Derived Categories

Introduction

We first discuss triangulated categories and localization in triangulated categories. Next, we prove that the homotopy category of complexes in an additive category is a triangulated category. Once this is done we define the derived category of an abelian category as the localization of the homotopy category with respect to quasi-isomorphisms. A good reference is Verdier’s thesis [Verdier].

Triangulated categories

Triangulated categories are a convenient tool to describe the type of structure inherent in the derived category of an abelian category. Some references are [Verdier], [KS], and [Neeman].

The definition of a triangulated category

In this section we collect most of the definitions concerning triangulated and pre-triangulated categories.

Definition

Let \(\mathcal{D}\) be an additive category. Let \([1] : \mathcal{D} \to \mathcal{D}\), \(E \mapsto E[1]\) be an additive functor which is an auto-equivalence of \(\mathcal{D}\).

1. A triangle is a sextuple \((X, Y, Z, f, g, h)\) where \(X, Y, Z \in \Ob(\mathcal{D})\) and \(f : X \to Y\), \(g : Y \to Z\) and \(h : Z \to X[1]\) are morphisms of \(\mathcal{D}\).

2. A morphism of triangles \((X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')\) is given by morphisms \(a : X \to X'\), \(b : Y \to Y'\) and \(c : Z \to Z'\) of \(\mathcal{D}\) such that \(b \circ f = f' \circ a\), \(c \circ g = g' \circ b\) and \(a[1] \circ h = h' \circ c\).

A morphism of triangles is visualized by the following commutative diagram \[\xymatrix{ X \ar[r] \ar[d]^a & Y \ar[r] \ar[d]^b & Z \ar[r] \ar[d]^c & X[1] \ar[d]^{a[1]} \\ X' \ar[r] & Y' \ar[r] & Z' \ar[r] & X'[1] }\] In the setting of Definition 0144, we write \([0] = \text{id}\), for \(n > 0\) we denote \([n]\) the \(n\)-fold composition of \([1]\), we choose a quasi-inverse \([-1]\) of \([1]\), and we set \([-n]\) equal to the \(n\)-fold composition of \([-1]\). Then \(\{[n]\}_{n \in \mathbf{Z}}\) is a collection of additive auto-equivalences of \(\mathcal{D}\) indexed by \(n \in \mathbf{Z}\) such that we are given isomorphisms of functors \([n] \circ [m] \cong [n + m]\).

Here is the definition of a triangulated category as given in Verdier’s thesis.

Definition

A triangulated category consists of a triple \((\mathcal{D}, \{[n]\}_{n\in \mathbf{Z}}, \mathcal{T})\) where

1. \(\mathcal{D}\) is an additive category,

2. \([1] : \mathcal{D} \to \mathcal{D}\), \(E \mapsto E[1]\) is an additive auto-equivalence and \([n]\) for \(n \in \mathbf{Z}\) is as discussed above, and

3. \(\mathcal{T}\) is a set of triangles (Definition 0144) called the distinguished triangles

subject to the following conditions

1. Any triangle isomorphic to a distinguished triangle is a distinguished triangle. Any triangle of the form \((X, X, 0, \text{id}, 0, 0)\) is distinguished. For any morphism \(f : X \to Y\) of \(\mathcal{D}\) there exists a distinguished triangle of the form \((X, Y, Z, f, g, h)\).

2. The triangle \((X, Y, Z, f, g, h)\) is distinguished if and only if the triangle \((Y, Z, X[1], g, h, -f[1])\) is.

3. Given a solid diagram \[\xymatrix{ X \ar[r]^f \ar[d]^a & Y \ar[r]^g \ar[d]^b & Z \ar[r]^h \ar@{-->}[d] & X[1] \ar[d]^{a[1]} \\ X' \ar[r]^{f'} & Y' \ar[r]^{g'} & Z' \ar[r]^{h'} & X'[1] }\] whose rows are distinguished triangles and which satisfies \(b \circ f = f' \circ a\), there exists a morphism \(c : Z \to Z'\) such that \((a, b, c)\) is a morphism of triangles.

4. Given objects \(X\), \(Y\), \(Z\) of \(\mathcal{D}\), and morphisms \(f : X \to Y\), \(g : Y \to Z\), and distinguished triangles \((X, Y, Q_1, f, p_1, d_1)\), \((X, Z, Q_2, g \circ f, p_2, d_2)\), and \((Y, Z, Q_3, g, p_3, d_3)\), there exist morphisms \(a : Q_1 \to Q_2\) and \(b : Q_2 \to Q_3\) such that

   1. \((Q_1, Q_2, Q_3, a, b, p_1[1] \circ d_3)\) is a distinguished triangle,

   2. the triple \((\text{id}_X, g, a)\) is a morphism of triangles \((X, Y, Q_1, f, p_1, d_1) \to (X, Z, Q_2, g \circ f, p_2, d_2)\), and

   3. the triple \((f, \text{id}_Z, b)\) is a morphism of triangles \((X, Z, Q_2, g \circ f, p_2, d_2) \to (Y, Z, Q_3, g, p_3, d_3)\).

We will call \((\mathcal{D}, [\ ], \mathcal{T})\) a pre-triangulated category if TR1, TR2 and TR3 hold.¹

The explanation of TR4 is that if you think of \(Q_1\) as \(Y/X\), \(Q_2\) as \(Z/X\) and \(Q_3\) as \(Z/Y\), then TR4(a) expresses the isomorphism \((Z/X)/(Y/X) \cong Z/Y\) and TR4(b) and TR4(c) express that we can compare the triangles \(X \to Y \to Q_1 \to X[1]\) etc with morphisms of triangles. For a more precise reformulation of this idea see the proof of Lemma 014R.

The sign in TR2 means that if \((X, Y, Z, f, g, h)\) is a distinguished triangle then in the long sequence [05QL]\[\begin{equation} \ldots \to Z[-1] \xrightarrow{-h[-1]} X \xrightarrow{f} Y \xrightarrow{g} Z \xrightarrow{h} X[1] \xrightarrow{-f[1]} Y[1] \xrightarrow{-g[1]} Z[1] \to \ldots \end{equation}\] each four term sequence gives a distinguished triangle.

As usual we abuse notation and we simply speak of a (pre-)triangulated category \(\mathcal{D}\) without explicitly introducing notation for the additional data. The notion of a pre-triangulated category is useful in finding statements equivalent to TR4.

We have the following definition of a triangulated functor.

Definition

Let \(\mathcal{D}\), \(\mathcal{D}'\) be pre-triangulated categories. An exact functor, or a triangulated functor from \(\mathcal{D}\) to \(\mathcal{D}'\) is a functor \(F : \mathcal{D} \to \mathcal{D}'\) together with given functorial isomorphisms \(\xi_X : F(X[1]) \to F(X)[1]\) such that for every distinguished triangle \((X, Y, Z, f, g, h)\) of \(\mathcal{D}\) the triangle \((F(X), F(Y), F(Z), F(f), F(g), \xi_X \circ F(h))\) is a distinguished triangle of \(\mathcal{D}'\).

An exact functor is additive, see Lemma 05QY. When we say two triangulated categories are equivalent we mean that they are equivalent in the \(2\)-category of triangulated categories. A \(2\)-morphism \(a : (F, \xi) \to (F', \xi')\) in this \(2\)-category is simply a transformation of functors \(a : F \to F'\) which is compatible with \(\xi\) and \(\xi'\), i.e., \[\xymatrix{ F \circ [1] \ar[r]_\xi \ar[d]_{a \star 1} & [1] \circ F \ar[d]^{1 \star a} \\ F' \circ [1] \ar[r]^{\xi'} & [1] \circ F' }\] commutes.

Definition

Let \((\mathcal{D}, [\ ], \mathcal{T})\) be a pre-triangulated category. A pre-triangulated subcategory² is a pair \((\mathcal{D}', \mathcal{T}')\) such that

1. \(\mathcal{D}'\) is an additive subcategory of \(\mathcal{D}\) which is preserved under \([1]\) and such that \([1] : \mathcal{D}' \to \mathcal{D}'\) is an auto-equivalence,

2. \(\mathcal{T}' \subset \mathcal{T}\) is a subset such that for every \((X, Y, Z, f, g, h) \in \mathcal{T}'\) we have \(X, Y, Z \in \Ob(\mathcal{D}')\) and \(f, g, h \in \text{Arrows}(\mathcal{D}')\), and

3. \((\mathcal{D}', [\ ], \mathcal{T}')\) is a pre-triangulated category.

If \(\mathcal{D}\) is a triangulated category, then we say \((\mathcal{D}', \mathcal{T}')\) is a triangulated subcategory if it is a pre-triangulated subcategory and \((\mathcal{D}', [\ ], \mathcal{T}')\) is a triangulated category.

In this situation the inclusion functor \(\mathcal{D}' \to \mathcal{D}\) is an exact functor with \(\xi_X : X[1] \to X[1]\) given by the identity on \(X[1]\).

We will see in Lemma 0146 that for a distinguished triangle \((X, Y, Z, f, g, h)\) in a pre-triangulated category the composition \(g \circ f : X \to Z\) is zero. Thus the sequence (05QL) is a complex. A homological functor is one that turns this complex into a long exact sequence.

Definition

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(\mathcal{A}\) be an abelian category. An additive functor \(H : \mathcal{D} \to \mathcal{A}\) is called homological if for every distinguished triangle \((X, Y, Z, f, g, h)\) the sequence \[H(X) \to H(Y) \to H(Z)\] is exact in the abelian category \(\mathcal{A}\). An additive functor \(H : \mathcal{D}^{opp} \to \mathcal{A}\) is called cohomological if the corresponding functor \(\mathcal{D} \to \mathcal{A}^{opp}\) is homological.

If \(H : \mathcal{D} \to \mathcal{A}\) is a homological functor we often write \(H^n(X) = H(X[n])\) so that \(H(X) = H^0(X)\). Our discussion of TR2 above implies that a distinguished triangle \((X, Y, Z, f, g, h)\) determines a long exact sequence [0148]\[\begin{equation} \xymatrix@C=3pc{ H^{-1}(Z) \ar[r]^{H(h[-1])} & H^0(X) \ar[r]^{H(f)} & H^0(Y) \ar[r]^{H(g)} & H^0(Z) \ar[r]^{H(h)} & H^1(X) } \end{equation}\] This will be called the long exact sequence associated to the distinguished triangle and the homological functor. As indicated we will not use any signs for the morphisms in the long exact sequence. This has the side effect that maps in the long exact sequence associated to the rotation (TR2) of a distinguished triangle differ from the maps in the sequence above by some signs.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(\mathcal{D}\) be a triangulated category. A \(\delta\)-functor from \(\mathcal{A}\) to \(\mathcal{D}\) is given by a functor \(G : \mathcal{A} \to \mathcal{D}\) and a rule which assigns to every short exact sequence \[0 \to A \xrightarrow{a} B \xrightarrow{b} C \to 0\] a morphism \(\delta = \delta_{A \to B \to C} : G(C) \to G(A)[1]\) such that

1. the triangle \((G(A), G(B), G(C), G(a), G(b), \delta_{A \to B \to C})\) is a distinguished triangle of \(\mathcal{D}\) for any short exact sequence as above, and

2. for every morphism \((A \to B \to C) \to (A' \to B' \to C')\) of short exact sequences the diagram \[\xymatrix{ G(C) \ar[d] \ar[rr]_{\delta_{A \to B \to C}} & & G(A)[1] \ar[d] \\ G(C') \ar[rr]^{\delta_{A' \to B' \to C'}} & & G(A')[1] }\] is commutative.

In this situation we call \((G(A), G(B), G(C), G(a), G(b), \delta_{A \to B \to C})\) the image of the short exact sequence under the given \(\delta\)-functor.

Note how a \(\delta\)-functor comes equipped with additional structure. Strictly speaking it does not make sense to say that a given functor \(\mathcal{A} \to \mathcal{D}\) is a \(\delta\)-functor, but we will often do so anyway.

Elementary results on triangulated categories

Most of the results in this section are proved for pre-triangulated categories and a fortiori hold in any triangulated category.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \((X, Y, Z, f, g, h)\) be a distinguished triangle. Then \(g \circ f = 0\), \(h \circ g = 0\) and \(f[1] \circ h = 0\).

Proof

By TR1 we know \((X, X, 0, 1, 0, 0)\) is a distinguished triangle. Apply TR3 to \[\xymatrix{ X \ar[r] \ar[d]^1 & X \ar[r] \ar[d]^f & 0 \ar[r] \ar@{-->}[d] & X[1] \ar[d]^{1[1]} \\ X \ar[r]^f & Y \ar[r]^g & Z \ar[r]^h & X[1] }\] Of course the dotted arrow is the zero map. Hence the commutativity of the diagram implies that \(g \circ f = 0\). For the other cases rotate the triangle, i.e., apply TR2.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. For any object \(W\) of \(\mathcal{D}\) the functor \(\Hom_\mathcal{D}(W, -)\) is homological, and the functor \(\Hom_\mathcal{D}(-, W)\) is cohomological.

Proof

Consider a distinguished triangle \((X, Y, Z, f, g, h)\). We have already seen that \(g \circ f = 0\), see Lemma 0146. Suppose \(a : W \to Y\) is a morphism such that \(g \circ a = 0\). Then we get a commutative diagram \[\xymatrix{ W \ar[r]_1 \ar@{..>}[d]^b & W \ar[r] \ar[d]^a & 0 \ar[r] \ar[d]^0 & W[1] \ar@{..>}[d]^{b[1]} \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] }\] Both rows are distinguished triangles (use TR1 for the top row). Hence we can fill the dotted arrow \(b\) (first rotate using TR2, then apply TR3, and then rotate back). This proves the lemma.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \[(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')\] be a morphism of distinguished triangles. If two among \(a, b, c\) are isomorphisms so is the third.

Proof

Assume that \(a\) and \(c\) are isomorphisms. For any object \(W\) of \(\mathcal{D}\) write \(H_W( - ) = \Hom_\mathcal{D}(W, -)\). Then we get a commutative diagram of abelian groups \[\xymatrix{ H_W(Z[-1]) \ar[r] \ar[d] & H_W(X) \ar[r] \ar[d] & H_W(Y) \ar[r] \ar[d] & H_W(Z) \ar[r] \ar[d] & H_W(X[1]) \ar[d] \\ H_W(Z'[-1]) \ar[r] & H_W(X') \ar[r] & H_W(Y') \ar[r] & H_W(Z') \ar[r] & H_W(X'[1]) }\] By assumption the right two and left two vertical arrows are bijective. As \(H_W\) is homological by Lemma 0149 and the five lemma (Homology, Lemma 05QB) it follows that the middle vertical arrow is an isomorphism. Hence by Yoneda’s lemma, see Categories, Lemma 001P we see that \(b\) is an isomorphism. This implies the other cases by rotating (using TR2).

Remark

Let \(\mathcal{D}\) be an additive category with translation functors \([n]\) as in Definition 0144. Let us call a triangle \((X, Y, Z, f, g, h)\) special³ if for every object \(W\) of \(\mathcal{D}\) the long sequence of abelian groups \[\ldots \to \Hom_\mathcal{D}(W, X) \to \Hom_\mathcal{D}(W, Y) \to \Hom_\mathcal{D}(W, Z) \to \Hom_\mathcal{D}(W, X[1]) \to \ldots\] is exact. The proof of Lemma 014A shows that if \[(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')\] is a morphism of special triangles and if two among \(a, b, c\) are isomorphisms so is the third. There is a dual statement for co-special triangles, i.e., triangles which turn into long exact sequences on applying the functor \(\Hom_\mathcal{D}(-, W)\). Thus distinguished triangles are special and co-special, but in general there are many more (co-)special triangles, than there are distinguished triangles.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \[(0, b, 0), (0, b', 0) : (X, Y, Z, f, g, h) \to (X, Y, Z, f, g, h)\] be endomorphisms of a distinguished triangle. Then \(bb' = 0\).

Proof

Picture \[\xymatrix{ X \ar[r] \ar[d]^0 & Y \ar[r] \ar[d]^{b, b'} \ar@{..>}[ld]^\alpha & Z \ar[r] \ar[d]^0 \ar@{..>}[ld]^\beta & X[1] \ar[d]^0 \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] }\] Applying Lemma 0149 we find dotted arrows \(\alpha\) and \(\beta\) such that \(b' = f \circ \alpha\) and \(b = \beta \circ g\). Then \(bb' = \beta \circ g \circ f \circ \alpha = 0\) as \(g \circ f = 0\) by Lemma 0146.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \((X, Y, Z, f, g, h)\) be a distinguished triangle. If \[\xymatrix{ Z \ar[r]_h \ar[d]_c & X[1] \ar[d]^{a[1]} \\ Z \ar[r]^h & X[1] }\] is commutative and \(a^2 = a\), \(c^2 = c\), then there exists a morphism \(b : Y \to Y\) with \(b^2 = b\) such that \((a, b, c)\) is an endomorphism of the triangle \((X, Y, Z, f, g, h)\).

Proof

By TR3 there exists a morphism \(b'\) such that \((a, b', c)\) is an endomorphism of \((X, Y, Z, f, g, h)\). Then \((0, (b')^2 - b', 0)\) is also an endomorphism. By Lemma 05QP we see that \((b')^2 - b'\) has square zero. Set \(b = b' - (2b' - 1)((b')^2 - b') = 3(b')^2 - 2(b')^3\). A computation shows that \((a, b, c)\) is an endomorphism and that \(b^2 - b = (4(b')^2 - 4b' - 3)((b')^2 - b')^2 = 0\).

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(f : X \to Y\) be a morphism of \(\mathcal{D}\). There exists a distinguished triangle \((X, Y, Z, f, g, h)\) which is unique up to (nonunique) isomorphism of triangles. More precisely, given a second such distinguished triangle \((X, Y, Z', f, g', h')\) there exists an isomorphism \[(1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Z', f, g', h')\]

Proof

Existence by TR1. Uniqueness up to isomorphism by TR3 and Lemma 014A.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \[(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')\] be a morphism of distinguished triangles. If one of the following conditions holds

1. \(\Hom(Y, X') = 0\),

2. \(\Hom(Z, Y') = 0\),

3. \(\Hom(X, X') = \Hom(Z, X') = 0\),

4. \(\Hom(Z, X') = \Hom(Z, Z') = 0\), or

5. \(\Hom(X[1], Z') = \Hom(Z, X') = 0\)

then \(b\) is the unique morphism from \(Y \to Y'\) such that \((a, b, c)\) is a morphism of triangles.

Proof

If we have a second morphism of triangles \((a, b', c)\) then \((0, b - b', 0)\) is a morphism of triangles. Hence we have to show: the only morphism \(b : Y \to Y'\) such that \(X \to Y \to Y'\) and \(Y \to Y' \to Z'\) are zero is \(0\). We will use Lemma 0149 without further mention. In particular, condition (3) implies (1). Given condition (1) if the composition \(g' \circ b : Y \to Y' \to Z'\) is zero, then \(b\) lifts to a morphism \(Y \to X'\) which has to be zero. This proves (1).

The proof of (2) and (4) are dual to this argument.

Assume (5). Consider the diagram \[\xymatrix{ X \ar[r]_f \ar[d]^0 & Y \ar[r]_g \ar[d]^b & Z \ar[r]_h \ar[d]^0 \ar@{..>}[ld]^\epsilon & X[1] \ar[d]^0 \\ X' \ar[r]^{f'} & Y' \ar[r]^{g'} & Z' \ar[r]^{h'} & X'[1] }\] We may choose \(\epsilon\) such that \(b = \epsilon \circ g\). Then \(g' \circ \epsilon \circ g = 0\) which implies that \(g' \circ \epsilon = \delta \circ h\) for some \(\delta \in \Hom(X[1], Z')\). Since \(\Hom(X[1], Z') = 0\) we conclude that \(g' \circ \epsilon = 0\). Hence \(\epsilon = f' \circ \gamma\) for some \(\gamma \in \Hom(Z, X')\). Since \(\Hom(Z, X') = 0\) we conclude that \(\epsilon = 0\) and hence \(b = 0\) as desired.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(f : X \to Y\) be a morphism of \(\mathcal{D}\). The following are equivalent

1. \(f\) is an isomorphism,

2. \((X, Y, 0, f, 0, 0)\) is a distinguished triangle, and

3. for any distinguished triangle \((X, Y, Z, f, g, h)\) we have \(Z = 0\).

Proof

By TR1 the triangle \((X, X, 0, 1, 0, 0)\) is distinguished. Let \((X, Y, Z, f, g, h)\) be a distinguished triangle. By TR3 there is a map of distinguished triangles \((1, f, 0) : (X, X, 0) \to (X, Y, Z)\). If \(f\) is an isomorphism, then \((1, f, 0)\) is an isomorphism of triangles by Lemma 014A and \(Z = 0\). Conversely, if \(Z = 0\), then \((1, f, 0)\) is an isomorphism of triangles as well, hence \(f\) is an isomorphism.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \((X, Y, Z, f, g, h)\) and \((X', Y', Z', f', g', h')\) be triangles. The following are equivalent

1. \((X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')\) is a distinguished triangle,

2. both \((X, Y, Z, f, g, h)\) and \((X', Y', Z', f', g', h')\) are distinguished triangles.

Proof

Assume (2). By TR1 we may choose a distinguished triangle \((X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')\). By TR3 we can find morphisms of distinguished triangles \((X, Y, Z, f, g, h) \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')\) and \((X', Y', Z', f', g', h') \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')\). Taking the direct sum of these morphisms we obtain a morphism of triangles \[\xymatrix{ (X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \ar[d]^{(1, 1, c)} \\ (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h''). }\] In the terminology of Remark 09WA this is a map of special triangles (because a direct sum of special triangles is special) and we conclude that \(c\) is an isomorphism. Thus (1) holds.

Assume (1). We will show that \((X, Y, Z, f, g, h)\) is a distinguished triangle. First observe that \((X, Y, Z, f, g, h)\) is a special triangle (terminology from Remark 09WA) as a direct summand of the distinguished hence special triangle \((X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')\). Using TR1 let \((X, Y, Q, f, g'', h'')\) be a distinguished triangle. By TR3 there exists a morphism of distinguished triangles \((X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \to (X, Y, Q, f, g'', h'')\). Composing this with the inclusion map we get a morphism of triangles \[(1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Q, f, g'', h'')\] By Remark 09WA we find that \(c\) is an isomorphism and we conclude that (2) holds.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \((X, Y, Z, f, g, h)\) be a distinguished triangle.

1. If \(h = 0\), then there exists a right inverse \(s : Z \to Y\) to \(g\).

2. For any right inverse \(s : Z \to Y\) of \(g\) the map \(f \oplus s : X \oplus Z \to Y\) is an isomorphism.

3. For any objects \(X', Z'\) of \(\mathcal{D}\) the triangle \((X', X' \oplus Z', Z', (1, 0), (0, 1), 0)\) is distinguished.

Proof

To see (1) use that \(\Hom_\mathcal{D}(Z, Y) \to \Hom_\mathcal{D}(Z, Z) \to \Hom_\mathcal{D}(Z, X[1])\) is exact by Lemma 0149. By the same token, if \(s\) is as in (2), then \(h = 0\) and the sequence \[0 \to \Hom_\mathcal{D}(W, X) \to \Hom_\mathcal{D}(W, Y) \to \Hom_\mathcal{D}(W, Z) \to 0\] is split exact (split by \(s : Z \to Y\)). Hence by Yoneda’s lemma we see that \(X \oplus Z \to Y\) is an isomorphism. The last assertion follows from TR1 and Lemma 05QS.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(f : X \to Y\) be a morphism of \(\mathcal{D}\). The following are equivalent

1. \(f\) has a kernel,

2. \(f\) has a cokernel,

3. \(f\) is the isomorphic to a composition \(K \oplus Z \to Z \to Z \oplus Q\) of a projection and coprojection for some objects \(K, Z, Q\) of \(\mathcal{D}\).

Proof

Any morphism isomorphic to a map of the form \(X' \oplus Z \to Z \oplus Y'\) has both a kernel and a cokernel. Hence (3) \(\Rightarrow\) (1), (2). Next we prove (1) \(\Rightarrow\) (3). Suppose first that \(f : X \to Y\) is a monomorphism, i.e., its kernel is zero. By TR1 there exists a distinguished triangle \((X, Y, Z, f, g, h)\). By Lemma 0146 the composition \(f \circ h[-1] = 0\). As \(f\) is a monomorphism we see that \(h[-1] = 0\) and hence \(h = 0\). Then Lemma 05QT implies that \(Y = X \oplus Z\), i.e., we see that (3) holds. Next, assume \(f\) has a kernel \(K\). As \(K \to X\) is a monomorphism we conclude \(X = K \oplus X'\) and \(f|_{X'} : X' \to Y\) is a monomorphism. Hence \(Y = X' \oplus Y'\) and we win. The implication (2) \(\Rightarrow\) (3) is dual to this.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(I\) be a set.

1. Let \(X_i\), \(i \in I\) be a family of objects of \(\mathcal{D}\).

   1. If \(\prod X_i\) exists, then \((\prod X_i)[1] = \prod X_i[1]\).

   2. If \(\bigoplus X_i\) exists, then \((\bigoplus X_i)[1] = \bigoplus X_i[1]\).

2. Let \(X_i \to Y_i \to Z_i \to X_i[1]\) be a family of distinguished triangles of \(\mathcal{D}\).

   1. If \(\prod X_i\), \(\prod Y_i\), \(\prod Z_i\) exist, then \(\prod X_i \to \prod Y_i \to \prod Z_i \to \prod X_i[1]\) is a distinguished triangle.

   2. If \(\bigoplus X_i\), \(\bigoplus Y_i\), \(\bigoplus Z_i\) exist, then \(\bigoplus X_i \to \bigoplus Y_i \to \bigoplus Z_i \to \bigoplus X_i[1]\) is a distinguished triangle.

Proof

Part (1) is true because \([1]\) is an autoequivalence of \(\mathcal{D}\) and because direct sums and products are defined in terms of the category structure. Let us prove (2)(a). Choose a distinguished triangle \(\prod X_i \to \prod Y_i \to Z \to \prod X_i[1]\). For each \(j\) we can use TR3 to choose a morphism \(p_j : Z \to Z_j\) fitting into a morphism of distinguished triangles with the projection maps \(\prod X_i \to X_j\) and \(\prod Y_i \to Y_j\). Using the definition of products we obtain a map \(\prod p_i : Z \to \prod Z_i\) fitting into a morphism of triangles from the distinguished triangle to the triangle made out of the products. Observe that the “product” triangle \(\prod X_i \to \prod Y_i \to \prod Z_i \to \prod X_i[1]\) is special in the terminology of Remark 09WA because products of exact sequences of abelian groups are exact. Hence Remark 09WA shows that the morphism of triangles is an isomorphism and we conclude by TR1. The proof of (2)(b) is dual.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. If \(\mathcal{D}\) has countable products, then \(\mathcal{D}\) is Karoubian. If \(\mathcal{D}\) has countable coproducts, then \(\mathcal{D}\) is Karoubian.

Proof

Assume \(\mathcal{D}\) has countable products. By Homology, Lemma 05QV it suffices to check that morphisms which have a right inverse have kernels. Any morphism which has a right inverse is an epimorphism, hence has a kernel by Lemma 05QU. The second statement is dual to the first.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(\operatorname{Ar}(\mathcal{D})\) be the category whose objects are morphisms of \(\mathcal{D}\) and whose morphisms are commutative squares. Let \(\operatorname{DTri}(\mathcal{D})\) be the category whose objects are distinguished triangles and whose morphisms are morphisms of triangles. Let \[U : \operatorname{DTri}(\mathcal{D}) \longrightarrow \operatorname{Ar}(\mathcal{D})\] be the functor which sends a distinguished triangle to its first arrow. Assume \(U\) has a section. If \(\mathcal{D}\) has countable products or countable coproducts, then every morphism of \(\mathcal{D}\) is isomorphic to a composition \[K \oplus Z \longrightarrow Z \longrightarrow Z \oplus Q\] of a projection and a coprojection. In particular, \(\mathcal{D}\) has kernels and cokernels.

Proof

By Lemma 05QW, the category \(\mathcal{D}\) is Karoubian. Let \(T\) be a section of \(U\), let \(f : X \to Y\), and write \[T(f) = (X, Y, Z_f, f, g_f, h_f).\] Set \(d_f = -h_f[-1] : Z_f[-1] \to X\). Exactness of the sequence obtained from this triangle gives a natural epimorphism \[q_f : \Hom_\mathcal{D}(-, Z_f[-1]) \longrightarrow \mathcal{K}_f, \qquad \mathcal{K}_f(W) = \{v : W \to X \mid f \circ v = 0\},\] given by composition with \(d_f\).

Write \(T(W \to 0) = (W, 0, Z_W, 0, 0, h_W)\). By two rotations and Lemma 05QR, the map \(h_W : Z_W \to W[1]\) is an isomorphism. Given \(v \in \mathcal{K}_f(W)\), applying \(T\) to the commutative square \((v, 0) : (W \to 0) \to (X \to Y)\) gives a morphism of triangles \((v, 0, c_v)\). Define \[s_{f,W}(v) = -c_v[-1] \circ h_W[-1]^{-1}.\] Functoriality of \(T\) shows that these maps are natural in \(W\). The last square in the morphism of triangles gives \(h_f[-1] \circ c_v[-1] = v \circ h_W[-1]\); hence the chosen signs give \(q_f \circ s_f = 1_{\mathcal{K}_f}\).

Thus \(e_f = s_f \circ q_f\) is an idempotent natural endomorphism of \(\Hom_\mathcal{D}(-, Z_f[-1])\). By Yoneda it comes from an idempotent \(e : Z_f[-1] \to Z_f[-1]\). Since \(\mathcal{D}\) is Karoubian, \(e\) splits, and the image of this splitting represents \(\mathcal{K}_f\). Thus \(f\) has a kernel. The conclusion now follows from Lemma 05QU.

The following lemma makes it slightly easier to prove that a pre-triangulated category is triangulated.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. In order to prove TR4 it suffices to show that given any pair of composable morphisms \(f : X \to Y\) and \(g : Y \to Z\) there exist

1. isomorphisms \(i : X' \to X\), \(j : Y' \to Y\) and \(k : Z' \to Z\), and then setting \(f' = j^{-1}fi : X' \to Y'\) and \(g' = k^{-1}gj : Y' \to Z'\) there exist

2. distinguished triangles \((X', Y', Q_1, f', p_1, d_1)\), \((X', Z', Q_2, g' \circ f', p_2, d_2)\) and \((Y', Z', Q_3, g', p_3, d_3)\), such that the assertion of TR4 holds.

Proof

The replacement of \(X, Y, Z\) by \(X', Y', Z'\) is harmless by our definition of distinguished triangles and their isomorphisms. The lemma follows from the fact that the distinguished triangles \((X', Y', Q_1, f', p_1, d_1)\), \((X', Z', Q_2, g' \circ f', p_2, d_2)\) and \((Y', Z', Q_3, g', p_3, d_3)\) are unique up to isomorphism by Lemma 014B.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Assume that \(\mathcal{D}'\) is an additive full subcategory of \(\mathcal{D}\). The following are equivalent

1. there exists a set of triangles \(\mathcal{T}'\) such that \((\mathcal{D}', \mathcal{T}')\) is a pre-triangulated subcategory of \(\mathcal{D}\),

2. \(\mathcal{D}'\) is preserved under \([1]\) and \([1] : \mathcal{D}' \to \mathcal{D}'\) is an auto-equivalence and given any morphism \(f : X \to Y\) in \(\mathcal{D}'\) there exists a distinguished triangle \((X, Y, Z, f, g, h)\) in \(\mathcal{D}\) such that \(Z\) is isomorphic to an object of \(\mathcal{D}'\).

In this case \(\mathcal{T}'\) as in (1) is the set of distinguished triangles \((X, Y, Z, f, g, h)\) of \(\mathcal{D}\) such that \(X, Y, Z \in \Ob(\mathcal{D}')\). Finally, if \(\mathcal{D}\) is a triangulated category, then (1) and (2) are also equivalent to

1. \(\mathcal{D}'\) is a triangulated subcategory.

Proof

Omitted.

Lemma

An exact functor of pre-triangulated categories is additive.

Proof

Let \(F : \mathcal{D} \to \mathcal{D}'\) be an exact functor of pre-triangulated categories. Since \((0, 0, 0, 1_0, 1_0, 0)\) is a distinguished triangle of \(\mathcal{D}\) the triangle \[(F(0), F(0), F(0), 1_{F(0)}, 1_{F(0)}, F(0))\] is distinguished in \(\mathcal{D}'\). This implies that \(1_{F(0)} \circ 1_{F(0)}\) is zero, see Lemma 0146. Hence \(F(0)\) is the zero object of \(\mathcal{D}'\). This also implies that \(F\) applied to any zero morphism is zero (since a morphism in an additive category is zero if and only if it factors through the zero object). Next, using that \((X, X \oplus Y, Y, (1, 0), (0, 1), 0)\) is a distinguished triangle by Lemma 05QT part (3), we see that \((F(X), F(X \oplus Y), F(Y), F(1, 0), F(0, 1), 0)\) is one too. This implies that the map \(F(X) \oplus F(Y) \to F(X \oplus Y)\) is an isomorphism by Lemma 05QT part (2). To finish we apply Homology, Lemma 0DLP.

Lemma

Let \(F : \mathcal{D} \to \mathcal{D}'\) be a fully faithful exact functor of pre-triangulated categories. Then a triangle \((X, Y, Z, f, g, h)\) of \(\mathcal{D}\) is distinguished if and only if \((F(X), F(Y), F(Z), F(f), F(g), F(h))\) is distinguished in \(\mathcal{D}'\).

Proof

The “only if” part is clear. Assume \((F(X), F(Y), F(Z))\) is distinguished in \(\mathcal{D}'\). Pick a distinguished triangle \((X, Y, Z', f, g', h')\) in \(\mathcal{D}\). By Lemma 014B there exists an isomorphism of triangles \[(1, 1, c') : (F(X), F(Y), F(Z)) \longrightarrow (F(X), F(Y), F(Z')).\] Since \(F\) is fully faithful, there exists a morphism \(c : Z \to Z'\) such that \(F(c) = c'\). Then \((1, 1, c)\) is an isomorphism between \((X, Y, Z)\) and \((X, Y, Z')\). Hence \((X, Y, Z)\) is distinguished by TR1.

Lemma

Let \(\mathcal{D}, \mathcal{D}', \mathcal{D}''\) be pre-triangulated categories. Let \(F : \mathcal{D} \to \mathcal{D}'\) and \(F' : \mathcal{D}' \to \mathcal{D}''\) be exact functors. Then \(F' \circ F\) is an exact functor.

Proof

Omitted.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(\mathcal{A}\) be an abelian category. Let \(H : \mathcal{D} \to \mathcal{A}\) be a homological functor.

1. Let \(\mathcal{D}'\) be a pre-triangulated category. Let \(F : \mathcal{D}' \to \mathcal{D}\) be an exact functor. Then the composition \(H \circ F\) is a homological functor as well.

2. Let \(\mathcal{A}'\) be an abelian category. Let \(G : \mathcal{A} \to \mathcal{A}'\) be an exact functor. Then \(G \circ H\) is a homological functor as well.

Proof

Omitted.

Lemma

Let \(\mathcal{D}\) be a triangulated category. Let \(\mathcal{A}\) be an abelian category. Let \(G : \mathcal{A} \to \mathcal{D}\) be a \(\delta\)-functor.

1. Let \(\mathcal{D}'\) be a triangulated category. Let \(F : \mathcal{D} \to \mathcal{D}'\) be an exact functor. Then the composition \(F \circ G\) is a \(\delta\)-functor as well.

2. Let \(\mathcal{A}'\) be an abelian category. Let \(H : \mathcal{A}' \to \mathcal{A}\) be an exact functor. Then \(G \circ H\) is a \(\delta\)-functor as well.

Proof

Omitted.

Lemma

Let \(\mathcal{D}\) be a triangulated category. Let \(\mathcal{A}\) and \(\mathcal{B}\) be abelian categories. Let \(G : \mathcal{A} \to \mathcal{D}\) be a \(\delta\)-functor. Let \(H : \mathcal{D} \to \mathcal{B}\) be a homological functor. Assume that \(H^{-1}(G(A)) = 0\) for all \(A\) in \(\mathcal{A}\). Then the collection \[\{H^n \circ G, H^n(\delta_{A \to B \to C})\}_{n \geq 0}\] is a \(\delta\)-functor from \(\mathcal{A} \to \mathcal{B}\), see Homology, Definition 010Q.

Proof

The notation signifies the following. If \(0 \to A \xrightarrow{a} B \xrightarrow{b} C \to 0\) is a short exact sequence in \(\mathcal{A}\), then \[\delta = \delta_{A \to B \to C} : G(C) \to G(A)[1]\] is a morphism in \(\mathcal{D}\) such that \((G(A), G(B), G(C), a, b, \delta)\) is a distinguished triangle, see Definition 0150. Then \(H^n(\delta) : H^n(G(C)) \to H^n(G(A)[1]) = H^{n + 1}(G(A))\) is clearly functorial in the short exact sequence. Finally, the long exact cohomology sequence (0148) combined with the vanishing of \(H^{-1}(G(C))\) gives a long exact sequence \[0 \to H^0(G(A)) \to H^0(G(B)) \to H^0(G(C)) \xrightarrow{H^0(\delta)} H^1(G(A)) \to \ldots\] in \(\mathcal{B}\) as desired.

The proof of the following result uses TR4.

Proposition

Let \(\mathcal{D}\) be a triangulated category. Any commutative diagram \[\xymatrix{ X \ar[r] \ar[d] & Y \ar[d] \\ X' \ar[r] & Y' }\] can be extended to a diagram \[\xymatrix{ X \ar[r] \ar[d] & Y \ar[r] \ar[d] & Z \ar[r] \ar[d] & X[1] \ar[d] \\ X' \ar[r] \ar[d] & Y' \ar[r] \ar[d] & Z' \ar[r] \ar[d] & X'[1] \ar[d] \\ X'' \ar[r] \ar[d] & Y'' \ar[r] \ar[d] & Z'' \ar[r] \ar[d] & X''[1] \ar[d] \\ X[1] \ar[r] & Y[1] \ar[r] & Z[1] \ar[r] & X[2] }\] where all the squares are commutative, except for the lower right square which is anticommutative. Moreover, each of the rows and columns are distinguished triangles. Finally, the morphisms on the bottom row (resp. right column) are obtained from the morphisms of the top row (resp. left column) by applying \([1]\).

Proof

During this proof we avoid writing the arrows in order to make the proof legible. Choose distinguished triangles \((X, Y, Z)\), \((X', Y', Z')\), \((X, X', X'')\), \((Y, Y', Y'')\), and \((X, Y', A)\). Note that the morphism \(X \to Y'\) is both equal to the composition \(X \to Y \to Y'\) and equal to the composition \(X \to X' \to Y'\). Hence, we can find morphisms

1. \(a : Z \to A\) and \(b : A \to Y''\), and

2. \(a' : X'' \to A\) and \(b' : A \to Z'\)

as in TR4. Denote \(c : Y'' \to Z[1]\) the composition \(Y'' \to Y[1] \to Z[1]\) and denote \(c' : Z' \to X''[1]\) the composition \(Z' \to X'[1] \to X''[1]\). The conclusion of our application TR4 are that

1. \((Z, A, Y'', a, b, c)\), \((X'', A, Z', a', b', c')\) are distinguished triangles,

2. \((X, Y, Z) \to (X, Y', A)\), \((X, Y', A) \to (Y, Y', Y'')\), \((X, X', X'') \to (X, Y', A)\), \((X, Y', A) \to (X', Y', Z')\) are morphisms of triangles.

First using that \((X, X', X'') \to (X, Y', A)\) and \((X, Y', A) \to (Y, Y', Y'')\). are morphisms of triangles we see the first of the diagrams \[\vcenter{ \xymatrix{ X' \ar[r] \ar[d] & Y' \ar[d] \\ X'' \ar[r]^{b \circ a'} \ar[d] & Y'' \ar[d] \\ X[1] \ar[r] & Y[1] } } \quad\text{and}\quad \vcenter{ \xymatrix{ Y \ar[r] \ar[d] & Z \ar[d]^{b' \circ a} \ar[r] & X[1] \ar[d] \\ Y' \ar[r] & Z' \ar[r] & X'[1] } }\] is commutative. The second is commutative too using that \((X, Y, Z) \to (X, Y', A)\) and \((X, Y', A) \to (X', Y', Z')\) are morphisms of triangles. At this point we choose a distinguished triangle \((X'', Y'' , Z'')\) starting with the map \(b \circ a' : X'' \to Y''\).

Next we apply TR4 one more time to the morphisms \(X'' \to A \to Y''\) and the triangles \((X'', A, Z', a', b', c')\), \((X'', Y'', Z'')\), and \((A, Y'', Z[1], b, c , -a[1])\) to get morphisms \(a'' : Z' \to Z''\) and \(b'' : Z'' \to Z[1]\). Then \((Z', Z'', Z[1], a'', b'', - b'[1] \circ a[1])\) is a distinguished triangle, hence also \((Z, Z', Z'', -b' \circ a, a'', -b'')\) and hence also \((Z, Z', Z'', b' \circ a, a'', b'')\). Moreover, \((X'', A, Z') \to (X'', Y'', Z'')\) and \((X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])\) are morphisms of triangles. At this point we have defined all the distinguished triangles and all the morphisms, and all that’s left is to verify some commutativity relations.

To see that the middle square in the diagram commutes, note that the arrow \(Y' \to Z'\) factors as \(Y' \to A \to Z'\) because \((X, Y', A) \to (X', Y', Z')\) is a morphism of triangles. Similarly, the morphism \(Y' \to Y''\) factors as \(Y' \to A \to Y''\) because \((X, Y', A) \to (Y, Y', Y'')\) is a morphism of triangles. Hence the middle square commutes because the square with sides \((A, Z', Z'', Y'')\) commutes as \((X'', A, Z') \to (X'', Y'', Z'')\) is a morphism of triangles (by TR4). The square with sides \((Y'', Z'', Y[1], Z[1])\) commutes because \((X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])\) is a morphism of triangles and \(c : Y'' \to Z[1]\) is the composition \(Y'' \to Y[1] \to Z[1]\). The square with sides \((Z', X'[1], X''[1], Z'')\) is commutative because \((X'', A, Z') \to (X'', Y'', Z'')\) is a morphism of triangles and \(c' : Z' \to X''[1]\) is the composition \(Z' \to X'[1] \to X''[1]\). Finally, we have to show that the square with sides \((Z'', X''[1], Z[1], X[2])\) anticommutes. This holds because \((X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])\) is a morphism of triangles and we’re done.

