Exact complexes, signs and lifting covers

Written by GPT-6.1 Sol (OpenAI), October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Original text public domain (CC0); referenced Stacks proofs retain GFDL-1.2.

An exact sequence does not promise that every map into its middle term lifts through the preceding term. It promises such a lift after replacing the test object by an object mapping epimorphically onto it. This distinction makes diagram arguments work in an arbitrary abelian category, without projective objects. It also identifies precisely which parts of two rows must be exact to force a comparison map to be monic or epic.

We use Formal linear combinations and finite sums, Section 1, for biproduct maps, and Trace coreflections and balanced nonabelian categories, Section 3, for agreement between the ordinary and additive image constructions. Testing exactness with doubled objects, Section 1, proves the full Hom tests. Work in a locally small additive category; whenever exactness or cohomology is mentioned, assume it is abelian. All diagram indices are small in the chosen universe. No generator or supply of projectives is assumed.

1. Which positions in a complex are being tested?

A cohomological complex has objects \(X^n\) for \(n\in\mathbb Z\), maps \(d^n:X^n\to X^{n+1}\), and equations \(d^{n+1}d^n=0\). A finite string is a complex when its successive composites vanish wherever both arrows are drawn. It can be extended by zero objects and zero maps on either side. This extension is a complex automatically. Exactness at a newly exposed endpoint is a further condition.

For example, in an abelian category the finite complex \(A\xrightarrow f B\xrightarrow g C\) tests exactness at \(B\). Adding an initial zero tests monicity of \(f\); adding a final zero tests epicity of \(g\). A short exact sequence tests all three conditions. The kernel and cokernel universal properties identify its two maps as a kernel followed by its cokernel. Every morphism has the two short exact factorizations

\[ \begin{gathered} 0\to\ker f\to A\to\operatorname{im}f\to0,\\ 0\to\operatorname{im}f\to B\to\operatorname{coker}f\to0. \end{gathered} \tag{1.1} \]

Here the specified coimage-to-image isomorphism supplies the quotient map in the first row. These are the abelian conventions of Stacks, Section 12.5. The Hom tests in the prerequisite prove the endpoint recognition, including the universal maps.

A rectangular diagram of complexes requires every row and every column to be a complex. In an abelian category an exact diagram requires all these rows and columns to be exact at their drawn interior positions. Commutativity of squares is specified separately. For a square with top arrow \(a\), bottom arrow \(b\), left arrow \(r\) and right arrow \(s\), an \(\varepsilon\)-commutative square satisfies

\[ sa=\varepsilon br,\qquad\varepsilon\in\{1,-1\}. \tag{1.2} \]

The choice \(-1\) is called anticommutativity. The sign is multiplication in an abelian Hom group; it makes sense even when the category is not linear over a field. Exercise 3 shows how these square signs control a total differential.

2. One cohomology object with several universal descriptions

Fix a complex \(A\xrightarrow f B\xrightarrow g C\). Put \(K=\ker g\), \(Q=\operatorname{coker}f\), \(I=\operatorname{im}f\), and \(J=\operatorname{im}g\). Write \(k:K\to B\), \(q:B\to Q\), and \(j:J\to C\) for their structural maps. Since \(gf=0\), the map \(g\) factors through \(q\), so kills its kernel \(I\to B\). That inclusion therefore factors uniquely through \(K\), giving a monomorphism \(i:I\to K\). The image quotient \(e:B\to J\) kills \(f\), because \(jef=gf=0\) and \(j\) is monic. Thus it gives \(p:Q\to J\) with \(pq=e\). Define \(u=qk\).

Proposition 2.1. There are canonical identifications

\[ \begin{gathered} H=\operatorname{im}u\simeq\operatorname{coker}i,\\ H\simeq\operatorname{coker}(A\to K),\\ H\simeq\ker p\simeq\ker(Q\to C). \end{gathered} \tag{2.1} \]

The complex is exact at \(B\) if and only if any, hence all, of the following hold: \(H=0\), \(u=0\), \(i\) is invertible, \(A\to K\) is epic, \(p\) is invertible, or \(Q\to C\) is monic.

Proof. We first identify the kernel of \(u\). If \(t:T\to K\) satisfies \(ut=0\), then \(qkt=0\). The inclusion \(I\to B\) is the kernel of \(q\), so \(kt\) factors uniquely through \(I\). Since \(k\) is monic, that factor gives \(t=i\bar t\). Conversely \(ui=0\). Thus \(i\) is the kernel of \(u\), and \(\operatorname{coim}u=\operatorname{coker}i\).