Localization of triangulated categories

In order to construct the derived category starting from the homotopy category of complexes, we will use a localization process.

Definition

Let \(\mathcal{D}\) be a pre-triangulated category. We say a multiplicative system \(S\) is compatible with the triangulated structure if the following two conditions hold:

1. For a morphism \(f\) of \(\mathcal{D}\) we have \(f \in S \Leftrightarrow f[1] \in S\)⁴.

2. Given a solid commutative square \[\xymatrix{ X \ar[r] \ar[d]^s & Y \ar[r] \ar[d]^{s'} & Z \ar[r] \ar@{-->}[d] & X[1] \ar[d]^{s[1]} \\ X' \ar[r] & Y' \ar[r] & Z' \ar[r] & X'[1] }\] whose rows are distinguished triangles with \(s, s' \in S\) there exists a morphism \(s'' : Z \to Z'\) in \(S\) such that \((s, s', s'')\) is a morphism of triangles.

It turns out that these axioms are not independent of the axioms defining multiplicative systems.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(S \subset \text{Arrows}(\mathcal{D})\).

1. If \(S\) contains all identities and MS6 holds (Definition 05R2), then every isomorphism of \(\mathcal{D}\) is in \(S\).

2. If MS1, MS5 (Categories, Definition 04VC) and MS6 hold, then MS2 holds.

Proof

Assume \(S\) contains all identities and MS6 holds. Let \(f : X \to Y\) be an isomorphism of \(\mathcal{D}\). Consider the diagram \[\xymatrix{ 0 \ar[r] \ar[d]^1 & X \ar[r]_1 \ar[d]^1 & X \ar[r] \ar@{-->}[d] & 0[1] \ar[d]^{1[1]} \\ 0 \ar[r] & X \ar[r]^f & Y \ar[r] & 0[1] }\] The rows are distinguished triangles by Lemma 05QR. By MS6 we see that the dotted arrow exists and is in \(S\), so \(f\) is in \(S\).

Assume MS1, MS5, MS6. Suppose that \(f : X \to Y\) is a morphism of \(\mathcal{D}\) and \(t : X \to X'\) an element of \(S\). Choose a distinguished triangle \((X, Y, Z, f, g, h)\). Next, choose a distinguished triangle \((X', Y', Z, f', g', t[1] \circ h)\) (here we use TR1 and TR2). By MS5, MS6 (and TR2 to rotate) we can find the dotted arrow in the commutative diagram \[\xymatrix{ X \ar[r] \ar[d]^t & Y \ar[r] \ar@{..>}[d]^{s'} & Z \ar[r] \ar[d]^1 & X[1] \ar[d]^{t[1]} \\ X' \ar[r] & Y' \ar[r] & Z \ar[r] & X'[1] }\] with moreover \(s' \in S\). This proves LMS2. The proof of RMS2 is dual.

Remark

In the presence of MS1 and MS6, condition MS5 is equivalent to asking \(s[n] \in S\) for all \(s \in S\) and \(n \in \mathbf{Z}\). For example, suppose MS5 holds, we have \(s \in S\), and we want to show \(s[-1] \in S\). This isn’t immediate because \(s[-1][1]\) is not equal to \(s\), only isomorphic to \(s\) as an arrow of \(\mathcal{D}\). Still, this does imply that \(s[-1][1] = f \circ s \circ g\) for isomorphisms \(f\), \(g\). By Lemma 05R3 (1) we find \(f, g \in S\), hence \(s[-1][1] \in S\) by MS1, hence \(s[-1] \in S\) by MS5. We leave a complete proof to the reader as an exercise.

Lemma

Let \(F : \mathcal{D} \to \mathcal{D}'\) be an exact functor of pre-triangulated categories. Let \[S = \{f \in \text{Arrows}(\mathcal{D}) \mid F(f)\text{ is an isomorphism}\}\] Then \(S\) is a saturated (see Categories, Definition 05Q8) multiplicative system compatible with the triangulated structure on \(\mathcal{D}\).

Proof

We have to prove axioms MS1 – MS6, see Categories, Definitions 04VC and 05Q8 and Definition 05R2. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and Lemma 014A. By Lemma 05R3 we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let \(f, g : X \to Y\) be morphisms of \(\mathcal{D}\), and let \(t : Z \to X\) be an element of \(S\) such that \(f \circ t = g \circ t\). As \(\mathcal{D}\) is additive this simply means that \(a \circ t = 0\) with \(a = f - g\). Choose a distinguished triangle \((Z, X, Q, t, d, h)\) using TR1. Since \(a \circ t = 0\) we see by Lemma 0149 there exists a morphism \(i : Q \to Y\) such that \(i \circ d = a\). Finally, using TR1 again we can choose a triangle \((Q, Y, W, i, j, k)\). Here is a picture \[\xymatrix{ Z \ar[r]_t & X \ar[r]_d \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^j \\ & & W }\] OK, and now we apply the functor \(F\) to this diagram. Since \(t \in S\) we see that \(F(Q) = 0\), see Lemma 05QR. Hence \(F(j)\) is an isomorphism by the same lemma, i.e., \(j \in S\). Finally, \(j \circ a = j \circ i \circ d = 0\) as \(j \circ i = 0\). Thus \(j \circ f = j \circ g\) and we see that LMS3 holds. The proof of RMS3 is dual.

Lemma

Let \(H : \mathcal{D} \to \mathcal{A}\) be a homological functor between a pre-triangulated category and an abelian category. Let \[S = \{f \in \text{Arrows}(\mathcal{D}) \mid H^i(f)\text{ is an isomorphism for all }i \in \mathbf{Z}\}\] Then \(S\) is a saturated (see Categories, Definition 05Q8) multiplicative system compatible with the triangulated structure on \(\mathcal{D}\).

Proof

We have to prove axioms MS1 – MS6, see Categories, Definitions 04VC and 05Q8 and Definition 05R2. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and the long exact cohomology sequence (0148). By Lemma 05R3 we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let \(f, g : X \to Y\) be morphisms of \(\mathcal{D}\), and let \(t : Z \to X\) be an element of \(S\) such that \(f \circ t = g \circ t\). As \(\mathcal{D}\) is additive this simply means that \(a \circ t = 0\) with \(a = f - g\). Choose a distinguished triangle \((Z, X, Q, t, u, v)\) using TR1 and TR2. Since \(a \circ t = 0\) we see by Lemma 0149 there exists a morphism \(i : Q \to Y\) such that \(i \circ u = a\). Finally, using TR1 again we can choose a triangle \((Q, Y, W, i, j, k)\). Here is a picture \[\xymatrix{ Z \ar[r]_t & X \ar[r]_u \ar[d]^1 & Q \ar[r]_v \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^j \\ & & W }\] OK, and now we apply the functors \(H^i\) to this diagram. Since \(t \in S\) we see that \(H^i(Q) = 0\) by the long exact cohomology sequence (0148). Hence \(H^i(j)\) is an isomorphism for all \(i\) by the same argument, i.e., \(j \in S\). Finally, \(j \circ a = j \circ i \circ u = 0\) as \(j \circ i = 0\). Thus \(j \circ f = j \circ g\) and we see that LMS3 holds. The proof of RMS3 is dual.

Proposition

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(S\) be a multiplicative system compatible with the triangulated structure. Then there exists a unique structure of a pre-triangulated category on \(S^{-1}\mathcal{D}\) such that \([1] \circ Q = Q \circ [1]\) and the localization functor \(Q : \mathcal{D} \to S^{-1}\mathcal{D}\) is exact. Moreover, if \(\mathcal{D}\) is a triangulated category, so is \(S^{-1}\mathcal{D}\).

Proof

We have seen that \(S^{-1}\mathcal{D}\) is an additive category and that the localization functor \(Q\) is additive in Homology, Lemma 05QE. It follows from MS5 that there is a unique additive auto-equivalence \([1] : S^{-1}\mathcal{D} \to S^{-1}\mathcal{D}\) such that \(Q \circ [1] = [1] \circ Q\) (equality of functors); we omit the details. We say a triangle of \(S^{-1}\mathcal{D}\) is distinguished if it is isomorphic to the image of a distinguished triangle under the localization functor \(Q\).

Proof of TR1. The only thing to prove here is that if \(a : Q(X) \to Q(Y)\) is a morphism of \(S^{-1}\mathcal{D}\), then \(a\) fits into a distinguished triangle. Write \(a = Q(s)^{-1} \circ Q(f)\) for some \(s : Y \to Y'\) in \(S\) and \(f : X \to Y'\). Choose a distinguished triangle \((X, Y', Z, f, g, h)\) in \(\mathcal{D}\). Then we see that \((Q(X), Q(Y), Q(Z), a, Q(g) \circ Q(s), Q(h))\) is a distinguished triangle of \(S^{-1}\mathcal{D}\).

Proof of TR2. This is immediate from the definitions.

Proof of TR3. Note that the existence of the dotted arrow which is required to exist may be proven after replacing the two triangles by isomorphic triangles. Hence we may assume given distinguished triangles \((X, Y, Z, f, g, h)\) and \((X', Y', Z', f', g', h')\) of \(\mathcal{D}\) and a commutative diagram \[\xymatrix{ Q(X) \ar[r]_{Q(f)} \ar[d]_a & Q(Y) \ar[d]^b \\ Q(X') \ar[r]^{Q(f')} & Q(Y') }\] in \(S^{-1}\mathcal{D}\). Now we apply Categories, Lemma 05Q3 to find a morphism \(f'' : X'' \to Y''\) in \(\mathcal{D}\) and a commutative diagram \[\xymatrix{ X \ar[d]_f \ar[r]_k & X'' \ar[d]^{f''} & X' \ar[d]^{f'} \ar[l]^s \\ Y \ar[r]^l & Y'' & Y' \ar[l]_t }\] in \(\mathcal{D}\) with \(s, t \in S\) and \(a = s^{-1}k\), \(b = t^{-1}l\). At this point we can use TR3 for \(\mathcal{D}\) and MS6 to find a commutative diagram \[\xymatrix{ X \ar[r] \ar[d]^k & Y \ar[r] \ar[d]^l & Z \ar[r] \ar[d]^m & X[1] \ar[d]^{k[1]} \\ X'' \ar[r] & Y'' \ar[r] & Z'' \ar[r] & X''[1] \\ X' \ar[r] \ar[u]_s & Y' \ar[r] \ar[u]_t & Z' \ar[r] \ar[u]_r & X'[1] \ar[u]_{s[1]} }\] with \(r \in S\). It follows that setting \(c = Q(r)^{-1}Q(m)\) we obtain the desired morphism of triangles \[\xymatrix{ (Q(X), Q(Y), Q(Z), Q(f), Q(g), Q(h)) \ar[d]^{(a, b, c)} \\ (Q(X'), Q(Y'), Q(Z'), Q(f'), Q(g'), Q(h')) }\]

This proves the first statement of the proposition. If \(\mathcal{D}\) is also a triangulated category, then we still have to prove TR4 in order to show that \(S^{-1}\mathcal{D}\) is triangulated as well. To do this we reduce by Lemma 014C to the following statement: Given composable morphisms \(a : Q(X) \to Q(Y)\) and \(b : Q(Y) \to Q(Z)\) we have to produce an octahedron after possibly replacing \(Q(X), Q(Y), Q(Z)\) by isomorphic objects. To do this we may first replace \(Y\) by an object such that \(a = Q(f)\) for some morphism \(f : X \to Y\) in \(\mathcal{D}\). (More precisely, write \(a = s^{-1}f\) with \(s : Y \to Y'\) in \(S\) and \(f : X \to Y'\). Then replace \(Y\) by \(Y'\).) After this we similarly replace \(Z\) by an object such that \(b = Q(g)\) for some morphism \(g : Y \to Z\). Now we can find distinguished triangles \((X, Y, Q_1, f, p_1, d_1)\), \((X, Z, Q_2, g \circ f, p_2, d_2)\), and \((Y, Z, Q_3, g, p_3, d_3)\) in \(\mathcal{D}\) (by TR1), and morphisms \(a : Q_1 \to Q_2\) and \(b : Q_2 \to Q_3\) as in TR4. Then it is immediately verified that applying the functor \(Q\) to all these data gives a corresponding structure in \(S^{-1}\mathcal{D}\).

The universal property of the localization of a triangulated category is as follows (we formulate this for pre-triangulated categories, hence it holds a fortiori for triangulated categories).

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(S\) be a multiplicative system compatible with the triangulated structure. Let \(Q : \mathcal{D} \to S^{-1}\mathcal{D}\) be the localization functor, see Proposition 05R6.

1. If \(H : \mathcal{D} \to \mathcal{A}\) is a homological functor into an abelian category \(\mathcal{A}\) such that \(H(s)\) is an isomorphism for all \(s \in S\), then the unique factorization \(H' : S^{-1}\mathcal{D} \to \mathcal{A}\) such that \(H = H' \circ Q\) (see Categories, Lemma 04VG) is a homological functor too.

2. If \(F : \mathcal{D} \to \mathcal{D}'\) is an exact functor into a pre-triangulated category \(\mathcal{D}'\) such that \(F(s)\) is an isomorphism for all \(s \in S\), then the unique factorization \(F' : S^{-1}\mathcal{D} \to \mathcal{D}'\) such that \(F = F' \circ Q\) (see Categories, Lemma 04VG) is an exact functor too.

Proof

This lemma proves itself. Details omitted.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category and let \(\mathcal{D}' \subset \mathcal{D}\) be a full, pre-triangulated subcategory. Let \(S\) be a saturated multiplicative system of \(\mathcal{D}\) compatible with the triangulated structure. Assume that for each \(X\) in \(\mathcal{D}\) there exists an \(s : X' \to X\) in \(S\) such that \(X'\) is an object of \(\mathcal{D}'\). Then \(S' = S \cap \text{Arrows}(\mathcal{D}')\) is a saturated multiplicative system compatible with the triangulated structure and the functor \[(S')^{-1}\mathcal{D}' \longrightarrow S^{-1}\mathcal{D}\] is an equivalence of pre-triangulated categories.

Proof

Consider the quotient functor \(Q : \mathcal{D} \to S^{-1}\mathcal{D}\) of Proposition 05R6. Since \(S\) is saturated we have that a morphism \(f : X \to Y\) is in \(S\) if and only if \(Q(f)\) is invertible, see Categories, Lemma 05Q9. Thus \(S'\) is the collection of arrows which are turned into isomorphisms by the composition \(\mathcal{D}' \to \mathcal{D} \to S^{-1}\mathcal{D}\). Hence \(S'\) is a saturated multiplicative system compatible with the triangulated structure by Lemma 05R4. By Lemma 05R7 we obtain the exact functor \((S')^{-1}\mathcal{D}' \to S^{-1}\mathcal{D}\) of pre-triangulated categories. By assumption this functor is essentially surjective. Let \(X', Y'\) be objects of \(\mathcal{D}'\). By Categories, Remark 05Q4 we have \[\Mor_{S^{-1}\mathcal{D}}(X', Y') = \colim_{s : X \to X'\text{ in }S} \Mor_\mathcal{D}(X, Y')\] Our assumption implies that for any \(s : X \to X'\) in \(S\) we can find a morphism \(s' : X'' \to X\) in \(S\) with \(X''\) in \(\mathcal{D}'\). Then \(s \circ s' : X'' \to X'\) is in \(S'\). Hence the colimit above is equal to \[\colim_{s'' : X'' \to X'\text{ in }S'} \Mor_{\mathcal{D}'}(X'', Y') = \Mor_{(S')^{-1}\mathcal{D}'}(X', Y')\] This proves our functor is also fully faithful and the proof is complete.

The following lemma describes the kernel (see Definition 05RF) of the localization functor.

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(S\) be a multiplicative system compatible with the triangulated structure. Let \(Z\) be an object of \(\mathcal{D}\). The following are equivalent

1. \(Q(Z) = 0\) in \(S^{-1}\mathcal{D}\),

2. there exists \(Z' \in \Ob(\mathcal{D})\) such that \(0 : Z \to Z'\) is an element of \(S\),

3. there exists \(Z' \in \Ob(\mathcal{D})\) such that \(0 : Z' \to Z\) is an element of \(S\), and

4. there exists an object \(Z'\) and a distinguished triangle \((X, Y, Z \oplus Z', f, g, h)\) such that \(f \in S\).

If \(S\) is saturated, then these are also equivalent to

1. the morphism \(0 \to Z\) is an element of \(S\),

2. the morphism \(Z \to 0\) is an element of \(S\),

3. there exists a distinguished triangle \((X, Y, Z, f, g, h)\) such that \(f \in S\).

Proof

The equivalence of (1), (2), and (3) is Homology, Lemma 05QF. If (2) holds, then \((Z'[-1], Z'[-1] \oplus Z, Z, (1, 0), (0, 1), 0)\) is a distinguished triangle (see Lemma 05QT) with “\(0 \in S\)”. By rotating we conclude that (4) holds. If \((X, Y, Z \oplus Z', f, g, h)\) is a distinguished triangle with \(f \in S\) then \(Q(f)\) is an isomorphism hence \(Q(Z \oplus Z') = 0\) hence \(Q(Z) = 0\). Thus (1) – (4) are all equivalent.

Next, assume that \(S\) is saturated. Note that each of (5), (6), (7) implies one of the equivalent conditions (1) – (4). Suppose that \(Q(Z) = 0\). Then \(0 \to Z\) is a morphism of \(\mathcal{D}\) which becomes an isomorphism in \(S^{-1}\mathcal{D}\). According to Categories, Lemma 05Q9 the fact that \(S\) is saturated implies that \(0 \to Z\) is in \(S\). Hence (1) \(\Rightarrow\) (5). Dually (1) \(\Rightarrow\) (6). Finally, if \(0 \to Z\) is in \(S\), then the triangle \((0, Z, Z, 0, \text{id}_Z, 0)\) is distinguished by TR1 and TR2 and is a triangle as in (7).

Lemma

Let \(\mathcal{D}\) be a pre-triangulated category. Let \(S\) be a saturated multiplicative system in \(\mathcal{D}\) that is compatible with the triangulated structure. Let \((X, Y, Z, f, g, h)\) be a distinguished triangle in \(\mathcal{D}\). Consider the category of morphisms of triangles \[\mathcal{I} = \{(s, s', s'') : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h') \mid s, s', s'' \in S\}\] Then \(\mathcal{I}\) is a filtered category and the functors \(\mathcal{I} \to X/S\), \(\mathcal{I} \to Y/S\), and \(\mathcal{I} \to Z/S\) are cofinal.

Proof

We strongly suggest the reader skip the proof of this lemma and instead work it out on a napkin.

The first remark is that using rotation of distinguished triangles (TR2) gives an equivalence of categories between \(\mathcal{I}\) and the corresponding category for the distinguished triangle \((Y, Z, X[1], g, h, -f[1])\). Using this we see for example that if we prove the functor \(\mathcal{I} \to X/S\) is cofinal, then the same thing is true for the functors \(\mathcal{I} \to Y/S\) and \(\mathcal{I} \to Z/S\).

Note that if \(s : X \to X'\) is a morphism of \(S\), then using MS2 we can find \(s' : Y \to Y'\) and \(f' : X' \to Y'\) such that \(f' \circ s = s' \circ f\), whereupon we can use MS6 to complete this into an object of \(\mathcal{I}\). Hence the functor \(\mathcal{I} \to X/S\) is surjective on objects. Using rotation as above this implies the same thing is true for the functors \(\mathcal{I} \to Y/S\) and \(\mathcal{I} \to Z/S\).

Suppose given objects \(s_1 : X \to X_1\) and \(s_2 : X \to X_2\) in \(X/S\) and a morphism \(a : X_1 \to X_2\) in \(X/S\). Since \(S\) is saturated, we see that \(a \in S\), see Categories, Lemma 05Q9. By the argument of the previous paragraph we can complete \(s_1 : X \to X_1\) to an object \((s_1, s'_1, s''_1) : (X, Y, Z, f, g, h) \to (X_1, Y_1, Z_1, f_1, g_1, h_1)\) in \(\mathcal{I}\). Then we can repeat and find \((a, b, c) : (X_1, Y_1, Z_1, f_1, g_1, h_1) \to (X_2, Y_2, Z_2, f_2, g_2, h_2)\) with \(a, b, c \in S\) completing the given \(a : X_1 \to X_2\). But then \((a, b, c)\) is a morphism in \(\mathcal{I}\). In this way we conclude that the functor \(\mathcal{I} \to X/S\) is also surjective on arrows. Using rotation as above, this implies the same thing is true for the functors \(\mathcal{I} \to Y/S\) and \(\mathcal{I} \to Z/S\).

The category \(\mathcal{I}\) is nonempty as the identity provides an object. This proves the condition (1) of the definition of a filtered category, see Categories, Definition 002V.

We check condition (2) of Categories, Definition 002V for the category \(\mathcal{I}\). Suppose given objects \((s_1, s'_1, s''_1) : (X, Y, Z, f, g, h) \to (X_1, Y_1, Z_1, f_1, g_1, h_1)\) and \((s_2, s'_2, s''_2) : (X, Y, Z, f, g, h) \to (X_2, Y_2, Z_2, f_2, g_2, h_2)\) in \(\mathcal{I}\). We want to find an object of \(\mathcal{I}\) which is the target of an arrow from both \((X_1, Y_1, Z_1, f_1, g_1, h_1)\) and \((X_2, Y_2, Z_2, f_2, g_2, h_2)\). By Categories, Remark 05Q0 the categories \(X/S\), \(Y/S\), \(Z/S\) are filtered. Thus we can find \(X \to X_3\) in \(X/S\) and morphisms \(s : X_2 \to X_3\) and \(a : X_1 \to X_3\). By the above we can find a morphism \((s, s', s'') : (X_2, Y_2, Z_2, f_2, g_2, h_2) \to (X_3, Y_3, Z_3, f_3, g_3, h_3)\) with \(s', s'' \in S\). After replacing \((X_2, Y_2, Z_2)\) by \((X_3, Y_3, Z_3)\) we may assume that there exists a morphism \(a : X_1 \to X_2\) in \(X/S\). Repeating the argument for \(Y\) and \(Z\) (by rotating as above) we may assume there is a morphism \(a : X_1 \to X_2\) in \(X/S\), \(b : Y_1 \to Y_2\) in \(Y/S\), and \(c : Z_1 \to Z_2\) in \(Z/S\). However, these morphisms do not necessarily give rise to a morphism of distinguished triangles. On the other hand, the necessary diagrams do commute in \(S^{-1}\mathcal{D}\). Hence we see (for example) that there exists a morphism \(s'_2 : Y_2 \to Y_3\) in \(S\) such that \(s'_2 \circ f_2 \circ a = s'_2 \circ b \circ f_1\). Another replacement of \((X_2, Y_2, Z_2)\) as above then gets us to the situation where \(f_2 \circ a = b \circ f_1\). Rotating and applying the same argument two more times we see that we may assume \((a, b, c)\) is a morphism of triangles. This proves condition (2).

Next we check condition (3) of Categories, Definition 002V. Suppose \((s_1, s_1', s_1'') : (X, Y, Z) \to (X_1, Y_1, Z_1)\) and \((s_2, s_2', s_2'') : (X, Y, Z) \to (X_2, Y_2, Z_2)\) are objects of \(\mathcal{I}\), and suppose \((a, b, c), (a', b', c')\) are two morphisms between them. Since \(a \circ s_1 = a' \circ s_1\) there exists a morphism \(s_3 : X_2 \to X_3\) such that \(s_3 \circ a = s_3 \circ a'\). Using the surjectivity statement we can complete this to a morphism of triangles \((s_3, s_3', s_3'') : (X_2, Y_2, Z_2) \to (X_3, Y_3, Z_3)\) with \(s_3, s_3', s_3'' \in S\). Thus \((s_3 \circ s_2, s_3' \circ s_2', s_3'' \circ s_2'') : (X, Y, Z) \to (X_3, Y_3, Z_3)\) is also an object of \(\mathcal{I}\) and after composing the maps \((a, b, c), (a', b', c')\) with \((s_3, s_3', s_3'')\) we obtain \(a = a'\). By rotating we may do the same to get \(b = b'\) and \(c = c'\).

Finally, we check that \(\mathcal{I} \to X/S\) is cofinal, see Categories, Definition 04E6. The first condition is true as the functor is surjective. Suppose that we have an object \(s : X \to X'\) in \(X/S\) and two objects \((s_1, s'_1, s''_1) : (X, Y, Z, f, g, h) \to (X_1, Y_1, Z_1, f_1, g_1, h_1)\) and \((s_2, s'_2, s''_2) : (X, Y, Z, f, g, h) \to (X_2, Y_2, Z_2, f_2, g_2, h_2)\) in \(\mathcal{I}\) as well as morphisms \(t_1 : X' \to X_1\) and \(t_2 : X' \to X_2\) in \(X/S\). By property (2) of \(\mathcal{I}\) proved above we can find morphisms \((s_3, s'_3, s''_3) : (X_1, Y_1, Z_1, f_1, g_1, h_1) \to (X_3, Y_3, Z_3, f_3, g_3, h_3)\) and \((s_4, s'_4, s''_4) : (X_2, Y_2, Z_2, f_2, g_2, h_2) \to (X_3, Y_3, Z_3, f_3, g_3, h_3)\) in \(\mathcal{I}\). We would be done if the compositions \(X' \to X_1 \to X_3\) and \(X' \to X_2 \to X_3\) were equal (see displayed equation in Categories, Definition 04E6). If not, then, because \(X/S\) is filtered, we can choose a morphism \(X_3 \to X_4\) in \(X/S\) such that the compositions \(X' \to X_1 \to X_3 \to X_4\) and \(X' \to X_2 \to X_3 \to X_4\) are equal. Then we finally complete \(X_3 \to X_4\) to a morphism \((X_3, Y_3, Z_3) \to (X_4, Y_4, Z_4)\) in \(\mathcal{I}\) and compose with that morphism to see that the result is true.

Quotients of triangulated categories

Given a triangulated category and a triangulated subcategory we can construct another triangulated category by taking the “quotient”. The construction uses a localization. This is similar to the quotient of an abelian category by a Serre subcategory, see Homology, Section 02MN. Before we do the actual construction we briefly discuss kernels of exact functors.

Definition

Let \(\mathcal{D}\) be a pre-triangulated category. We say a full pre-triangulated subcategory \(\mathcal{D}'\) of \(\mathcal{D}\) is saturated if whenever \(X \oplus Y\) is isomorphic to an object of \(\mathcal{D}'\) then both \(X\) and \(Y\) are isomorphic to objects of \(\mathcal{D}'\).

A saturated triangulated subcategory is sometimes called a thick triangulated subcategory. In some references, this is only used for strictly full triangulated subcategories (and sometimes the definition is written such that it implies strictness). There is another notion, that of an épaisse triangulated subcategory. The definition is that given a commutative diagram \[\xymatrix{ & S \ar[rd] \\ X \ar[ru] \ar[rr] & & Y \ar[r] & T \ar[r] & X[1] }\] where the second line is a distinguished triangle and \(S\) and \(T\) are isomorphic to objects of \(\mathcal{D}'\), then also \(X\) and \(Y\) are isomorphic to objects of \(\mathcal{D}'\). It turns out that this is equivalent to being saturated (this is elementary and can be found in [Rickard-derived]) and the notion of a saturated category is easier to work with.

Lemma

Let \(F : \mathcal{D} \to \mathcal{D}'\) be an exact functor of pre-triangulated categories. Let \(\mathcal{D}''\) be the full subcategory of \(\mathcal{D}\) with objects \[\Ob(\mathcal{D}'') = \{X \in \Ob(\mathcal{D}) \mid F(X) = 0\}\] Then \(\mathcal{D}''\) is a strictly full saturated pre-triangulated subcategory of \(\mathcal{D}\). If \(\mathcal{D}\) is a triangulated category, then \(\mathcal{D}''\) is a triangulated subcategory.

Proof

It is clear that \(\mathcal{D}''\) is preserved under \([1]\) and \([-1]\). If \((X, Y, Z, f, g, h)\) is a distinguished triangle of \(\mathcal{D}\) and \(F(X) = F(Y) = 0\), then also \(F(Z) = 0\) as \((F(X), F(Y), F(Z), F(f), F(g), F(h))\) is distinguished. Hence we may apply Lemma 05QX to see that \(\mathcal{D}''\) is a pre-triangulated subcategory (respectively a triangulated subcategory if \(\mathcal{D}\) is a triangulated category). The final assertion of being saturated follows from \(F(X) \oplus F(Y) = 0 \Rightarrow F(X) = F(Y) = 0\).

Lemma

Let \(H : \mathcal{D} \to \mathcal{A}\) be a homological functor of a pre-triangulated category into an abelian category. Let \(\mathcal{D}'\) be the full subcategory of \(\mathcal{D}\) with objects \[\Ob(\mathcal{D}') = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \in \mathbf{Z}\}\] Then \(\mathcal{D}'\) is a strictly full saturated pre-triangulated subcategory of \(\mathcal{D}\). If \(\mathcal{D}\) is a triangulated category, then \(\mathcal{D}'\) is a triangulated subcategory.

Proof

It is clear that \(\mathcal{D}'\) is preserved under \([1]\) and \([-1]\). If \((X, Y, Z, f, g, h)\) is a distinguished triangle of \(\mathcal{D}\) and \(H(X[n]) = H(Y[n]) = 0\) for all \(n\), then also \(H(Z[n]) = 0\) for all \(n\) by the long exact sequence (0148). Hence we may apply Lemma 05QX to see that \(\mathcal{D}'\) is a pre-triangulated subcategory (respectively a triangulated subcategory if \(\mathcal{D}\) is a triangulated category). The assertion of being saturated follows from \[\begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*}\] for all \(n \in \mathbf{Z}\).

Lemma

Let \(H : \mathcal{D} \to \mathcal{A}\) be a homological functor of a pre-triangulated category into an abelian category. Let \(\mathcal{D}_H^{+}, \mathcal{D}_H^{-}, \mathcal{D}_H^b\) be the full subcategory of \(\mathcal{D}\) with objects \[\begin{matrix} \Ob(\mathcal{D}_H^{+}) = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \ll 0\} \\ \Ob(\mathcal{D}_H^{-}) = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \gg 0\} \\ \Ob(\mathcal{D}_H^b) = \{X \in \Ob(\mathcal{D}) \mid H(X[n]) = 0\text{ for all }|n| \gg 0\} \end{matrix}\] Each of these is a strictly full saturated pre-triangulated subcategory of \(\mathcal{D}\). If \(\mathcal{D}\) is a triangulated category, then each is a triangulated subcategory.

Proof

Let us prove this for \(\mathcal{D}_H^{+}\). It is clear that it is preserved under \([1]\) and \([-1]\). If \((X, Y, Z, f, g, h)\) is a distinguished triangle of \(\mathcal{D}\) and \(H(X[n]) = H(Y[n]) = 0\) for all \(n \ll 0\), then also \(H(Z[n]) = 0\) for all \(n \ll 0\) by the long exact sequence (0148). Hence we may apply Lemma 05QX to see that \(\mathcal{D}_H^{+}\) is a pre-triangulated subcategory (respectively a triangulated subcategory if \(\mathcal{D}\) is a triangulated category). The assertion of being saturated follows from \[\begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*}\] for all \(n \in \mathbf{Z}\).

Definition

Let \(\mathcal{D}\) be a (pre-)triangulated category.

1. Let \(F : \mathcal{D} \to \mathcal{D}'\) be an exact functor. The kernel of \(F\) is the strictly full saturated (pre-)triangulated subcategory described in Lemma 05RC.

2. Let \(H : \mathcal{D} \to \mathcal{A}\) be a homological functor. The kernel of \(H\) is the strictly full saturated (pre-)triangulated subcategory described in Lemma 05RD.

These are sometimes denoted \(\Ker(F)\) or \(\Ker(H)\).

The proof of the following lemma uses TR4.

Lemma

Let \(\mathcal{D}\) be a triangulated category. Let \(\mathcal{D}' \subset \mathcal{D}\) be a full triangulated subcategory. Set [05RH]\[\begin{equation} S = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }\\ (X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with } Z\text{ isomorphic to an object of }\mathcal{D}' \end{matrix} \right\} \end{equation}\] Then \(S\) is a multiplicative system compatible with the triangulated structure on \(\mathcal{D}\). In this situation the following are equivalent

1. \(S\) is a saturated multiplicative system,

2. \(\mathcal{D}'\) is a saturated triangulated subcategory.

Proof

To prove the first assertion we have to prove that MS1, MS2, MS3 and MS5, MS6 hold.

Proof of MS1. It is clear that identities are in \(S\) because \((X, X, 0, 1, 0, 0)\) is distinguished for every object \(X\) of \(\mathcal{D}\) and because \(0\) is an object of \(\mathcal{D}'\). Let \(f : X \to Y\) and \(g : Y \to Z\) be composable morphisms contained in \(S\). Choose distinguished triangles \((X, Y, Q_1, f, p_1, d_1)\), \((X, Z, Q_2, g \circ f, p_2, d_2)\), and \((Y, Z, Q_3, g, p_3, d_3)\). By assumption we know that \(Q_1\) and \(Q_3\) are isomorphic to objects of \(\mathcal{D}'\). By TR4 we know there exists a distinguished triangle \((Q_1, Q_2, Q_3, a, b, c)\). Since \(\mathcal{D}'\) is a triangulated subcategory we conclude that \(Q_2\) is isomorphic to an object of \(\mathcal{D}'\). Hence \(g \circ f \in S\).

Proof of MS3. Let \(a : X \to Y\) be a morphism and let \(t : Z \to X\) be an element of \(S\) such that \(a \circ t = 0\). To prove LMS3 it suffices to find an \(s \in S\) such that \(s \circ a = 0\), compare with the proof of Lemma 05R4. Choose a distinguished triangle \((Z, X, Q, t, g, h)\) using TR1 and TR2. Since \(a \circ t = 0\) we see by Lemma 0149 there exists a morphism \(i : Q \to Y\) such that \(i \circ g = a\). Finally, using TR1 again we can choose a triangle \((Q, Y, W, i, s, k)\). Here is a picture \[\xymatrix{ Z \ar[r]_t & X \ar[r]_g \ar[d]^1 & Q \ar[r] \ar[d]^i & Z[1] \\ & X \ar[r]_a & Y \ar[d]^s \\ & & W }\] Since \(t \in S\) we see that \(Q\) is isomorphic to an object of \(\mathcal{D}'\). Hence \(s \in S\). Finally, \(s \circ a = s \circ i \circ g = 0\) as \(s \circ i = 0\) by Lemma 0146. We conclude that LMS3 holds. The proof of RMS3 is dual.

Proof of MS5. Follows as distinguished triangles and \(\mathcal{D}'\) are stable under translations

Proof of MS6. Suppose given a commutative diagram \[\xymatrix{ X \ar[r] \ar[d]^s & Y \ar[d]^{s'} \\ X' \ar[r] & Y' }\] with \(s, s' \in S\). By Proposition 05R0 we can extend this to a nine square diagram. As \(s, s'\) are elements of \(S\) we see that \(X'', Y''\) are isomorphic to objects of \(\mathcal{D}'\). Since \(\mathcal{D}'\) is a full triangulated subcategory we see that \(Z''\) is also isomorphic to an object of \(\mathcal{D}'\). Whence the morphism \(Z \to Z'\) is an element of \(S\). This proves MS6.

MS2 is a formal consequence of MS1, MS5, and MS6, see Lemma 05R3. This finishes the proof of the first assertion of the lemma.

Let’s assume that \(S\) is saturated. (In the following we will use rotation of distinguished triangles without further mention.) Let \(X \oplus Y\) be an object isomorphic to an object of \(\mathcal{D}'\). Consider the morphism \(f : 0 \to X\). The composition \(0 \to X \to X \oplus Y\) is an element of \(S\) as \((0, X \oplus Y, X \oplus Y, 0, 1, 0)\) is a distinguished triangle. The composition \(Y[-1] \to 0 \to X\) is an element of \(S\) as \((X, X \oplus Y, Y, (1, 0), (0, 1), 0)\) is a distinguished triangle, see Lemma 05QT. Hence \(0 \to X\) is an element of \(S\) (as \(S\) is saturated). Thus \(X\) is isomorphic to an object of \(\mathcal{D}'\) as desired.

Finally, assume \(\mathcal{D}'\) is a saturated triangulated subcategory. Let \[W \xrightarrow{h} X \xrightarrow{g} Y \xrightarrow{f} Z\] be composable morphisms of \(\mathcal{D}\) such that \(fg, gh \in S\). We will build up a picture of objects as in the diagram below. \[\xymatrix{ & & Q_{12} \ar[rd] & & Q_{23} \ar[rd] \\ & Q_1 \ar[ld]_{\! + \! 1} \ar[ru] & & Q_2 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \ar[ru] & & Q_3 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \\ W \ar[rr] & & X \ar[lu] \ar[rr] & & Y \ar[lu] \ar[rr] & & Z \ar[lu] }\] First choose distinguished triangles \((W, X, Q_1)\), \((X, Y, Q_2)\), \((Y, Z, Q_3)\) \((W, Y, Q_{12})\), and \((X, Z, Q_{23})\). Denote \(s : Q_2 \to Q_1[1]\) the composition \(Q_2 \to X[1] \to Q_1[1]\). Denote \(t : Q_3 \to Q_2[1]\) the composition \(Q_3 \to Y[1] \to Q_2[1]\). By TR4 applied to the composition \(W \to X \to Y\) and the composition \(X \to Y \to Z\) there exist distinguished triangles \((Q_1, Q_{12}, Q_2)\) and \((Q_2, Q_{23}, Q_3)\) which use the morphisms \(s\) and \(t\). The objects \(Q_{12}\) and \(Q_{23}\) are isomorphic to objects of \(\mathcal{D}'\) as \(W \to Y\) and \(X \to Z\) are assumed in \(S\). Hence also \(s[1]t\) is an element of \(S\) as \(S\) is closed under compositions and shifts. Note that \(s[1]t = 0\) as \(Y[1] \to Q_2[1] \to X[2]\) is zero, see Lemma 0146. Hence \(Q_3[1] \oplus Q_1[2]\) is isomorphic to an object of \(\mathcal{D}'\), see Lemma 05QT. By assumption on \(\mathcal{D}'\) we conclude that \(Q_3\) and \(Q_1\) are isomorphic to objects of \(\mathcal{D}'\). Looking at the distinguished triangle \((Q_1, Q_{12}, Q_2)\) we conclude that \(Q_2\) is also isomorphic to an object of \(\mathcal{D}'\). Looking at the distinguished triangle \((X, Y, Q_2)\) we finally conclude that \(g \in S\). (It also follows that \(h, f \in S\), but we don’t need this.)

Definition

Let \(\mathcal{D}\) be a triangulated category. Let \(\mathcal{B}\) be a full triangulated subcategory. We define the quotient category \(\mathcal{D}/\mathcal{B}\) by the formula \(\mathcal{D}/\mathcal{B} = S^{-1}\mathcal{D}\), where \(S\) is the multiplicative system of \(\mathcal{D}\) associated to \(\mathcal{B}\) via Lemma 05RG. The localization functor \(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}\) is called the quotient functor in this case.

Note that the quotient functor \(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}\) is an exact functor of triangulated categories, see Proposition 05R6. The universal property of this construction is the following.

Lemma

Let \(\mathcal{D}\) be a triangulated category. Let \(\mathcal{B}\) be a full triangulated subcategory of \(\mathcal{D}\). Let \(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}\) be the quotient functor.

1. If \(H : \mathcal{D} \to \mathcal{A}\) is a homological functor into an abelian category \(\mathcal{A}\) such that \(\mathcal{B} \subset \Ker(H)\) then there exists a unique factorization \(H' : \mathcal{D}/\mathcal{B} \to \mathcal{A}\) such that \(H = H' \circ Q\) and \(H'\) is a homological functor too.

2. If \(F : \mathcal{D} \to \mathcal{D}'\) is an exact functor into a pre-triangulated category \(\mathcal{D}'\) such that \(\mathcal{B} \subset \Ker(F)\) then there exists a unique factorization \(F' : \mathcal{D}/\mathcal{B} \to \mathcal{D}'\) such that \(F = F' \circ Q\) and \(F'\) is an exact functor too.

Proof

This lemma follows from Lemma 05R7. Namely, if \(f : X \to Y\) is a morphism of \(\mathcal{D}\) such that for some distinguished triangle \((X, Y, Z, f, g, h)\) the object \(Z\) is isomorphic to an object of \(\mathcal{B}\), then \(H(f)\), resp. \(F(f)\) is an isomorphism under the assumptions of (1), resp. (2). Details omitted.

The kernel of the quotient functor can be described as follows.

Lemma

Let \(\mathcal{D}\) be a triangulated category. Let \(\mathcal{B}\) be a full triangulated subcategory. The kernel of the quotient functor \(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}\) is the strictly full subcategory of \(\mathcal{D}\) whose objects are \[\Ob(\Ker(Q)) = \left\{ \begin{matrix} Z \in \Ob(\mathcal{D}) \text{ such that there exists a }Z' \in \Ob(\mathcal{D}) \\ \text{ such that }Z \oplus Z'\text{ is isomorphic to an object of }\mathcal{B} \end{matrix} \right\}\] In other words it is the smallest strictly full saturated triangulated subcategory of \(\mathcal{D}\) containing \(\mathcal{B}\).

Proof

First note that the kernel is automatically a strictly full triangulated subcategory containing summands of any of its objects, see Lemma 05RC. The description of its objects follows from the definitions and Lemma 05R8 part (4).

Let \(\mathcal{D}\) be a triangulated category. At this point we have constructions which induce order preserving maps between

1. the partially ordered set of multiplicative systems \(S\) in \(\mathcal{D}\) compatible with the triangulated structure, and

2. the partially ordered set of full triangulated subcategories \(\mathcal{B} \subset \mathcal{D}\).

Namely, the constructions are given by \(S \mapsto \mathcal{B}(S) = \Ker(Q : \mathcal{D} \to S^{-1}\mathcal{D})\) and \(\mathcal{B} \mapsto S(\mathcal{B})\) where \(S(\mathcal{B})\) is the multiplicative set of (05RH), i.e., \[S(\mathcal{B}) = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }\\ (X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with } Z\text{ isomorphic to an object of }\mathcal{B} \end{matrix} \right\}\] Note that it is not the case that these operations are mutually inverse.

Lemma

Let \(\mathcal{D}\) be a triangulated category. The operations described above have the following properties

1. \(S(\mathcal{B}(S))\) is the “saturation” of \(S\), i.e., it is the smallest saturated multiplicative system in \(\mathcal{D}\) containing \(S\), and

2. \(\mathcal{B}(S(\mathcal{B}))\) is the “saturation” of \(\mathcal{B}\), i.e., it is the smallest strictly full saturated triangulated subcategory of \(\mathcal{D}\) containing \(\mathcal{B}\).

In particular, the constructions define mutually inverse maps between the (partially ordered) set of saturated multiplicative systems in \(\mathcal{D}\) compatible with the triangulated structure on \(\mathcal{D}\) and the (partially ordered) set of strictly full saturated triangulated subcategories of \(\mathcal{D}\).

Proof

First, let’s start with a full triangulated subcategory \(\mathcal{B}\). Then \(\mathcal{B}(S(\mathcal{B})) = \Ker(Q : \mathcal{D} \to \mathcal{D}/\mathcal{B})\) and hence (2) is the content of Lemma 05RK.

Next, suppose that \(S\) is a multiplicative system in \(\mathcal{D}\) compatible with the triangulation on \(\mathcal{D}\). Then \(\mathcal{B}(S) = \Ker(Q : \mathcal{D} \to S^{-1}\mathcal{D})\). Hence (using Lemma 05QR in the localized category) \[\begin{align*} S(\mathcal{B}(S)) & = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished}\\ \text{triangle }(X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with }Q(Z) = 0 \end{matrix} \right\}. \\ & = \{f \in \text{Arrows}(\mathcal{D}) \mid Q(f)\text{ is an isomorphism}\} \\ & = \hat S = S' \end{align*}\] in the notation of Categories, Lemma 05Q9. The final statement of that lemma finishes the proof.

Lemma

Let \(H : \mathcal{D} \to \mathcal{A}\) be a homological functor from a triangulated category \(\mathcal{D}\) to an abelian category \(\mathcal{A}\), see Definition 0147. The subcategory \(\Ker(H)\) of \(\mathcal{D}\) is a strictly full saturated triangulated subcategory of \(\mathcal{D}\) whose corresponding saturated multiplicative system (see Lemma 05RL) is the set \[S = \{f \in \text{Arrows}(\mathcal{D}) \mid H^i(f)\text{ is an isomorphism for all }i \in \mathbf{Z}\}.\] The functor \(H\) factors through the quotient functor \(Q : \mathcal{D} \to \mathcal{D}/\Ker(H)\).

Proof

The category \(\Ker(H)\) is a strictly full saturated triangulated subcategory of \(\mathcal{D}\) by Lemma 05RD. The set \(S\) is a saturated multiplicative system compatible with the triangulated structure by Lemma 05R5. Recall that the multiplicative system corresponding to \(\Ker(H)\) is the set \[\left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }\\ (X, Y, Z, f, g, h)\text{ with } H^i(Z) = 0 \text{ for all }i \end{matrix} \right\}.\] By the long exact cohomology sequence, see (0148), it is clear that \(f\) is an element of this set if and only if \(f\) is an element of \(S\). Finally, the factorization of \(H\) through \(Q\) is a consequence of Lemma 05RJ.

Adjoints for exact functors

Results on adjoint functors between triangulated categories.

Lemma

Let \(F : \mathcal{D} \to \mathcal{D}'\) be an exact functor between triangulated categories. If \(F\) admits a right adjoint \(G: \mathcal{D'} \to \mathcal{D}\), then \(G\) is also an exact functor.

Proof

Let \(\xi_X : F(X[1]) \to F(X)[1]\) be as in Definition 014V. Let \(\epsilon_A : F(G(A)) \to A\) be the adjunction map. Consider the composition \[F(G(A)[1]) \xrightarrow{\xi_{G(A)}} F(G(A))[1] \xrightarrow{\epsilon_A[1]} A[1]\] This map is adjoint to a map \(G(A)[1] \to G(A[1])\) which we claim to be an isomorphism. To see this, by the Yoneda lemma it suffices to show that we get a bijection on applying \(\Mor_\mathcal{D}(X[1], -)\) for every object \(X\) of \(\mathcal{D}\). Now, every morphism \(f : X[1] \to G(A)[1]\) is of the form \(g[1]\) for a unique \(g : X \to G(A)\) and every \(g\) is adjoint to a unique \(h : F(X) \to A\) and in turn \(h[1] \circ \xi_X\) is adjoint to a unique \(i : X[1] \to G(A[1])\). The rule sending \(f\) to \(i\) gives a bijection between \(\Mor_\mathcal{D}(X[1], G(A)[1])\) and \(\Mor_\mathcal{D}(X[1], G(A[1]))\). That this bijection is given by the morphism above follows from the discussion in Categories, Section 0036 and the following calculation \[\begin{align*} \epsilon_A[1] \circ \xi_{G(A)} \circ F(f) & = \epsilon_A[1] \circ \xi_{G(A)} \circ F(g[1]) \\ & = \epsilon_A[1] \circ F(g)[1] \circ \xi_X \\ & = (\epsilon_A \circ F(g))[1] \circ \xi_X \\ & = h[1] \circ \xi_X \end{align*}\] Some details omitted. We will show that \(G\) is an exact functor using the inverse of the isomorphisms \(G(A)[1] \to G(A[1])\). These isomorphisms are functorial in \(A\) (details omitted).

Let \(A \to B \to C \to A[1]\) be a distinguished triangle in \(\mathcal{D}'\). Choose a distinguished triangle \[G(A) \to G(B) \to X \to G(A)[1]\] in \(\mathcal{D}\). Then \(F(G(A)) \to F(G(B)) \to F(X) \to F(G(A))[1]\) is a distinguished triangle in \(\mathcal{D}'\). By TR3 we can choose a morphism of distinguished triangles \[\xymatrix{ F(G(A)) \ar[r] \ar[d]^{\epsilon_A} & F(G(B)) \ar[r] \ar[d]^{\epsilon_B} & F(X) \ar[r] \ar[d] & F(G(A))[1] \ar[d]^{\epsilon_A[1]} \\ A \ar[r] & B \ar[r] & C \ar[r] & A[1] }\] Recall that the arrow \(F(X) \to F(G(A))[1]\) is the composition of \(F(X) \to F(G(A)[1])\) with \(\xi_{G(A)}\). Hence using that \(G\) is the right adjoint of \(F\) and our discussion above, we conclude the existence of a morphism \(X \to G(C)\) such that the diagram \[\xymatrix{ G(A) \ar[r] \ar[d] & G(B) \ar[r] \ar[d] & X \ar[r] \ar[d] & G(A)[1] \ar[d] \\ G(A) \ar[r] & G(B) \ar[r] & G(C) \ar[r] & G(A[1]) }\] commutes and the right vertical arrow is the morphism constructed above. Applying the homological functor \(\Hom_{\mathcal{D}}(W, -)\) for an object \(W\) of \(\mathcal{D}\) we deduce from the \(5\) lemma that \[\Hom_{\mathcal{D}}(W, X) \to \Hom_{\mathcal{D}}(W, G(C))\] is a bijection and using the Yoneda lemma once more we conclude that \(X \to G(C)\) is an isomorphism. Hence we conclude that \(G(A) \to G(B) \to G(C) \to G(A)[1]\) is a distinguished triangle which is what we wanted to show.

Lemma

Let \(\mathcal{D}\), \(\mathcal{D}'\) be triangulated categories. Let \(F : \mathcal{D} \to \mathcal{D}'\) and \(G : \mathcal{D}' \to \mathcal{D}\) be functors. Assume that

1. \(F\) and \(G\) are exact functors,

2. \(F\) is fully faithful,

3. \(G\) is a right adjoint to \(F\), and

4. the kernel of \(G\) is zero.

Then \(F\) is an equivalence of categories.

Proof

Since \(F\) is fully faithful the adjunction map \(\text{id} \to G \circ F\) is an isomorphism (Categories, Lemma 07RB). Let \(X\) be an object of \(\mathcal{D}'\). Choose a distinguished triangle \[F(G(X)) \to X \to Y \to F(G(X))[1]\] in \(\mathcal{D}'\). Applying \(G\) and using that \(G(F(G(X))) = G(X)\) we find a distinguished triangle \[G(X) \to G(X) \to G(Y) \to G(X)[1]\] Hence \(G(Y) = 0\). Thus \(Y = 0\). Thus \(F(G(X)) \to X\) is an isomorphism.

The homotopy category

Let \(\mathcal{A}\) be an additive category. The homotopy category \(K(\mathcal{A})\) of \(\mathcal{A}\) is the category of complexes of \(\mathcal{A}\) with morphisms given by morphisms of complexes up to homotopy. Here is the formal definition.

Definition

Let \(\mathcal{A}\) be an additive category.

1. We let \(\text{Comp}(\mathcal{A}) = \text{CoCh}(\mathcal{A})\) be the category of (cochain) complexes.

2. A complex \(K^\bullet\) is said to be bounded below if \(K^n = 0\) for all \(n \ll 0\).

3. A complex \(K^\bullet\) is said to be bounded above if \(K^n = 0\) for all \(n \gg 0\).

4. A complex \(K^\bullet\) is said to be bounded if \(K^n = 0\) for all \(|n| \gg 0\).

5. We let \(\text{Comp}^{+}(\mathcal{A})\), \(\text{Comp}^{-}(\mathcal{A})\), resp. \(\text{Comp}^b(\mathcal{A})\) be the full subcategory of \(\text{Comp}(\mathcal{A})\) whose objects are the complexes which are bounded below, bounded above, resp. bounded.

6. We let \(K(\mathcal{A})\) be the category with the same objects as \(\text{Comp}(\mathcal{A})\) but as morphisms homotopy classes of maps of complexes (see Homology, Lemma 0112).

7. We let \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), resp. \(K^b(\mathcal{A})\) be the full subcategory of \(K(\mathcal{A})\) whose objects are bounded below, bounded above, resp. bounded complexes of \(\mathcal{A}\).

It will turn out that the categories \(K(\mathcal{A})\), \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), and \(K^b(\mathcal{A})\) are triangulated categories. To prove this we first develop some machinery related to cones and split exact sequences.

Cones and termwise split sequences

Let \(\mathcal{A}\) be an additive category, and let \(K(\mathcal{A})\) denote the category of complexes of \(\mathcal{A}\) with morphisms given by morphisms of complexes up to homotopy. Note that the shift functors \([n]\) on complexes, see Homology, Definition 011G, give rise to functors \([n] : K(\mathcal{A}) \to K(\mathcal{A})\) such that \([n] \circ [m] = [n + m]\) and \([0] = \text{id}\).

Definition

Let \(\mathcal{A}\) be an additive category. Let \(f : K^\bullet \to L^\bullet\) be a morphism of complexes of \(\mathcal{A}\). The cone of \(f\) is the complex \(C(f)^\bullet\) given by \(C(f)^n = L^n \oplus K^{n + 1}\) and differential \[d_{C(f)}^n = \left( \begin{matrix} d^n_L & f^{n + 1} \\ 0 & -d_K^{n + 1} \end{matrix} \right)\] It comes equipped with canonical morphisms of complexes \(i : L^\bullet \to C(f)^\bullet\) and \(p : C(f)^\bullet \to K^\bullet[1]\) induced by the obvious maps \(L^n \to C(f)^n \to K^{n + 1}\).

In other words \((K, L, C(f), f, i, p)\) forms a triangle: \[K^\bullet \to L^\bullet \to C(f)^\bullet \to K^\bullet[1]\] The formation of this triangle is functorial in the following sense.

Lemma

Suppose that \[\xymatrix{ K_1^\bullet \ar[r]_{f_1} \ar[d]_a & L_1^\bullet \ar[d]^b \\ K_2^\bullet \ar[r]^{f_2} & L_2^\bullet }\] is a diagram of morphisms of complexes which is commutative up to homotopy. Then there exists a morphism \(c : C(f_1)^\bullet \to C(f_2)^\bullet\) which gives rise to a morphism of triangles \((a, b, c) : (K_1^\bullet, L_1^\bullet, C(f_1)^\bullet, f_1, i_1, p_1) \to (K_2^\bullet, L_2^\bullet, C(f_2)^\bullet, f_2, i_2, p_2)\) of \(K(\mathcal{A})\).

Proof

Let \(h^n : K_1^n \to L_2^{n - 1}\) be a family of morphisms such that \(b \circ f_1 - f_2 \circ a= d \circ h + h \circ d\). Define \(c^n\) by the matrix \[c^n = \left( \begin{matrix} b^n & h^{n + 1} \\ 0 & a^{n + 1} \end{matrix} \right) : L_1^n \oplus K_1^{n + 1} \to L_2^n \oplus K_2^{n + 1}\] A matrix computation shows that \(c\) is a morphism of complexes. It is trivial that \(c \circ i_1 = i_2 \circ b\), and it is trivial also to check that \(p_2 \circ c = a \circ p_1\).

Note that the morphism \(c : C(f_1)^\bullet \to C(f_2)^\bullet\) constructed in the proof of Lemma 014F in general depends on the chosen homotopy \(h\) between \(f_2 \circ a\) and \(b \circ f_1\).

Lemma

Suppose that \(f: K^\bullet \to L^\bullet\) and \(g : L^\bullet \to M^\bullet\) are morphisms of complexes such that \(g \circ f\) is homotopic to zero. Then

1. \(g\) factors through a morphism \(C(f)^\bullet \to M^\bullet\), and

2. \(f\) factors through a morphism \(K^\bullet \to C(g)^\bullet[-1]\).

Proof

The assumptions say that the diagram \[\xymatrix{ K^\bullet \ar[r]_f \ar[d] & L^\bullet \ar[d]^g \\ 0 \ar[r] & M^\bullet }\] commutes up to homotopy. Since the cone on \(0 \to M^\bullet\) is \(M^\bullet\) the map \(C(f)^\bullet \to C(0 \to M^\bullet) = M^\bullet\) of Lemma 014F is the map in (1). The cone on \(K^\bullet \to 0\) is \(K^\bullet[1]\) and applying Lemma 014F gives a map \(K^\bullet[1] \to C(g)^\bullet\). Applying \([-1]\) we obtain the map in (2).

Note that the morphisms \(C(f)^\bullet \to M^\bullet\) and \(K^\bullet \to C(g)^\bullet[-1]\) constructed in the proof of Lemma 08RI in general depend on the chosen homotopy.

Definition

Let \(\mathcal{A}\) be an additive category. A termwise split injection \(\alpha : A^\bullet \to B^\bullet\) is a morphism of complexes such that each \(A^n \to B^n\) is isomorphic to the inclusion of a direct summand. A termwise split surjection \(\beta : B^\bullet \to C^\bullet\) is a morphism of complexes such that each \(B^n \to C^n\) is isomorphic to the projection onto a direct summand.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \[\xymatrix{ A^\bullet \ar[r]_f \ar[d]_a & B^\bullet \ar[d]^b \\ C^\bullet \ar[r]^g & D^\bullet }\] be a diagram of morphisms of complexes commuting up to homotopy. If \(f\) is a termwise split injection, then \(b\) is homotopic to a morphism which makes the diagram commute. If \(g\) is a termwise split surjection, then \(a\) is homotopic to a morphism which makes the diagram commute.

Proof

Let \(h^n : A^n \to D^{n - 1}\) be a collection of morphisms such that \(bf - ga = dh + hd\). Suppose that \(\pi^n : B^n \to A^n\) are morphisms splitting the morphisms \(f^n\). Take \(b' = b - dh\pi - h\pi d\). Suppose \(s^n : D^n \to C^n\) are morphisms splitting the morphisms \(g^n : C^n \to D^n\). Take \(a' = a + dsh + shd\). Computations omitted.

The following lemma can be used to replace a morphism of complexes by a morphism where in each degree the map is the injection of a direct summand.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(\alpha : K^\bullet \to L^\bullet\) be a morphism of complexes of \(\mathcal{A}\). There exists a factorization \[\xymatrix{ K^\bullet \ar[r]^{\tilde \alpha} \ar@/_1pc/[rr]_\alpha & \tilde L^\bullet \ar[r]^\pi & L^\bullet }\] such that

1. \(\tilde \alpha\) is a termwise split injection (see Definition 014G),

2. there is a map of complexes \(s : L^\bullet \to \tilde L^\bullet\) such that \(\pi \circ s = \text{id}_{L^\bullet}\) and such that \(s \circ \pi\) is homotopic to \(\text{id}_{\tilde L^\bullet}\).

Moreover, if both \(K^\bullet\) and \(L^\bullet\) are in \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), or \(K^b(\mathcal{A})\), then so is \(\tilde L^\bullet\).

Proof

We set \[\tilde L^n = L^n \oplus K^n \oplus K^{n + 1}\] and we define \[d^n_{\tilde L} = \left( \begin{matrix} d^n_L & 0 & 0 \\ 0 & d^n_K & \text{id}_{K^{n + 1}} \\ 0 & 0 & -d^{n + 1}_K \end{matrix} \right)\] In other words, \(\tilde L^\bullet = L^\bullet \oplus C(1_{K^\bullet})\). Moreover, we set \[\tilde \alpha = \left( \begin{matrix} \alpha \\ \text{id}_{K^n} \\ 0 \end{matrix} \right)\] which is clearly a split injection. It is also clear that it defines a morphism of complexes. We define \[\pi = \left( \begin{matrix} \text{id}_{L^n} & 0 & 0 \end{matrix} \right)\] so that clearly \(\pi \circ \tilde \alpha = \alpha\). We set \[s = \left( \begin{matrix} \text{id}_{L^n} \\ 0 \\ 0 \end{matrix} \right)\] so that \(\pi \circ s = \text{id}_{L^\bullet}\). Finally, let \(h^n : \tilde L^n \to \tilde L^{n - 1}\) be the map which maps the summand \(K^n\) of \(\tilde L^n\) via the identity morphism to the summand \(K^n\) of \(\tilde L^{n - 1}\). Then it is a trivial matter (see computations in remark below) to prove that \[\text{id}_{\tilde L^\bullet} - s \circ \pi = d \circ h + h \circ d\] which finishes the proof of the lemma.

Remark

To see the last displayed equality in the proof above we can argue with elements as follows. We have \(s\pi(l, k, k^{+}) = (l, 0, 0)\). Hence the morphism of the left hand side maps \((l, k, k^{+})\) to \((0, k, k^{+})\). On the other hand \(h(l, k, k^{+}) = (0, 0, k)\) and \(d(l, k, k^{+}) = (dl, dk + k^{+}, -dk^{+})\). Hence \((dh + hd)(l, k, k^{+}) = d(0, 0, k) + h(dl, dk + k^{+}, -dk^{+}) = (0, k, -dk) + (0, 0, dk + k^{+}) = (0, k, k^{+})\) as desired.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(\alpha : K^\bullet \to L^\bullet\) be a morphism of complexes of \(\mathcal{A}\). There exists a factorization \[\xymatrix{ K^\bullet \ar[r]^i \ar@/_1pc/[rr]_\alpha & \tilde K^\bullet \ar[r]^{\tilde \alpha} & L^\bullet }\] such that

1. \(\tilde \alpha\) is a termwise split surjection (see Definition 014G),

2. there is a map of complexes \(s : \tilde K^\bullet \to K^\bullet\) such that \(s \circ i = \text{id}_{K^\bullet}\) and such that \(i \circ s\) is homotopic to \(\text{id}_{\tilde K^\bullet}\).

Moreover, if both \(K^\bullet\) and \(L^\bullet\) are in \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), or \(K^b(\mathcal{A})\), then so is \(\tilde K^\bullet\).

Proof

Dual to Lemma 013N. Take \[\tilde K^n = K^n \oplus L^{n - 1} \oplus L^n\] and we define \[d^n_{\tilde K} = \left( \begin{matrix} d^n_K & 0 & 0 \\ 0 & - d^{n - 1}_L & \text{id}_{L^n} \\ 0 & 0 & d^n_L \end{matrix} \right)\] in other words \(\tilde K^\bullet = K^\bullet \oplus C(1_{L^\bullet[-1]})\). Moreover, we set \[\tilde \alpha = \left( \begin{matrix} \alpha & 0 & \text{id}_{L^n} \end{matrix} \right)\] which is clearly a split surjection. It is also clear that it defines a morphism of complexes. We define \[i = \left( \begin{matrix} \text{id}_{K^n} \\ 0 \\ 0 \end{matrix} \right)\] so that clearly \(\tilde \alpha \circ i = \alpha\). We set \[s = \left( \begin{matrix} \text{id}_{K^n} & 0 & 0 \end{matrix} \right)\] so that \(s \circ i = \text{id}_{K^\bullet}\). Finally, let \(h^n : \tilde K^n \to \tilde K^{n - 1}\) be the map which maps the summand \(L^{n - 1}\) of \(\tilde K^n\) via the identity morphism to the summand \(L^{n - 1}\) of \(\tilde K^{n - 1}\). Then it is a trivial matter to prove that \[\text{id}_{\tilde K^\bullet} - i \circ s = d \circ h + h \circ d\] which finishes the proof of the lemma.

Definition

Let \(\mathcal{A}\) be an additive category. A termwise split exact sequence of complexes of \(\mathcal{A}\) is a complex of complexes \[0 \to A^\bullet \xrightarrow{\alpha} B^\bullet \xrightarrow{\beta} C^\bullet \to 0\] together with given direct sum decompositions \(B^n = A^n \oplus C^n\) compatible with \(\alpha^n\) and \(\beta^n\). We often write \(s^n : C^n \to B^n\) and \(\pi^n : B^n \to A^n\) for the maps induced by the direct sum decompositions. According to Homology, Lemma 011J we get an associated morphism of complexes \[\delta : C^\bullet \longrightarrow A^\bullet[1]\] which in degree \(n\) is the map \(\pi^{n + 1} \circ d_B^n \circ s^n\). In other words \((A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta)\) forms a triangle \[A^\bullet \to B^\bullet \to C^\bullet \to A^\bullet[1]\] This will be the triangle associated to the termwise split sequence of complexes.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) be a termwise split exact sequence as in Definition 014I. Let \((\pi')^n\), \((s')^n\) be a second collection of splittings. Denote \(\delta' : C^\bullet \longrightarrow A^\bullet[1]\) the morphism associated to this second set of splittings. Then \[(1, 1, 1) : (A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta) \longrightarrow (A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta')\] is an isomorphism of triangles in \(K(\mathcal{A})\).

Proof

The statement simply means that \(\delta\) and \(\delta'\) are homotopic maps of complexes. This is Homology, Lemma 011L.

Remark

Let \(\mathcal{A}\) be an additive category. Let \(0 \to A_i^\bullet \to B_i^\bullet \to C_i^\bullet \to 0\), \(i = 1, 2\) be termwise split exact sequences. Suppose that \(a : A_1^\bullet \to A_2^\bullet\), \(b : B_1^\bullet \to B_2^\bullet\), and \(c : C_1^\bullet \to C_2^\bullet\) are morphisms of complexes such that \[\xymatrix{ A_1^\bullet \ar[d]_a \ar[r] & B_1^\bullet \ar[r] \ar[d]_b & C_1^\bullet \ar[d]_c \\ A_2^\bullet \ar[r] & B_2^\bullet \ar[r] & C_2^\bullet }\] commutes in \(K(\mathcal{A})\). In general, there does not exist a morphism \(b' : B_1^\bullet \to B_2^\bullet\) which is homotopic to \(b\) such that the diagram above commutes in the category of complexes. Namely, consider Examples, Equation (087K). If we could replace the middle map there by a homotopic one such that the diagram commutes, then we would have additivity of traces which we do not.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(0 \to A_i^\bullet \to B_i^\bullet \to C_i^\bullet \to 0\), \(i = 1, 2, 3\) be termwise split exact sequences of complexes. Let \(b : B_1^\bullet \to B_2^\bullet\) and \(b' : B_2^\bullet \to B_3^\bullet\) be morphisms of complexes such that \[\vcenter{ \xymatrix{ A_1^\bullet \ar[d]_0 \ar[r] & B_1^\bullet \ar[r] \ar[d]_b & C_1^\bullet \ar[d]_0 \\ A_2^\bullet \ar[r] & B_2^\bullet \ar[r] & C_2^\bullet } } \quad\text{and}\quad \vcenter{ \xymatrix{ A_2^\bullet \ar[d]^0 \ar[r] & B_2^\bullet \ar[r] \ar[d]^{b'} & C_2^\bullet \ar[d]^0 \\ A_3^\bullet \ar[r] & B_3^\bullet \ar[r] & C_3^\bullet } }\] commute in \(K(\mathcal{A})\). Then \(b' \circ b = 0\) in \(K(\mathcal{A})\).

Proof

By Lemma 014H we can replace \(b\) and \(b'\) by homotopic maps such that the right square of the left diagram commutes and the left square of the right diagram commutes. In other words, we have \(\Im(b^n) \subset \Im(A_2^n \to B_2^n)\) and \(\Ker((b')^n) \supset \Im(A_2^n \to B_2^n)\). Then \(b' \circ b = 0\) as a map of complexes.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(f_1 : K_1^\bullet \to L_1^\bullet\) and \(f_2 : K_2^\bullet \to L_2^\bullet\) be morphisms of complexes. Let \[(a, b, c) : (K_1^\bullet, L_1^\bullet, C(f_1)^\bullet, f_1, i_1, p_1) \longrightarrow (K_2^\bullet, L_2^\bullet, C(f_2)^\bullet, f_2, i_2, p_2)\] be any morphism of triangles of \(K(\mathcal{A})\). If \(a\) and \(b\) are homotopy equivalences then so is \(c\).

Proof

Let \(a^{-1} : K_2^\bullet \to K_1^\bullet\) be a morphism of complexes which is inverse to \(a\) in \(K(\mathcal{A})\). Let \(b^{-1} : L_2^\bullet \to L_1^\bullet\) be a morphism of complexes which is inverse to \(b\) in \(K(\mathcal{A})\). Let \(c' : C(f_2)^\bullet \to C(f_1)^\bullet\) be the morphism from Lemma 014F applied to \(f_1 \circ a^{-1} = b^{-1} \circ f_2\). If we can show that \(c \circ c'\) and \(c' \circ c\) are isomorphisms in \(K(\mathcal{A})\) then we win. Hence it suffices to prove the following: Given a morphism of triangles \((1, 1, c) : (K^\bullet, L^\bullet, C(f)^\bullet, f, i, p)\) in \(K(\mathcal{A})\) the morphism \(c\) is an isomorphism in \(K(\mathcal{A})\). By assumption the two squares in the diagram \[\xymatrix{ L^\bullet \ar[r] \ar[d]_1 & C(f)^\bullet \ar[r] \ar[d]_c & K^\bullet[1] \ar[d]_1 \\ L^\bullet \ar[r] & C(f)^\bullet \ar[r] & K^\bullet[1] }\] commute up to homotopy. By construction of \(C(f)^\bullet\) the rows form termwise split sequences of complexes. Thus we see that \((c - 1)^2 = 0\) in \(K(\mathcal{A})\) by Lemma 086L. Hence \(c\) is an isomorphism in \(K(\mathcal{A})\) with inverse \(2 - c\).

Hence if \(a\) and \(b\) are homotopy equivalences then the resulting morphism of triangles is an isomorphism of triangles in \(K(\mathcal{A})\). It turns out that the collection of triangles of \(K(\mathcal{A})\) given by cones and the collection of triangles of \(K(\mathcal{A})\) given by termwise split sequences of complexes are the same up to isomorphisms, at least up to sign!

Lemma

Let \(\mathcal{A}\) be an additive category.

1. Given a termwise split sequence of complexes \((\alpha : A^\bullet \to B^\bullet, \beta : B^\bullet \to C^\bullet, s^n, \pi^n)\) there exists a homotopy equivalence \(C(\alpha)^\bullet \to C^\bullet\) such that the diagram \[\xymatrix{ A^\bullet \ar[r] \ar[d] & B^\bullet \ar[d] \ar[r] & C(\alpha)^\bullet \ar[r]_{-p} \ar[d] & A^\bullet[1] \ar[d] \\ A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r]^\delta & A^\bullet[1] }\] defines an isomorphism of triangles in \(K(\mathcal{A})\).

2. Given a morphism of complexes \(f : K^\bullet \to L^\bullet\) there exists an isomorphism of triangles \[\xymatrix{ K^\bullet \ar[r] \ar[d] & \tilde L^\bullet \ar[d] \ar[r] & M^\bullet \ar[r]_{\delta} \ar[d] & K^\bullet[1] \ar[d] \\ K^\bullet \ar[r] & L^\bullet \ar[r] & C(f)^\bullet \ar[r]^{-p} & K^\bullet[1] }\] where the upper triangle is the triangle associated to a termwise split exact sequence \(K^\bullet \to \tilde L^\bullet \to M^\bullet\).

Proof

Proof of (1). We have \(C(\alpha)^n = B^n \oplus A^{n + 1}\) and we simply define \(C(\alpha)^n \to C^n\) via the projection onto \(B^n\) followed by \(\beta^n\). This defines a morphism of complexes because the compositions \(A^{n + 1} \to B^{n + 1} \to C^{n + 1}\) are zero. To get a homotopy inverse we take \(C^\bullet \to C(\alpha)^\bullet\) given by \((s^n , -\delta^n)\) in degree \(n\). This is a morphism of complexes because the morphism \(\delta^n\) can be characterized as the unique morphism \(C^n \to A^{n + 1}\) such that \(d \circ s^n - s^{n + 1} \circ d = \alpha \circ \delta^n\), see proof of Homology, Lemma 011J. The composition \(C^\bullet \to C(\alpha)^\bullet \to C^\bullet\) is the identity. The composition \(C(\alpha)^\bullet \to C^\bullet \to C(\alpha)^\bullet\) is equal to the morphism \[\left( \begin{matrix} s^n \circ \beta^n & 0 \\ -\delta^n \circ \beta^n & 0 \end{matrix} \right)\] To see that this is homotopic to the identity map use the homotopy \(h^n : C(\alpha)^n \to C(\alpha)^{n - 1}\) given by the matrix \[\left( \begin{matrix} 0 & 0 \\ \pi^n & 0 \end{matrix} \right) : C(\alpha)^n = B^n \oplus A^{n + 1} \to B^{n - 1} \oplus A^n = C(\alpha)^{n - 1}\] It is trivial to verify that \[\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) - \left( \begin{matrix} s^n \\ -\delta^n \end{matrix} \right) \left( \begin{matrix} \beta^n & 0 \end{matrix} \right) = \left( \begin{matrix} d & \alpha^n \\ 0 & -d \end{matrix} \right) \left( \begin{matrix} 0 & 0 \\ \pi^n & 0 \end{matrix} \right) + \left( \begin{matrix} 0 & 0 \\ \pi^{n + 1} & 0 \end{matrix} \right) \left( \begin{matrix} d & \alpha^{n + 1} \\ 0 & -d \end{matrix} \right)\] To finish the proof of (1) we have to show that the morphisms \(-p : C(\alpha)^\bullet \to A^\bullet[1]\) (see Definition 014E) and \(C(\alpha)^\bullet \to C^\bullet \to A^\bullet[1]\) agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse \((s, -\delta) : C^\bullet \to C(\alpha)^\bullet\) and check instead that the two maps \(C^\bullet \to A^\bullet[1]\) agree. And note that \(p \circ (s, -\delta) = -\delta\) as desired.

Proof of (2). We let \(\tilde f : K^\bullet \to \tilde L^\bullet\), \(s : L^\bullet \to \tilde L^\bullet\) and \(\pi : \tilde L^\bullet \to L^\bullet\) be as in Lemma 013N. By Lemmas 014F and 014K the triangles \((K^\bullet, L^\bullet, C(f), f, i, p)\) and \((K^\bullet, \tilde L^\bullet, C(\tilde f), \tilde f, \tilde i, \tilde p)\) are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace \(L^\bullet\) by \(\tilde L^\bullet\) and \(f\) by \(\tilde f\). In other words we may assume that \(f\) is a termwise split injection. In this case the result follows from part (1).

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(A_1^\bullet \to A_2^\bullet \to \ldots \to A_n^\bullet\) be a sequence of composable morphisms of complexes. There exists a commutative diagram \[\xymatrix{ A_1^\bullet \ar[r] & A_2^\bullet \ar[r] & \ldots \ar[r] & A_n^\bullet \\ B_1^\bullet \ar[r] \ar[u] & B_2^\bullet \ar[r] \ar[u] & \ldots \ar[r] & B_n^\bullet \ar[u] }\] such that each morphism \(B_i^\bullet \to B_{i + 1}^\bullet\) is a termwise split injection and each \(B_i^\bullet \to A_i^\bullet\) is a homotopy equivalence. Moreover, if all \(A_i^\bullet\) are in \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), or \(K^b(\mathcal{A})\), then so are the \(B_i^\bullet\).

Proof

The case \(n = 1\) is without content. Lemma 013N is the case \(n = 2\). Suppose we have constructed the diagram except for \(B_n^\bullet\). Apply Lemma 013N to the composition \(B_{n - 1}^\bullet \to A_{n - 1}^\bullet \to A_n^\bullet\). The result is a factorization \(B_{n - 1}^\bullet \to B_n^\bullet \to A_n^\bullet\) as desired.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \((\alpha : A^\bullet \to B^\bullet, \beta : B^\bullet \to C^\bullet, s^n, \pi^n)\) be a termwise split sequence of complexes. Let \((A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta)\) be the associated triangle. Then the triangle \((C^\bullet[-1], A^\bullet, B^\bullet, \delta[-1], \alpha, \beta)\) is isomorphic to the triangle \((C^\bullet[-1], A^\bullet, C(\delta[-1])^\bullet, \delta[-1], i, p)\).

Proof

We write \(B^n = A^n \oplus C^n\) and we identify \(\alpha^n\) and \(\beta^n\) with the natural inclusion and projection maps. By construction of \(\delta\) we have \[d_B^n = \left( \begin{matrix} d_A^n & \delta^n \\ 0 & d_C^n \end{matrix} \right)\] On the other hand the cone of \(\delta[-1] : C^\bullet[-1] \to A^\bullet\) is given as \(C(\delta[-1])^n = A^n \oplus C^n\) with differential identical with the matrix above! Whence the lemma.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(f : K^\bullet \to L^\bullet\) be a morphism of complexes. The triangle \((L^\bullet, C(f)^\bullet, K^\bullet[1], i, p, f[1])\) is the triangle associated to the termwise split sequence \[0 \to L^\bullet \to C(f)^\bullet \to K^\bullet[1] \to 0\] coming from the definition of the cone of \(f\).

Proof

Immediate from the definitions.

Distinguished triangles in the homotopy category

Since we want our boundary maps in long exact sequences of cohomology to be given by the maps in the snake lemma without signs we define distinguished triangles in the homotopy category as follows.

Definition

Let \(\mathcal{A}\) be an additive category. A triangle \((X, Y, Z, f, g, h)\) of \(K(\mathcal{A})\) is called a distinguished triangle of \(K(\mathcal{A})\) if it is isomorphic to the triangle associated to a termwise split exact sequence of complexes, see Definition 014I. Same definition for \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), and \(K^b(\mathcal{A})\).

Note that according to Lemma 014L a triangle of the form \((K^\bullet, L^\bullet, C(f)^\bullet, f, i, -p)\) is a distinguished triangle. This does indeed lead to a triangulated category, see Proposition 014S. Before we can prove the proposition we need one more lemma in order to be able to prove TR4.

Lemma

Let \(\mathcal{A}\) be an additive category. Suppose that \(\alpha : A^\bullet \to B^\bullet\) and \(\beta : B^\bullet \to C^\bullet\) are split injections of complexes. Then there exist distinguished triangles \((A^\bullet, B^\bullet, Q_1^\bullet, \alpha, p_1, d_1)\), \((A^\bullet, C^\bullet, Q_2^\bullet, \beta \circ \alpha, p_2, d_2)\) and \((B^\bullet, C^\bullet, Q_3^\bullet, \beta, p_3, d_3)\) for which TR4 holds.

Proof

Say \(\pi_1^n : B^n \to A^n\), and \(\pi_3^n : C^n \to B^n\) are the splittings. Then also \(A^\bullet \to C^\bullet\) is a split injection with splittings \(\pi_2^n = \pi_1^n \circ \pi_3^n\). Let us write \(Q_1^\bullet\), \(Q_2^\bullet\) and \(Q_3^\bullet\) for the “quotient” complexes. In other words, \(Q_1^n = \Ker(\pi_1^n)\), \(Q_3^n = \Ker(\pi_3^n)\) and \(Q_2^n = \Ker(\pi_2^n)\). Note that the kernels exist. Then \(B^n = A^n \oplus Q_1^n\) and \(C^n = B^n \oplus Q_3^n\), where we think of \(A^n\) as a subobject of \(B^n\) and so on. This implies \(C^n = A^n \oplus Q_1^n \oplus Q_3^n\). Note that \(\pi_2^n = \pi_1^n \circ \pi_3^n\) is zero on both \(Q_1^n\) and \(Q_3^n\). Hence \(Q_2^n = Q_1^n \oplus Q_3^n\). Consider the commutative diagram \[\begin{matrix} 0 & \to & A^\bullet & \to & B^\bullet & \to & Q_1^\bullet & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & A^\bullet & \to & C^\bullet & \to & Q_2^\bullet & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & \\ 0 & \to & B^\bullet & \to & C^\bullet & \to & Q_3^\bullet & \to & 0 \end{matrix}\] The rows of this diagram are termwise split exact sequences, and hence determine distinguished triangles by definition. Moreover downward arrows in the diagram above are compatible with the chosen splittings and hence define morphisms of triangles \[(A^\bullet \to B^\bullet \to Q_1^\bullet \to A^\bullet[1]) \longrightarrow (A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet[1])\] and \[(A^\bullet \to C^\bullet \to Q_2^\bullet \to A^\bullet[1]) \longrightarrow (B^\bullet \to C^\bullet \to Q_3^\bullet \to B^\bullet[1]).\] Note that the splittings \(Q_3^n \to C^n\) of the bottom split sequence in the diagram provides a splitting for the split sequence \(0 \to Q_1^\bullet \to Q_2^\bullet \to Q_3^\bullet \to 0\) upon composing with \(C^n \to Q_2^n\). It follows easily from this that the morphism \(\delta : Q_3^\bullet \to Q_1^\bullet[1]\) in the corresponding distinguished triangle \[(Q_1^\bullet \to Q_2^\bullet \to Q_3^\bullet \to Q_1^\bullet[1])\] is equal to the composition \(Q_3^\bullet \to B^\bullet[1] \to Q_1^\bullet[1]\). Hence we get a structure as in the conclusion of axiom TR4.

Proposition

Let \(\mathcal{A}\) be an additive category. The category \(K(\mathcal{A})\) of complexes up to homotopy with its natural translation functors and distinguished triangles as defined above is a triangulated category.

Proof

Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Also, any triangle \((A^\bullet, A^\bullet, 0, 1, 0, 0)\) is distinguished since \(0 \to A^\bullet \to A^\bullet \to 0 \to 0\) is a termwise split sequence of complexes. Finally, given any morphism of complexes \(f : K^\bullet \to L^\bullet\) the triangle \((K, L, C(f), f, i, -p)\) is distinguished by Lemma 014L.

Proof of TR2. Let \((X, Y, Z, f, g, h)\) be a triangle. Assume \((Y, Z, X[1], g, h, -f[1])\) is distinguished. Then there exists a termwise split sequence of complexes \(A^\bullet \to B^\bullet \to C^\bullet\) such that the associated triangle \((A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta)\) is isomorphic to \((Y, Z, X[1], g, h, -f[1])\). Rotating back we see that \((X, Y, Z, f, g, h)\) is isomorphic to \((C^\bullet[-1], A^\bullet, B^\bullet, -\delta[-1], \alpha, \beta)\). It follows from Lemma 014N that the triangle \((C^\bullet[-1], A^\bullet, B^\bullet, \delta[-1], \alpha, \beta)\) is isomorphic to \((C^\bullet[-1], A^\bullet, C(\delta[-1])^\bullet, \delta[-1], i, p)\). Precomposing the previous isomorphism of triangles with \(-1\) on \(Y\) it follows that \((X, Y, Z, f, g, h)\) is isomorphic to \((C^\bullet[-1], A^\bullet, C(\delta[-1])^\bullet, \delta[-1], i, -p)\). Hence it is distinguished by Lemma 014L. On the other hand, suppose that \((X, Y, Z, f, g, h)\) is distinguished. By Lemma 014L this means that it is isomorphic to a triangle of the form \((K^\bullet, L^\bullet, C(f), f, i, -p)\) for some morphism of complexes \(f\). Then the rotated triangle \((Y, Z, X[1], g, h, -f[1])\) is isomorphic to \((L^\bullet, C(f), K^\bullet[1], i, -p, -f[1])\) which is isomorphic to the triangle \((L^\bullet, C(f), K^\bullet[1], i, p, f[1])\). By Lemma 014O this triangle is distinguished. Hence \((Y, Z, X[1], g, h, -f[1])\) is distinguished as desired.

Proof of TR3. Let \((X, Y, Z, f, g, h)\) and \((X', Y', Z', f', g', h')\) be distinguished triangles of \(K(\mathcal{A})\) and let \(a : X \to X'\) and \(b : Y \to Y'\) be morphisms such that \(f' \circ a = b \circ f\). By Lemma 014L we may assume that \((X, Y, Z, f, g, h) = (X, Y, C(f), f, i, -p)\) and \((X', Y', Z', f', g', h') = (X', Y', C(f'), f', i', -p')\). At this point we simply apply Lemma 014F to the commutative diagram given by \(f, f', a, b\).

Proof of TR4. At this point we know that \(K(\mathcal{A})\) is a pre-triangulated category. Hence we can use Lemma 014C. Let \(A^\bullet \to B^\bullet\) and \(B^\bullet \to C^\bullet\) be composable morphisms of \(K(\mathcal{A})\). By Lemma 014M we may assume that \(A^\bullet \to B^\bullet\) and \(B^\bullet \to C^\bullet\) are split injective morphisms. In this case the result follows from Lemma 014R.

Remark

Let \(\mathcal{A}\) be an additive category. Exactly the same proof as the proof of Proposition 014S shows that the categories \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), and \(K^b(\mathcal{A})\) are triangulated categories. Namely, the cone of a morphism between bounded (above, below) is bounded (above, below). But we prove below that these are triangulated subcategories of \(K(\mathcal{A})\) which gives another proof.

Lemma

Let \(\mathcal{A}\) be an additive category. The categories \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), and \(K^b(\mathcal{A})\) are full triangulated subcategories of \(K(\mathcal{A})\).