Next \(p\) is epic: if two maps agree after \(p\), they agree after \(pq=e\), which is epic. Also \(pu=ek=0\). If \(r:Q\to T\) kills \(u\), then \(rq\) kills \(k\). The map \(e\) is the cokernel of \(k\), by the image factorization of \(g\), so \(rq=se\) for a unique \(s:J\to T\). Consequently \(rq=spq\); canceling \(q\) gives \(r=sp\). Uniqueness also follows from epicity of \(p\). Therefore \(p\) is a cokernel of \(u\), and \(\operatorname{im}u=\ker p\).

The abelian coimage-to-image isomorphism now identifies \(\operatorname{coker}i\) with \(\ker p\). The induced map \(A\to K\) is the composite of the image quotient \(A\twoheadrightarrow I\) and \(i\); killing this composite is equivalent to killing \(i\). Their cokernels are therefore canonically the same. Finally \(Q\to C\) is \(jp\), and monicity of \(j\) identifies its kernel with that of \(p\). These arguments preserve every structural map in (2.1).

Since \(i\) is monic, its cokernel is zero exactly when it is invertible. Its image is all of \(K\) exactly when \(A\to K\) is epic. Since \(p\) is epic, its kernel is zero exactly when it is invertible. The last kernel description gives the monicity test. An image is zero precisely when its original map is zero, which gives the condition on \(u\). These are exactly the equality of the incoming image and outgoing kernel that defines middle exactness. \(\square\)

Call \(H\) the cohomology at \(B\). Reversing arrows exchanges kernels with cokernels and images with coimages, as proved in Stacks, Lemma 12.5.2. For the reversed complex in the opposite category, the map corresponding to \(u\) is \(u^{\mathrm{op}}\). Its image is the opposite of \(\operatorname{coim}u\), hence is canonically \(H^{\mathrm{op}}\). Thus exactness is unchanged by passing to the opposite category.

3. Retain the complete lifting results

Three precise open results provide the covering step in the next proof.

The full proofs remain at these exact references. The objects and arrows in the last assertion are in the original abelian category. Applying it to \(A\twoheadrightarrow B\to0\) gives a lift through any epimorphism after an epimorphic change of test object. It does not assert a lift on \(S\) itself. Exercise 1 gives an elementary obstruction to such a global lift.

4. A four-arrow test with only two exact positions

Consider two complexes and a commutative comparison:

\[ \begin{gathered} X_0\xrightarrow{a_0}X_1\xrightarrow{a_1}X_2\xrightarrow{a_2}X_3,\\ Y_0\xrightarrow{b_0}Y_1\xrightarrow{b_1}Y_2\xrightarrow{b_2}Y_3,\\ f_i:X_i\to Y_i\quad(0\le i\le3),\\ f_{i+1}a_i=b_if_i\quad(0\le i<3). \end{gathered} \tag{4.1} \]

Assume the upper row is exact at \(X_2\) and the lower row is exact at \(Y_1\). No other interior exactness is required.

Theorem 4.1. If \(f_0\) is epic and \(f_1,f_3\) are monic, then \(f_2\) is monic. If \(f_3\) is monic and \(f_0,f_2\) are epic, then \(f_1\) is epic.

Proof of monicity. Let \(h:S\to X_2\) satisfy \(f_2h=0\). Then \(f_3a_2h=b_2f_2h=0\), so monicity of \(f_3\) gives \(a_2h=0\). Exactness at \(X_2\), through Tag 08N5, supplies an epimorphism \(e_1:S_1\to S\) and \(t:S_1\to X_1\) such that \(a_1t=he_1\).

The map \(f_1t\) is killed by \(b_1\), since

\[ b_1f_1t=f_2a_1t=f_2he_1=0. \tag{4.2} \]

Exactness at \(Y_1\) supplies an epimorphism \(e_2:S_2\to S_1\) and \(y:S_2\to Y_0\) with \(b_0y=f_1te_2\). Pull back the epimorphism \(f_0\) along \(y\). Tag 08N4 supplies an epimorphism \(e_3:S_3\to S_2\) and \(x:S_3\to X_0\) with \(f_0x=ye_3\). Now

\[ \begin{gathered} f_1a_0x=b_0f_0x=b_0ye_3,\\ b_0ye_3=f_1te_2e_3. \end{gathered} \tag{4.3} \]

Monicity of \(f_1\) gives \(a_0x=te_2e_3\). Hence

\[ he_1e_2e_3=a_1te_2e_3=a_1a_0x=0. \tag{4.4} \]

The last equality uses the complex condition at \(X_1\), not exactness there. The composite of the three epimorphisms is epic, so \(h=0\). For any pair \(r,s:S\to X_2\) with \(f_2r=f_2s\), apply the argument to \(h=r-s\). Thus \(r=s\), proving monicity.