Proof

Each of the categories mentioned is a full additive subcategory. We use the criterion of Lemma 05QX to show that they are triangulated subcategories. It is clear that each of the categories \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), and \(K^b(\mathcal{A})\) is preserved under the shift functors \([1], [-1]\). Finally, suppose that \(f : A^\bullet \to B^\bullet\) is a morphism in \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), or \(K^b(\mathcal{A})\). Then \((A^\bullet, B^\bullet, C(f)^\bullet, f, i, -p)\) is a distinguished triangle of \(K(\mathcal{A})\) with \(C(f)^\bullet \in K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), or \(K^b(\mathcal{A})\) as is clear from the construction of the cone. Thus the lemma is proved. (Alternatively, \(K^\bullet \to L^\bullet\) is isomorphic to a termwise split injection of complexes in \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), or \(K^b(\mathcal{A})\), see Lemma 013N and then one can directly take the associated distinguished triangle.)

Lemma

Let \(\mathcal{A}\), \(\mathcal{B}\) be additive categories. Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor. The induced functors \[\begin{matrix} F : K(\mathcal{A}) \longrightarrow K(\mathcal{B}) \\ F : K^{+}(\mathcal{A}) \longrightarrow K^{+}(\mathcal{B}) \\ F : K^{-}(\mathcal{A}) \longrightarrow K^{-}(\mathcal{B}) \\ F : K^b(\mathcal{A}) \longrightarrow K^b(\mathcal{B}) \end{matrix}\] are exact functors of triangulated categories.

Proof

Suppose \(A^\bullet \to B^\bullet \to C^\bullet\) is a termwise split sequence of complexes of \(\mathcal{A}\) with splittings \((s^n, \pi^n)\) and associated morphism \(\delta : C^\bullet \to A^\bullet[1]\), see Definition 014I. Then \(F(A^\bullet) \to F(B^\bullet) \to F(C^\bullet)\) is a termwise split sequence of complexes with splittings \((F(s^n), F(\pi^n))\) and associated morphism \(F(\delta) : F(C^\bullet) \to F(A^\bullet)[1]\). Thus \(F\) transforms distinguished triangles into distinguished triangles.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \((A^\bullet, B^\bullet, C^\bullet, a, b, c)\) be a distinguished triangle in \(K(\mathcal{A})\). Then there exists an isomorphic distinguished triangle \((A^\bullet, (B')^\bullet, C^\bullet, a', b', c)\) such that \(0 \to A^n \to (B')^n \to C^n \to 0\) is a split short exact sequence for all \(n\).

Proof

We will use that \(K(\mathcal{A})\) is a triangulated category by Proposition 014S. Let \(W^\bullet\) be the cone on \(c : C^\bullet \to A^\bullet[1]\) with its maps \(i : A^\bullet[1] \to W^\bullet\) and \(p : W^\bullet \to C^\bullet[1]\). Then \((C^\bullet, A^\bullet[1], W^\bullet, c, i, -p)\) is a distinguished triangle by Lemma 014L. Rotating backwards twice we see that \((A^\bullet, W^\bullet[-1], C^\bullet, -i[-1], p[-1], c)\) is a distinguished triangle. By TR3 there is a morphism of distinguished triangles \((\text{id}, \beta, \text{id}) : (A^\bullet, B^\bullet, C^\bullet, a, b, c) \to (A^\bullet, W^\bullet[-1], C^\bullet, -i[-1], p[-1], c)\) which must be an isomorphism by Lemma 014A. This finishes the proof because \(0 \to A^\bullet \to W^\bullet[-1] \to C^\bullet \to 0\) is a termwise split short exact sequence of complexes by the very construction of cones in Section 014D.

Remark

Let \(\mathcal{A}\) be an additive category with countable direct sums. Let \(\text{DoubleComp}(\mathcal{A})\) denote the category of double complexes in \(\mathcal{A}\), see Homology, Section 0FNB. We can use this category to construct two triangulated categories.

1. We can consider an object \(A^{\bullet, \bullet}\) of \(\text{DoubleComp}(\mathcal{A})\) as a complex of complexes as follows \[\ldots \to A^{\bullet, -1} \to A^{\bullet, 0} \to A^{\bullet, 1} \to \ldots\] and take the homotopy category \(K_{first}(\text{DoubleComp}(\mathcal{A}))\) with the corresponding triangulated structure given by Proposition 014S. By Homology, Remark 0G6A the functor \[\text{Tot} : K_{first}(\text{DoubleComp}(\mathcal{A})) \longrightarrow K(\mathcal{A})\] is an exact functor of triangulated categories.

2. We can consider an object \(A^{\bullet, \bullet}\) of \(\text{DoubleComp}(\mathcal{A})\) as a complex of complexes as follows \[\ldots \to A^{-1, \bullet} \to A^{0, \bullet} \to A^{1, \bullet} \to \ldots\] and take the homotopy category \(K_{second}(\text{DoubleComp}(\mathcal{A}))\) with the corresponding triangulated structure given by Proposition 014S. By Homology, Remark 0G6B the functor \[\text{Tot} : K_{second}(\text{DoubleComp}(\mathcal{A})) \longrightarrow K(\mathcal{A})\] is an exact functor of triangulated categories.

Remark

Let \(\mathcal{A}\), \(\mathcal{B}\), \(\mathcal{C}\) be additive categories and assume \(\mathcal{C}\) has countable direct sums. Suppose that \[\otimes : \mathcal{A} \times \mathcal{B} \longrightarrow \mathcal{C}, \quad (X, Y) \longmapsto X \otimes Y\] is a functor which is bilinear on morphisms. This determines a functor \[\text{Comp}(\mathcal{A}) \times \text{Comp}(\mathcal{B}) \longrightarrow \text{DoubleComp}(\mathcal{C}), \quad (X^\bullet, Y^\bullet) \longmapsto X^\bullet \otimes Y^\bullet\] See Homology, Example 0A5J.

1. For a fixed object \(X^\bullet\) of \(\text{Comp}(\mathcal{A})\) the functor \[K(\mathcal{B}) \longrightarrow K(\mathcal{C}), \quad Y^\bullet \longmapsto \text{Tot}(X^\bullet \otimes Y^\bullet)\] is an exact functor of triangulated categories.

2. For a fixed object \(Y^\bullet\) of \(\text{Comp}(\mathcal{B})\) the functor \[K(\mathcal{A}) \longrightarrow K(\mathcal{C}), \quad X^\bullet \longmapsto \text{Tot}(X^\bullet \otimes Y^\bullet)\] is an exact functor of triangulated categories.

This follows from Remark 0G6D since the functors \(\text{Comp}(\mathcal{A}) \to \text{DoubleComp}(\mathcal{C})\), \(Y^\bullet \mapsto X^\bullet \otimes Y^\bullet\) and \(\text{Comp}(\mathcal{B}) \to \text{DoubleComp}(\mathcal{C})\), \(X^\bullet \mapsto X^\bullet \otimes Y^\bullet\) are immediately seen to be compatible with homotopies and termwise split short exact sequences and hence induce exact functors of triangulated categories \[K(\mathcal{B}) \to K_{first}(\text{DoubleComp}(\mathcal{C})) \quad\text{and}\quad K(\mathcal{A}) \to K_{second}(\text{DoubleComp}(\mathcal{C}))\] Observe that for the first of the two the isomorphism \[\text{Tot}(X^\bullet \otimes Y^\bullet[1]) \cong \text{Tot}(X^\bullet \otimes Y^\bullet)[1]\] involves signs (this goes back to the signs chosen in Homology, Remark 0FLG).

Derived categories

In this section we construct the derived category of an abelian category \(\mathcal{A}\) by inverting the quasi-isomorphisms in \(K(\mathcal{A})\). Before we do this recall that the functors \(H^i : \text{Comp}(\mathcal{A}) \to \mathcal{A}\) factor through \(K(\mathcal{A})\), see Homology, Lemma 0116. Moreover, in Homology, Definition 011H we have defined identifications \(H^i(K^\bullet[n]) = H^{i + n}(K^\bullet)\). At this point it makes sense to redefine \[H^i(K^\bullet) = H^0(K^\bullet[i])\] in order to avoid confusion and possible sign errors.

Lemma

Let \(\mathcal{A}\) be an abelian category. The functor \[H^0 : K(\mathcal{A}) \longrightarrow \mathcal{A}\] is homological.

Proof

Because \(H^0\) is a functor, and by our definition of distinguished triangles it suffices to prove that given a termwise split short exact sequence of complexes \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) the sequence \(H^0(A^\bullet) \to H^0(B^\bullet) \to H^0(C^\bullet)\) is exact. This follows from Homology, Lemma 0117.

In particular, this lemma implies that a distinguished triangle \((X, Y, Z, f, g, h)\) in \(K(\mathcal{A})\) gives rise to a long exact cohomology sequence [05ST]\[\begin{equation} \xymatrix{ \ldots \ar[r] & H^i(X) \ar[r]^{H^i(f)} & H^i(Y) \ar[r]^{H^i(g)} & H^i(Z) \ar[r]^{H^i(h)} & H^{i + 1}(X) \ar[r] & \ldots } \end{equation}\] see (0148). Moreover, there is a compatibility with the long exact sequence of cohomology associated to a short exact sequence of complexes. For example, if \((A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta)\) is the distinguished triangle associated to a termwise split exact sequence of complexes (see Definition 014I), then the cohomology sequence above agrees with the one defined using the snake lemma, see Homology, Lemma 0117 and for agreement of sequences, see Homology, Lemma 011K.

Recall that a complex \(K^\bullet\) is acyclic if \(H^i(K^\bullet) = 0\) for all \(i \in \mathbf{Z}\). Moreover, recall that a morphism of complexes \(f : K^\bullet \to L^\bullet\) is a quasi-isomorphism if and only if \(H^i(f)\) is an isomorphism for all \(i\). See Homology, Definition 0115.

Lemma

Let \(\mathcal{A}\) be an abelian category. The full subcategory \(\text{Ac}(\mathcal{A})\) of \(K(\mathcal{A})\) consisting of acyclic complexes is a strictly full saturated triangulated subcategory of \(K(\mathcal{A})\). The corresponding saturated multiplicative system (see Lemma 05RL) of \(K(\mathcal{A})\) is the set \(\text{Qis}(\mathcal{A})\) of quasi-isomorphisms. In particular, the kernel of the localization functor \(Q : K(\mathcal{A}) \to \text{Qis}(\mathcal{A})^{-1}K(\mathcal{A})\) is \(\text{Ac}(\mathcal{A})\) and the functor \(H^0\) factors through \(Q\).

Proof

We know that \(H^0\) is a homological functor by Lemma 05RS. Thus this lemma is a special case of Lemma 05RM.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(\text{Ac}(\mathcal{A})\) and \(\text{Qis}(\mathcal{A})\) be as in Lemma 05RT. The derived category of \(\mathcal{A}\) is the triangulated category \[D(\mathcal{A}) = K(\mathcal{A})/\text{Ac}(\mathcal{A}) = \text{Qis}(\mathcal{A})^{-1} K(\mathcal{A}).\] We denote by \(H^0 : D(\mathcal{A}) \to \mathcal{A}\) the unique functor whose composition with the quotient functor gives back the functor \(H^0\) defined above. Using Lemma 05RE we introduce the strictly full saturated triangulated subcategories \(D^{+}(\mathcal{A}), D^{-}(\mathcal{A}), D^b(\mathcal{A})\) whose sets of objects are \[\begin{matrix} \Ob(D^{+}(\mathcal{A})) = \{X \in \Ob(D(\mathcal{A})) \mid H^n(X) = 0\text{ for all }n \ll 0\} \\ \Ob(D^{-}(\mathcal{A})) = \{X \in \Ob(D(\mathcal{A})) \mid H^n(X) = 0\text{ for all }n \gg 0\} \\ \Ob(D^b(\mathcal{A})) = \{X \in \Ob(D(\mathcal{A})) \mid H^n(X) = 0\text{ for all }|n| \gg 0\} \end{matrix}\] The category \(D^b(\mathcal{A})\) is called the bounded derived category of \(\mathcal{A}\).

If \(K^\bullet\) and \(L^\bullet\) are complexes of \(\mathcal{A}\) then we sometimes say “\(K^\bullet\) is quasi-isomorphic to \(L^\bullet\)” to indicate that \(K^\bullet\) and \(L^\bullet\) are isomorphic objects of \(D(\mathcal{A})\).

Remark

In this chapter, we consistently work with “small” abelian categories (as is the convention in the Stacks project). For a “big” abelian category \(\mathcal{A}\), it isn’t clear that the derived category \(D(\mathcal{A})\) exists, because it isn’t clear that morphisms in the derived category are sets. In fact, in general they aren’t, see Examples, Lemma 07JT. However, if \(\mathcal{A}\) is a Grothendieck abelian category, and given \(K^\bullet, L^\bullet\) in \(K(\mathcal{A})\), then by Injectives, Theorem 079P there exists a quasi-isomorphism \(L^\bullet \to I^\bullet\) to a K-injective complex \(I^\bullet\) and Lemma 070I shows that \[\Hom_{D(\mathcal{A})}(K^\bullet, L^\bullet) = \Hom_{K(\mathcal{A})}(K^\bullet, I^\bullet)\] which is a set. Some examples of Grothendieck abelian categories are the category of modules over a ring, or more generally the category of sheaves of modules on a ringed site.

Each of the variants \(D^{+}(\mathcal{A}), D^{-}(\mathcal{A}), D^b(\mathcal{A})\) can be constructed as a localization of the corresponding homotopy category. This relies on the following simple lemma.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet\) be a complex.

1. If \(H^n(K^\bullet) = 0\) for all \(n \ll 0\), then there exists a quasi-isomorphism \(K^\bullet \to L^\bullet\) with \(L^\bullet\) bounded below.

2. If \(H^n(K^\bullet) = 0\) for all \(n \gg 0\), then there exists a quasi-isomorphism \(M^\bullet \to K^\bullet\) with \(M^\bullet\) bounded above.

3. If \(H^n(K^\bullet) = 0\) for all \(|n| \gg 0\), then there exists a commutative diagram of morphisms of complexes \[\xymatrix{ K^\bullet \ar[r] & L^\bullet \\ M^\bullet \ar[u] \ar[r] & N^\bullet \ar[u] }\] where all the arrows are quasi-isomorphisms, \(L^\bullet\) bounded below, \(M^\bullet\) bounded above, and \(N^\bullet\) a bounded complex.

Proof

Pick \(a \ll 0 \ll b\) and set \(M^\bullet = \tau_{\leq b}K^\bullet\), \(L^\bullet = \tau_{\geq a}K^\bullet\), and \(N^\bullet = \tau_{\leq b}L^\bullet = \tau_{\geq a}M^\bullet\). See Homology, Section 0118 for the truncation functors.

To state the following lemma denote \(\text{Ac}^{+}(\mathcal{A})\), \(\text{Ac}^{-}(\mathcal{A})\), resp. \(\text{Ac}^b(\mathcal{A})\) the intersection of \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), resp. \(K^b(\mathcal{A})\) with \(\text{Ac}(\mathcal{A})\). Denote \(\text{Qis}^{+}(\mathcal{A})\), \(\text{Qis}^{-}(\mathcal{A})\), resp. \(\text{Qis}^b(\mathcal{A})\) the intersection of \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), resp. \(K^b(\mathcal{A})\) with \(\text{Qis}(\mathcal{A})\).

Lemma

Let \(\mathcal{A}\) be an abelian category. The subcategories \(\text{Ac}^{+}(\mathcal{A})\), \(\text{Ac}^{-}(\mathcal{A})\), resp. \(\text{Ac}^b(\mathcal{A})\) are strictly full saturated triangulated subcategories of \(K^{+}(\mathcal{A})\), \(K^{-}(\mathcal{A})\), resp. \(K^b(\mathcal{A})\). The corresponding saturated multiplicative systems (see Lemma 05RL) are the sets \(\text{Qis}^{+}(\mathcal{A})\), \(\text{Qis}^{-}(\mathcal{A})\), resp. \(\text{Qis}^b(\mathcal{A})\).

1. The kernel of the functor \(K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})\) is \(\text{Ac}^{+}(\mathcal{A})\) and this induces an equivalence of triangulated categories \[K^{+}(\mathcal{A})/\text{Ac}^{+}(\mathcal{A}) = \text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{A})\]

2. The kernel of the functor \(K^{-}(\mathcal{A}) \to D^{-}(\mathcal{A})\) is \(\text{Ac}^{-}(\mathcal{A})\) and this induces an equivalence of triangulated categories \[K^{-}(\mathcal{A})/\text{Ac}^{-}(\mathcal{A}) = \text{Qis}^{-}(\mathcal{A})^{-1}K^{-}(\mathcal{A}) \longrightarrow D^{-}(\mathcal{A})\]

3. The kernel of the functor \(K^b(\mathcal{A}) \to D^b(\mathcal{A})\) is \(\text{Ac}^b(\mathcal{A})\) and this induces an equivalence of triangulated categories \[K^b(\mathcal{A})/\text{Ac}^b(\mathcal{A}) = \text{Qis}^b(\mathcal{A})^{-1}K^b(\mathcal{A}) \longrightarrow D^b(\mathcal{A})\]

Proof

The initial statements follow from Lemma 05RM by considering the restriction of the homological functor \(H^0\). The statement on kernels in (1), (2), (3) is a consequence of the definitions in each case. Each of the functors is essentially surjective by Lemma 05RV. To finish the proof we have to show the functors are fully faithful. We first do this for the bounded below version.

Suppose that \(K^\bullet, L^\bullet\) are bounded below complexes. A morphism between these in \(D(\mathcal{A})\) is of the form \(s^{-1}f\) for a pair \(f : K^\bullet \to (L')^\bullet\), \(s : L^\bullet \to (L')^\bullet\) where \(s\) is a quasi-isomorphism. This implies that \((L')^\bullet\) has cohomology bounded below. Hence by Lemma 05RV we can choose a quasi-isomorphism \(s' : (L')^\bullet \to (L'')^\bullet\) with \((L'')^\bullet\) bounded below. Then the pair \((s' \circ f, s' \circ s)\) defines a morphism in \(\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})\). Hence the functor is “full”. Finally, suppose that the pair \(f : K^\bullet \to (L')^\bullet\), \(s : L^\bullet \to (L')^\bullet\) defines a morphism in \(\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})\) which is zero in \(D(\mathcal{A})\). This means that there exists a quasi-isomorphism \(s' : (L')^\bullet \to (L'')^\bullet\) such that \(s' \circ f = 0\). Using Lemma 05RV once more we obtain a quasi-isomorphism \(s'' : (L'')^\bullet \to (L''')^\bullet\) with \((L''')^\bullet\) bounded below. Thus we see that \(s'' \circ s' \circ f = 0\) which implies that \(s^{-1}f\) is zero in \(\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})\). This finishes the proof that the functor in (1) is an equivalence.

The proof of (2) is dual to the proof of (1). To prove (3) we may use the result of (2). Hence it suffices to prove that the functor \(\text{Qis}^b(\mathcal{A})^{-1}K^b(\mathcal{A}) \to \text{Qis}^{-}(\mathcal{A})^{-1}K^{-}(\mathcal{A})\) is fully faithful. The argument given in the previous paragraph applies directly to show this where we consistently work with complexes which are already bounded above.

The canonical delta-functor

The derived category should be the receptacle for the universal cohomology functor. In order to state the result we use the notion of a \(\delta\)-functor from an abelian category into a triangulated category, see Definition 0150.

Consider the functor \(\text{Comp}(\mathcal{A}) \to K(\mathcal{A})\). This functor is not a \(\delta\)-functor in general. The easiest way to see this is to consider a nonsplit short exact sequence \(0 \to A \to B \to C \to 0\) of objects of \(\mathcal{A}\). Since \(\Hom_{K(\mathcal{A})}(C[0], A[1]) = 0\) we see that any distinguished triangle arising from this short exact sequence would look like \((A[0], B[0], C[0], a, b, 0)\). But the existence of such a distinguished triangle in \(K(\mathcal{A})\) implies that the extension is split. A contradiction.

It turns out that the functor \(\text{Comp}(\mathcal{A}) \to D(\mathcal{A})\) is a \(\delta\)-functor. In order to see this we have to define the morphisms \(\delta\) associated to a short exact sequence \[0 \to A^\bullet \xrightarrow{a} B^\bullet \xrightarrow{b} C^\bullet \to 0\] of complexes in the abelian category \(\mathcal{A}\). Consider the cone \(C(a)^\bullet\) of the morphism \(a\). We have \(C(a)^n = B^n \oplus A^{n + 1}\) and we define \(q^n : C(a)^n \to C^n\) via the projection to \(B^n\) followed by \(b^n\). Hence a morphism of complexes \[q : C(a)^\bullet \longrightarrow C^\bullet.\] It is clear that \(q \circ i = b\) where \(i\) is as in Definition 014E. Note that, as \(a^\bullet\) is injective in each degree, the kernel of \(q\) is identified with the cone of \(\text{id}_{A^\bullet}\) which is acyclic. Hence we see that \(q\) is a quasi-isomorphism. According to Lemma 014L the triangle \[(A, B, C(a), a, i, -p)\] is a distinguished triangle in \(K(\mathcal{A})\). As the localization functor \(K(\mathcal{A}) \to D(\mathcal{A})\) is exact we see that \((A, B, C(a), a, i, -p)\) is a distinguished triangle in \(D(\mathcal{A})\). Since \(q\) is a quasi-isomorphism we see that \(q\) is an isomorphism in \(D(\mathcal{A})\). Hence we deduce that \[(A, B, C, a, b, -p \circ q^{-1})\] is a distinguished triangle of \(D(\mathcal{A})\). This suggests the following lemma.

Lemma

Let \(\mathcal{A}\) be an abelian category. The functor \(\text{Comp}(\mathcal{A}) \to D(\mathcal{A})\) defined above has the natural structure of a \(\delta\)-functor, with \[\delta_{A^\bullet \to B^\bullet \to C^\bullet} = - p \circ q^{-1}\] with \(p\) and \(q\) as explained above. The same construction turns the functors \(\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})\), \(\text{Comp}^{-}(\mathcal{A}) \to D^{-}(\mathcal{A})\), and \(\text{Comp}^b(\mathcal{A}) \to D^b(\mathcal{A})\) into \(\delta\)-functors.

Proof

We have already seen that this choice leads to a distinguished triangle whenever given a short exact sequence of complexes. We have to show that given a commutative diagram \[\xymatrix{ 0 \ar[r] & A^\bullet \ar[r]_a \ar[d]_f & B^\bullet \ar[r]_b \ar[d]_g & C^\bullet \ar[r] \ar[d]_h & 0 \\ 0 \ar[r] & (A')^\bullet \ar[r]^{a'} & (B')^\bullet \ar[r]^{b'} & (C')^\bullet \ar[r] & 0 }\] we get the desired commutative diagram of Definition 0150 (2). By Lemma 014F the pair \((f, g)\) induces a canonical morphism \(c : C(a)^\bullet \to C(a')^\bullet\). It is a simple computation to show that \(q' \circ c = h \circ q\) and \(f[1] \circ p = p' \circ c\). From this the result follows directly.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \[\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & D^\bullet \ar[r] & E^\bullet \ar[r] & F^\bullet \ar[r] & 0 }\] be a commutative diagram of morphisms of complexes such that the rows are short exact sequences of complexes, and the vertical arrows are quasi-isomorphisms. The \(\delta\)-functor of Lemma 0152 above maps the short exact sequences \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) and \(0 \to D^\bullet \to E^\bullet \to F^\bullet \to 0\) to isomorphic distinguished triangles.

Proof

Trivial from the fact that \(K(\mathcal{A}) \to D(\mathcal{A})\) transforms quasi-isomorphisms into isomorphisms and that the associated distinguished triangles are functorial.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \[\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r] & 0 }\] be a short exact sequence of complexes. Assume this short exact sequence is termwise split. Let \((A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta)\) be the distinguished triangle of \(K(\mathcal{A})\) associated to the sequence. The \(\delta\)-functor of Lemma 0152 above maps the short exact sequence \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) to a triangle isomorphic to the distinguished triangle \[(A^\bullet, B^\bullet, C^\bullet, \alpha, \beta, \delta).\]

Proof

Follows from Lemma 014L.

Remark

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet\) be a complex of \(\mathcal{A}\). Let \(a \in \mathbf{Z}\). We claim there is a canonical distinguished triangle \[\tau_{\leq a}K^\bullet \to K^\bullet \to \tau_{\geq a + 1}K^\bullet \to (\tau_{\leq a}K^\bullet)[1]\] in \(D(\mathcal{A})\). Here we have used the canonical truncation functors \(\tau\) from Homology, Section 0118. Namely, we first take the distinguished triangle associated by our \(\delta\)-functor (Lemma 0152) to the short exact sequence of complexes \[0 \to \tau_{\leq a}K^\bullet \to K^\bullet \to K^\bullet/\tau_{\leq a}K^\bullet \to 0\] Next, we use that the map \(K^\bullet \to \tau_{\geq a + 1}K^\bullet\) factors through a quasi-isomorphism \(K^\bullet/\tau_{\leq a}K^\bullet \to \tau_{\geq a + 1}K^\bullet\) by the description of cohomology groups in Homology, Section 0118. In a similar way we obtain canonical distinguished triangles \[\tau_{\leq a}K^\bullet \to \tau_{\leq a + 1}K^\bullet \to H^{a + 1}(K^\bullet)[-a-1] \to (\tau_{\leq a}K^\bullet)[1]\] and \[H^a(K^\bullet)[-a] \to \tau_{\geq a}K^\bullet \to \tau_{\geq a + 1}K^\bullet \to H^a(K^\bullet)[-a + 1]\]

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \[K_0^\bullet \to K_1^\bullet \to \ldots \to K_n^\bullet\] be maps of complexes such that

1. \(H^i(K_0^\bullet) = 0\) for \(i > 0\),

2. \(H^{-j}(K_j^\bullet) \to H^{-j}(K_{j + 1}^\bullet)\) is zero.

Then the composition \(K_0^\bullet \to K_n^\bullet\) factors through \(\tau_{\leq -n}K_n^\bullet \to K_n^\bullet\) in \(D(\mathcal{A})\). Dually, given maps of complexes \[K_n^\bullet \to K_{n - 1}^\bullet \to \ldots \to K_0^\bullet\] such that

1. \(H^i(K_0^\bullet) = 0\) for \(i < 0\),

2. \(H^j(K_{j + 1}^\bullet) \to H^j(K_j^\bullet)\) is zero,

then the composition \(K_n^\bullet \to K_0^\bullet\) factors through \(K_n^\bullet \to \tau_{\geq n}K_n^\bullet\) in \(D(\mathcal{A})\).

Proof

The case \(n = 1\). Since \(\tau_{\leq 0}K_0^\bullet = K_0^\bullet\) in \(D(\mathcal{A})\) we can replace \(K_0^\bullet\) by \(\tau_{\leq 0}K_0^\bullet\) and \(K_1^\bullet\) by \(\tau_{\leq 0}K_1^\bullet\). Consider the distinguished triangle \[\tau_{\leq -1}K_1^\bullet \to K_1^\bullet \to H^0(K_1^\bullet)[0] \to (\tau_{\leq -1}K_1^\bullet)[1]\] (Remark 08J5). The composition \(K_0^\bullet \to K_1^\bullet \to H^0(K_1^\bullet)[0]\) is zero as it is equal to \(K_0^\bullet \to H^0(K_0^\bullet)[0] \to H^0(K_1^\bullet)[0]\) which is zero by assumption. The fact that \(\Hom_{D(\mathcal{A})}(K_0^\bullet, -)\) is a homological functor (Lemma 0149), allows us to find the desired factorization. For \(n = 2\) we get a factorization \(K_0^\bullet \to \tau_{\leq -1}K_1^\bullet\) by the case \(n = 1\) and we can apply the case \(n = 1\) to the map of complexes \(\tau_{\leq -1}K_1^\bullet \to \tau_{\leq -1}K_2^\bullet\) to get a factorization \(\tau_{\leq -1}K_1^\bullet \to \tau_{\leq -2}K_2^\bullet\). The general case is proved in exactly the same manner.

Filtered derived categories

A reference for this section is [cotangent, I, Chapter V]. Let \(\mathcal{A}\) be an abelian category. In this section we will define the filtered derived category \(DF(\mathcal{A})\) of \(\mathcal{A}\). In short, we will define it as the derived category of the exact category of objects of \(\mathcal{A}\) endowed with a finite filtration. (Thus our construction is a special case of a more general construction of the derived category of an exact category, see for example [Buhler], [Keller].) Illusie’s filtered derived category is the full subcategory of ours consisting of those objects whose filtration is finite. (In our category the filtration is still finite in each degree, but may not be uniformly bounded.) The rationale for our choice is that it is not harder and it allows us to apply the discussion to the spectral sequences of Lemma 015J, see also Remark 015K.

We will use the notation regarding filtered objects introduced in Homology, Section 0120. The category of filtered objects of \(\mathcal{A}\) is denoted \(\text{Fil}(\mathcal{A})\). All filtrations will be decreasing by fiat.

Definition

Let \(\mathcal{A}\) be an abelian category. The category of finite filtered objects of \(\mathcal{A}\) is the category of filtered objects \((A, F)\) of \(\mathcal{A}\) whose filtration \(F\) is finite. We denote it \(\text{Fil}^f(\mathcal{A})\).

Thus \(\text{Fil}^f(\mathcal{A})\) is a full subcategory of \(\text{Fil}(\mathcal{A})\). For each \(p \in \mathbf{Z}\) there is a functor \(\text{gr}^p : \text{Fil}^f(\mathcal{A}) \to \mathcal{A}\). There is a functor \[\text{gr} = \bigoplus\nolimits_{p \in \mathbf{Z}} \text{gr}^p : \text{Fil}^f(\mathcal{A}) \to \text{Gr}(\mathcal{A})\] where \(\text{Gr}(\mathcal{A})\) is the category of graded objects of \(\mathcal{A}\), see Homology, Definition 0125. Finally, there is a functor \[(\text{forget }F) : \text{Fil}^f(\mathcal{A}) \longrightarrow \mathcal{A}\] which associates to the filtered object \((A, F)\) the underlying object of \(\mathcal{A}\). The category \(\text{Fil}^f(\mathcal{A})\) is an additive category, but not abelian in general, see Homology, Example 0108.

Because the functors \(\text{gr}^p\), \(\text{gr}\), \((\text{forget }F)\) are additive they induce exact functors of triangulated categories \[\text{gr}^p, (\text{forget }F) : K(\text{Fil}^f(\mathcal{A})) \to K(\mathcal{A}) \quad\text{and}\quad \text{gr} : K(\text{Fil}^f(\mathcal{A})) \to K(\text{Gr}(\mathcal{A}))\] by Lemma 014X. By analogy with the case of the homotopy category of an abelian category we make the following definitions.

Definition

Let \(\mathcal{A}\) be an abelian category.

1. Let \(\alpha : K^\bullet \to L^\bullet\) be a morphism of \(K(\text{Fil}^f(\mathcal{A}))\). We say that \(\alpha\) is a filtered quasi-isomorphism if the morphism \(\text{gr}(\alpha)\) is a quasi-isomorphism.

2. Let \(K^\bullet\) be an object of \(K(\text{Fil}^f(\mathcal{A}))\). We say that \(K^\bullet\) is filtered acyclic if the complex \(\text{gr}(K^\bullet)\) is acyclic.

Note that \(\alpha : K^\bullet \to L^\bullet\) is a filtered quasi-isomorphism if and only if each \(\text{gr}^p(\alpha)\) is a quasi-isomorphism. Similarly a complex \(K^\bullet\) is filtered acyclic if and only if each \(\text{gr}^p(K^\bullet)\) is acyclic.

Lemma

Let \(\mathcal{A}\) be an abelian category.

1. The functor \(K(\text{Fil}^f(\mathcal{A})) \longrightarrow \text{Gr}(\mathcal{A})\), \(K^\bullet \longmapsto H^0(\text{gr}(K^\bullet))\) is homological.

2. The functor \(K(\text{Fil}^f(\mathcal{A})) \rightarrow \mathcal{A}\), \(K^\bullet \longmapsto H^0(\text{gr}^p(K^\bullet))\) is homological.

3. The functor \(K(\text{Fil}^f(\mathcal{A})) \longrightarrow \mathcal{A}\), \(K^\bullet \longmapsto H^0((\text{forget }F)K^\bullet)\) is homological.

Proof

This follows from the fact that \(H^0 : K(\mathcal{A}) \to \mathcal{A}\) is homological, see Lemma 05RS and the fact that the functors \(\text{gr}, \text{gr}^p, (\text{forget }F)\) are exact functors of triangulated categories. See Lemma 05QZ.

Lemma

Let \(\mathcal{A}\) be an abelian category. The full subcategory \(\text{FAc}(\mathcal{A})\) of \(K(\text{Fil}^f(\mathcal{A}))\) consisting of filtered acyclic complexes is a strictly full saturated triangulated subcategory of \(K(\text{Fil}^f(\mathcal{A}))\). The corresponding saturated multiplicative system (see Lemma 05RL) of \(K(\text{Fil}^f(\mathcal{A}))\) is the set \(\text{FQis}(\mathcal{A})\) of filtered quasi-isomorphisms. In particular, the kernel of the localization functor \[Q : K(\text{Fil}^f(\mathcal{A})) \longrightarrow \text{FQis}(\mathcal{A})^{-1}K(\text{Fil}^f(\mathcal{A}))\] is \(\text{FAc}(\mathcal{A})\) and the functor \(H^0 \circ \text{gr}\) factors through \(Q\).

Proof

We know that \(H^0 \circ \text{gr}\) is a homological functor by Lemma 05S0. Thus this lemma is a special case of Lemma 05RM.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(\text{FAc}(\mathcal{A})\) and \(\text{FQis}(\mathcal{A})\) be as in Lemma 05S1. The filtered derived category of \(\mathcal{A}\) is the triangulated category \[DF(\mathcal{A}) = K(\text{Fil}^f(\mathcal{A}))/\text{FAc}(\mathcal{A}) = \text{FQis}(\mathcal{A})^{-1} K(\text{Fil}^f(\mathcal{A})).\]

Lemma

The functors \(\text{gr}^p, \text{gr}, (\text{forget }F)\) induce canonical exact functors \[\text{gr}^p, (\text{forget }F): DF(\mathcal{A}) \longrightarrow D(\mathcal{A})\] and \[\text{gr}: DF(\mathcal{A}) \longrightarrow D(\text{Gr}(\mathcal{A}))\] which commute with the localization functors.

Proof

This follows from the universal property of localization, see Lemma 05R7, provided we can show that a filtered quasi-isomorphism is turned into a quasi-isomorphism by each of the functors \(\text{gr}^p, \text{gr}, (\text{forget }F)\). This is true by definition for the first two. For the last one we have to do a little bit of work. Let \(f : K^\bullet \to L^\bullet\) be a filtered quasi-isomorphism in \(K(\text{Fil}^f(\mathcal{A}))\). Choose a distinguished triangle \((K^\bullet, L^\bullet, M^\bullet, f, g, h)\) which contains \(f\). Then \(M^\bullet\) is filtered acyclic, see Lemma 05S1. Hence by the corresponding lemma for \(K(\mathcal{A})\) it suffices to show that a filtered acyclic complex is an acyclic complex if we forget the filtration. This follows from Homology, Lemma 05QH.

Definition

Let \(\mathcal{A}\) be an abelian category. The bounded filtered derived category \(DF^b(\mathcal{A})\) is the full subcategory of \(DF(\mathcal{A})\) with objects those \(X\) such that \(\text{gr}(X) \in D^b(\mathcal{A})\). Similarly for the bounded below filtered derived category \(DF^{+}(\mathcal{A})\) and the bounded above filtered derived category \(DF^{-}(\mathcal{A})\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet \in K(\text{Fil}^f(\mathcal{A}))\).

1. If \(H^n(\text{gr}(K^\bullet)) = 0\) for all \(n < a\), then there exists a filtered quasi-isomorphism \(K^\bullet \to L^\bullet\) with \(L^n = 0\) for all \(n < a\).

2. If \(H^n(\text{gr}(K^\bullet)) = 0\) for all \(n > b\), then there exists a filtered quasi-isomorphism \(M^\bullet \to K^\bullet\) with \(M^n = 0\) for all \(n > b\).

3. If \(H^n(\text{gr}(K^\bullet)) = 0\) for all \(|n| \gg 0\), then there exists a commutative diagram of morphisms of complexes \[\xymatrix{ K^\bullet \ar[r] & L^\bullet \\ M^\bullet \ar[u] \ar[r] & N^\bullet \ar[u] }\] where all the arrows are filtered quasi-isomorphisms, \(L^\bullet\) bounded below, \(M^\bullet\) bounded above, and \(N^\bullet\) a bounded complex.

Proof

Suppose that \(H^n(\text{gr}(K^\bullet)) = 0\) for all \(n < a\). By Homology, Lemma 05QH the sequence \[K^{a - 2} \xrightarrow{d^{a - 2}} K^{a - 1} \xrightarrow{d^{a - 1}} K^a\] is an exact sequence of objects of \(\mathcal{A}\) and the morphisms \(d^{a - 2}\) and \(d^{a - 1}\) are strict. Hence \(\Coim(d^{a - 1}) = \Im(d^{a - 1})\) in \(\text{Fil}^f(\mathcal{A})\) and the map \(\text{gr}(\Im(d^{a - 1})) \to \text{gr}(K^a)\) is injective with image equal to the image of \(\text{gr}(K^{a - 1}) \to \text{gr}(K^a)\), see Homology, Lemma 0127. This means that the map \(K^\bullet \to \tau_{\geq a}K^\bullet\) into the truncation \[\tau_{\geq a}K^\bullet = (\ldots \to 0 \to K^a/\Im(d^{a - 1}) \to K^{a + 1} \to \ldots)\] is a filtered quasi-isomorphism. This proves (1). The proof of (2) is dual to the proof of (1). Part (3) follows formally from (1) and (2).

To state the following lemma denote \(\text{FAc}^{+}(\mathcal{A})\), \(\text{FAc}^{-}(\mathcal{A})\), resp. \(\text{FAc}^b(\mathcal{A})\) the intersection of \(K^{+}(\text{Fil}^f\mathcal{A})\), \(K^{-}(\text{Fil}^f\mathcal{A})\), resp. \(K^b(\text{Fil}^f\mathcal{A})\) with \(\text{FAc}(\mathcal{A})\). Denote \(\text{FQis}^{+}(\mathcal{A})\), \(\text{FQis}^{-}(\mathcal{A})\), resp. \(\text{FQis}^b(\mathcal{A})\) the intersection of \(K^{+}(\text{Fil}^f\mathcal{A})\), \(K^{-}(\text{Fil}^f\mathcal{A})\), resp. \(K^b(\text{Fil}^f\mathcal{A})\) with \(\text{FQis}(\mathcal{A})\).

Lemma

Let \(\mathcal{A}\) be an abelian category. The subcategories \(\text{FAc}^{+}(\mathcal{A})\), \(\text{FAc}^{-}(\mathcal{A})\), resp. \(\text{FAc}^b(\mathcal{A})\) are strictly full saturated triangulated subcategories of \(K^{+}(\text{Fil}^f\mathcal{A})\), \(K^{-}(\text{Fil}^f\mathcal{A})\), resp. \(K^b(\text{Fil}^f\mathcal{A})\). The corresponding saturated multiplicative systems (see Lemma 05RL) are the sets \(\text{FQis}^{+}(\mathcal{A})\), \(\text{FQis}^{-}(\mathcal{A})\), resp. \(\text{FQis}^b(\mathcal{A})\).

1. The kernel of the functor \(K^{+}(\text{Fil}^f\mathcal{A}) \to DF^{+}(\mathcal{A})\) is \(\text{FAc}^{+}(\mathcal{A})\) and this induces an equivalence of triangulated categories \[K^{+}(\text{Fil}^f\mathcal{A})/\text{FAc}^{+}(\mathcal{A}) = \text{FQis}^{+}(\mathcal{A})^{-1}K^{+}(\text{Fil}^f\mathcal{A}) \longrightarrow DF^{+}(\mathcal{A})\]

2. The kernel of the functor \(K^{-}(\text{Fil}^f\mathcal{A}) \to DF^{-}(\mathcal{A})\) is \(\text{FAc}^{-}(\mathcal{A})\) and this induces an equivalence of triangulated categories \[K^{-}(\text{Fil}^f\mathcal{A})/\text{FAc}^{-}(\mathcal{A}) = \text{FQis}^{-}(\mathcal{A})^{-1}K^{-}(\text{Fil}^f\mathcal{A}) \longrightarrow DF^{-}(\mathcal{A})\]

3. The kernel of the functor \(K^b(\text{Fil}^f\mathcal{A}) \to DF^b(\mathcal{A})\) is \(\text{FAc}^b(\mathcal{A})\) and this induces an equivalence of triangulated categories \[K^b(\text{Fil}^f\mathcal{A})/\text{FAc}^b(\mathcal{A}) = \text{FQis}^b(\mathcal{A})^{-1}K^b(\text{Fil}^f\mathcal{A}) \longrightarrow DF^b(\mathcal{A})\]

Proof

This follows from the results above, in particular Lemma 05S5, by exactly the same arguments as used in the proof of Lemma 05RW.

Derived functors in general

A reference for this section is Deligne’s exposé XVII in [SGA4]. A very general notion of right and left derived functors exists where we have an exact functor between triangulated categories, a multiplicative system in the source category and we want to find the “correct” extension of the exact functor to the localized category.

Situation

Here \(F : \mathcal{D} \to \mathcal{D}'\) is an exact functor of triangulated categories and \(S\) is a saturated multiplicative system in \(\mathcal{D}\) compatible with the structure of triangulated category on \(\mathcal{D}\).

Let \(X \in \Ob(\mathcal{D})\). Recall from Categories, Remark 05Q0 the filtered category \(X/S\) of arrows \(s : X \to X'\) in \(S\) with source \(X\). Dually, in Categories, Remark 05Q4 we defined the cofiltered category \(S/X\) of arrows \(s : X' \to X\) in \(S\) with target \(X\).

Definition

Assumptions and notation as in Situation 05S8. Let \(X \in \Ob(\mathcal{D})\).

1. we say the right derived functor \(RF\) is defined at \(X\) if the ind-object \[(X/S) \longrightarrow \mathcal{D}', \quad (s : X \to X') \longmapsto F(X')\] is essentially constant⁵; in this case the value \(Y\) in \(\mathcal{D}'\) is called the value of \(RF\) at \(X\).

2. we say the left derived functor \(LF\) is defined at \(X\) if the pro-object \[(S/X) \longrightarrow \mathcal{D}', \quad (s: X' \to X) \longmapsto F(X')\] is essentially constant; in this case the value \(Y\) in \(\mathcal{D}'\) is called the value of \(LF\) at \(X\).

By abuse of notation we often denote the values simply \(RF(X)\) or \(LF(X)\).

It will turn out that the full subcategory of \(\mathcal{D}\) consisting of objects where \(RF\) is defined is a triangulated subcategory, and \(RF\) will define a functor on this subcategory which transforms morphisms of \(S\) into isomorphisms.

Lemma

Assumptions and notation as in Situation 05S8. Let \(f : X \to Y\) be a morphism of \(\mathcal{D}\).

1. If \(RF\) is defined at \(X\) and \(Y\) then there exists a unique morphism \(RF(f) : RF(X) \to RF(Y)\) between the values such that for any commutative diagram \[\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }\] with \(s, s' \in S\) the diagram \[\xymatrix{ F(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & RF(X) \ar[d] \\ F(Y) \ar[r] & F(Y') \ar[r] & RF(Y) }\] commutes.

2. If \(LF\) is defined at \(X\) and \(Y\) then there exists a unique morphism \(LF(f) : LF(X) \to LF(Y)\) between the values such that for any commutative diagram \[\xymatrix{ X' \ar[d]_{f'} \ar[r]_s & X \ar[d]^f \\ Y' \ar[r]^{s'} & Y }\] with \(s, s'\) in \(S\) the diagram \[\xymatrix{ LF(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & F(X) \ar[d] \\ LF(Y) \ar[r] & F(Y') \ar[r] & F(Y) }\] commutes.

Proof

Part (1) holds if we only assume that the colimits \[RF(X) = \colim_{s : X \to X'} F(X') \quad\text{and}\quad RF(Y) = \colim_{s' : Y \to Y'} F(Y')\] exist. Namely, to give a morphism \(RF(X) \to RF(Y)\) between the colimits is the same thing as giving for each \(s : X \to X'\) in \(\Ob(X/S)\) a morphism \(F(X') \to RF(Y)\) compatible with morphisms in the category \(X/S\). To get the morphism we choose a commutative diagram \[\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }\] with \(s, s'\) in \(S\) as is possible by MS2 and we set \(F(X') \to RF(Y)\) equal to the composition \(F(X') \to F(Y') \to RF(Y)\). To see that this is independent of the choice of the diagram above use MS3. Details omitted. The proof of (2) is dual.

Lemma

Assumptions and notation as in Situation 05S8. Let \(s : X \to Y\) be an element of \(S\).

1. \(RF\) is defined at \(X\) if and only if it is defined at \(Y\). In this case the map \(RF(s) : RF(X) \to RF(Y)\) between values is an isomorphism.

2. \(LF\) is defined at \(X\) if and only if it is defined at \(Y\). In this case the map \(LF(s) : LF(X) \to LF(Y)\) between values is an isomorphism.

Proof

Omitted.

Lemma

Assumptions and notation as in Situation 05S8. Let \(X\) be an object of \(\mathcal{D}\) and \(n \in \mathbf{Z}\).

1. \(RF\) is defined at \(X\) if and only if it is defined at \(X[n]\). In this case there is a canonical isomorphism \(RF(X)[n]= RF(X[n])\) between values.

2. \(LF\) is defined at \(X\) if and only if it is defined at \(X[n]\). In this case there is a canonical isomorphism \(LF(X)[n] \to LF(X[n])\) between values.

Proof

Omitted.

Lemma

Assumptions and notation as in Situation 05S8. Let \((X, Y, Z, f, g, h)\) be a distinguished triangle of \(\mathcal{D}\). If \(RF\) is defined at two out of three of \(X, Y, Z\), then it is defined at the third. Moreover, in this case \[(RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h))\] is a distinguished triangle in \(\mathcal{D}'\). Similarly for \(LF\).

Proof

Say \(RF\) is defined at \(X, Y\) with values \(A, B\). Let \(RF(f) : A \to B\) be the induced morphism, see Lemma 05SA. We may choose a distinguished triangle \((A, B, C, RF(f), b, c)\) in \(\mathcal{D}'\). We claim that \(C\) is a value of \(RF\) at \(Z\).

To see this pick \(s : X \to X'\) in \(S\) such that there exists a morphism \(\alpha : A \to F(X')\) as in Categories, Definition 05PU. We may choose a commutative diagram \[\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }\] with \(s' \in S\) by MS2. Using that \(Y/S\) is filtered we can (after replacing \(s'\) by some \(s'' : Y \to Y''\) in \(S\)) assume that there exists a morphism \(\beta : B \to F(Y')\) as in Categories, Definition 05PU. Picture \[\xymatrix{ A \ar[d]_{RF(f)} \ar[r]_-\alpha & F(X') \ar[r] \ar[d]^{F(f')} & A \ar[d]^{RF(f)} \\ B \ar[r]^-\beta & F(Y') \ar[r] & B }\] It may not be true that the left square commutes, but the outer and right squares commute. The assumption that the ind-object \(\{F(Y')\}_{s' : Y \to Y'}\) is essentially constant means that there exists a \(s'' : Y \to Y''\) in \(S\) and a morphism \(h : Y' \to Y''\) such that \(s'' = h \circ s'\) and such that \(F(h)\) equal to \(F(Y') \to B \to F(Y') \to F(Y'')\). Hence after replacing \(Y'\) by \(Y''\) and \(\beta\) by \(F(h) \circ \beta\) the diagram will commute (by direct computation with arrows).

Using MS6 choose a morphism of triangles \[(s, s', s'') : (X, Y, Z, f, g, h) \longrightarrow (X', Y', Z', f', g', h')\] with \(s'' \in S\). By TR3 choose a morphism of triangles \[(\alpha, \beta, \gamma) : (A, B, C, RF(f), b, c) \longrightarrow (F(X'), F(Y'), F(Z'), F(f'), F(g'), F(h'))\]

By Lemma 05SB it suffices to prove that \(RF(Z')\) is defined and that the arrow \(\gamma : C \to F(Z')\) induces an isomorphism \(C \to RF(Z')\). Namely, then we will get an isomorphism \[(A, B, C, RF(f), b, c) \longrightarrow (RF(X'), RF(Y'), RF(Z'), RF(f'), RF(g'), RF(h'))\] of triangles and by TR1 we conclude that the target is a distinguished triangle. Consider the category \(\mathcal{I}\) of Lemma 05R9 of triangles \[\mathcal{I} = \{(t, t', t'') : (X', Y', Z', f', g', h') \to (X'', Y'', Z'', f'', g'', h'') \mid (t, t', t'') \in S\}\] To show that the system \(F(Z'')\) is essentially constant over the category \(Z'/S\) is equivalent to showing that the system of \(F(Z'')\) is essentially constant over \(\mathcal{I}\) because \(\mathcal{I} \to Z'/S\) is cofinal, see Categories, Lemma 0A1S (cofinality is proven in Lemma 05R9). For any object \(W\) in \(\mathcal{D}'\) we consider the diagram \[\xymatrix{ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(X'')) \ar[d] & \Mor_{\mathcal{D}'}(W, A) \ar[l] \ar[d] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Y'')) \ar[d] & \Mor_{\mathcal{D}'}(W, B) \ar[d] \ar[l] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Z'')) \ar[d] & \Mor_{\mathcal{D}'}(W, C) \ar[d] \ar[l] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(X''[1])) \ar[d] & \Mor_{\mathcal{D}'}(W, A[1]) \ar[d] \ar[l] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Y''[1])) & \Mor_{\mathcal{D}'}(W, B[1]) \ar[l] }\] where the horizontal arrows are given by composing with \((\alpha, \beta, \gamma)\). Since filtered colimits are exact (Algebra, Lemma 00DB) the left column is an exact sequence. Thus the \(5\) lemma (Homology, Lemma 05QB) tells us the map \[\colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Z'')) \longrightarrow \Mor_{\mathcal{D}'}(W, C)\] is bijective. We conclude that \(F(Z'')\) is essentially constant over \(\mathcal{I}\) with value \(C\) by part (4) of Categories, Lemma 05PY.

Lemma

Assumptions and notation as in Situation 05S8. Let \(X, Y\) be objects of \(\mathcal{D}\).

1. If \(RF\) is defined at \(X\) and \(Y\), then \(RF\) is defined at \(X \oplus Y\).

2. If \(\mathcal{D}'\) is Karoubian and \(RF\) is defined at \(X \oplus Y\), then \(RF\) is defined at both \(X\) and \(Y\).

In either case we have \(RF(X \oplus Y) = RF(X) \oplus RF(Y)\). Similarly for \(LF\).

Proof

If \(RF\) is defined at \(X\) and \(Y\), then the distinguished triangle \(X \to X \oplus Y \to Y \to X[1]\) (Lemma 05QT) and Lemma 05SC shows that \(RF\) is defined at \(X \oplus Y\) and that we have a distinguished triangle \(RF(X) \to RF(X \oplus Y) \to RF(Y) \to RF(X)[1]\). Applying Lemma 05QT to this once more we find that \(RF(X \oplus Y) = RF(X) \oplus RF(Y)\). This proves (1) and the final assertion.

Conversely, assume that \(RF\) is defined at \(X \oplus Y\) and that \(\mathcal{D}'\) is Karoubian. Since \(S\) is a saturated system \(S\) is the set of arrows which become invertible under the additive localization functor \(Q : \mathcal{D} \to S^{-1}\mathcal{D}\), see Categories, Lemma 05Q9. Thus for any \(s : X \to X'\) and \(s' : Y \to Y'\) in \(S\) the morphism \(s \oplus s' : X \oplus Y \to X' \oplus Y'\) is an element of \(S\). In this way we obtain a functor \[X/S \times Y/S \longrightarrow (X \oplus Y)/S\] Recall that the categories \(X/S, Y/S, (X \oplus Y)/S\) are filtered (Categories, Remark 05Q0). By Categories, Lemma 0A2C \(X/S \times Y/S\) is filtered and \(F|_{X/S} : X/S \to \mathcal{D}'\) (resp. \(F|_{Y/S} : Y/S \to \mathcal{D}'\)) is essentially constant if and only if \(F|_{X/S} \circ \text{pr}_1 : X/S \times Y/S \to \mathcal{D}'\) (resp. \(F|_{Y/S} \circ \text{pr}_2 : X/S \times Y/S \to \mathcal{D}'\)) is essentially constant. Below we will show that the displayed functor is cofinal, hence by Categories, Lemma 0A1S, we see that \(F|_{(X \oplus Y)/S}\) is essentially constant implies that \(F|_{X/S} \circ \text{pr}_1 \oplus F|_{Y/S} \circ \text{pr}_2 : X/S \times Y/S \to \mathcal{D}'\) is essentially constant. By Homology, Lemma 0A2G (and this is where we use that \(\mathcal{D}'\) is Karoubian) we see that \(F|_{X/S} \circ \text{pr}_1 \oplus F|_{Y/S} \circ \text{pr}_2\) being essentially constant implies \(F|_{X/S} \circ \text{pr}_1\) and \(F|_{Y/S} \circ \text{pr}_2\) are essentially constant proving that \(RF\) is defined at \(X\) and \(Y\).

Proof that the displayed functor is cofinal. To do this pick any \(t : X \oplus Y \to Z\) in \(S\). Using MS2 we can find morphisms \(Z \to X'\), \(Z \to Y'\) and \(s : X \to X'\), \(s' : Y \to Y'\) in \(S\) such that \[\xymatrix{ X \ar[d]^s & X \oplus Y \ar[d] \ar[l] \ar[r] & Y \ar[d]_{s'} \\ X' & Z \ar[l] \ar[r] & Y' }\] commutes. This proves there is a map \(Z \to X' \oplus Y'\) in \((X \oplus Y)/S\), i.e., we get part (1) of Categories, Definition 04E6. To prove part (2) it suffices to prove that given \(t : X \oplus Y \to Z\) and morphisms \(s_i \oplus s'_i : Z \to X'_i \oplus Y'_i\), \(i = 1, 2\) in \((X \oplus Y)/S\) we can find morphisms \(a : X'_1 \to X'\), \(b : X'_2 \to X'\), \(c : Y'_1 \to Y'\), \(d : Y'_2 \to Y'\) in \(S\) such that \(a \circ s_1 = b \circ s_2\) and \(c \circ s'_1 = d \circ s'_2\). To do this we first choose any \(X'\) and \(Y'\) and maps \(a, b, c, d\) in \(S\); this is possible as \(X/S\) and \(Y/S\) are filtered. Then the two maps \(a \circ s_1, b \circ s_2 : Z \to X'\) become equal in \(S^{-1}\mathcal{D}\). Hence we can find a morphism \(X' \to X''\) in \(S\) equalizing them. Similarly we find \(Y' \to Y''\) in \(S\) equalizing \(c \circ s'_1\) and \(d \circ s'_2\). Replacing \(X'\) by \(X''\) and \(Y'\) by \(Y''\) we get \(a \circ s_1 = b \circ s_2\) and \(c \circ s'_1 = d \circ s'_2\).

The proof of the corresponding statements for \(LF\) are dual.

Proposition

Assumptions and notation as in Situation 05S8.

1. The full subcategory \(\mathcal{E}\) of \(\mathcal{D}\) consisting of objects at which \(RF\) is defined is a strictly full triangulated subcategory of \(\mathcal{D}\).

2. We obtain an exact functor \(RF : \mathcal{E} \longrightarrow \mathcal{D}'\) of triangulated categories.

3. Elements of \(S\) with either source or target in \(\mathcal{E}\) are morphisms of \(\mathcal{E}\).

4. Any element of \(S_\mathcal{E} = \text{Arrows}(\mathcal{E}) \cap S\) is mapped to an isomorphism by \(RF\).

5. The set \(S_\mathcal{E}\) is a saturated multiplicative system in \(\mathcal{E}\) compatible with the triangulated structure.

6. The functor \(S_\mathcal{E}^{-1}\mathcal{E} \to S^{-1}\mathcal{D}\) is a fully faithful exact functor of triangulated categories.

7. We obtain an exact functor \[RF : S_\mathcal{E}^{-1}\mathcal{E} \longrightarrow \mathcal{D}'.\]

8. If \(\mathcal{D}'\) is Karoubian, then \(\mathcal{E}\) is a saturated triangulated subcategory of \(\mathcal{D}\).

A similar result holds for \(LF\).

Proof

Since \(S\) is saturated it contains all isomorphisms (see remark following Categories, Definition 05Q8). Hence (1) follows from Lemmas 05SB, 05SC, and 05SU. We get (2) from Lemmas 05SA, 05SU, and 05SC. We get (3) from Lemma 05SB. Part (4) follows from Lemma 05SB. Part (5) follows from the definitions and part (3). The fully faithfulness in (6) follows from (3) and the definitions. The fact that \(S_\mathcal{E}^{-1}\mathcal{E} \to S^{-1}\mathcal{D}\) is exact follows from the fact that a triangle in \(S_\mathcal{E}^{-1}\mathcal{E}\) is distinguished if and only if it is isomorphic to the image of a distinguished triangle in \(\mathcal{E}\), see proof of Proposition 05R6. The factorization of \(RF : \mathcal{E} \to \mathcal{D}'\) through an exact functor \(S_\mathcal{E}^{-1}\mathcal{E} \to \mathcal{D}'\) follows from Lemma 05R7. Finally, part (8) follows from Lemma 05SD.

Proposition 05SE tells us that \(RF\) lives on a maximal strictly full triangulated subcategory of \(S^{-1}\mathcal{D}\) and is an exact functor on this triangulated category. Picture: \[\xymatrix{ \mathcal{D} \ar[d]_Q \ar[rrr]_F & & & \mathcal{D}' \\ S^{-1}\mathcal{D} & & S_\mathcal{E}^{-1}\mathcal{E} \ar[ll]_{\text{fully faithful}}^{\text{exact}} \ar[ur]_{RF} }\]

Definition

In Situation 05S8. We say \(F\) is right derivable, or that \(RF\) everywhere defined if \(RF\) is defined at every object of \(\mathcal{D}\). We say \(F\) is left derivable, or that \(LF\) everywhere defined if \(LF\) is defined at every object of \(\mathcal{D}\).

In this case we obtain a right (resp. left) derived functor [05SW]\[\begin{equation} RF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}', \quad\text{(resp. } LF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}'), \end{equation}\] see Proposition 05SE. In most interesting situations it is not the case that \(RF \circ Q\) is equal to \(F\). In fact, it might happen that the canonical map \(F(X) \to RF(X)\) is never an isomorphism. In practice this does not happen, because in practice we only know how to prove \(F\) is right derivable by showing that \(RF\) can be computed by evaluating \(F\) at judiciously chosen objects of the triangulated category \(\mathcal{D}\). This warrants a definition.

Definition

In Situation 05S8.

1. An object \(X\) of \(\mathcal{D}\) computes \(RF\) if \(RF\) is defined at \(X\) and the canonical map \(F(X) \to RF(X)\) is an isomorphism.

2. An object \(X\) of \(\mathcal{D}\) computes \(LF\) if \(LF\) is defined at \(X\) and the canonical map \(LF(X) \to F(X)\) is an isomorphism.

Lemma

Assumptions and notation as in Situation 05S8. Let \(X\) be an object of \(\mathcal{D}\) and \(n \in \mathbf{Z}\).

1. \(X\) computes \(RF\) if and only if \(X[n]\) computes \(RF\).

2. \(X\) computes \(LF\) if and only if \(X[n]\) computes \(LF\).

Proof

Omitted.

Lemma

Assumptions and notation as in Situation 05S8. Let \((X, Y, Z, f, g, h)\) be a distinguished triangle of \(\mathcal{D}\). If \(X, Y\) compute \(RF\) then so does \(Z\). Similar for \(LF\).

Proof

By Lemma 05SC we know that \(RF\) is defined at \(Z\) and that \(RF\) applied to the triangle produces a distinguished triangle. Consider the morphism of distinguished triangles \[\xymatrix{ (F(X), F(Y), F(Z), F(f), F(g), F(h)) \ar[d] \\ (RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h)) }\] Two out of three maps are isomorphisms, hence so is the third.

Lemma

Assumptions and notation as in Situation 05S8. Let \(X, Y\) be objects of \(\mathcal{D}\). If \(X \oplus Y\) computes \(RF\), then \(X\) and \(Y\) compute \(RF\). Similarly for \(LF\).

Proof

If \(X \oplus Y\) computes \(RF\), then \(RF(X \oplus Y) = F(X) \oplus F(Y)\). In the proof of Lemma 05SD we have seen that the functor \(X/S \times Y/S \to (X \oplus Y)/S\), \((s, s') \mapsto s \oplus s'\) is cofinal. Thus by Categories, Lemma 0A1S and by characterization (4) of Categories, Lemma 05PY we know that for any object \(W\) in \(\mathcal{D}'\) the map \[\Hom_{\mathcal{D}'}(F(X \oplus Y), W) \longrightarrow \colim_{s : X \to X', s' : Y \to Y'} \Hom_{\mathcal{D}'}(F(X' \oplus Y'), W)\] is bijective. Since this arrow is clearly compatible with direct sum decompositions on both sides, we conclude that the map \[\Hom_{\mathcal{D}'}(F(X), W) \longrightarrow \colim_{s : X \to X'} \Hom_{\mathcal{D}'}(F(X'), W)\] is bijective (minor detail omitted). Thus by Categories, Lemma 05PY we conclude \(RF\) is defined at \(X\) with value \(F(X)\). Similarly for \(Y\).

Lemma

Assumptions and notation as in Situation 05S8.

1. If for every object \(X \in \Ob(\mathcal{D})\) there exists an arrow \(s : X \to X'\) in \(S\) such that \(X'\) computes \(RF\), then \(RF\) is everywhere defined.

2. If for every object \(X \in \Ob(\mathcal{D})\) there exists an arrow \(s : X' \to X\) in \(S\) such that \(X'\) computes \(LF\), then \(LF\) is everywhere defined.

Proof

This is clear from the definitions.

Lemma

Assumptions and notation as in Situation 05S8. If there exists a subset \(\mathcal{I} \subset \Ob(\mathcal{D})\) such that

1. for all \(X \in \Ob(\mathcal{D})\) there exists \(s : X \to X'\) in \(S\) with \(X' \in \mathcal{I}\), and

2. for every arrow \(s : X \to X'\) in \(S\) with \(X, X' \in \mathcal{I}\) the map \(F(s) : F(X) \to F(X')\) is an isomorphism,

then \(RF\) is everywhere defined and every \(X \in \mathcal{I}\) computes \(RF\). Dually, if there exists a subset \(\mathcal{P} \subset \Ob(\mathcal{D})\) such that

1. for all \(X \in \Ob(\mathcal{D})\) there exists \(s : X' \to X\) in \(S\) with \(X' \in \mathcal{P}\), and

2. for every arrow \(s : X \to X'\) in \(S\) with \(X, X' \in \mathcal{P}\) the map \(F(s) : F(X) \to F(X')\) is an isomorphism,

then \(LF\) is everywhere defined and every \(X \in \mathcal{P}\) computes \(LF\).

Proof

Let \(X\) be an object of \(\mathcal{D}\). Assumption (1) implies that the arrows \(s : X \to X'\) in \(S\) with \(X' \in \mathcal{I}\) are cofinal in the category \(X/S\). Assumption (2) implies that \(F\) is constant on this cofinal subcategory. Clearly this implies that \(F : (X/S) \to \mathcal{D}'\) is essentially constant with value \(F(X')\) for any \(s : X \to X'\) in \(S\) with \(X' \in \mathcal{I}\).

Lemma

Let \(\mathcal{A}, \mathcal{B}, \mathcal{C}\) be triangulated categories. Let \(S\), resp. \(S'\) be a saturated multiplicative system in \(\mathcal{A}\), resp. \(\mathcal{B}\) compatible with the triangulated structure. Let \(F : \mathcal{A} \to \mathcal{B}\) and \(G : \mathcal{B} \to \mathcal{C}\) be exact functors. Denote \(F' : \mathcal{A} \to (S')^{-1}\mathcal{B}\) the composition of \(F\) with the localization functor.

1. If \(RF'\), \(RG\), \(R(G \circ F)\) are everywhere defined, then there is a canonical transformation of functors \(t : R(G \circ F) \longrightarrow RG \circ RF'\).

2. If \(LF'\), \(LG\), \(L(G \circ F)\) are everywhere defined, then there is a canonical transformation of functors \(t : LG \circ LF' \to L(G \circ F)\).

Proof

In this proof we try to be careful. Hence let us think of the derived functors as the functors \[RF' : S^{-1}\mathcal{A} \to (S')^{-1}\mathcal{B}, \quad R(G \circ F) : S^{-1}\mathcal{A} \to \mathcal{C}, \quad RG : (S')^{-1}\mathcal{B} \to \mathcal{C}.\] Let us denote \(Q_A : \mathcal{A} \to S^{-1}\mathcal{A}\) and \(Q_B : \mathcal{B} \to (S')^{-1}\mathcal{B}\) the localization functors. Then \(F' = Q_B \circ F\). Note that for every object \(Y\) of \(\mathcal{B}\) there is a canonical map \[G(Y) \longrightarrow RG(Q_B(Y))\] in other words, there is a transformation of functors \(t' : G \to RG \circ Q_B\). Let \(X\) be an object of \(\mathcal{A}\). We have \[\begin{align*} R(G \circ F)(Q_A(X)) & = \colim_{s : X \to X' \in S} G(F(X')) \\ & \xrightarrow{t'} \colim_{s : X \to X' \in S} RG(Q_B(F(X'))) \\ & = \colim_{s : X \to X' \in S} RG(F'(X')) \\ & = RG(\colim_{s : X \to X' \in S} F'(X')) \\ & = RG(RF'(X)). \end{align*}\] The system \(F'(X')\) is essentially constant in the category \((S')^{-1}\mathcal{B}\). Hence we may pull the colimit inside the functor \(RG\) in the third equality of the diagram above, see Categories, Lemma 05SH and its proof. We omit the proof this defines a transformation of functors. The case of left derived functors is similar.

Derived functors on derived categories

In practice derived functors come about most often when given an additive functor between abelian categories.

Situation

Here \(F : \mathcal{A} \to \mathcal{B}\) is an additive functor between abelian categories. This induces exact functors \[F : K(\mathcal{A}) \to K(\mathcal{B}), \quad K^{+}(\mathcal{A}) \to K^{+}(\mathcal{B}), \quad K^{-}(\mathcal{A}) \to K^{-}(\mathcal{B}).\] See Lemma 014X. We also denote \(F\) the composition \(K(\mathcal{A}) \to D(\mathcal{B})\), \(K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\), and \(K^{-}(\mathcal{A}) \to D^-(\mathcal{B})\) of \(F\) with the localization functor \(K(\mathcal{B}) \to D(\mathcal{B})\), etc. This situation leads to four derived functors we will consider in the following.

1. The right derived functor of \(F : K(\mathcal{A}) \to D(\mathcal{B})\) relative to the multiplicative system \(\text{Qis}(\mathcal{A})\).

2. The right derived functor of \(F : K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) relative to the multiplicative system \(\text{Qis}^{+}(\mathcal{A})\).

3. The left derived functor of \(F : K(\mathcal{A}) \to D(\mathcal{B})\) relative to the multiplicative system \(\text{Qis}(\mathcal{A})\).

4. The left derived functor of \(F : K^{-}(\mathcal{A}) \to D^{-}(\mathcal{B})\) relative to the multiplicative system \(\text{Qis}^-(\mathcal{A})\).

Each of these cases is an example of Situation 05S8.

Some of the ambiguity that may arise is alleviated by the following.

Lemma

In Situation 05T4.

1. Let \(X\) be an object of \(K^{+}(\mathcal{A})\). The right derived functor of \(K(\mathcal{A}) \to D(\mathcal{B})\) is defined at \(X\) if and only if the right derived functor of \(K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is defined at \(X\). Moreover, the values are canonically isomorphic.

2. Let \(X\) be an object of \(K^{+}(\mathcal{A})\). Then \(X\) computes the right derived functor of \(K(\mathcal{A}) \to D(\mathcal{B})\) if and only if \(X\) computes the right derived functor of \(K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\).

3. Let \(X\) be an object of \(K^{-}(\mathcal{A})\). The left derived functor of \(K(\mathcal{A}) \to D(\mathcal{B})\) is defined at \(X\) if and only if the left derived functor of \(K^{-}(\mathcal{A}) \to D^{-}(\mathcal{B})\) is defined at \(X\). Moreover, the values are canonically isomorphic.

4. Let \(X\) be an object of \(K^{-}(\mathcal{A})\). Then \(X\) computes the left derived functor of \(K(\mathcal{A}) \to D(\mathcal{B})\) if and only if \(X\) computes the left derived functor of \(K^{-}(\mathcal{A}) \to D^{-}(\mathcal{B})\).

Proof

Let \(X\) be an object of \(K^{+}(\mathcal{A})\). Consider a quasi-isomorphism \(s : X \to X'\) in \(K(\mathcal{A})\). By Lemma 05RV there exists a quasi-isomorphism \(X' \to X''\) with \(X''\) bounded below. Hence we see that \(X/\text{Qis}^+(\mathcal{A})\) is cofinal in \(X/\text{Qis}(\mathcal{A})\). Thus it is clear that (1) holds. Part (2) follows directly from part (1). Parts (3) and (4) are dual to parts (1) and (2).

Given an object \(A\) of an abelian category \(\mathcal{A}\) we get a complex \[A[0] = ( \ldots \to 0 \to A \to 0 \to \ldots )\] where \(A\) is placed in degree zero. Hence a functor \(\mathcal{A} \to K(\mathcal{A})\), \(A \mapsto A[0]\). Let us temporarily say that a partial functor is one that is defined on a subcategory.

Definition

In Situation 05T4.

1. The right derived functors of \(F\) are the partial functors \(RF\) associated to cases (1) and (2) of Situation 05T4.

2. The left derived functors of \(F\) are the partial functors \(LF\) associated to cases (3) and (4) of Situation 05T4.

3. An object \(A\) of \(\mathcal{A}\) is said to be right acyclic for \(F\), or acyclic for \(RF\) if \(A[0]\) computes \(RF\).

4. An object \(A\) of \(\mathcal{A}\) is said to be left acyclic for \(F\), or acyclic for \(LF\) if \(A[0]\) computes \(LF\).

The following few lemmas give some criteria for the existence of enough acyclics.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(\mathcal{P} \subset \Ob(\mathcal{A})\) be a subset containing \(0\) such that every object of \(\mathcal{A}\) is a quotient of an element of \(\mathcal{P}\). Let \(a \in \mathbf{Z}\).

1. Given \(K^\bullet\) with \(K^n = 0\) for \(n > a\) there exists a quasi-isomorphism \(P^\bullet \to K^\bullet\) with \(P^n \in \mathcal{P}\) and \(P^n \to K^n\) surjective for all \(n\) and \(P^n = 0\) for \(n > a\).

2. Given \(K^\bullet\) with \(H^n(K^\bullet) = 0\) for \(n > a\) there exists a quasi-isomorphism \(P^\bullet \to K^\bullet\) with \(P^n \in \mathcal{P}\) for all \(n\) and \(P^n = 0\) for \(n > a\).

Proof

Proof of part (1). Consider the following induction hypothesis \(IH_n\): There are \(P^j \in \mathcal{P}\), \(j \geq n\), with \(P^j = 0\) for \(j > a\), maps \(d^j : P^j \to P^{j + 1}\) for \(j \geq n\), and surjective maps \(\alpha^j : P^j \to K^j\) for \(j \geq n\) such that the diagram \[\xymatrix{ & & P^n \ar[d]^\alpha \ar[r] & P^{n + 1} \ar[d]^\alpha \ar[r] & P^{n + 2} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^n \ar[r] & K^{n + 1} \ar[r] & K^{n + 2} \ar[r] & \ldots }\] is commutative, such that \(d^{j + 1} \circ d^j = 0\) for \(j \geq n\), such that \(\alpha\) induces isomorphisms \(\Ker(d^j)/\Im(d^{j - 1}) \to H^j(K^\bullet)\) for \(j > n\), and such that \(\alpha : \Ker(d^n) \to \Ker(d_K^n)\) is surjective. Then we choose a surjection \[P^{n - 1} \longrightarrow K^{n - 1} \times_{K^n} \Ker(d^n) = K^{n - 1} \times_{\Ker(d_K^n)} \Ker(d^n)\] with \(P^{n - 1}\) in \(\mathcal{P}\). This allows us to extend the diagram above to \[\xymatrix{ & P^{n - 1} \ar[d]^\alpha \ar[r] & P^n \ar[d]^\alpha \ar[r] & P^{n + 1} \ar[d]^\alpha \ar[r] & P^{n + 2} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^n \ar[r] & K^{n + 1} \ar[r] & K^{n + 2} \ar[r] & \ldots }\] The reader easily checks that \(IH_{n - 1}\) holds with this choice.

We finish the proof of (1) as follows. First we note that \(IH_n\) is true for \(n = a + 1\) since we can just take \(P^j = 0\) for \(j > a\). Hence we see that proceeding by descending induction we produce a complex \(P^\bullet\) with \(P^n = 0\) for \(n > a\) consisting of objects from \(\mathcal{P}\), and a termwise surjective quasi-isomorphism \(\alpha : P^\bullet \to K^\bullet\) as desired.

Proof of part (2). The assumption implies that the morphism \(\tau_{\leq a}K^\bullet \to K^\bullet\) (Homology, Section 0118) is a quasi-isomorphism. Apply part (1) to find \(P^\bullet \to \tau_{\leq a}K^\bullet\). The composition \(P^\bullet \to K^\bullet\) is the desired quasi-isomorphism.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(\mathcal{I} \subset \Ob(\mathcal{A})\) be a subset containing \(0\) such that every object of \(\mathcal{A}\) is a subobject of an element of \(\mathcal{I}\). Let \(a \in \mathbf{Z}\).

1. Given \(K^\bullet\) with \(K^n = 0\) for \(n < a\) there exists a quasi-isomorphism \(K^\bullet \to I^\bullet\) with \(K^n \to I^n\) injective and \(I^n \in \mathcal{I}\) for all \(n\) and \(I^n = 0\) for \(n < a\),

2. Given \(K^\bullet\) with \(H^n(K^\bullet) = 0\) for \(n < a\) there exists a quasi-isomorphism \(K^\bullet \to I^\bullet\) with \(I^n \in \mathcal{I}\) and \(I^n = 0\) for \(n < a\).

Proof

This lemma is dual to Lemma 05T7.

Lemma

In Situation 05T4. Let \(\mathcal{I} \subset \Ob(\mathcal{A})\) be a subset with the following properties:

1. every object of \(\mathcal{A}\) is a subobject of an element of \(\mathcal{I}\),

2. for any short exact sequence \(0 \to P \to Q \to R \to 0\) of \(\mathcal{A}\) with \(P, Q \in \mathcal{I}\), then \(R \in \mathcal{I}\), and \(0 \to F(P) \to F(Q) \to F(R) \to 0\) is exact.

Then every object of \(\mathcal{I}\) is acyclic for \(RF\).

Proof

Pick \(A \in \mathcal{I}\). Let \(A[0] \to K^\bullet\) be a quasi-isomorphism with \(K^\bullet\) bounded below. Then we can find a quasi-isomorphism \(K^\bullet \to I^\bullet\) with \(I^\bullet\) bounded below and each \(I^n \in \mathcal{I}\), see Lemma 05T6⁶. Hence we see that these resolutions are cofinal in the category \(A[0]/\text{Qis}^{+}(\mathcal{A})\). To finish the proof it therefore suffices to show that for any quasi-isomorphism \(A[0] \to I^\bullet\) with \(I^\bullet\) bounded below and \(I^n \in \mathcal{I}\) we have \(F(A)[0] \to F(I^\bullet)\) is a quasi-isomorphism. To see this suppose that \(I^n = 0\) for \(n < n_0\). Of course we may assume that \(n_0 < 0\). Starting with \(n = n_0\) we prove inductively that \(\Im(d^{n - 1}) = \Ker(d^n)\) and \(\Im(d^{-1})\) are elements of \(\mathcal{I}\) using property (2) and the exact sequences \[0 \to \Ker(d^n) \to I^n \to \Im(d^n) \to 0.\] Moreover, property (2) also guarantees that the complex \[0 \to F(I^{n_0}) \to F(I^{n_0 + 1}) \to \ldots \to F(I^{-1}) \to F(\Im(d^{-1})) \to 0\] is exact. The exact sequence \(0 \to \Im(d^{-1}) \to I^0 \to I^0/\Im(d^{-1}) \to 0\) implies that \(I^0/\Im(d^{-1})\) is an element of \(\mathcal{I}\). The exact sequence \(0 \to A \to I^0/\Im(d^{-1}) \to \Im(d^0) \to 0\) then implies that \(\Im(d^0) = \Ker(d^1)\) is an element of \(\mathcal{I}\) and from then on one continues as before to show that \(\Im(d^{n - 1}) = \Ker(d^n)\) is an element of \(\mathcal{I}\) for all \(n > 0\). Applying \(F\) to each of the short exact sequences mentioned above and using (2) we observe that \(F(A)[0] \to F(I^\bullet)\) is a quasi-isomorphism as desired.

Lemma

In Situation 05T4. Let \(\mathcal{P} \subset \Ob(\mathcal{A})\) be a subset with the following properties:

1. every object of \(\mathcal{A}\) is a quotient of an element of \(\mathcal{P}\),

2. for any short exact sequence \(0 \to P \to Q \to R \to 0\) of \(\mathcal{A}\) with \(Q, R \in \mathcal{P}\), then \(P \in \mathcal{P}\), and \(0 \to F(P) \to F(Q) \to F(R) \to 0\) is exact.

Then every object of \(\mathcal{P}\) is acyclic for \(LF\).

Proof

Dual to the proof of Lemma 05T8.

Higher derived functors

The following simple lemma shows that right derived functors “move to the right”.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor between abelian categories. Let \(K^\bullet\) be a complex of \(\mathcal{A}\) and \(a \in \mathbf{Z}\).

1. If \(H^i(K^\bullet) = 0\) for all \(i < a\) and \(RF\) is defined at \(K^\bullet\), then \(H^i(RF(K^\bullet)) = 0\) for all \(i < a\).

2. If \(RF\) is defined at \(K^\bullet\) and \(\tau_{\leq a}K^\bullet\), then \(H^i(RF(\tau_{\leq a}K^\bullet)) = H^i(RF(K^\bullet))\) for all \(i \leq a\).

Proof

Assume \(K^\bullet\) satisfies the assumptions of (1). Let \(s : K^\bullet \to L^\bullet\) be any quasi-isomorphism. Then it is also true that \(K^\bullet \to \tau_{\geq a}L^\bullet\) is a quasi-isomorphism by our assumption on \(K^\bullet\). Hence in the category \(K^\bullet/\text{Qis}^{+}(\mathcal{A})\) the quasi-isomorphisms \(s : K^\bullet \to L^\bullet\) with \(L^n = 0\) for \(n < a\) are cofinal. From Categories, Lemma 0A1S we deduce that \(RF\) is the value of the essentially constant ind-object \(F(L^\bullet)\) for these \(s\). This means that \(\text{id} : RF(K^\bullet) \to RF(K^\bullet)\) factors through \(F(L^\bullet)\) for some complex \(L^\bullet\) with \(L^n = 0\) for \(n < a\). It follows that \(H^i(RF(K^\bullet)) = 0\) for \(i < a\).

To prove (2) we use the distinguished triangle \[\tau_{\leq a}K^\bullet \to K^\bullet \to \tau_{\geq a + 1}K^\bullet \to (\tau_{\leq a}K^\bullet)[1]\] of Remark 08J5 to conclude via Lemma 05SC that \(RF\) is defined at \(\tau_{\geq a + 1}K^\bullet\) as well and that we have a distinguished triangle \[RF(\tau_{\leq a}K^\bullet) \to RF(K^\bullet) \to RF(\tau_{\geq a + 1}K^\bullet) \to RF(\tau_{\leq a}K^\bullet)[1]\] in \(D(\mathcal{B})\). By part (1) we see that \(RF(\tau_{\geq a + 1}K^\bullet)\) has vanishing cohomology in degrees \(< a + 1\). The long exact cohomology sequence of this distinguished triangle then shows what we want.

Definition

Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor between abelian categories. Assume \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is everywhere defined. Let \(i \in \mathbf{Z}\). The \(i\)th right derived functor \(R^iF\) of \(F\) is the functor \[R^iF = H^i \circ RF : \mathcal{A} \longrightarrow \mathcal{B}\]

The following lemma shows that it really does not make a lot of sense to take the right derived functor unless the functor is left exact.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor between abelian categories and assume \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is everywhere defined.

1. We have \(R^iF = 0\) for \(i < 0\),

2. \(R^0F\) is left exact,

3. the map \(F \to R^0F\) is an isomorphism if and only if \(F\) is left exact.

Proof

Let \(A\) be an object of \(\mathcal{A}\). By Lemma 05TC we have \(H^i(RF(A[0])) = 0\) for \(i < 0\). This proves (1).

Let \(0 \to A \to B \to C \to 0\) be a short exact sequence of \(\mathcal{A}\). By Lemma 0152 we obtain a distinguished triangle \((A[0], B[0], C[0], a, b, c)\) in \(D^{+}(\mathcal{A})\). From the long exact cohomology sequence (and the vanishing for \(i < 0\) proved above) we deduce that \(0 \to R^0F(A) \to R^0F(B) \to R^0F(C)\) is exact. Hence \(R^0F\) is left exact. Of course this also proves that if \(F \to R^0F\) is an isomorphism, then \(F\) is left exact.

Assume \(F\) is left exact. Recall that \(RF(A[0])\) is the value of the essentially constant system \(F(K^\bullet)\) for \(s : A[0] \to K^\bullet\) quasi-isomorphisms. It follows that \(R^0F(A)\) is the value of the essentially constant system \(H^0(F(K^\bullet))\) for \(s : A[0] \to K^\bullet\) quasi-isomorphisms, see Categories, Lemma 05SH. But if \(s : A[0] \to K^\bullet\) is a quasi-isomorphism, then \(A[0] \to \tau_{\geq 0}K^\bullet\) is a quasi-isomorphism. Hence in the category \(A[0]/\text{Qis}^{+}(\mathcal{A})\) the quasi-isomorphisms \(s : A[0] \to K^\bullet\) with \(K^n = 0\) for \(n < 0\) are cofinal. It follows from Categories, Lemma 0A1S that we may restrict to such \(s\). Moreover, for such an \(s\) the sequence \[0 \to A \to K^0 \to K^1\] is exact. Since \(F\) is left exact we see that \(0 \to F(A) \to F(K^0) \to F(K^1)\) is exact as well. It follows that \(F(A) \to H^0(F(K^\bullet))\) is an isomorphism and the system is actually constant with value \(F(A)\). We conclude \(R^0F = F\) as desired.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor between abelian categories and assume \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is everywhere defined. Let \(A\) be an object of \(\mathcal{A}\).

1. \(A\) is right acyclic for \(F\) if and only if \(F(A) \to R^0F(A)\) is an isomorphism and \(R^iF(A) = 0\) for all \(i > 0\),

2. if \(F\) is left exact, then \(A\) is right acyclic for \(F\) if and only if \(R^iF(A) = 0\) for all \(i > 0\).

Proof

If \(A\) is right acyclic for \(F\), then \(RF(A[0]) = F(A)[0]\) and in particular \(F(A) \to R^0F(A)\) is an isomorphism and \(R^iF(A) = 0\) for \(i \not = 0\). Conversely, if \(F(A) \to R^0F(A)\) is an isomorphism and \(R^iF(A) = 0\) for all \(i > 0\) then \(F(A[0]) \to RF(A[0])\) is a quasi-isomorphism by Lemma 05TD part (1) and hence \(A\) is acyclic. If \(F\) is left exact then \(F = R^0F\), see Lemma 05TD.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be a left exact functor between abelian categories and assume \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is everywhere defined. Let \(0 \to A \to B \to C \to 0\) be a short exact sequence of \(\mathcal{A}\).

1. If \(A\) and \(C\) are right acyclic for \(F\) then so is \(B\).

2. If \(A\) and \(B\) are right acyclic for \(F\) then so is \(C\).

3. If \(B\) and \(C\) are right acyclic for \(F\) and \(F(B) \to F(C)\) is surjective then \(A\) is right acyclic for \(F\).

In each of the three cases \[0 \to F(A) \to F(B) \to F(C) \to 0\] is a short exact sequence of \(\mathcal{B}\).

Proof

By Lemma 0152 we obtain a distinguished triangle \((A[0], B[0], C[0], a, b, c)\) in \(D^{+}(\mathcal{A})\). As \(RF\) is an exact functor and since \(R^iF = 0\) for \(i < 0\) and \(R^0F = F\) (Lemma 05TD) we obtain an exact cohomology sequence \[0 \to F(A) \to F(B) \to F(C) \to R^1F(A) \to \ldots\] in the abelian category \(\mathcal{B}\). Thus the lemma follows from the characterization of acyclic objects in Lemma 015C.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor between abelian categories and assume \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is everywhere defined.

1. The functors \(R^iF\), \(i \geq 0\) come equipped with a canonical structure of a \(\delta\)-functor from \(\mathcal{A} \to \mathcal{B}\), see Homology, Definition 010Q.

2. If every object of \(\mathcal{A}\) is a subobject of a right acyclic object for \(F\), then \(\{R^iF, \delta\}_{i \geq 0}\) is a universal \(\delta\)-functor, see Homology, Definition 010S.

Proof

The functor \(\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})\), \(A \mapsto A[0]\) is exact. The functor \(\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})\) is a \(\delta\)-functor, see Lemma 0152. The functor \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is exact. Finally, the functor \(H^0 : D^{+}(\mathcal{B}) \to \mathcal{B}\) is a homological functor, see Definition 05RU. Hence we get the structure of a \(\delta\)-functor from Lemma 05SR and Lemma 0151. Part (2) follows from Homology, Lemma 010T and the description of acyclics in Lemma 015C.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor between abelian categories. Let \(A^\bullet\) be a bounded below complex of right \(F\)-acyclic objects such that \(RF\) is defined at \(A^\bullet\)⁷. The canonical map \[F(A^\bullet) \longrightarrow RF(A^\bullet)\] is an isomorphism in \(D^{+}(\mathcal{B})\), i.e., \(A^\bullet\) computes \(RF\).

Proof

Let \(A^\bullet\) be a bounded complex of right \(F\)-acyclic objects. We claim that \(RF\) is defined at \(A^\bullet\) and that \(F(A^\bullet) \to RF(A^\bullet)\) is an isomorphism in \(D^+(\mathcal{B})\). Namely, it holds for complexes with at most one nonzero right \(F\)-acyclic object by Definition 0157. Next, suppose that \(A^n = 0\) for \(n \not \in [a, b]\). Using the “stupid” truncations we obtain a termwise split short exact sequence of complexes \[0 \to \sigma_{\geq a + 1} A^\bullet \to A^\bullet \to \sigma_{\leq a} A^\bullet \to 0\] see Homology, Section 0118. Thus a distinguished triangle \((\sigma_{\geq a + 1} A^\bullet, A^\bullet, \sigma_{\leq a} A^\bullet)\). By induction hypothesis \(RF\) is defined for the two outer complexes and these complexes compute \(RF\). Then the same is true for the middle one by Lemma 05SZ.

Suppose that \(A^\bullet\) is a bounded below complex of acyclic objects such that \(RF\) is defined at \(A^\bullet\). To show that \(F(A^\bullet) \to RF(A^\bullet)\) is an isomorphism in \(D^{+}(\mathcal{B})\) it suffices to show that \(H^i(F(A^\bullet)) \to H^i(RF(A^\bullet))\) is an isomorphism for all \(i\). Pick \(i\). Consider the termwise split short exact sequence of complexes \[0 \to \sigma_{\geq i + 2} A^\bullet \to A^\bullet \to \sigma_{\leq i + 1} A^\bullet \to 0.\] Note that this induces a termwise split short exact sequence \[0 \to \sigma_{\geq i + 2} F(A^\bullet) \to F(A^\bullet) \to \sigma_{\leq i + 1} F(A^\bullet) \to 0.\] Hence we get distinguished triangles \[(\sigma_{\geq i + 2} A^\bullet, A^\bullet, \sigma_{\leq i + 1} A^\bullet) \quad\text{and}\quad (\sigma_{\geq i + 2} F(A^\bullet), F(A^\bullet), \sigma_{\leq i + 1} F(A^\bullet))\] Since \(RF\) is defined at \(A^\bullet\) (by assumption) and at \(\sigma_{\leq i + 1}A^\bullet\) (by the first paragraph) we see that \(RF\) is defined at \(\sigma_{\geq i + 2}A^\bullet\) and we get a distinguished triangle \[(RF(\sigma_{\geq i + 2} A^\bullet), RF(A^\bullet), RF(\sigma_{\leq i + 1} A^\bullet))\] See Lemma 05SC. Using these distinguished triangles we obtain a map of exact sequences \[\xymatrix{ H^i(\sigma_{\geq i + 2} F(A^\bullet)) \ar[r] \ar[d] & H^i(F(A^\bullet)) \ar[r] \ar[d]^\alpha & H^i(\sigma_{\leq i + 1} F(A^\bullet)) \ar[r] \ar[d]^\beta & H^{i + 1}(\sigma_{\geq i + 2} F(A^\bullet)) \ar[d] \\ H^i(RF(\sigma_{\geq i + 2} A^\bullet)) \ar[r] & H^i(RF(A^\bullet)) \ar[r] & H^i(RF(\sigma_{\leq i + 1} A^\bullet)) \ar[r] & H^{i + 1}(RF(\sigma_{\geq i + 2} A^\bullet)) }\] By the results of the first paragraph the map \(\beta\) is an isomorphism. By inspection the objects on the upper left and the upper right are zero. Hence to finish the proof it suffices to show that \(H^i(RF(\sigma_{\geq i + 2} A^\bullet)) = 0\) and \(H^{i + 1}(RF(\sigma_{\geq i + 2} A^\bullet)) = 0\). This follows immediately from Lemma 05TC.

Proposition

Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor of abelian categories.

1. If every object of \(\mathcal{A}\) injects into an object acyclic for \(RF\), then \(RF\) is defined on all of \(K^{+}(\mathcal{A})\) and we obtain an exact functor \[RF : D^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{B})\] see (05SW). Moreover, any bounded below complex \(A^\bullet\) whose terms are acyclic for \(RF\) computes \(RF\).

2. If every object of \(\mathcal{A}\) is quotient of an object acyclic for \(LF\), then \(LF\) is defined on all of \(K^{-}(\mathcal{A})\) and we obtain an exact functor \[LF : D^{-}(\mathcal{A}) \longrightarrow D^{-}(\mathcal{B})\] see (05SW). Moreover, any bounded above complex \(A^\bullet\) whose terms are acyclic for \(LF\) computes \(LF\).

Proof

Assume every object of \(\mathcal{A}\) injects into an object acyclic for \(RF\). Let \(\mathcal{I}\) be the set of objects acyclic for \(RF\). Let \(K^\bullet\) be a bounded below complex in \(\mathcal{A}\). By Lemma 05T6 there exists a quasi-isomorphism \(\alpha : K^\bullet \to I^\bullet\) with \(I^\bullet\) bounded below and \(I^n \in \mathcal{I}\). Hence in order to prove (1) it suffices to show that \(F(I^\bullet) \to F((I')^\bullet)\) is a quasi-isomorphism when \(s : I^\bullet \to (I')^\bullet\) is a quasi-isomorphism of bounded below complexes of objects from \(\mathcal{I}\), see Lemma 06XN. Note that the cone \(C(s)^\bullet\) is an acyclic bounded below complex all of whose terms are in \(\mathcal{I}\). Hence it suffices to show: given an acyclic bounded below complex \(I^\bullet\) all of whose terms are in \(\mathcal{I}\) the complex \(F(I^\bullet)\) is acyclic.

Say \(I^n = 0\) for \(n < n_0\). Setting \(J^n = \Im(d^n)\) we break \(I^\bullet\) into short exact sequences \(0 \to J^n \to I^{n + 1} \to J^{n + 1} \to 0\) for \(n \geq n_0\). These sequences induce distinguished triangles \((J^n, I^{n + 1}, J^{n + 1})\) in \(D^+(\mathcal{A})\) by Lemma 0152. For each \(k \in \mathbf{Z}\) denote \(H_k\) the assertion: For all \(n \leq k\) the object \(J^n\) is in \(\mathcal{I}\). Then \(H_k\) holds trivially for \(k < n_0\). If \(H_n\) holds, then Lemma 05SZ shows that \(J^{n + 1}\) is in \(\mathcal{I}\) and we have \(H_{n + 1}\). By Proposition 05SE we have a distinguished triangle \((RF(J^n), RF(I^{n + 1}), RF(J^{n + 1}))\). Since \(J^n, I^{n + 1}, J^{n + 1}\) are in \(\mathcal{I}\) the long exact cohomology sequence (05ST) associated to this distinguished triangle collapses to an exact sequence \[0 \to F(J^n) \to F(I^{n + 1}) \to F(J^{n + 1}) \to 0\] This in turn proves that \(F(I^\bullet)\) is exact.

The proof in the case of \(LF\) is dual.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be an exact functor of abelian categories. Then

1. every object of \(\mathcal{A}\) is right acyclic for \(F\),

2. \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is everywhere defined,

3. \(RF : D(\mathcal{A}) \to D(\mathcal{B})\) is everywhere defined,

4. every complex computes \(RF\), in other words, the canonical map \(F(K^\bullet) \to RF(K^\bullet)\) is an isomorphism for all complexes, and

5. \(R^iF = 0\) for \(i \not = 0\).

Proof

This is true because \(F\) transforms acyclic complexes into acyclic complexes and quasi-isomorphisms into quasi-isomorphisms. Details omitted.

Triangulated subcategories of the derived category

Let \(\mathcal{A}\) be an abelian category. In this section we look at certain strictly full saturated triangulated subcategories \(\mathcal{D}' \subset D(\mathcal{A})\).

Let \(\mathcal{B} \subset \mathcal{A}\) be a weak Serre subcategory, see Homology, Definition 02MO and Lemma 0754. We let \(D_\mathcal{B}(\mathcal{A})\) be the full subcategory of \(D(\mathcal{A})\) whose objects are \[\Ob(D_\mathcal{B}(\mathcal{A})) = \{X \in \Ob(D(\mathcal{A})) \mid H^n(X) \text{ is an object of }\mathcal{B}\text{ for all }n\}\] We also define \(D^{+}_\mathcal{B}(\mathcal{A}) = D^{+}(\mathcal{A}) \cap D_\mathcal{B}(\mathcal{A})\) and similarly for the other bounded versions.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(\mathcal{B} \subset \mathcal{A}\) be a weak Serre subcategory. The category \(D_\mathcal{B}(\mathcal{A})\) is a strictly full saturated triangulated subcategory of \(D(\mathcal{A})\). Similarly for the bounded versions.

Proof

It is clear that \(D_\mathcal{B}(\mathcal{A})\) is an additive subcategory preserved under the translation functors. If \(X \oplus Y\) is in \(D_\mathcal{B}(\mathcal{A})\), then both \(H^n(X)\) and \(H^n(Y)\) are kernels of maps between objects of \(\mathcal{B}\) as \(H^n(X \oplus Y) = H^n(X) \oplus H^n(Y)\). Hence both \(X\) and \(Y\) are in \(D_\mathcal{B}(\mathcal{A})\). By Lemma 05QX it therefore suffices to show that given a distinguished triangle \((X, Y, Z, f, g, h)\) such that \(X\) and \(Y\) are in \(D_\mathcal{B}(\mathcal{A})\) then \(Z\) is an object of \(D_\mathcal{B}(\mathcal{A})\). The long exact cohomology sequence (05ST) and the definition of a weak Serre subcategory (see Homology, Definition 02MO) show that \(H^n(Z)\) is an object of \(\mathcal{B}\) for all \(n\). Thus \(Z\) is an object of \(D_\mathcal{B}(\mathcal{A})\).

We continue to assume that \(\mathcal{B}\) is a weak Serre subcategory of the abelian category \(\mathcal{A}\). Then \(\mathcal{B}\) is an abelian category and the inclusion functor \(\mathcal{B} \to \mathcal{A}\) is exact. Hence we obtain a derived functor \(D(\mathcal{B}) \to D(\mathcal{A})\), see Lemma 015F. Clearly the functor \(D(\mathcal{B}) \to D(\mathcal{A})\) factors through a canonical exact functor [06UR]\[\begin{equation} D(\mathcal{B}) \longrightarrow D_\mathcal{B}(\mathcal{A}) \end{equation}\] After all a complex made from objects of \(\mathcal{B}\) certainly gives rise to an object of \(D_\mathcal{B}(\mathcal{A})\) and as distinguished triangles in \(D_\mathcal{B}(\mathcal{A})\) are exactly the distinguished triangles of \(D(\mathcal{A})\) whose vertices are in \(D_\mathcal{B}(\mathcal{A})\) we see that the functor is exact since \(D(\mathcal{B}) \to D(\mathcal{A})\) is exact. Similarly we obtain functors \(D^+(\mathcal{B}) \to D^+_\mathcal{B}(\mathcal{A})\), \(D^-(\mathcal{B}) \to D^-_\mathcal{B}(\mathcal{A})\), and \(D^b(\mathcal{B}) \to D^b_\mathcal{B}(\mathcal{A})\) for the bounded versions. A key question in many cases is whether the displayed functor is an equivalence.

Now, suppose that \(\mathcal{B}\) is a Serre subcategory of \(\mathcal{A}\). In this case we have the quotient functor \(\mathcal{A} \to \mathcal{A}/\mathcal{B}\), see Homology, Lemma 02MS. In this case \(D_\mathcal{B}(\mathcal{A})\) is the kernel of the functor \(D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B})\). Thus we obtain a canonical functor \[D(\mathcal{A})/D_\mathcal{B}(\mathcal{A}) \longrightarrow D(\mathcal{A}/\mathcal{B})\] by Lemma 05RJ. Similarly for the bounded versions.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(\mathcal{B} \subset \mathcal{A}\) be a Serre subcategory. Then \(D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B})\) is essentially surjective.

Proof

We will use the description of the category \(\mathcal{A}/\mathcal{B}\) in the proof of Homology, Lemma 02MS. Let \((X^\bullet, d^\bullet)\) be a complex of \(\mathcal{A}/\mathcal{B}\). This means that \(X^i\) is an object of \(\mathcal{A}\) and \(d^i : X^i \to X^{i + 1}\) is a morphism in \(\mathcal{A}/\mathcal{B}\) such that \(d^i \circ d^{i - 1} = 0\) in \(\mathcal{A}/\mathcal{B}\).

For \(i \geq 0\) we may write \(d^i = (s^i, f^i)\) where \(s^i : Y^i \to X^i\) is a morphism of \(\mathcal{A}\) whose kernel and cokernel are in \(\mathcal{B}\) (equivalently \(s^i\) becomes an isomorphism in the quotient category) and \(f^i : Y^i \to X^{i + 1}\) is a morphism of \(\mathcal{A}\). By induction we will construct a commutative diagram \[\xymatrix{ & (X')^1 \ar@{..>}[r] & (X')^2 \ar@{..>}[r] & \ldots \\ X^0 \ar@{..>}[ru] & X^1 \ar@{..>}[u] & X^2 \ar@{..>}[u] & \ldots \\ Y^0 \ar[u]_{s^0} \ar[ru]_{f^0} & Y^1 \ar[u]_{s^1} \ar[ru]_{f^1} & Y^2 \ar[u]_{s^2} \ar[ru]_{f^2} & \ldots }\] where the vertical arrows \(X^i \to (X')^i\) become isomorphisms in the quotient category. Namely, we first let \((X')^1 = \Coker(Y^0 \to X^0 \oplus X^1)\) (or rather the pushout of the diagram with arrows \(s^0\) and \(f^0\)) which gives the first commutative diagram. Next, we take \((X')^2 = \Coker(Y^1 \to (X')^1 \oplus X^2)\). And so on. Setting additionally \((X')^n = X^n\) for \(n \leq 0\) we see that the map \((X^\bullet, d^\bullet) \to ((X')^\bullet, (d')^\bullet)\) is an isomorphism of complexes in \(\mathcal{A}/\mathcal{B}\). Hence we may assume \(d^n : X^n \to X^{n + 1}\) is given by a map \(X^n \to X^{n + 1}\) in \(\mathcal{A}\) for \(n \geq 0\).

Dually, for \(i < 0\) we may write \(d^i = (g^i, t^{i + 1})\) where \(t^{i + 1} : X^{i + 1} \to Z^{i + 1}\) is an isomorphism in the quotient category and \(g^i : X^i \to Z^{i + 1}\) is a morphism. By induction we will construct a commutative diagram \[\xymatrix{ \ldots & Z^{-2} & Z^{-1} & Z^0 \\ \ldots & X^{-2} \ar[u]_{t_{-2}} \ar[ru]_{g_{-2}} & X^{-1} \ar[u]_{t_{-1}} \ar[ru]_{g_{-1}} & X^0 \ar[u]_{t^0} \\ \ldots & (X')^{-2} \ar@{..>}[u] \ar@{..>}[r] & (X')^{-1} \ar@{..>}[u] \ar@{..>}[ru] }\] where the vertical arrows \((X')^i \to X^i\) become isomorphisms in the quotient category. Namely, we take \((X')^{-1} = X^{-1} \times_{Z^0} X^0\). Then we take \((X')^{-2} = X^{-2} \times_{Z^{-1}} (X')^{-1}\). And so on. Setting additionally \((X')^n = X^n\) for \(n \geq 0\) we see that the map \(((X')^\bullet, (d')^\bullet) \to (X^\bullet, d^\bullet)\) is an isomorphism of complexes in \(\mathcal{A}/\mathcal{B}\). Hence we may assume \(d^n : X^n \to X^{n + 1}\) is given by a map \(d^n : X^n \to X^{n + 1}\) in \(\mathcal{A}\) for all \(n \in \mathbf{Z}\).

In this case we know the compositions \(d^n \circ d^{n - 1}\) are zero in \(\mathcal{A}/\mathcal{B}\). If for \(n > 0\) we replace \(X^n\) by \[(X')^n = X^n/\sum\nolimits_{0 < k \leq n} \Im(\Im(X^{k - 2} \to X^k) \to X^n)\] then the compositions \(d^n \circ d^{n - 1}\) are zero for \(n \geq 0\). (Similarly to the second paragraph above we obtain an isomorphism of complexes \((X^\bullet, d^\bullet) \to ((X')^\bullet, (d')^\bullet)\).) Finally, for \(n < 0\) we replace \(X^n\) by \[(X')^n = \bigcap\nolimits_{n \leq k < 0} (X^n \to X^k)^{-1}\Ker(X^k \to X^{k + 2})\] and we argue in the same manner to get a complex in \(\mathcal{A}\) whose image in \(\mathcal{A}/\mathcal{B}\) is isomorphic to the given one.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(\mathcal{B} \subset \mathcal{A}\) be a Serre subcategory. Suppose that the functor \(v : \mathcal{A} \to \mathcal{A}/\mathcal{B}\) has a left adjoint \(u : \mathcal{A}/\mathcal{B} \to \mathcal{A}\) such that \(vu \cong \text{id}\). Then \[D(\mathcal{A})/D_\mathcal{B}(\mathcal{A}) = D(\mathcal{A}/\mathcal{B})\] and similarly for the bounded versions.

Proof

The functor \(D(v) : D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B})\) is essentially surjective by Lemma 06XL. For an object \(X\) of \(D(\mathcal{A})\) the adjunction mapping \(c_X : uvX \to X\) maps to an isomorphism in \(D(\mathcal{A}/\mathcal{B})\) because \(vuv \cong v\) by the assumption that \(vu \cong \text{id}\). Thus in a distinguished triangle \((uvX, X, Z, c_X, g, h)\) the object \(Z\) is an object of \(D_\mathcal{B}(\mathcal{A})\) as we see by looking at the long exact cohomology sequence. Hence \(c_X\) is an element of the multiplicative system used to define the quotient category \(D(\mathcal{A})/D_\mathcal{B}(\mathcal{A})\). Thus \(uvX \cong X\) in \(D(\mathcal{A})/D_\mathcal{B}(\mathcal{A})\). For \(X, Y \in \Ob(\mathcal{A})\) the map \[\Hom_{D(\mathcal{A})/D_\mathcal{B}(\mathcal{A})}(X, Y) \longrightarrow \Hom_{D(\mathcal{A}/\mathcal{B})}(vX, vY)\] is bijective because \(u\) gives an inverse (by the remarks above).

For certain Serre subcategories \(\mathcal{B} \subset \mathcal{A}\) we can prove that the functor \(D(\mathcal{B}) \to D_\mathcal{B}(\mathcal{A})\) is fully faithful.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(\mathcal{B} \subset \mathcal{A}\) be a Serre subcategory. Assume that for every surjection \(X \to Y\) with \(X \in \Ob(\mathcal{A})\) and \(Y \in \Ob(\mathcal{B})\) there exists \(X' \subset X\), \(X' \in \Ob(\mathcal{B})\) which surjects onto \(Y\). Then the functor \(D^-(\mathcal{B}) \to D^-_\mathcal{B}(\mathcal{A})\) of (06UR) is an equivalence.

Proof

Let \(X^\bullet\) be a bounded above complex of \(\mathcal{A}\) such that \(H^i(X^\bullet) \in \Ob(\mathcal{B})\) for all \(i \in \mathbf{Z}\). Moreover, suppose we are given \(B^i \subset X^i\), \(B^i \in \Ob(\mathcal{B})\) for all \(i \in \mathbf{Z}\). Claim: there exists a subcomplex \(Y^\bullet \subset X^\bullet\) such that

1. \(Y^\bullet \to X^\bullet\) is a quasi-isomorphism,

2. \(Y^i \in \Ob(\mathcal{B})\) for all \(i \in \mathbf{Z}\), and

3. \(B^i \subset Y^i\) for all \(i \in \mathbf{Z}\).

To prove the claim, using the assumption of the lemma we first choose \(C^i \subset \Ker(d^i : X^i \to X^{i + 1})\), \(C^i \in \Ob(\mathcal{B})\) surjecting onto \(H^i(X^\bullet)\). Setting \(D^i = C^i + d^{i - 1}(B^{i - 1}) + B^i\) we find a subcomplex \(D^\bullet\) satisfying (2) and (3) such that \(H^i(D^\bullet) \to H^i(X^\bullet)\) is surjective for all \(i \in \mathbf{Z}\). For any choice of \(E^i \subset X^i\) with \(E^i \in \Ob(\mathcal{B})\) and \(d^i(E^i) \subset D^{i + 1} + E^{i + 1}\) we see that setting \(Y^i = D^i + E^i\) gives a subcomplex whose terms are in \(\mathcal{B}\) and whose cohomology surjects onto the cohomology of \(X^\bullet\). Clearly, if \(d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \Im(d^i)\) then we see that the map on cohomology is also injective. For \(n \gg 0\) we can take \(E^n\) equal to \(0\). By descending induction we can choose \(E^i\) for all \(i\) with the desired property. Namely, given \(E^{i + 1}, E^{i + 2}, \ldots\) we choose \(E^i \subset X^i\) such that \(d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \Im(d^i)\). This is possible by our assumption in the lemma combined with the fact that \((D^{i + 1} + E^{i + 1}) \cap \Im(d^i)\) is in \(\mathcal{B}\) as \(\mathcal{B}\) is a Serre subcategory of \(\mathcal{A}\).

The claim above implies the lemma. Essential surjectivity is immediate from the claim. Let us prove faithfulness. Namely, suppose we have a morphism \(f : U^\bullet \to V^\bullet\) of bounded above complexes of \(\mathcal{B}\) whose image in \(D(\mathcal{A})\) is zero. Then there exists a quasi-isomorphism \(s : V^\bullet \to X^\bullet\) into a bounded above complex of \(\mathcal{A}\) such that \(s \circ f\) is homotopic to zero. Choose a homotopy \(h^i : U^i \to X^{i - 1}\) between \(0\) and \(s \circ f\). Apply the claim with \(B^i = h^{i + 1}(U^{i + 1}) + s^i(V^i)\). The resulting map \(s' : V^\bullet \to Y^\bullet\) is a quasi-isomorphism as well and \(s' \circ f\) is homotopic to zero as is clear from the fact that \(h^i\) factors through \(Y^{i - 1}\). This proves faithfulness. Fullness is proved in the exact same manner.

Injective resolutions

In this section we prove some lemmas regarding the existence of injective resolutions in abelian categories having enough injectives.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(A \in \Ob(\mathcal{A})\). An injective resolution of \(A\) is a complex \(I^\bullet\) together with a map \(A \to I^0\) such that:

1. We have \(I^n = 0\) for \(n < 0\).

2. Each \(I^n\) is an injective object of \(\mathcal{A}\).

3. The map \(A \to I^0\) is an isomorphism onto \(\Ker(d^0)\).

4. We have \(H^i(I^\bullet) = 0\) for \(i > 0\).

Hence \(A[0] \to I^\bullet\) is a quasi-isomorphism. In other words the complex \[\ldots \to 0 \to A \to I^0 \to I^1 \to \ldots\] is acyclic. Let \(K^\bullet\) be a complex in \(\mathcal{A}\). An injective resolution of \(K^\bullet\) is a complex \(I^\bullet\) together with a map \(\alpha : K^\bullet \to I^\bullet\) of complexes such that

1. We have \(I^n = 0\) for \(n \ll 0\), i.e., \(I^\bullet\) is bounded below.

2. Each \(I^n\) is an injective object of \(\mathcal{A}\).

3. The map \(\alpha : K^\bullet \to I^\bullet\) is a quasi-isomorphism.

In other words an injective resolution \(K^\bullet \to I^\bullet\) gives rise to a diagram \[\xymatrix{ \ldots \ar[r] & K^{n - 1} \ar[d] \ar[r] & K^n \ar[d] \ar[r] & K^{n + 1} \ar[d] \ar[r] & \ldots \\ \ldots \ar[r] & I^{n - 1} \ar[r] & I^n \ar[r] & I^{n + 1} \ar[r] & \ldots }\] which induces an isomorphism on cohomology objects in each degree. An injective resolution of an object \(A\) of \(\mathcal{A}\) is almost the same thing as an injective resolution of the complex \(A[0]\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet\) be a complex of \(\mathcal{A}\).

1. If \(K^\bullet\) has an injective resolution then \(H^n(K^\bullet) = 0\) for \(n \ll 0\).

2. If \(H^n(K^\bullet) = 0\) for all \(n \ll 0\) then there exists a quasi-isomorphism \(K^\bullet \to L^\bullet\) with \(L^\bullet\) bounded below.

Proof

Omitted. For the second statement use \(L^\bullet = \tau_{\geq n}K^\bullet\) for some \(n \ll 0\). See Homology, Section 0118 for the definition of the truncation \(\tau_{\geq n}\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Assume \(\mathcal{A}\) has enough injectives.

1. Any object of \(\mathcal{A}\) has an injective resolution.

2. If \(H^n(K^\bullet) = 0\) for all \(n \ll 0\) then \(K^\bullet\) has an injective resolution.

3. If \(K^\bullet\) is a complex with \(K^n = 0\) for \(n < a\), then there exists an injective resolution \(\alpha : K^\bullet \to I^\bullet\) with \(I^n = 0\) for \(n < a\) such that each \(\alpha^n : K^n \to I^n\) is injective.

Proof

Proof of (1). First choose an injection \(A \to I^0\) of \(A\) into an injective object of \(\mathcal{A}\). Next, choose an injection \(I^0/A \to I^1\) into an injective object of \(\mathcal{A}\). Denote \(d^0\) the induced map \(I^0 \to I^1\). Next, choose an injection \(I^1/\Im(d^0) \to I^2\) into an injective object of \(\mathcal{A}\). Denote \(d^1\) the induced map \(I^1 \to I^2\). And so on. By Lemma 013J part (2) follows from part (3). Part (3) is a special case of Lemma 05T6.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet\) be an acyclic complex. Let \(I^\bullet\) be bounded below and consisting of injective objects. Any morphism \(K^\bullet \to I^\bullet\) is homotopic to zero.

Proof

Let \(\alpha : K^\bullet \to I^\bullet\) be a morphism of complexes. Note that \(\alpha^j = 0\) for \(j \ll 0\) as \(I^\bullet\) is bounded below. In particular, we can find an \(n\) such that there exist \(h^j : K^j \to I^{j - 1}\) for \(j \leq n\) such that \(\alpha^j = d^{j - 1} \circ h^j + h^{j + 1} \circ d^j\) for \(j < n\). We will show that there exists a morphism \(h^{n + 1} : K^{n + 1} \to I^n\) such that \(\alpha^n = d^{n - 1} \circ h^n + h^{n + 1} \circ d^n\). Note that \[\begin{align*} (\alpha^n - d^{n - 1} \circ h^n) \circ d^{n - 1} & = \alpha^n \circ d^{n - 1} - d^{n - 1} \circ h^n \circ d^{n - 1} \\ & = d^{n - 1} \circ \alpha^{n - 1} - d^{n - 1} \circ h^n \circ d^{n - 1} \\ & = d^{n - 1} \circ (d^{n - 2} \circ h^{n - 1} + h^n \circ d^{n - 1}) - d^{n - 1} \circ h^n \circ d^{n - 1} \\ & = 0 \end{align*}\] Since \(K^\bullet\) is acyclic we have \(d^{n - 1}(K^{n - 1}) = \Ker(K^n \to K^{n + 1})\). Hence we can think of \(\alpha^n - d^{n - 1} \circ h^n\) as a map into \(I^n\) defined on the subobject \(\Im(K^n \to K^{n + 1})\) of \(K^{n + 1}\). By injectivity of the object \(I^n\) we can extend this to a map \(h^{n + 1} : K^{n + 1} \to I^n\). With this choice the reader checks that we indeed have \(\alpha^n = d^{n - 1} \circ h^n + h^{n + 1} \circ d^n\).

By induction on \(n\) we conclude we can find \(h = (h^j)_{j \in \mathbf{Z}}\) which forms a homotopy between \(\alpha\) and \(0\) as desired.

Remark

Let \(\mathcal{A}\) be an abelian category. Using the fact that \(K(\mathcal{A})\) is a triangulated category we may use Lemma 013R to obtain proofs of some of the lemmas below which are usually proved by chasing through diagrams. Namely, suppose that \(\alpha : K^\bullet \to L^\bullet\) is a quasi-isomorphism of complexes. Then \[(K^\bullet, L^\bullet, C(\alpha)^\bullet, \alpha, i, -p)\] is a distinguished triangle in \(K(\mathcal{A})\) (Lemma 014L) and \(C(\alpha)^\bullet\) is an acyclic complex (Lemma 05RT). Next, let \(I^\bullet\) be a bounded below complex of injective objects. Then \[\xymatrix{ \Hom_{K(\mathcal{A})}(C(\alpha)^\bullet, I^\bullet) \ar[r] & \Hom_{K(\mathcal{A})}(L^\bullet, I^\bullet) \ar[r] & \Hom_{K(\mathcal{A})}(K^\bullet, I^\bullet) \ar[lld] \\ \Hom_{K(\mathcal{A})}(C(\alpha)^\bullet[-1], I^\bullet) }\] is an exact sequence of abelian groups, see Lemma 0149. At this point Lemma 013R guarantees that the outer two groups are zero and hence \(\Hom_{K(\mathcal{A})}(L^\bullet, I^\bullet) = \Hom_{K(\mathcal{A})}(K^\bullet, I^\bullet)\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Consider a solid diagram \[\xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_\gamma & L^\bullet \ar@{-->}[dl]^\beta \\ I^\bullet }\] where \(I^\bullet\) is bounded below and consists of injective objects, and \(\alpha\) is a quasi-isomorphism.

1. There exists a map of complexes \(\beta\) making the diagram commute up to homotopy.

2. If \(\alpha\) is injective in every degree then we can find a \(\beta\) which makes the diagram commute.

Proof

The “correct” proof of part (1) is explained in Remark 05TF. We also give a direct proof here.

We first show that (2) implies (1). Namely, let \(\tilde \alpha : K^\bullet \to \tilde L^\bullet\), \(\pi\), \(s\) be as in Lemma 013N. Since \(\tilde \alpha\) is injective by (2) there exists a morphism \(\tilde \beta : \tilde L^\bullet \to I^\bullet\) such that \(\gamma = \tilde \beta \circ \tilde \alpha\). Set \(\beta = \tilde \beta \circ s\). Then we have \[\beta \circ \alpha = \tilde \beta \circ s \circ \pi \circ \tilde \alpha \sim \tilde \beta \circ \tilde \alpha = \gamma\] as desired.

Assume that \(\alpha : K^\bullet \to L^\bullet\) is injective. Suppose we have already defined \(\beta\) in all degrees \(\leq n - 1\) compatible with differentials and such that \(\gamma^j = \beta^j \circ \alpha^j\) for all \(j \leq n - 1\). Consider the commutative solid diagram \[\xymatrix{ K^{n - 1} \ar[r] \ar@/_2pc/[dd]_\gamma \ar[d]^\alpha & K^n \ar@/^2pc/[dd]^\gamma \ar[d]^\alpha \\ L^{n - 1} \ar[r] \ar[d]^\beta & L^n \ar@{-->}[d] \\ I^{n - 1} \ar[r] & I^n }\] Thus we see that the dotted arrow is prescribed on the subobjects \(\alpha(K^n)\) and \(d^{n - 1}(L^{n - 1})\). Moreover, these two arrows agree on \(\alpha(d^{n - 1}(K^{n - 1}))\). Hence if [013Q]\[\begin{equation} \alpha(d^{n - 1}(K^{n - 1})) = \alpha(K^n) \cap d^{n - 1}(L^{n - 1}) \end{equation}\] then these morphisms glue to a morphism \(\alpha(K^n) + d^{n - 1}(L^{n - 1}) \to I^n\) and, using the injectivity of \(I^n\), we can extend this to a morphism from all of \(L^n\) into \(I^n\). After this by induction we get the morphism \(\beta\) for all \(n\) simultaneously (note that we can set \(\beta^n = 0\) for all \(n \ll 0\) since \(I^\bullet\) is bounded below – in this way starting the induction).

It remains to prove the equality (013Q). The reader is encouraged to argue this for themselves with a suitable diagram chase. Nonetheless here is our argument. Note that the inclusion \(\alpha(d^{n - 1}(K^{n - 1})) \subset \alpha(K^n) \cap d^{n - 1}(L^{n - 1})\) is obvious. Take an object \(T\) of \(\mathcal{A}\) and a morphism \(x : T \to L^n\) whose image is contained in the subobject \(\alpha(K^n) \cap d^{n - 1}(L^{n - 1})\). Since \(\alpha\) is injective we see that \(x = \alpha \circ x'\) for some \(x' : T \to K^n\). Moreover, since \(x\) lies in \(d^{n - 1}(L^{n - 1})\) we see that \(d^n \circ x = 0\). Hence using injectivity of \(\alpha\) again we see that \(d^n \circ x' = 0\). Thus \(x'\) gives a morphism \([x'] : T \to H^n(K^\bullet)\). On the other hand the corresponding map \([x] : T \to H^n(L^\bullet)\) induced by \(x\) is zero by assumption. Since \(\alpha\) is a quasi-isomorphism we conclude that \([x'] = 0\). This of course means exactly that the image of \(x'\) is contained in \(d^{n - 1}(K^{n - 1})\) and we win.

Lemma

Let \(\mathcal{A}\) be an abelian category. Consider a solid diagram \[\xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_\gamma & L^\bullet \ar@{-->}[dl]^{\beta_i} \\ I^\bullet }\] where \(I^\bullet\) is bounded below and consists of injective objects, and \(\alpha\) is a quasi-isomorphism. Any two morphisms \(\beta_1, \beta_2\) making the diagram commute up to homotopy are homotopic.

Proof

This follows from Remark 05TF. We also give a direct argument here.

Let \(\tilde \alpha : K^\bullet \to \tilde L^\bullet\), \(\pi\), \(s\) be as in Lemma 013N. If we can show that \(\beta_1 \circ\pi\) is homotopic to \(\beta_2 \circ \pi\), then we deduce that \(\beta_1 \sim \beta_2\) because \(\pi \circ s\) is the identity. Hence we may assume \(\alpha^n : K^n \to L^n\) is the inclusion of a direct summand for all \(n\). Thus we get a short exact sequence of complexes \[0 \to K^\bullet \to L^\bullet \to M^\bullet \to 0\] which is termwise split and such that \(M^\bullet\) is acyclic. We choose splittings \(L^n = K^n \oplus M^n\), so we have \(\beta_i^n : K^n \oplus M^n \to I^n\) and \(\gamma^n : K^n \to I^n\). In this case the condition on \(\beta_i\) is that there are morphisms \(h_i^n : K^n \to I^{n - 1}\) such that \[\gamma^n - \beta_i^n|_{K^n} = d \circ h_i^n + h_i^{n + 1} \circ d\] Thus we see that \[\beta_1^n|_{K^n} - \beta_2^n|_{K^n} = d \circ (h_1^n - h_2^n) + (h_1^{n + 1} - h_2^{n + 1}) \circ d\] Consider the map \(h^n : K^n \oplus M^n \to I^{n - 1}\) which equals \(h_1^n - h_2^n\) on the first summand and zero on the second. Then we see that \[\beta_1^n - \beta_2^n - (d \circ h^n + h^{n + 1} \circ d)\] is a morphism of complexes \(L^\bullet \to I^\bullet\) which is identically zero on the subcomplex \(K^\bullet\). Hence it factors as \(L^\bullet \to M^\bullet \to I^\bullet\). Thus the result of the lemma follows from Lemma 013R.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(I^\bullet\) be bounded below complex consisting of injective objects. Let \(L^\bullet \in K(\mathcal{A})\). Then \[\Mor_{K(\mathcal{A})}(L^\bullet, I^\bullet) = \Mor_{D(\mathcal{A})}(L^\bullet, I^\bullet).\]

Proof

Let \(a\) be an element of the right hand side. We may represent \(a = \gamma\alpha^{-1}\) where \(\alpha : K^\bullet \to L^\bullet\) is a quasi-isomorphism and \(\gamma : K^\bullet \to I^\bullet\) is a map of complexes. By Lemma 013P we can find a morphism \(\beta : L^\bullet \to I^\bullet\) such that \(\beta \circ \alpha\) is homotopic to \(\gamma\). This proves that the map is surjective. Let \(b\) be an element of the left hand side which maps to zero in the right hand side. Then \(b\) is the homotopy class of a morphism \(\beta : L^\bullet \to I^\bullet\) such that there exists a quasi-isomorphism \(\alpha : K^\bullet \to L^\bullet\) with \(\beta \circ \alpha\) homotopic to zero. Then Lemma 013S shows that \(\beta\) is homotopic to zero also, i.e., \(b = 0\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Assume \(\mathcal{A}\) has enough injectives. For any short exact sequence \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) of \(\text{Comp}^{+}(\mathcal{A})\) there exists a commutative diagram in \(\text{Comp}^{+}(\mathcal{A})\) \[\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_1^\bullet \ar[r] & I_2^\bullet \ar[r] & I_3^\bullet \ar[r] & 0 }\] where the vertical arrows are injective resolutions and the rows are short exact sequences of complexes. Additionally,

1. given any injective resolution \(A^\bullet \to I^\bullet\) we may assume \(I_1^\bullet = I^\bullet\),

2. if \(A^n = B^n = C^n = 0\) for \(n < 0\), then we may assume \(I_j^n = 0\) for \(n < 0\),

3. we can combine (1) and (2) if also \(I^n = 0\) for \(n < 0\).

Proof

Step 1. Choose an injective resolution \(A^\bullet \to I^\bullet\) (see Lemma 013K) or use the given one. Recall that \(\text{Comp}^{+}(\mathcal{A})\) is an abelian category, see Homology, Lemma 0114. Hence we may form the pushout along the map \(A^\bullet \to I^\bullet\) to get \[\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I^\bullet \ar[r] & E^\bullet \ar[r] & C^\bullet \ar[r] & 0 }\] Because of the \(5\)-lemma and the last assertion of Homology, Lemma 0117 the map \(B^\bullet \to E^\bullet\) is a quasi-isomorphism. Note that the lower short exact sequence is termwise split, see Homology, Lemma 0136. Hence it suffices to prove the lemma when \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) is termwise split.

Step 2. Choose splittings. In other words, write \(B^n = A^n \oplus C^n\). Denote \(\delta : C^\bullet \to A^\bullet[1]\) the morphism as in Homology, Lemma 011J. Choose injective resolutions \(f_1 : A^\bullet \to I_1^\bullet\) and \(f_3 : C^\bullet \to I_3^\bullet\). (If \(A^\bullet\) is a complex of injectives, then use \(I_1^\bullet = A^\bullet\). If \(A^n = C^n = 0\) for \(n < 0\), then pick these such that \(I_1^n = I_3^n = 0\) for \(n < 0\), see Lemma 013K.) We may assume \(f_3\) is injective in every degree. By Lemma 013P we may find a morphism \(\delta' : I_3^\bullet \to I_1^\bullet[1]\) such that \(\delta' \circ f_3 = f_1[1] \circ \delta\) (equality of morphisms of complexes). Set \(I_2^n = I_1^n \oplus I_3^n\). Define \[d_{I_2}^n = \left( \begin{matrix} d_{I_1}^n & (\delta')^n \\ 0 & d_{I_3}^n \end{matrix} \right)\] and define the maps \(B^n \to I_2^n\) to be given as the sum of the maps \(A^n \to I_1^n\) and \(C^n \to I_3^n\). Everything is clear.

Projective resolutions

This section is dual to Section 013G. We give definitions and state results, but we do not reprove the lemmas.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(A \in \Ob(\mathcal{A})\). A projective resolution of \(A\) is a complex \(P^\bullet\) together with a map \(P^0 \to A\) such that:

1. We have \(P^n = 0\) for \(n > 0\).

2. Each \(P^n\) is a projective object of \(\mathcal{A}\).

3. The map \(P^0 \to A\) induces an isomorphism \(\Coker(d^{-1}) \to A\).

4. We have \(H^i(P^\bullet) = 0\) for \(i < 0\).

Hence \(P^\bullet \to A[0]\) is a quasi-isomorphism. In other words the complex \[\ldots \to P^{-1} \to P^0 \to A \to 0 \to \ldots\] is acyclic. Let \(K^\bullet\) be a complex in \(\mathcal{A}\). A projective resolution of \(K^\bullet\) is a complex \(P^\bullet\) together with a map \(\alpha : P^\bullet \to K^\bullet\) of complexes such that

1. We have \(P^n = 0\) for \(n \gg 0\), i.e., \(P^\bullet\) is bounded above.

2. Each \(P^n\) is a projective object of \(\mathcal{A}\).

3. The map \(\alpha : P^\bullet \to K^\bullet\) is a quasi-isomorphism.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet\) be a complex of \(\mathcal{A}\).

1. If \(K^\bullet\) has a projective resolution then \(H^n(K^\bullet) = 0\) for \(n \gg 0\).

2. If \(H^n(K^\bullet) = 0\) for \(n \gg 0\) then there exists a quasi-isomorphism \(L^\bullet \to K^\bullet\) with \(L^\bullet\) bounded above.

Proof

Dual to Lemma 013J.

Lemma

Let \(\mathcal{A}\) be an abelian category. Assume \(\mathcal{A}\) has enough projectives.

1. Any object of \(\mathcal{A}\) has a projective resolution.

2. If \(H^n(K^\bullet) = 0\) for all \(n \gg 0\) then \(K^\bullet\) has a projective resolution.

3. If \(K^\bullet\) is a complex with \(K^n = 0\) for \(n > a\), then there exists a projective resolution \(\alpha : P^\bullet \to K^\bullet\) with \(P^n = 0\) for \(n > a\) such that each \(\alpha^n : P^n \to K^n\) is surjective.

Proof

Dual to Lemma 013K.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet\) be an acyclic complex. Let \(P^\bullet\) be bounded above and consisting of projective objects. Any morphism \(P^\bullet \to K^\bullet\) is homotopic to zero.

Proof

Dual to Lemma 013R.

Remark

Let \(\mathcal{A}\) be an abelian category. Suppose that \(\alpha : K^\bullet \to L^\bullet\) is a quasi-isomorphism of complexes. Let \(P^\bullet\) be a bounded above complex of projectives. Then \[\Hom_{K(\mathcal{A})}(P^\bullet, K^\bullet) \longrightarrow \Hom_{K(\mathcal{A})}(P^\bullet, L^\bullet)\] is an isomorphism. This is dual to Remark 05TF.

Lemma

Let \(\mathcal{A}\) be an abelian category. Consider a solid diagram \[\xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_\beta }\] where \(P^\bullet\) is bounded above and consists of projective objects, and \(\alpha\) is a quasi-isomorphism.

1. There exists a map of complexes \(\beta\) making the diagram commute up to homotopy.

2. If \(\alpha\) is surjective in every degree then we can find a \(\beta\) which makes the diagram commute.

Proof

Dual to Lemma 013P.

Lemma

Let \(\mathcal{A}\) be an abelian category. Consider a solid diagram \[\xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_{\beta_i} }\] where \(P^\bullet\) is bounded above and consists of projective objects, and \(\alpha\) is a quasi-isomorphism. Any two morphisms \(\beta_1, \beta_2\) making the diagram commute up to homotopy are homotopic.

Proof

Dual to Lemma 013S.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(P^\bullet\) be bounded above complex consisting of projective objects. Let \(L^\bullet \in K(\mathcal{A})\). Then \[\Mor_{K(\mathcal{A})}(P^\bullet, L^\bullet) = \Mor_{D(\mathcal{A})}(P^\bullet, L^\bullet).\]

Proof

Dual to Lemma 05TG.

Lemma

Let \(\mathcal{A}\) be an abelian category. Assume \(\mathcal{A}\) has enough projectives. For any short exact sequence \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) of \(\text{Comp}^{-}(\mathcal{A})\) there exists a commutative diagram in \(\text{Comp}^{-}(\mathcal{A})\) \[\xymatrix{ 0 \ar[r] & P_1^\bullet \ar[r] \ar[d] & P_2^\bullet \ar[r] \ar[d] & P_3^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r] & 0 }\] where the vertical arrows are projective resolutions and the rows are short exact sequences of complexes. In fact, given any projective resolution \(P^\bullet \to C^\bullet\) we may assume \(P_3^\bullet = P^\bullet\).

Proof

Dual to Lemma 013T.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(P^\bullet\), \(K^\bullet\) be complexes. Let \(n \in \mathbf{Z}\). Assume that

1. \(P^\bullet\) is a bounded complex consisting of projective objects,

2. \(P^i = 0\) for \(i < n\), and

3. \(H^i(K^\bullet) = 0\) for \(i \geq n\).

Then \(\Hom_{K(\mathcal{A})}(P^\bullet, K^\bullet) = \Hom_{D(\mathcal{A})}(P^\bullet, K^\bullet) = 0\).

Proof

The first equality follows from Lemma 064B. Note that there is a distinguished triangle \[(\tau_{\leq n - 1}K^\bullet, K^\bullet, \tau_{\geq n}K^\bullet, f, g, h)\] by Remark 08J5. Hence, by Lemma 0149 it suffices to prove \(\Hom_{K(\mathcal{A})}(P^\bullet, \tau_{\leq n - 1}K^\bullet) = 0\) and \(\Hom_{K(\mathcal{A})}(P^\bullet, \tau_{\geq n} K^\bullet) = 0\). The first vanishing is trivial and the second is Lemma 0647.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(\beta : P^\bullet \to L^\bullet\) and \(\alpha : E^\bullet \to L^\bullet\) be maps of complexes. Let \(n \in \mathbf{Z}\). Assume

1. \(P^\bullet\) is a bounded complex of projectives and \(P^i = 0\) for \(i < n\),

2. \(H^i(\alpha)\) is an isomorphism for \(i > n\) and surjective for \(i = n\).

Then there exists a map of complexes \(\gamma : P^\bullet \to E^\bullet\) such that \(\alpha \circ \gamma\) and \(\beta\) are homotopic.

Proof

Consider the cone \(C^\bullet = C(\alpha)^\bullet\) with map \(i : L^\bullet \to C^\bullet\). Note that \(i \circ \beta\) is zero by Lemma 064D. Hence we can lift \(\beta\) to \(E^\bullet\) by Lemma 0149.

Right derived functors and injective resolutions

At this point we can use the material above to define the right derived functors of an additive functor between an abelian category having enough injectives and a general abelian category.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(I \in \Ob(\mathcal{A})\) be an injective object. Let \(I^\bullet\) be a bounded below complex of injectives in \(\mathcal{A}\).

1. \(I^\bullet\) computes \(RF\) relative to \(\text{Qis}^{+}(\mathcal{A})\) for any exact functor \(F : K^{+}(\mathcal{A}) \to \mathcal{D}\) into any triangulated category \(\mathcal{D}\).

2. \(I\) is right acyclic for any additive functor \(F : \mathcal{A} \to \mathcal{B}\) into any abelian category \(\mathcal{B}\).

Proof

Part (2) is a direct consequence of part (1) and Definition 0157. To prove (1) let \(\alpha : I^\bullet \to K^\bullet\) be a quasi-isomorphism into a complex. By Lemma 013P we see that \(\alpha\) has a left inverse. Hence the category \(I^\bullet/\text{Qis}^{+}(\mathcal{A})\) is essentially constant with value \(\text{id} : I^\bullet \to I^\bullet\). Thus also the ind-object \[I^\bullet/\text{Qis}^{+}(\mathcal{A}) \longrightarrow \mathcal{D}, \quad (I^\bullet \to K^\bullet) \longmapsto F(K^\bullet)\] is essentially constant with value \(F(I^\bullet)\). This proves (1), see Definitions 05S9 and 05SX.

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives.

1. For any exact functor \(F : K^{+}(\mathcal{A}) \to \mathcal{D}\) into a triangulated category \(\mathcal{D}\) the right derived functor \[RF : D^{+}(\mathcal{A}) \longrightarrow \mathcal{D}\] is everywhere defined.

2. For any additive functor \(F : \mathcal{A} \to \mathcal{B}\) into an abelian category \(\mathcal{B}\) the right derived functor \[RF : D^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{B})\] is everywhere defined.

Proof

Combine Lemma 05TH and Proposition 05TA for the second assertion. To see the first assertion combine Lemma 013K, Lemma 05TH, Lemma 05T1, and Equation (05SW).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor.

1. The functor \(RF\) is an exact functor \(D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\).

2. The functor \(RF\) induces an exact functor \(K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\).

3. The functor \(RF\) induces a \(\delta\)-functor \(\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\).

4. The functor \(RF\) induces a \(\delta\)-functor \(\mathcal{A} \to D^{+}(\mathcal{B})\).

Proof

This lemma simply reviews some of the results obtained so far. Note that by Lemma 05TI \(RF\) is everywhere defined. Here are some references:

1. The derived functor is exact: This boils down to Lemma 05SC.

2. This is true because \(K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})\) is exact and compositions of exact functors are exact.

3. This is true because \(\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})\) is a \(\delta\)-functor, see Lemma 0152.

4. This is true because \(\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})\) is exact and precomposing a \(\delta\)-functor by an exact functor gives a \(\delta\)-functor.

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Let \(F : \mathcal{A} \to \mathcal{B}\) be a left exact functor.

1. For any short exact sequence \(0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0\) of complexes in \(\text{Comp}^{+}(\mathcal{A})\) there is an associated long exact sequence \[\ldots \to H^i(RF(A^\bullet)) \to H^i(RF(B^\bullet)) \to H^i(RF(C^\bullet)) \to H^{i + 1}(RF(A^\bullet)) \to \ldots\]

2. The functors \(R^iF : \mathcal{A} \to \mathcal{B}\) are zero for \(i < 0\). Also \(R^0F = F : \mathcal{A} \to \mathcal{B}\).

3. We have \(R^iF(I) = 0\) for \(i > 0\) and \(I\) injective.

4. The sequence \((R^iF, \delta)\) forms a universal \(\delta\)-functor (see Homology, Definition 010S) from \(\mathcal{A}\) to \(\mathcal{B}\).

Proof

This lemma simply reviews some of the results obtained so far. Note that by Lemma 05TI \(RF\) is everywhere defined. Here are some references:

1. This follows from Lemma 0159 part (3) combined with the long exact cohomology sequence (05ST) for \(D^{+}(\mathcal{B})\).

2. This is Lemma 05TD.

3. This is the fact that injective objects are acyclic.

4. This is Lemma 05TE.

Cartan-Eilenberg resolutions

This section can be expanded. The material can be generalized and applied in more cases. Resolutions need not use injectives and the method also works in the unbounded case in some situations.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet\) be a bounded below complex. A Cartan-Eilenberg resolution of \(K^\bullet\) is given by a double complex \(I^{\bullet, \bullet}\) and a morphism of complexes \(\epsilon : K^\bullet \to I^{\bullet, 0}\) with the following properties:

1. There exists an \(i \ll 0\) such that \(I^{p, q} = 0\) for all \(p < i\) and all \(q\).

2. We have \(I^{p, q} = 0\) if \(q < 0\).

3. The complex \(I^{p, \bullet}\) is an injective resolution of \(K^p\).

4. The complex \(\Ker(d_1^{p, \bullet})\) is an injective resolution of \(\Ker(d_K^p)\).

5. The complex \(\Im(d_1^{p, \bullet})\) is an injective resolution of \(\Im(d_K^p)\).

6. The complex \(H^p_I(I^{\bullet, \bullet})\) is an injective resolution of \(H^p(K^\bullet)\).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Let \(K^\bullet\) be a bounded below complex. There exists a Cartan-Eilenberg resolution of \(K^\bullet\).

Proof

Suppose that \(K^p = 0\) for \(p < n\). Decompose \(K^\bullet\) into short exact sequences as follows: Set \(Z^p = \Ker(d^p)\), \(B^p = \Im(d^{p - 1})\), \(H^p = Z^p/B^p\), and consider \[\begin{matrix} 0 \to Z^n \to K^n \to B^{n + 1} \to 0 \\ 0 \to B^{n + 1} \to Z^{n + 1} \to H^{n + 1} \to 0 \\ 0 \to Z^{n + 1} \to K^{n + 1} \to B^{n + 2} \to 0 \\ 0 \to B^{n + 2} \to Z^{n + 2} \to H^{n + 2} \to 0 \\ \ldots \end{matrix}\] Set \(I^{p, q} = 0\) for \(p < n\). Inductively we choose injective resolutions as follows:

1. Choose an injective resolution \(Z^n \to J_Z^{n, \bullet}\) with \(J_Z^{n, m} = 0\) for \(m < 0\), see Lemma 013K.

2. Using Lemma 013T choose injective resolutions \(K^n \to I^{n, \bullet}\), \(B^{n + 1} \to J_B^{n + 1, \bullet}\) with \(I^{n, m} = 0\) for \(m < 0\) and \(J_B^{n + 1, m} = 0\) for \(m < 0\), and an exact sequence of complexes \(0 \to J_Z^{n, \bullet} \to I^{n, \bullet} \to J_B^{n + 1, \bullet} \to 0\) compatible with the short exact sequence \(0 \to Z^n \to K^n \to B^{n + 1} \to 0\).

3. Using Lemma 013T choose injective resolutions \(Z^{n + 1} \to J_Z^{n + 1, \bullet}\), \(H^{n + 1} \to J_H^{n + 1, \bullet}\) with \(J_Z^{n + 1, m} = 0\) for \(m < 0\) and \(J_H^{n + 1, m} = 0\) for \(m < 0\) and an exact sequence of complexes \(0 \to J_B^{n + 1, \bullet} \to J_Z^{n + 1, \bullet} \to J_H^{n + 1, \bullet} \to 0\) compatible with the short exact sequence \(0 \to B^{n + 1} \to Z^{n + 1} \to H^{n + 1} \to 0\).

4. Etc.

Taking as maps \(d_1^\bullet : I^{p, \bullet} \to I^{p + 1, \bullet}\) the compositions \(I^{p, \bullet} \to J_B^{p + 1, \bullet} \to J_Z^{p + 1, \bullet} \to I^{p + 1, \bullet}\) everything is clear.

Lemma

Let \(F : \mathcal{A} \to \mathcal{B}\) be a left exact functor of abelian categories. Let \(K^\bullet\) be a bounded below complex of \(\mathcal{A}\). Let \(I^{\bullet, \bullet}\) be a Cartan-Eilenberg resolution for \(K^\bullet\). The spectral sequences \(({}'E_r, {}'d_r)_{r \geq 0}\) and \(({}''E_r, {}''d_r)_{r \geq 0}\) associated to the double complex \(F(I^{\bullet, \bullet})\) satisfy the relations \[{}'E_1^{p, q} = R^qF(K^p) \quad \text{and} \quad {}''E_2^{p, q} = R^pF(H^q(K^\bullet))\] Moreover, these spectral sequences are bounded, converge to \(H^*(RF(K^\bullet))\), and the associated induced filtrations on \(H^n(RF(K^\bullet))\) are finite.

Proof

We will use the following remarks without further mention:

1. As \(I^{p, \bullet}\) is an injective resolution of \(K^p\) we see that \(RF\) is defined at \(K^p[0]\) with value \(F(I^{p, \bullet})\).

2. As \(H^p_I(I^{\bullet, \bullet})\) is an injective resolution of \(H^p(K^\bullet)\) the derived functor \(RF\) is defined at \(H^p(K^\bullet)[0]\) with value \(F(H^p_I(I^{\bullet, \bullet}))\).

3. By Homology, Lemma 0133 the total complex \(\text{Tot}(I^{\bullet, \bullet})\) is an injective resolution of \(K^\bullet\). Hence \(RF\) is defined at \(K^\bullet\) with value \(F(\text{Tot}(I^{\bullet, \bullet}))\).

Consider the two spectral sequences associated to the double complex \(L^{\bullet, \bullet} = F(I^{\bullet, \bullet})\), see Homology, Lemma 0130. These are both bounded, converge to \(H^*(\text{Tot}(L^{\bullet, \bullet}))\), and induce finite filtrations on \(H^n(\text{Tot}(L^{\bullet, \bullet}))\), see Homology, Lemma 0132. Since \(\text{Tot}(L^{\bullet, \bullet}) = \text{Tot}(F(I^{\bullet, \bullet})) = F(\text{Tot}(I^{\bullet, \bullet}))\) computes \(RF(K^\bullet)\) we find the final assertion of the lemma holds true.

Computation of the first spectral sequence. We have \({}'E_1^{p, q} = H^q(L^{p, \bullet})\) in other words \[{}'E_1^{p, q} = H^q(F(I^{p, \bullet})) = R^qF(K^p)\] as desired. Observe for later use that the maps \({}'d_1^{p, q} : {}'E_1^{p, q} \to {}'E_1^{p + 1, q}\) are the maps \(R^qF(K^p) \to R^qF(K^{p + 1})\) induced by \(K^p \to K^{p + 1}\) and the fact that \(R^qF\) is a functor.

Computation of the second spectral sequence. We have \({}''E_1^{p, q} = H^q(L^{\bullet, p}) = H^q(F(I^{\bullet, p}))\). Note that the complex \(I^{\bullet, p}\) is bounded below, consists of injectives, and moreover each kernel, image, and cohomology group of the differentials is an injective object of \(\mathcal{A}\). Hence we can split the differentials, i.e., each differential is a split surjection onto a direct summand. It follows that the same is true after applying \(F\). Hence \({}''E_1^{p, q} = F(H^q(I^{\bullet, p})) = F(H^q_I(I^{\bullet, p}))\). The differentials on this are \((-1)^q\) times \(F\) applied to the differential of the complex \(H^p_I(I^{\bullet, \bullet})\) which is an injective resolution of \(H^p(K^\bullet)\). Hence the description of the \(E_2\) terms.

Remark

The spectral sequences of Lemma 015J are functorial in the complex \(K^\bullet\). This follows from functoriality properties of Cartan-Eilenberg resolutions. On the other hand, they are both examples of a more general spectral sequence which may be associated to a filtered complex of \(\mathcal{A}\). The functoriality will follow from its construction. We will return to this in the section on the filtered derived category, see Remark 015X.

Composition of right derived functors

Sometimes we can compute the right derived functor of a composition. Suppose that \(\mathcal{A}, \mathcal{B}, \mathcal{C}\) are abelian categories. Let \(F : \mathcal{A} \to \mathcal{B}\) and \(G : \mathcal{B} \to \mathcal{C}\) be left exact functors. Assume that the right derived functors \(RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\), \(RG : D^{+}(\mathcal{B}) \to D^{+}(\mathcal{C})\), and \(R(G \circ F) : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{C})\) are everywhere defined. Then there exists a canonical transformation \[t : R(G \circ F) \longrightarrow RG \circ RF\] of functors from \(D^{+}(\mathcal{A})\) to \(D^{+}(\mathcal{C})\), see Lemma 05T2. This transformation need not always be an isomorphism.

Lemma

Let \(\mathcal{A}, \mathcal{B}, \mathcal{C}\) be abelian categories. Let \(F : \mathcal{A} \to \mathcal{B}\) and \(G : \mathcal{B} \to \mathcal{C}\) be left exact functors. Assume \(\mathcal{A}\), \(\mathcal{B}\) have enough injectives. The following are equivalent

1. \(F(I)\) is right acyclic for \(G\) for each injective object \(I\) of \(\mathcal{A}\), and

2. the canonical map \[t : R(G \circ F) \longrightarrow RG \circ RF\] is an isomorphism of functors from \(D^{+}(\mathcal{A})\) to \(D^{+}(\mathcal{C})\).

Proof

If (2) holds, then (1) follows by evaluating the isomorphism \(t\) on \(RF(I) = F(I)\). Conversely, assume (1) holds. Let \(A^\bullet\) be a bounded below complex of \(\mathcal{A}\). Choose an injective resolution \(A^\bullet \to I^\bullet\). The map \(t\) is given (see proof of Lemma 05T2) by the maps \[R(G \circ F)(A^\bullet) = (G \circ F)(I^\bullet) = G(F(I^\bullet)) \to RG(F(I^\bullet)) = RG(RF(A^\bullet))\] where the arrow is an isomorphism by Lemma 015E.

Lemma

Assume the hypotheses of Lemma 015M and that the equivalent conditions (1) and (2) hold. Let \(X\) be an object of \(D^{+}(\mathcal{A})\). There exists a spectral sequence \((E_r, d_r)_{r \geq 0}\) consisting of bigraded objects \(E_r\) of \(\mathcal{C}\) with \(d_r\) of bidegree \((r, - r + 1)\) and with \[E_2^{p, q} = R^pG(H^q(RF(X)))\] Moreover, this spectral sequence is bounded, converges to \(H^*(R(G \circ F)(X))\), and induces a finite filtration on each \(H^n(R(G \circ F)(X))\).

For an object \(A\) of \(\mathcal{A}\) we get \(E_2^{p, q} = R^pG(R^qF(A))\) converging to \(R^{p + q}(G \circ F)(A)\).

Proof

We may represent \(X\) by a bounded below complex \(A^\bullet\). Choose an injective resolution \(A^\bullet \to I^\bullet\). Choose a Cartan-Eilenberg resolution \(F(I^\bullet) \to I^{\bullet, \bullet}\) using Lemma 015I. Apply the second spectral sequence of Lemma 015J.

Resolution functors

Let \(\mathcal{A}\) be an abelian category with enough injectives. Denote \(\mathcal{I}\) the full additive subcategory of \(\mathcal{A}\) whose objects are the injective objects of \(\mathcal{A}\). It turns out that \(K^{+}(\mathcal{I})\) and \(D^{+}(\mathcal{A})\) are equivalent in this case (see Proposition 013V). For many purposes it therefore makes sense to think of \(D^{+}(\mathcal{A})\) as the (easier to grok) category \(K^{+}(\mathcal{I})\) in this case.

Proposition

Let \(\mathcal{A}\) be an abelian category. Assume \(\mathcal{A}\) has enough injectives. Denote \(\mathcal{I} \subset \mathcal{A}\) the strictly full additive subcategory whose objects are the injective objects of \(\mathcal{A}\). The functor \[K^{+}(\mathcal{I}) \longrightarrow D^{+}(\mathcal{A})\] is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories.

Proof

It is clear that the functor is exact. It is essentially surjective by Lemma 013K. Full faithfulness is a consequence of Lemma 05TG.

Proposition 013V implies that we can find resolution functors. It turns out that we can prove resolution functors exist even in some cases where the abelian category \(\mathcal{A}\) is a “big” category, i.e., has a class of objects.

Definition

Let \(\mathcal{A}\) be an abelian category with enough injectives. A resolution functor⁸ for \(\mathcal{A}\) is given by the following data:

1. for all \(K^\bullet \in \Ob(K^{+}(\mathcal{A}))\) a bounded below complex of injectives \(j(K^\bullet)\), and

2. for all \(K^\bullet \in \Ob(K^{+}(\mathcal{A}))\) a quasi-isomorphism \(i_{K^\bullet} : K^\bullet \to j(K^\bullet)\).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Given a resolution functor \((j, i)\) there is a unique way to turn \(j\) into a functor and \(i\) into a \(2\)-isomorphism producing a \(2\)-commutative diagram \[\xymatrix{ K^{+}(\mathcal{A}) \ar[rd] \ar[rr]_j & & K^{+}(\mathcal{I}) \ar[ld] \\ & D^{+}(\mathcal{A}) }\] where \(\mathcal{I}\) is the full additive subcategory of \(\mathcal{A}\) consisting of injective objects.

Proof

For every morphism \(\alpha : K^\bullet \to L^\bullet\) of \(K^{+}(\mathcal{A})\) there is a unique morphism \(j(\alpha) : j(K^\bullet) \to j(L^\bullet)\) in \(K^{+}(\mathcal{I})\) such that \[\xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_{i_{K^\bullet}} & L^\bullet \ar[d]^{i_{L^\bullet}} \\ j(K^\bullet) \ar[r]^{j(\alpha)} & j(L^\bullet) }\] is commutative in \(K^{+}(\mathcal{A})\). To see this either use Lemmas 013P and 013S or the equivalent Lemma 05TG. The uniqueness implies that \(j\) is a functor, and the commutativity of the diagram implies that \(i\) gives a \(2\)-morphism which witnesses the \(2\)-commutativity of the diagram of categories in the statement of the lemma.

Lemma

Let \(\mathcal{A}\) be an abelian category. Assume \(\mathcal{A}\) has enough injectives. Then a resolution functor \(j\) exists and is unique up to unique isomorphism of functors.

Proof

Consider the set of all objects \(K^\bullet\) of \(K^{+}(\mathcal{A})\). (Recall that by our conventions any category has a set of objects unless mentioned otherwise.) By Lemma 013K every object has an injective resolution. By the axiom of choice we can choose for each \(K^\bullet\) an injective resolution \(i_{K^\bullet} : K^\bullet \to j(K^\bullet)\).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Any resolution functor \(j : K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})\) is exact.

Proof

Denote \(i_{K^\bullet} : K^\bullet \to j(K^\bullet)\) the canonical maps of Definition 013W. First we discuss the existence of the functorial isomorphism \(j(K^\bullet[1]) \to j(K^\bullet)[1]\). Consider the diagram \[\xymatrix{ K^\bullet[1] \ar[d]^{i_{K^\bullet[1]}} \ar@{=}[rr] & & K^\bullet[1] \ar[d]^{i_{K^\bullet}[1]} \\ j(K^\bullet[1]) \ar@{..>}[rr]^{\xi_{K^\bullet}} & & j(K^\bullet)[1] }\] By Lemmas 013P and 013S there exists a unique dotted arrow \(\xi_{K^\bullet}\) in \(K^{+}(\mathcal{I})\) making the diagram commute in \(K^{+}(\mathcal{A})\). We omit the verification that this gives a functorial isomorphism. (Hint: use Lemma 013S again.)

Let \((K^\bullet, L^\bullet, M^\bullet, f, g, h)\) be a distinguished triangle of \(K^{+}(\mathcal{A})\). We have to show that \((j(K^\bullet), j(L^\bullet), j(M^\bullet), j(f), j(g), \xi_{K^\bullet} \circ j(h))\) is a distinguished triangle of \(K^{+}(\mathcal{I})\). Note that we have a commutative diagram \[\xymatrix{ K^\bullet \ar[r]_f \ar[d] & L^\bullet \ar[r]_g \ar[d] & M^\bullet \ar[rr]_h \ar[d] & & K^\bullet[1] \ar[d] \\ j(K^\bullet) \ar[r]^{j(f)} & j(L^\bullet) \ar[r]^{j(g)} & j(M^\bullet) \ar[rr]^{\xi_{K^\bullet} \circ j(h)} & & j(K^\bullet)[1] }\] in \(K^{+}(\mathcal{A})\) whose vertical arrows are the quasi-isomorphisms \(i_K, i_L, i_M\). Hence we see that the image of \((j(K^\bullet), j(L^\bullet), j(M^\bullet), j(f), j(g), \xi_{K^\bullet} \circ j(h))\) in \(D^{+}(\mathcal{A})\) is isomorphic to a distinguished triangle and hence a distinguished triangle by TR1. Thus we see from Lemma 05SQ that \((j(K^\bullet), j(L^\bullet), j(M^\bullet), j(f), j(g), \xi_{K^\bullet} \circ j(h))\) is a distinguished triangle in \(K^{+}(\mathcal{I})\).

Lemma

Let \(\mathcal{A}\) be an abelian category which has enough injectives. Let \(j\) be a resolution functor. Write \(Q : K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})\) for the natural functor. Then \(j = j' \circ Q\) for a unique functor \(j' : D^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})\) which is quasi-inverse to the canonical functor \(K^{+}(\mathcal{I}) \to D^{+}(\mathcal{A})\).

Proof

The functor \(Q\) is a localization by Lemma 05RW. To prove the existence of \(j'\) it suffices to show that any element of \(\text{Qis}^{+}(\mathcal{A})\) is mapped to an isomorphism under the functor \(j\), see Lemma 05R7. Consider the commutative square in proof of Lemma 014W. In this square, if \(\alpha\) is a quasi-isomorphism, then \(i_{L^\bullet}\circ\alpha=j(\alpha)\circ i_{K^\bullet}\) is a quasi-isomorphism too, hence so is \(j(\alpha)\). Thus by Proposition 013V the morphism \(j(\alpha)\) is an isomorphism in \(K^+(\mathcal{I})\). We omit the verification that \(j'\) is quasi-inverse to \(K^{+}(\mathcal{I}) \to D^{+}(\mathcal{A})\).

Remark

Suppose that \(\mathcal{A}\) is a “big” abelian category with enough injectives such as the category of abelian groups. In this case we have to be slightly more careful in constructing our resolution functor since we cannot use the axiom of choice with a quantifier ranging over a class. But note that the proof of the lemma does show that any two resolution functors are canonically isomorphic. Namely, given quasi-isomorphisms \(i : K^\bullet \to I^\bullet\) and \(i' : K^\bullet \to J^\bullet\) of a bounded below complex \(K^\bullet\) into bounded below complexes of injectives there exists a unique(!) morphism \(a : I^\bullet \to J^\bullet\) in \(K^{+}(\mathcal{I})\) such that \(i' = a \circ i\) as morphisms in \(K^{+}(\mathcal{A})\). Hence the only issue is existence, and we will see how to deal with this in the next section.

Functorial injective embeddings and resolution functors

In this section we redo the construction of a resolution functor \(K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})\) in case the category \(\mathcal{A}\) has functorial injective embeddings. There are two reasons for this: (1) the proof is easier and (2) the construction also works if \(\mathcal{A}\) is a “big” abelian category. See Remark 0142 below.

Let \(\mathcal{A}\) be an abelian category. As before denote \(\mathcal{I}\) the additive full subcategory of \(\mathcal{A}\) consisting of injective objects. Consider the category \(\text{InjRes}(\mathcal{A})\) of arrows \(\alpha : K^\bullet \to I^\bullet\) where \(K^\bullet\) is a bounded below complex of \(\mathcal{A}\), \(I^\bullet\) is a bounded below complex of injectives of \(\mathcal{A}\) and \(\alpha\) is a quasi-isomorphism. In other words, \(\alpha\) is an injective resolution and \(K^\bullet\) is bounded below. There is an obvious functor \[s : \text{InjRes}(\mathcal{A}) \longrightarrow \text{Comp}^{+}(\mathcal{A})\] defined by \((\alpha : K^\bullet \to I^\bullet) \mapsto K^\bullet\). There is also a functor \[t : \text{InjRes}(\mathcal{A}) \longrightarrow K^{+}(\mathcal{I})\] defined by \((\alpha : K^\bullet \to I^\bullet) \mapsto I^\bullet\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Assume \(\mathcal{A}\) has functorial injective embeddings, see Homology, Definition 0139.

1. There exists a functor \(inj : \text{Comp}^{+}(\mathcal{A}) \to \text{InjRes}(\mathcal{A})\) such that \(s \circ inj = \text{id}\).

2. For any functor \(inj : \text{Comp}^{+}(\mathcal{A}) \to \text{InjRes}(\mathcal{A})\) such that \(s \circ inj = \text{id}\) we obtain a resolution functor, see Definition 013W.

Proof

Let \(A \mapsto (A \to J(A))\) be a functorial injective embedding, see Homology, Definition 0139. We first note that we may assume \(J(0) = 0\). Namely, if not then for any object \(A\) we have \(0 \to A \to 0\) which gives a direct sum decomposition \(J(A) = J(0) \oplus \Ker(J(A) \to J(0))\). Note that the functorial morphism \(A \to J(A)\) has to map into the second summand. Hence we can replace our functor by \(J'(A) = \Ker(J(A) \to J(0))\) if needed.

Let \(K^\bullet\) be a bounded below complex of \(\mathcal{A}\). Say \(K^p = 0\) if \(p < B\). We are going to construct a double complex \(I^{\bullet, \bullet}\) of injectives, together with a map \(\alpha : K^\bullet \to I^{\bullet, 0}\) such that \(\alpha\) induces a quasi-isomorphism of \(K^\bullet\) with the associated total complex of \(I^{\bullet, \bullet}\). First we set \(I^{p, q} = 0\) whenever \(q < 0\). Next, we set \(I^{p, 0} = J(K^p)\) and \(\alpha^p : K^p \to I^{p, 0}\) the functorial embedding. Since \(J\) is a functor we see that \(I^{\bullet, 0}\) is a complex and that \(\alpha\) is a morphism of complexes. Each \(\alpha^p\) is injective. And \(I^{p, 0} = 0\) for \(p < B\) because \(J(0) = 0\). Next, we set \(I^{p, 1} = J(\Coker(K^p \to I^{p, 0}))\). Again by functoriality we see that \(I^{\bullet, 1}\) is a complex. And again we get that \(I^{p, 1} = 0\) for \(p < B\). It is also clear that \(K^p\) maps isomorphically onto \(\Ker(I^{p, 0} \to I^{p, 1})\). As our third step we take \(I^{p, 2} = J(\Coker(I^{p, 0} \to I^{p, 1}))\). And so on and so forth.

At this point we can apply Homology, Lemma 0133 to get that the map \[\alpha : K^\bullet \longrightarrow \text{Tot}(I^{\bullet, \bullet})\] is a quasi-isomorphism. To prove we get a functor \(inj\) it remains to show that the construction above is functorial. This verification is omitted.

Suppose we have a functor \(inj\) such that \(s \circ inj = \text{id}\). For every object \(K^\bullet\) of \(\text{Comp}^{+}(\mathcal{A})\) we can write \[inj(K^\bullet) = (i_{K^\bullet} : K^\bullet \to j(K^\bullet))\] This provides us with a resolution functor as in Definition 013W.

Remark

Suppose \(inj\) is a functor such that \(s \circ inj = \text{id}\) as in part (2) of Lemma 0141. Write \(inj(K^\bullet) = (i_{K^\bullet} : K^\bullet \to j(K^\bullet))\) as in the proof of that lemma. Suppose \(\alpha : K^\bullet \to L^\bullet\) is a map of bounded below complexes. Consider the map \(inj(\alpha)\) in the category \(\text{InjRes}(\mathcal{A})\). It induces a commutative diagram \[\xymatrix{ K^\bullet \ar[rr]^-{\alpha} \ar[d]_{i_K} & & L^\bullet \ar[d]^{i_L} \\ j(K^\bullet) \ar[rr]^-{inj(\alpha)} & & j(L^\bullet) }\] of morphisms of complexes. Hence, looking at the proof of Lemma 05TJ we see that the functor \(j : K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})\) is given by the rule \[j(\alpha\text{ up to homotopy}) = inj(\alpha)\text{ up to homotopy}\in \Hom_{K^{+}(\mathcal{I})}(j(K^\bullet), j(L^\bullet))\] Hence we see that \(j\) matches \(t \circ inj\) in this case, i.e., the diagram \[\xymatrix{ \text{Comp}^{+}(\mathcal{A}) \ar[rr]_{t \circ inj} \ar[rd] & & K^{+}(\mathcal{I}) \\ & K^{+}(\mathcal{A}) \ar[ru]_j }\] is commutative.

Remark

Let \(\textit{Mod}(\mathcal{O}_X)\) be the category of \(\mathcal{O}_X\)-modules on a ringed space \((X, \mathcal{O}_X)\) (or more generally on a ringed site). We will see later that \(\textit{Mod}(\mathcal{O}_X)\) has enough injectives and in fact functorial injective embeddings, see Injectives, Theorem 01DU. Note that the proof of Lemma 013X does not apply to \(\textit{Mod}(\mathcal{O}_X)\). But the proof of Lemma 0141 does apply to \(\textit{Mod}(\mathcal{O}_X)\). Thus we obtain \[j : K^{+}(\textit{Mod}(\mathcal{O}_X)) \longrightarrow K^{+}(\mathcal{I})\] which is a resolution functor where \(\mathcal{I}\) is the additive category of injective \(\mathcal{O}_X\)-modules. This argument also works in the following cases:

1. The category \(\text{Mod}_R\) of \(R\)-modules over a ring \(R\).

2. The category \(\textit{PMod}(\mathcal{O})\) of presheaves of \(\mathcal{O}\)-modules on a site endowed with a presheaf of rings.

3. The category \(\textit{Mod}(\mathcal{O})\) of sheaves of \(\mathcal{O}\)-modules on a ringed site.

Right derived functors via resolution functors

The content of the following lemma is that we can simply define \(RF(K^\bullet) = F(j(K^\bullet))\) if we are given a resolution functor \(j\).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Let \(F : \mathcal{A} \to \mathcal{B}\) be an additive functor into an abelian category. Let \((j, i)\) be a resolution functor, see Definition 013W. The right derived functor \(RF\) of \(F\) fits into the following \(2\)-commutative diagram \[\xymatrix{ D^{+}(\mathcal{A}) \ar[rd]_{RF} \ar[rr]^{j'} & & K^{+}(\mathcal{I}) \ar[ld]^F \\ & D^{+}(\mathcal{B}) }\] where \(j'\) is the functor from Lemma 05TK.

Proof

By Lemma 05TH we have \(RF(K^\bullet) = F(j(K^\bullet))\).

Remark

In the situation of Lemma 05TN we see that we have actually lifted the right derived functor to an exact functor \(F \circ j' : D^{+}(\mathcal{A}) \to K^{+}(\mathcal{B})\). It is occasionally useful to use such a factorization.

Filtered derived category and injective resolutions

Let \(\mathcal{A}\) be an abelian category. In this section we will show that if \(\mathcal{A}\) has enough injectives, then so does the category \(\text{Fil}^f(\mathcal{A})\) in some sense. One can use this observation to compute in the filtered derived category of \(\mathcal{A}\).

The category \(\text{Fil}^f(\mathcal{A})\) is an example of an exact category, see Injectives, Remark 05SF. A special role is played by the strict morphisms, see Homology, Definition 0123, i.e., the morphisms \(f\) such that \(\Coim(f) = \Im(f)\). We will say that a complex \(A \to B \to C\) in \(\text{Fil}^f(\mathcal{A})\) is exact if the sequence \(\text{gr}(A) \to \text{gr}(B) \to \text{gr}(C)\) is exact in \(\mathcal{A}\). This implies that \(A \to B\) and \(B \to C\) are strict morphisms, see Homology, Lemma 05QH.

Definition

Let \(\mathcal{A}\) be an abelian category. We say an object \(I\) of \(\text{Fil}^f(\mathcal{A})\) is filtered injective if each \(\text{gr}^p(I)\) is an injective object of \(\mathcal{A}\).

Lemma

Let \(\mathcal{A}\) be an abelian category. An object \(I\) of \(\text{Fil}^f(\mathcal{A})\) is filtered injective if and only if there exist \(a \leq b\), injective objects \(I_n\), \(a \leq n \leq b\) of \(\mathcal{A}\) and an isomorphism \(I \cong \bigoplus_{a \leq n \leq b} I_n\) such that \(F^pI = \bigoplus_{n \geq p} I_n\).

Proof

Follows from the fact that any injection \(J \to M\) of \(\mathcal{A}\) is split if \(J\) is an injective object. Details omitted.

Lemma

Let \(\mathcal{A}\) be an abelian category. Any strict monomorphism \(u : I \to A\) of \(\text{Fil}^f(\mathcal{A})\) where \(I\) is a filtered injective object is a split injection.

Proof

Let \(p\) be the largest integer such that \(F^pI \not = 0\). In particular \(\text{gr}^p(I) = F^pI\). Let \(I'\) be the object of \(\text{Fil}^f(\mathcal{A})\) whose underlying object of \(\mathcal{A}\) is \(F^pI\) and with filtration given by \(F^nI' = 0\) for \(n > p\) and \(F^nI' = I' = F^pI\) for \(n \leq p\). Note that \(I' \to I\) is a strict monomorphism too. The fact that \(u\) is a strict monomorphism implies that \(F^pI \to A/F^{p + 1}(A)\) is injective, see Homology, Lemma 0127. Choose a splitting \(s : A/F^{p + 1}A \to F^pI\) in \(\mathcal{A}\). The induced morphism \(s' : A \to I'\) is a strict morphism of filtered objects splitting the composition \(I' \to I \to A\). Hence we can write \(A = I' \oplus \Ker(s')\) and \(I = I' \oplus \Ker(s'|_I)\). Note that \(\Ker(s'|_I) \to \Ker(s')\) is a strict monomorphism and that \(\Ker(s'|_I)\) is a filtered injective object. By induction on the length of the filtration on \(I\) the map \(\Ker(s'|_I) \to \Ker(s')\) is a split injection. Thus we win.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(u : A \to B\) be a strict monomorphism of \(\text{Fil}^f(\mathcal{A})\) and \(f : A \to I\) a morphism from \(A\) into a filtered injective object in \(\text{Fil}^f(\mathcal{A})\). Then there exists a morphism \(g : B \to I\) such that \(f = g \circ u\).

Proof

The pushout \(f' : I \to I \amalg_A B\) of \(f\) by \(u\) is a strict monomorphism, see Homology, Lemma 05SM. Hence the result follows formally from Lemma 05TQ.

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. For any object \(A\) of \(\text{Fil}^f(\mathcal{A})\) there exists a strict monomorphism \(A \to I\) where \(I\) is a filtered injective object.

Proof

Pick \(a \leq b\) such that \(\text{gr}^p(A) = 0\) unless \(p \in \{a, a + 1, \ldots, b\}\). For each \(n \in \{a, a + 1, \ldots, b\}\) choose an injection \(u_n : A/F^{n + 1}A \to I_n\) with \(I_n\) an injective object. Set \(I = \bigoplus_{a \leq n \leq b} I_n\) with filtration \(F^pI = \bigoplus_{n \geq p} I_n\) and set \(u : A \to I\) equal to the direct sum of the maps \(u_n\).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. For any object \(A\) of \(\text{Fil}^f(\mathcal{A})\) there exists a filtered quasi-isomorphism \(A[0] \to I^\bullet\) where \(I^\bullet\) is a complex of filtered injective objects with \(I^n = 0\) for \(n < 0\).

Proof

First choose a strict monomorphism \(u_0 : A \to I^0\) of \(A\) into a filtered injective object, see Lemma 05TS. Next, choose a strict monomorphism \(u_1 : \Coker(u_0) \to I^1\) into a filtered injective object of \(\text{Fil}^f(\mathcal{A})\). Denote \(d^0\) the induced map \(I^0 \to I^1\). Next, choose a strict monomorphism \(u_2 : \Coker(u_1) \to I^2\) into a filtered injective object of \(\text{Fil}^f(\mathcal{A})\). Denote \(d^1\) the induced map \(I^1 \to I^2\). And so on. This works because each of the sequences \[0 \to \Coker(u_n) \to I^{n + 1} \to \Coker(u_{n + 1}) \to 0\] is short exact, i.e., induces a short exact sequence on applying \(\text{gr}\). To see this use Homology, Lemma 0127.

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Let \(f : A \to B\) be a morphism of \(\text{Fil}^f(\mathcal{A})\). Given filtered quasi-isomorphisms \(A[0] \to I^\bullet\) and \(B[0] \to J^\bullet\) where \(I^\bullet, J^\bullet\) are complexes of filtered injective objects with \(I^n = J^n = 0\) for \(n < 0\), then there exists a commutative diagram \[\xymatrix{ A[0] \ar[r] \ar[d] & B[0] \ar[d] \\ I^\bullet \ar[r] & J^\bullet }\]

Proof

As \(A[0] \to I^\bullet\) and \(B[0] \to J^\bullet\) are filtered quasi-isomorphisms we conclude that \(a : A \to I^0\), \(b : B \to J^0\) and all the morphisms \(d_I^n\), \(d_J^n\) are strict, see Homology, Lemma 05QH. We will inductively construct the maps \(f^n\) in the following commutative diagram \[\xymatrix{ A \ar[r]_a \ar[d]_f & I^0 \ar[r] \ar[d]^{f^0} & I^1 \ar[r] \ar[d]^{f^1} & I^2 \ar[r] \ar[d]^{f^2} & \ldots \\ B \ar[r]^b & J^0 \ar[r] & J^1 \ar[r] & J^2 \ar[r] & \ldots }\] Because \(A \to I^0\) is a strict monomorphism and because \(J^0\) is filtered injective, we can find a morphism \(f^0 : I^0 \to J^0\) such that \(f^0 \circ a = b \circ f\), see Lemma 05TR. The composition \(d_J^0 \circ b \circ f\) is zero, hence \(d_J^0 \circ f^0 \circ a = 0\), hence \(d_J^0 \circ f^0\) factors through a unique morphism \[\Coker(a) = \Coim(d_I^0) = \Im(d_I^0) \longrightarrow J^1.\] As \(\Im(d_I^0) \to I^1\) is a strict monomorphism we can extend the displayed arrow to a morphism \(f^1 : I^1 \to J^1\) by Lemma 05TR again. And so on.

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Let \(0 \to A \to B \to C \to 0\) be a short exact sequence in \(\text{Fil}^f(\mathcal{A})\). Given filtered quasi-isomorphisms \(A[0] \to I^\bullet\) and \(C[0] \to J^\bullet\) where \(I^\bullet, J^\bullet\) are complexes of filtered injective objects with \(I^n = J^n = 0\) for \(n < 0\), then there exists a commutative diagram \[\xymatrix{ 0 \ar[r] & A[0] \ar[r] \ar[d] & B[0] \ar[r] \ar[d] & C[0] \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I^\bullet \ar[r] & M^\bullet \ar[r] & J^\bullet \ar[r] & 0 }\] where the lower row is a termwise split sequence of complexes.

Proof

As \(A[0] \to I^\bullet\) and \(C[0] \to J^\bullet\) are filtered quasi-isomorphisms we conclude that \(a : A \to I^0\), \(c : C \to J^0\) and all the morphisms \(d_I^n\), \(d_J^n\) are strict, see Homology, Lemma 05QH. We are going to step by step construct the south-east and the south arrows in the following commutative diagram \[\xymatrix{ B \ar[r]_\beta \ar[rd]^b & C \ar[r]_c \ar[rd]^{\overline{b}} & J^0 \ar[d]^{\delta^0} \ar[r] & J^1 \ar[d]^{\delta^1} \ar[r] & \ldots \\ A \ar[u]^\alpha \ar[r]^a & I^0 \ar[r] & I^1 \ar[r] & I^2 \ar[r] & \ldots }\] As \(A \to B\) is a strict monomorphism, we can find a morphism \(b : B \to I^0\) such that \(b \circ \alpha = a\), see Lemma 05TR. As \(A\) is the kernel of the strict morphism \(I^0 \to I^1\) and \(\beta = \Coker(\alpha)\) we obtain a unique morphism \(\overline{b} : C \to I^1\) fitting into the diagram. As \(c\) is a strict monomorphism and \(I^1\) is filtered injective we can find \(\delta^0 : J^0 \to I^1\), see Lemma 05TR. Because \(B \to C\) is a strict epimorphism and because \(B \to I^0 \to I^1 \to I^2\) is zero, we see that \(C \to I^1 \to I^2\) is zero. Hence \(d_I^1 \circ \delta^0\) is zero on \(C \cong \Im(c)\). Hence \(d_I^1 \circ \delta^0\) factors through a unique morphism \[\Coker(c) = \Coim(d_J^0) = \Im(d_J^0) \longrightarrow I^2.\] As \(I^2\) is filtered injective and \(\Im(d_J^0) \to J^1\) is a strict monomorphism we can extend the displayed morphism to a morphism \(\delta^1 : J^1 \to I^2\), see Lemma 05TR. And so on. We set \(M^\bullet = I^\bullet \oplus J^\bullet\) with differential \[d_M^n = \left( \begin{matrix} d_I^n & (-1)^{n + 1}\delta^n \\ 0 & d_J^n \end{matrix} \right)\] Finally, the map \(B[0] \to M^\bullet\) is given by \(b \oplus c \circ \beta : B \to I^0 \oplus J^0\).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. For every \(K^\bullet \in K^{+}(\text{Fil}^f(\mathcal{A}))\) there exists a filtered quasi-isomorphism \(K^\bullet \to I^\bullet\) with \(I^\bullet\) bounded below, each \(I^n\) a filtered injective object, and each \(K^n \to I^n\) a strict monomorphism.

Proof

After replacing \(K^\bullet\) by a shift (which is harmless for the proof) we may assume that \(K^n = 0\) for \(n < 0\). Consider the short exact sequences \[\begin{matrix} 0 \to \Ker(d_K^0) \to K^0 \to \Coim(d_K^0) \to 0 \\ 0 \to \Ker(d_K^1) \to K^1 \to \Coim(d_K^1) \to 0 \\ 0 \to \Ker(d_K^2) \to K^2 \to \Coim(d_K^2) \to 0 \\ \ldots \end{matrix}\] of the exact category \(\text{Fil}^f(\mathcal{A})\) and the maps \(u_i : \Coim(d_K^i) \to \Ker(d_K^{i + 1})\). For each \(i \geq 0\) we may choose filtered quasi-isomorphisms \[\begin{matrix} \Ker(d_K^i)[0] \to I_{ker, i}^\bullet \\ \Coim(d_K^i)[0] \to I_{coim, i}^\bullet \end{matrix}\] with \(I_{ker, i}^n, I_{coim, i}^n\) filtered injective and zero for \(n < 0\), see Lemma 05TT. By Lemma 05TU we may lift \(u_i\) to a morphism of complexes \(u_i^\bullet : I_{coim, i}^\bullet \to I_{ker, i + 1}^\bullet\). Finally, for each \(i \geq 0\) we may complete the diagrams \[\xymatrix{ 0 \ar[r] & \Ker(d_K^i)[0] \ar[r] \ar[d] & K^i[0] \ar[r] \ar[d] & \Coim(d_K^i)[0] \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_{ker, i}^\bullet \ar[r]^{\alpha_i} & I_i^\bullet \ar[r]^{\beta_i} & I_{coim, i}^\bullet \ar[r] & 0 }\] with the lower sequence a termwise split exact sequence, see Lemma 05TV. For \(i \geq 0\) set \(d_i : I_i^\bullet \to I_{i + 1}^\bullet\) equal to \(d_i = \alpha_{i + 1} \circ u_i^\bullet \circ \beta_i\). Note that \(d_i \circ d_{i - 1} = 0\) because \(\beta_i \circ \alpha_i = 0\). Hence we have constructed a commutative diagram \[\xymatrix{ I_0^\bullet \ar[r] & I_1^\bullet \ar[r] & I_2^\bullet \ar[r] & \ldots \\ K^0[0] \ar[r] \ar[u] & K^1[0] \ar[r] \ar[u] & K^2[0] \ar[r] \ar[u] & \ldots }\] Here the vertical arrows are filtered quasi-isomorphisms. The upper row is a complex of complexes and each complex consists of filtered injective objects with no nonzero objects in degree \(< 0\). Thus we obtain a double complex by setting \(I^{a, b} = I_a^b\) and using \[d_1^{a, b} : I^{a, b} = I_a^b \to I_{a + 1}^b = I^{a + 1, b}\] the map \(d_a^b\) and using for \[d_2^{a, b} : I^{a, b} = I_a^b \to I_a^{b + 1} = I^{a, b + 1}\] the map \(d_{I_a}^b\). Denote by \(\text{Tot}(I^{\bullet, \bullet})\) the total complex associated to this double complex, see Homology, Definition 012Z. Observe that the maps \(K^n[0] \to I_n^\bullet\) come from maps \(K^n \to I^{n, 0}\) which give rise to a map of complexes \[K^\bullet \longrightarrow \text{Tot}(I^{\bullet, \bullet})\] We claim this is a filtered quasi-isomorphism. As \(\text{gr}(-)\) is an additive functor, we see that \(\text{gr}(\text{Tot}(I^{\bullet, \bullet})) = \text{Tot}(\text{gr}(I^{\bullet, \bullet}))\). Thus we can use Homology, Lemma 0133 to conclude that \(\text{gr}(K^\bullet) \to \text{gr}(\text{Tot}(I^{\bullet, \bullet}))\) is a quasi-isomorphism as desired.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K^\bullet, I^\bullet \in K(\text{Fil}^f(\mathcal{A}))\). Assume \(K^\bullet\) is filtered acyclic and \(I^\bullet\) bounded below and consisting of filtered injective objects. Any morphism \(K^\bullet \to I^\bullet\) is homotopic to zero: \(\Hom_{K(\text{Fil}^f(\mathcal{A}))}(K^\bullet, I^\bullet) = 0\).

Proof

Let \(\alpha : K^\bullet \to I^\bullet\) be a morphism of complexes. Assume that \(\alpha^j = 0\) for \(j < n\). We will show that there exists a morphism \(h : K^{n + 1} \to I^n\) such that \(\alpha^n = h \circ d\). Thus \(\alpha\) will be homotopic to the morphism of complexes \(\beta\) defined by \[\beta^j = \left\{ \begin{matrix} 0 & \text{if} & j \leq n \\ \alpha^{n + 1} - d \circ h & \text{if} & j = n + 1 \\ \alpha^j & \text{if} & j > n + 1 \end{matrix} \right.\] This will clearly prove the lemma (by induction). To prove the existence of \(h\) note that \(\alpha^n \circ d_K^{n - 1} = 0\) since \(\alpha^{n - 1} = 0\). Since \(K^\bullet\) is filtered acyclic we see that \(d_K^{n - 1}\) and \(d_K^n\) are strict and that \[0 \to \Im(d_K^{n - 1}) \to K^n \to \Im(d_K^n) \to 0\] is an exact sequence of the exact category \(\text{Fil}^f(\mathcal{A})\), see Homology, Lemma 05QH. Hence we can think of \(\alpha^n\) as a map into \(I^n\) defined on \(\Im(d_K^n)\). Using that \(\Im(d_K^n) \to K^{n + 1}\) is a strict monomorphism and that \(I^n\) is filtered injective we may lift this map to a map \(h : K^{n + 1} \to I^n\) as desired, see Lemma 05TR.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(I^\bullet \in K(\text{Fil}^f(\mathcal{A}))\) be a bounded below complex consisting of filtered injective objects.

1. Let \(\alpha : K^\bullet \to L^\bullet\) in \(K(\text{Fil}^f(\mathcal{A}))\) be a filtered quasi-isomorphism. Then the map \[\Hom_{K(\text{Fil}^f(\mathcal{A}))}(L^\bullet, I^\bullet) \to \Hom_{K(\text{Fil}^f(\mathcal{A}))}(K^\bullet, I^\bullet)\] is bijective.

2. Let \(L^\bullet \in K(\text{Fil}^f(\mathcal{A}))\). Then \[\Hom_{K(\text{Fil}^f(\mathcal{A}))}(L^\bullet, I^\bullet) = \Hom_{DF(\mathcal{A})}(L^\bullet, I^\bullet).\]

Proof

Proof of (1). Note that \[(K^\bullet, L^\bullet, C(\alpha)^\bullet, \alpha, i, -p)\] is a distinguished triangle in \(K(\text{Fil}^f(\mathcal{A}))\) (Lemma 014L) and \(C(\alpha)^\bullet\) is a filtered acyclic complex (Lemma 05S1). Then \[\xymatrix{ \Hom_{K(\text{Fil}^f(\mathcal{A}))}(C(\alpha)^\bullet, I^\bullet) \ar[r] & \Hom_{K(\text{Fil}^f(\mathcal{A}))}(L^\bullet, I^\bullet) \ar[r] & \Hom_{K(\text{Fil}^f(\mathcal{A}))}(K^\bullet, I^\bullet) \ar[lld] \\ \Hom_{K(\text{Fil}^f(\mathcal{A}))}(C(\alpha)^\bullet[-1], I^\bullet) }\] is an exact sequence of abelian groups, see Lemma 0149. At this point Lemma 05TX guarantees that the outer two groups are zero and hence \(\Hom_{K(\text{Fil}^f(\mathcal{A}))}(L^\bullet, I^\bullet) = \Hom_{K(\text{Fil}^f(\mathcal{A}))}(K^\bullet, I^\bullet)\).

Proof of (2). Let \(a\) be an element of the right hand side. We may represent \(a = \gamma\alpha^{-1}\) where \(\alpha : K^\bullet \to L^\bullet\) is a filtered quasi-isomorphism and \(\gamma : K^\bullet \to I^\bullet\) is a map of complexes. By part (1) we can find a morphism \(\beta : L^\bullet \to I^\bullet\) such that \(\beta \circ \alpha\) is homotopic to \(\gamma\). This proves that the map is surjective. Let \(b\) be an element of the left hand side which maps to zero in the right hand side. Then \(b\) is the homotopy class of a morphism \(\beta : L^\bullet \to I^\bullet\) such that there exists a filtered quasi-isomorphism \(\alpha : K^\bullet \to L^\bullet\) with \(\beta \circ \alpha\) homotopic to zero. Then part (1) shows that \(\beta\) is homotopic to zero also, i.e., \(b = 0\).

Lemma

Let \(\mathcal{A}\) be an abelian category with enough injectives. Let \(\mathcal{I}^f \subset \text{Fil}^f(\mathcal{A})\) denote the strictly full additive subcategory whose objects are the filtered injective objects. The canonical functor \[K^{+}(\mathcal{I}^f) \longrightarrow DF^{+}(\mathcal{A})\] is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories. Furthermore the diagrams \[\xymatrix{ K^{+}(\mathcal{I}^f) \ar[d]_{\text{gr}^p} \ar[r] & DF^{+}(\mathcal{A}) \ar[d]_{\text{gr}^p} \\ K^{+}(\mathcal{I}) \ar[r] & D^{+}(\mathcal{A}) } \quad \xymatrix{ K^{+}(\mathcal{I}^f) \ar[d]^{\text{forget }F} \ar[r] & DF^{+}(\mathcal{A}) \ar[d]^{\text{forget }F} \\ K^{+}(\mathcal{I}) \ar[r] & D^{+}(\mathcal{A}) }\] are commutative, where \(\mathcal{I} \subset \mathcal{A}\) is the strictly full additive subcategory whose objects are the injective objects.

Proof

The functor \(K^{+}(\mathcal{I}^f) \to DF^{+}(\mathcal{A})\) is essentially surjective by Lemma 05TW. It is fully faithful by Lemma 05TY. It is an exact functor by our definitions regarding distinguished triangles. The commutativity of the squares is immediate.

Remark

We can invert the arrow of the lemma only if \(\mathcal{A}\) is a category in our sense, namely if it has a set of objects. However, suppose given a big abelian category \(\mathcal{A}\) with enough injectives, such as \(\textit{Mod}(\mathcal{O}_X)\) for example. Then for any given set of objects \(\{A_i\}_{i\in I}\) there is an abelian subcategory \(\mathcal{A}' \subset \mathcal{A}\) containing all of them and having enough injectives, see Sets, Lemma 0010. Thus we may use the lemma above for \(\mathcal{A}'\). This essentially means that if we use a set’s worth of diagrams, etc then we will never run into trouble using the lemma.

Let \(\mathcal{A}, \mathcal{B}\) be abelian categories. Let \(T : \mathcal{A} \to \mathcal{B}\) be a left exact functor. (We cannot use the letter \(F\) for the functor since this would conflict too much with our use of the letter \(F\) to indicate filtrations.) Note that \(T\) induces an additive functor \[T : \text{Fil}^f(\mathcal{A}) \to \text{Fil}^f(\mathcal{B})\] by the rule \(T(A, F) = (T(A), F)\) where \(F^pT(A) = T(F^pA)\) which makes sense as \(T\) is left exact. (Warning: It may not be the case that \(\text{gr}(T(A)) = T(\text{gr}(A))\).) This induces functors of triangulated categories [05TZ]\[\begin{equation} T : K^{+}(\text{Fil}^f(\mathcal{A})) \longrightarrow K^{+}(\text{Fil}^f(\mathcal{B})) \end{equation}\] The filtered right derived functor of \(T\) is the right derived functor of Definition 05S9 for this exact functor composed with the exact functor \(K^{+}(\text{Fil}^f(\mathcal{B})) \to DF^{+}(\mathcal{B})\) and the multiplicative set \(\text{FQis}^{+}(\mathcal{A})\). Assume \(\mathcal{A}\) has enough injectives. At this point we can redo the discussion of Section 0156 to define the filtered right derived functors [015S]\[\begin{equation} RT : DF^{+}(\mathcal{A}) \longrightarrow DF^{+}(\mathcal{B}) \end{equation}\] of our functor \(T\).

However, instead we will proceed as in Section 05TM, and it will turn out that we can define \(RT\) even if \(T\) is just additive. Namely, we first choose a quasi-inverse \(j' : DF^{+}(\mathcal{A}) \to K^{+}(\mathcal{I}^f)\) of the equivalence of Lemma 015Q. By Lemma 05SQ we see that \(j'\) is an exact functor of triangulated categories. Next, we note that for a filtered injective object \(I\) we have a (noncanonical) decomposition [015T]\[\begin{equation} I \cong \bigoplus\nolimits_{p \in \mathbf{Z}} I_p, \quad\text{with}\quad F^pI = \bigoplus\nolimits_{q \geq p} I_q \end{equation}\] by Lemma 05TP. Hence if \(T\) is any additive functor \(T : \mathcal{A} \to \mathcal{B}\) then we get an additive functor [05U0]\[\begin{equation} T_{ext} : \mathcal{I}^f \to \text{Fil}^f(\mathcal{B}) \end{equation}\] by setting \(T_{ext}(I) = \bigoplus T(I_p)\) with \(F^pT_{ext}(I) = \bigoplus_{q \geq p} T(I_q)\). Note that we have the property \(\text{gr}(T_{ext}(I)) = T(\text{gr}(I))\) by construction. Hence we obtain a functor [05U1]\[\begin{equation} T_{ext} : K^{+}(\mathcal{I}^f) \to K^{+}(\text{Fil}^f(\mathcal{B})) \end{equation}\] which commutes with \(\text{gr}\). Then we define (015S) by the composition [05U2]\[\begin{equation} RT = T_{ext} \circ j'. \end{equation}\] Since \(RT : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\) is computed by injective resolutions as well, see Lemma 05TH, the commutation of \(T\) with \(\text{gr}\), and the commutative diagrams of Lemma 015Q imply that [015U]\[\begin{equation} \text{gr}^p \circ RT \cong RT \circ \text{gr}^p \end{equation}\] and [015V]\[\begin{equation} (\text{forget }F) \circ RT \cong RT \circ (\text{forget }F) \end{equation}\] as functors \(DF^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})\).

The filtered derived functor \(RT\) (015S) induces functors \[\begin{matrix} RT : \text{Fil}^f(\mathcal{A}) \to DF^{+}(\mathcal{B}), \\ RT : \text{Comp}^{+}(\text{Fil}^f(\mathcal{A})) \to DF^{+}(\mathcal{B}), \\ RT : KF^{+}(\mathcal{A}) \to DF^{+}(\mathcal{B}). \end{matrix}\] Note that since \(\text{Fil}^f(\mathcal{A})\), and \(\text{Comp}^{+}(\text{Fil}^f(\mathcal{A}))\) are no longer abelian it does not make sense to say that \(RT\) restricts to a \(\delta\)-functor on them. (This can be repaired by thinking of these categories as exact categories and formulating the notion of a \(\delta\)-functor from an exact category into a triangulated category.) But it does make sense, and it is true by construction, that \(RT\) is an exact functor on the triangulated category \(KF^{+}(\mathcal{A})\).

Lemma

Let \(\mathcal{A}, \mathcal{B}\) be abelian categories. Let \(T : \mathcal{A} \to \mathcal{B}\) be a left exact functor. Assume \(\mathcal{A}\) has enough injectives. Let \((K^\bullet, F)\) be an object of \(\text{Comp}^{+}(\text{Fil}^f(\mathcal{A}))\). There exists a spectral sequence \((E_r, d_r)_{r\geq 0}\) consisting of bigraded objects \(E_r\) of \(\mathcal{B}\) and \(d_r\) of bidegree \((r, - r + 1)\) and with \[E_1^{p, q} = R^{p + q}T(\text{gr}^p(K^\bullet))\] Moreover, this spectral sequence is bounded, converges to \(R^*T(K^\bullet)\), and induces a finite filtration on each \(R^nT(K^\bullet)\). The construction of this spectral sequence is functorial in the object \(K^\bullet\) of \(\text{Comp}^{+}(\text{Fil}^f(\mathcal{A}))\) and the terms \((E_r, d_r)\) for \(r \geq 1\) do not depend on any choices.

Proof

Choose a filtered quasi-isomorphism \(K^\bullet \to I^\bullet\) with \(I^\bullet\) a bounded below complex of filtered injective objects, see Lemma 05TW. Consider the complex \(RT(K^\bullet) = T_{ext}(I^\bullet)\), see (05U2). Thus we can consider the spectral sequence \((E_r, d_r)_{r \geq 0}\) associated to this as a filtered complex in \(\mathcal{B}\), see Homology, Section 012K. By Homology, Lemma 012M we have \(E_1^{p, q} = H^{p + q}(\text{gr}^p(T(I^\bullet)))\). By Equation (015T) we have \(E_1^{p, q} = H^{p + q}(T(\text{gr}^p(I^\bullet)))\), and by definition of a filtered injective resolution the map \(\text{gr}^p(K^\bullet) \to \text{gr}^p(I^\bullet)\) is an injective resolution. Hence \(E_1^{p, q} = R^{p + q}T(\text{gr}^p(K^\bullet))\).

On the other hand, each \(I^n\) has a finite filtration and hence each \(T(I^n)\) has a finite filtration. Thus we may apply Homology, Lemma 012W to conclude that the spectral sequence is bounded, converges to \(H^*(T(I^\bullet)) = R^*T(K^\bullet)\) and induces a finite filtration on each \(H^n(T(I^\bullet)) = R^nT(K^\bullet)\).

Suppose that \(K^\bullet \to L^\bullet\) is a morphism of \(\text{Comp}^{+}(\text{Fil}^f(\mathcal{A}))\). Choose a filtered quasi-isomorphism \(L^\bullet \to J^\bullet\) with \(J^\bullet\) a bounded below complex of filtered injective objects, see Lemma 05TW. By our results above, for example Lemma 05TY, there exists a diagram \[\xymatrix{ K^\bullet \ar[r] \ar[d] & L^\bullet \ar[d] \\ I^\bullet \ar[r] & J^\bullet }\] which commutes up to homotopy. Hence we get a morphism of filtered complexes \(T(I^\bullet) \to T(J^\bullet)\) which gives rise to the morphism of spectral sequences, see Homology, Lemma 012O. The last statement follows from this.

Remark

As promised in Remark 015K we discuss the connection of the lemma above with the constructions using Cartan-Eilenberg resolutions. Namely, let \(T : \mathcal{A} \to \mathcal{B}\) be a left exact functor of abelian categories, assume \(\mathcal{A}\) has enough injectives, and let \(K^\bullet\) be a bounded below complex of \(\mathcal{A}\). We give an alternative construction of the spectral sequences \({}'E\) and \({}''E\) of Lemma 015J.

First spectral sequence. Consider the “stupid” filtration on \(K^\bullet\) obtained by setting \(F^p(K^\bullet) = \sigma_{\geq p}(K^\bullet)\), see Homology, Section 0118. Note that this is stupid in the sense that \(d(F^p(K^\bullet)) \subset F^{p + 1}(K^\bullet)\), compare Homology, Lemma 012N. Note that \(\text{gr}^p(K^\bullet) = K^p[-p]\) with this filtration. According to Lemma 015W there is a spectral sequence with \(E_1\) term \[E_1^{p, q} = R^{p + q}T(K^p[-p]) = R^qT(K^p)\] as in the spectral sequence \({}'E_r\). Observe moreover that the differentials \(E_1^{p, q} \to E_1^{p + 1, q}\) agree with the differentials in \('{}E_1\), see Homology, Lemma 012N part (2) and the description of \({}'d_1\) in the proof of Lemma 015J.

Second spectral sequence. Consider the filtration on the complex \(K^\bullet\) obtained by setting \(F^p(K^\bullet) = \tau_{\leq -p}(K^\bullet)\), see Homology, Section 0118. The minus sign is necessary to get a decreasing filtration. Note that \(\text{gr}^p(K^\bullet)\) is quasi-isomorphic to \(H^{-p}(K^\bullet)[p]\) with this filtration. According to Lemma 015W there is a spectral sequence with \(E_1\) term \[E_1^{p, q} = R^{p + q}T(H^{-p}(K^\bullet)[p]) = R^{2p + q}T(H^{-p}(K^\bullet)) = {}''E_2^{i, j}\] with \(i = 2p + q\) and \(j = -p\). (This looks unnatural, but note that we could just as well have developed the whole theory of filtered complexes using increasing filtrations, with the end result that this then looks natural, but the other one doesn’t.) We leave it to the reader to see that the differentials match up.

Actually, given a Cartan-Eilenberg resolution \(K^\bullet \to I^{\bullet, \bullet}\) the induced morphism \(K^\bullet \to \text{Tot}(I^{\bullet, \bullet})\) into the associated total complex will be a filtered injective resolution for either filtration using suitable filtrations on \(\text{Tot}(I^{\bullet, \bullet})\). This can be used to match up the spectral sequences exactly.

Ext groups

In this section we start describing the Ext groups of objects of an abelian category. First we have the following very general definition.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(i \in \mathbf{Z}\). Let \(X, Y\) be objects of \(D(\mathcal{A})\). The \(i\)th extension group of \(X\) by \(Y\) is the group \[\Ext^i_\mathcal{A}(X, Y) = \Hom_{D(\mathcal{A})}(X, Y[i]) = \Hom_{D(\mathcal{A})}(X[-i], Y).\] If \(A, B \in \Ob(\mathcal{A})\) we set \(\Ext^i_\mathcal{A}(A, B) = \text{Ext}^i_\mathcal{A}(A[0], B[0])\).

Since \(\Hom_{D(\mathcal{A})}(X, -)\), resp. \(\Hom_{D(\mathcal{A})}(-, Y)\) is a homological, resp. cohomological functor, see Lemma 0149, we see that a distinguished triangle \((Y, Y', Y'')\), resp. \((X, X', X'')\) leads to a long exact sequence \[\ldots \to \Ext^i_\mathcal{A}(X, Y) \to \Ext^i_\mathcal{A}(X, Y') \to \Ext^i_\mathcal{A}(X, Y'') \to \Ext^{i + 1}_\mathcal{A}(X, Y) \to \ldots\] respectively \[\ldots \to \Ext^i_\mathcal{A}(X'', Y) \to \Ext^i_\mathcal{A}(X', Y) \to \Ext^i_\mathcal{A}(X, Y) \to \Ext^{i + 1}_\mathcal{A}(X'', Y) \to \ldots\] Note that since \(D^+(\mathcal{A})\), \(D^-(\mathcal{A})\), \(D^b(\mathcal{A})\) are full subcategories we may compute the Ext groups by Hom groups in these categories provided \(X\), \(Y\) are contained in them.

In case the category \(\mathcal{A}\) has enough injectives or enough projectives we can compute the Ext groups using injective or projective resolutions. To avoid confusion, recall that having an injective (resp. projective) resolution implies vanishing of homology in all low (resp. high) degrees, see Lemmas 013J and 0645.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(X^\bullet, Y^\bullet \in \Ob(K(\mathcal{A}))\).

1. Let \(Y^\bullet \to I^\bullet\) be an injective resolution (Definition 013I). Then \[\Ext^i_\mathcal{A}(X^\bullet, Y^\bullet) = \Hom_{K(\mathcal{A})}(X^\bullet, I^\bullet[i]).\]

2. Let \(P^\bullet \to X^\bullet\) be a projective resolution (Definition 0644). Then \[\Ext^i_\mathcal{A}(X^\bullet, Y^\bullet) = \Hom_{K(\mathcal{A})}(P^\bullet[-i], Y^\bullet).\]

Proof

Follows immediately from Lemma 05TG and Lemma 064B.

In the rest of this section we discuss extensions of objects of the abelian category itself. First we observe the following.

Lemma

Let \(\mathcal{A}\) be an abelian category.

1. Let \(X\), \(Y\) be objects of \(D(\mathcal{A})\). Given \(a, b \in \mathbf{Z}\) such that \(H^i(X) = 0\) for \(i > a\) and \(H^j(Y) = 0\) for \(j < b\), we have \(\Ext^n_\mathcal{A}(X, Y) = 0\) for \(n < b - a\) and \[\Ext^{b - a}_\mathcal{A}(X, Y) = \Hom_\mathcal{A}(H^a(X), H^b(Y))\]

2. Let \(A, B \in \Ob(\mathcal{A})\). For \(i < 0\) we have \(\Ext^i_\mathcal{A}(B, A) = 0\). We have \(\Ext^0_\mathcal{A}(B, A) = \Hom_\mathcal{A}(B, A)\).

Proof

Choose complexes \(X^\bullet\) and \(Y^\bullet\) representing \(X\) and \(Y\). Since \(Y^\bullet \to \tau_{\geq b}Y^\bullet\) is a quasi-isomorphism, we may assume that \(Y^j = 0\) for \(j < b\). Let \(L^\bullet \to X^\bullet\) be any quasi-isomorphism. Then \(\tau_{\leq a}L^\bullet \to X^\bullet\) is a quasi-isomorphism. Hence a morphism \(X \to Y[n]\) in \(D(\mathcal{A})\) can be represented as \(fs^{-1}\) where \(s : L^\bullet \to X^\bullet\) is a quasi-isomorphism, \(f : L^\bullet \to Y^\bullet[n]\) a morphism, and \(L^i = 0\) for \(i > a\). Note that \(f\) maps \(L^i\) to \(Y^{i + n}\). Thus \(f = 0\) if \(n < b - a\) because always either \(L^i\) or \(Y^{i + n}\) is zero. If \(n = b - a\), then \(f\) corresponds exactly to a morphism \(H^a(X) \to H^b(Y)\). Part (2) is a special case of (1).

Let \(\mathcal{A}\) be an abelian category. Suppose that \(0 \to A \to A' \to A'' \to 0\) is a short exact sequence of objects of \(\mathcal{A}\). Then \(0 \to A[0] \to A'[0] \to A''[0] \to 0\) leads to a distinguished triangle in \(D(\mathcal{A})\) (see Lemma 0152) hence a long exact sequence of Ext groups \[0 \to \Ext^0_\mathcal{A}(B, A) \to \Ext^0_\mathcal{A}(B, A') \to \Ext^0_\mathcal{A}(B, A'') \to \Ext^1_\mathcal{A}(B, A) \to \ldots\] Similarly, given a short exact sequence \(0 \to B \to B' \to B'' \to 0\) we obtain a long exact sequence of Ext groups \[0 \to \Ext^0_\mathcal{A}(B'', A) \to \Ext^0_\mathcal{A}(B', A) \to \Ext^0_\mathcal{A}(B, A) \to \Ext^1_\mathcal{A}(B'', A) \to \ldots\] We may view these Ext groups as an application of the construction of the derived category. It shows one can define Ext groups and construct the long exact sequence of Ext groups without needing the existence of enough injectives or projectives. There is an alternative construction of the Ext groups due to Yoneda which avoids the use of the derived category, see [Yoneda].

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(A, B \in \Ob(\mathcal{A})\). For \(i \geq 1\) a degree \(i\) Yoneda extension of \(B\) by \(A\) is an exact sequence \[E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0\] in \(\mathcal{A}\). We say two Yoneda extensions \(E\) and \(E'\) of the same degree are equivalent if there exists a commutative diagram \[\xymatrix{ 0 \ar[r] & A \ar[r] & Z_{i - 1} \ar[r] & \ldots \ar[r] & Z_0 \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & A \ar[r] \ar[u]^{\text{id}} \ar[d]_{\text{id}} & Z''_{i - 1} \ar[r] \ar[u] \ar[d] & \ldots \ar[r] & Z''_0 \ar[r] \ar[u] \ar[d] & B \ar[r] \ar[u]_{\text{id}} \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & A \ar[r] & Z'_{i - 1} \ar[r] & \ldots \ar[r] & Z'_0 \ar[r] & B \ar[r] & 0 }\] where the middle row is a Yoneda extension as well.

It is not immediately clear that the equivalence of the definition is an equivalence relation. Although it is instructive to prove this directly this will also follow from Lemma 06XU below.

Let \(\mathcal{A}\) be an abelian category with objects \(A\), \(B\). Given a Yoneda extension \(E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0\) we define an associated element \(\delta(E) \in \Ext^i_\mathcal{A}(B, A)\) as the morphism \(\delta(E) = fs^{-1} : B[0] \to A[i]\) where \(s\) is the quasi-isomorphism \[(\ldots \to 0 \to A \to Z_{i - 1} \to \ldots \to Z_0 \to 0 \to \ldots) \longrightarrow B[0]\] and \(f\) is the morphism of complexes \[(\ldots \to 0 \to A \to Z_{i - 1} \to \ldots \to Z_0 \to 0 \to \ldots) \longrightarrow A[i]\] We call \(\delta(E) = fs^{-1}\) the class of the Yoneda extension. It turns out that this class characterizes the equivalence class of the Yoneda extension.

Lemma

Let \(\mathcal{A}\) be an abelian category with objects \(A\), \(B\) and let \(i \geq 1\). Any element in \(\Ext^i_\mathcal{A}(B, A)\) is \(\delta(E)\) for some degree \(i\) Yoneda extension of \(B\) by \(A\). Given two Yoneda extensions \(E\), \(E'\) of the same degree then \(E\) is equivalent to \(E'\) if and only if \(\delta(E) = \delta(E')\).

Proof

Let \(\xi : B[0] \to A[i]\) be an element of \(\Ext^i_\mathcal{A}(B, A)\). We may write \(\xi = f s^{-1}\) for some quasi-isomorphism \(s : L^\bullet \to B[0]\) and map \(f : L^\bullet \to A[i]\). After replacing \(L^\bullet\) by \(\tau_{\leq 0}L^\bullet\) we may assume that \(L^j = 0\) for \(j > 0\). Picture \[\xymatrix{ L^{- i - 1} \ar[r] & L^{-i} \ar[r] \ar[d] & \ldots \ar[r] & L^0 \ar[r] & B \ar[r] & 0 \\ & A }\] Then setting \(Z_{i - 1} = (L^{- i + 1} \oplus A)/L^{-i}\) and \(Z_j = L^{-j}\) for \(j = i - 2, \ldots, 0\) we see that we obtain a degree \(i\) extension \(E\) of \(B\) by \(A\) whose class \(\delta(E)\) equals \(\xi\).

It is immediate from the definitions that equivalent Yoneda extensions have the same class. Suppose that \(E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0\) and \(E' : 0 \to A \to Z'_{i - 1} \to Z'_{i - 2} \to \ldots \to Z'_0 \to B \to 0\) are Yoneda extensions with the same class. By construction of \(D(\mathcal{A})\) as the localization of \(K(\mathcal{A})\) at the set of quasi-isomorphisms, this means there exists a complex \(L^\bullet\) and quasi-isomorphisms \[t : L^\bullet \to (\ldots \to 0 \to A \to Z_{i - 1} \to \ldots \to Z_0 \to 0 \to \ldots)\] and \[t' : L^\bullet \to (\ldots \to 0 \to A \to Z'_{i - 1} \to \ldots \to Z'_0 \to 0 \to \ldots)\] such that \(s \circ t = s' \circ t'\) and \(f \circ t = f' \circ t'\), see Categories, Section 04VB. Let \(E''\) be the degree \(i\) extension of \(B\) by \(A\) constructed from the pair \(L^\bullet \to B[0]\) and \(L^\bullet \to A[i]\) in the first paragraph of the proof. Then the reader sees readily that there exists “morphisms” of degree \(i\) Yoneda extensions \(E'' \to E\) and \(E'' \to E'\) as in the definition of equivalent Yoneda extensions (details omitted). This finishes the proof.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(A\), \(B\) be objects of \(\mathcal{A}\). Then \(\Ext^1_\mathcal{A}(B, A)\) is the group \(\Ext_\mathcal{A}(B, A)\) constructed in Homology, Definition 010K.

Proof

This is the case \(i = 1\) of Lemma 06XU.

Given an abelian category \(\mathcal{A}\) and objects \(X, Y, Z\) of \(D(\mathcal{A})\), there are bilinear and associative composition laws \[\Ext^j_\mathcal{A}(Y, Z) \times \Ext^i_\mathcal{A}(X, Y) \longrightarrow \Ext^{i + j}_\mathcal{A}(X, Z),\quad (\eta, \xi) \longmapsto \eta \circ \xi\] Namely, if \(\xi : X \to Y[i]\) and \(\eta : Y \to Z[j]\), then we define \(\eta \circ \xi\) as the composition of \(\xi\) by \(\eta[i] : Y[i] \to Z[i + j]\). If \(A, B, C\) are objects of \(\mathcal{A}\) and \(i, j \geq 1\), then this composition law \[\Ext^j_\mathcal{A}(B, C) \times \Ext^i_\mathcal{A}(A, B) \longrightarrow \Ext^{i + j}_\mathcal{A}(A, C)\] can be described in terms of Yoneda extensions as follows: the composition of \[0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0\] and \[0 \to B \to Z'_{j - 1} \to Z'_{j - 2} \to \ldots \to Z'_0 \to C \to 0\] is the Yoneda extension \[0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to Z'_{j - 1} \to Z'_{j - 2} \to \ldots \to Z'_0 \to C \to 0\]

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(0 \to A \to Z \to B \to 0\) and \(0 \to B \to Z' \to C \to 0\) be short exact sequences in \(\mathcal{A}\). Denote \([Z] \in \Ext^1_\mathcal{A}(B, A)\) and \([Z'] \in \Ext^1_\mathcal{A}(C, B)\) their classes. Then \([Z] \circ [Z'] \in \Ext^2_\mathcal{A}(C, A)\) is \(0\) if and only if there exists a commutative diagram \[\xymatrix{ & & 0 \ar[d] & 0 \ar[d] \\ 0 \ar[r] & A \ar[r] \ar[d]^1 & Z \ar[r] \ar[d] & B \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A \ar[r] & W \ar[r] \ar[d] & Z' \ar[r] \ar[d] & 0 \\ & & C \ar[r]^1 \ar[d]& C \ar[d]\\ & & 0 & 0 }\] with exact rows and columns in \(\mathcal{A}\).

Proof

Omitted. Hints: You can argue this using the result of Lemma 06XU and working out what it means for a \(2\)-extension class to be zero. Or you can use that if \([Z] \circ [Z'] \in \Ext^2_\mathcal{A}(C, A)\) is zero, then by the long exact cohomology sequence of \(\Ext\) the element \([Z] \in \Ext^1_\mathcal{A}(B, A)\) is the image of some element in \(\Ext^1_\mathcal{A}(Z', A)\).

Lemma

Let \(\mathcal{A}\) be an abelian category and let \(p \geq 0\). If \(\Ext^p_\mathcal{A}(B, A) = 0\) for any pair of objects \(A\), \(B\) of \(\mathcal{A}\), then \(\Ext^i_\mathcal{A}(B, A) = 0\) for \(i \geq p\) and any pair of objects \(A\), \(B\) of \(\mathcal{A}\).

Proof

For \(i > p\) write any class \(\xi\) as \(\delta(E)\) where \(E\) is a Yoneda extension \[E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0\] This is possible by Lemma 06XU. Set \(C = \Ker(Z_{p - 1} \to Z_{p - 2}) = \Im(Z_p \to Z_{p - 1})\). Then \(\delta(E)\) is the composition of \(\delta(E')\) and \(\delta(E'')\) where \[E' : 0 \to C \to Z_{p - 1} \to \ldots \to Z_0 \to B \to 0\] and \[E'' : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_p \to C \to 0\] Since \(\delta(E') \in \Ext^p_\mathcal{A}(B, C) = 0\) we conclude.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(K\) be an object of \(D^b(\mathcal{A})\) such that \(\Ext^p_\mathcal{A}(H^i(K), H^j(K)) = 0\) for all \(p \geq 2\) and \(i > j\). Then \(K\) is isomorphic to the direct sum of its cohomologies: \(K \cong \bigoplus H^i(K)[-i]\).

Proof

Choose \(a, b\) such that \(H^i(K) = 0\) for \(i \not \in [a, b]\). We will prove the lemma by induction on \(b - a\). If \(b - a \leq 0\), then the result is clear. If \(b - a > 0\), then we look at the distinguished triangle of truncations \[\tau_{\leq b - 1}K \to K \to H^b(K)[-b] \to (\tau_{\leq b - 1}K)[1]\] see Remark 08J5. By Lemma 05QT if the last arrow is zero, then \(K \cong \tau_{\leq b - 1}K \oplus H^b(K)[-b]\) and we win by induction. Again using induction we see that \[\Hom_{D(\mathcal{A})}(H^b(K)[-b], (\tau_{\leq b - 1}K)[1]) = \bigoplus\nolimits_{i < b} \Ext_\mathcal{A}^{b - i + 1}(H^b(K), H^i(K))\] By assumption the direct sum is zero and the proof is complete.

Lemma

Let \(\mathcal{A}\) be an abelian category. Assume \(\Ext^2_\mathcal{A}(B, A) = 0\) for any pair of objects \(A\), \(B\) of \(\mathcal{A}\). Then any object \(K\) of \(D^b(\mathcal{A})\) is isomorphic to the direct sum of its cohomologies: \(K \cong \bigoplus H^i(K)[-i]\).

Proof

The assumption implies that \(\Ext^i_\mathcal{A}(B, A) = 0\) for \(i \geq 2\) and any pair of objects \(A, B\) of \(\mathcal{A}\) by Lemma 0EWW. Hence this lemma is a special case of Lemma 0GM4.

K-groups

A tiny bit about \(K_0\) of a triangulated category.

Definition

Let \(\mathcal{D}\) be a triangulated category. We denote \(K_0(\mathcal{D})\) the zeroth \(K\)-group of \(\mathcal{D}\). It is the abelian group constructed as follows. Take the free abelian group on the objects on \(\mathcal{D}\) and for every distinguished triangle \(X \to Y \to Z\) impose the relation \([Y] - [X] - [Z] = 0\).