For the epic assertion, pass to the opposite category, reverse the rows and interchange them. The upper row is now \(Y_3^{\mathrm{op}}\to Y_2^{\mathrm{op}}\to Y_1^{\mathrm{op}}\to Y_0^{\mathrm{op}}\); the lower row is the reversed \(X\)-row. Their required exact positions are precisely the original exactness at \(Y_1\) and \(X_2\). The four vertical arrows, in order, are \(f_3^{\mathrm{op}},f_2^{\mathrm{op}},f_1^{\mathrm{op}},f_0^{\mathrm{op}}\). The monicity assertion therefore says that \(f_1^{\mathrm{op}}\) is monic under the stated hypotheses, which is epicity of \(f_1\). \(\square\)

For a comparison of five-term exact complexes, apply the monic test to the first four positions and the epic test to the last four. If the first comparison is epic, the second and fourth are invertible, and the fifth is monic, the middle comparison is both monic and epic, hence invertible. In particular, invertibility of the four outside comparisons suffices. Stacks, Lemma 12.5.20 gives the complete usual five-term result. Theorem 4.1 keeps the two-position hypothesis for the shorter rows.

5. Subobjects and the signs of a square

A subobject of \(B\) means an isomorphism class of monomorphisms into \(B\), where an isomorphism must commute with those maps. For subobjects \(U,V\subseteq B\), write

\[ \begin{gathered} U\cap V=U\times_B V,\\ U+V=\operatorname{im}(U\oplus V\to B). \end{gathered} \tag{5.1} \]

The map defining the sum restricts to the given inclusion on each summand. Iterating gives finite intersections and sums. The empty intersection is \(B\), and the empty sum is zero. The quotient \(B/U\) is the cokernel of the specified inclusion. For \(f:A\to B\), the inverse image of \(U\) is the pullback \(A\times_B U\); its map to \(A\) is monic. These constructions are invariant under the isomorphisms defining subobject classes, by their universal properties and uniqueness of image factorizations.

To recognize a square, let \(r:W\to A\), \(s:W\to B'\), \(f:A\to B\), and \(g:B'\to B\) satisfy \(fr=gs\). Put

\[ \begin{gathered} u=(r,s),\\ v=(f,-g),\\ vu=0. \end{gathered} \tag{5.2} \]

The square is a pullback exactly when \(0\to W\xrightarrow u A\oplus B'\xrightarrow v B\) is exact. It is a pushout exactly when \(W\xrightarrow u A\oplus B'\xrightarrow v B\to0\) is exact. Retain the complete proof of Stacks, Lemma 12.5.11, Tag 08N2, and the explicit signs in Exact squares and endpoint tests, Section 1. Stacks writes the pushout sequence using \((r,-s)\) followed by \((f,g)\). The involution \(\operatorname{diag}(1,-1)\) on the middle biproduct carries one sequence to the other. Thus the two sign conventions give the same exactness test.

An exact fork \(W\xrightarrow r A\rightrightarrows B\), with parallel maps \(f,g\), means that \(r\) is their equalizer. Additivity identifies this with the kernel of \(f-g\). Its recognition is therefore the exact sequence \(0\to W\xrightarrow r A\xrightarrow{f-g}B\). This specifies both the difference and the initial monomorphism.

6. Four graded exercises with full solutions

Exercise 1 (warm-up: a cover without a global section). In abelian groups, apply the lifting criterion to \(\mathbb Z\twoheadrightarrow\mathbb Z/2\to0\) and the identity of \(\mathbb Z/2\). Give a covering lift and prove that no lift from the original test object exists. Compute the intersection and sum of \(2\mathbb Z\) and \(3\mathbb Z\) in \(\mathbb Z\), and identify \((2\mathbb Z+3\mathbb Z)/3\mathbb Z\) with \(2\mathbb Z/(2\mathbb Z\cap3\mathbb Z)\) by an actual map.

Solution. Take \(S'=\mathbb Z\), let \(e:S'\to\mathbb Z/2\) be reduction modulo two, and let \(t:S'\to\mathbb Z\) be the identity. Reduction composed with \(t\) equals the given identity composed with \(e\). A global lift would send the class of one to an integer killed by two, hence to zero. Its reduction cannot be the class of one. Thus the covering lift exists but no global one does.

An integer in both subgroups is divisible by two and three, hence by six; conversely every multiple of six lies in both. The intersection is \(6\mathbb Z\). The sum is all of \(\mathbb Z\), since \(1=3-2\). Inclusion followed by reduction modulo three induces

\[ 2\mathbb Z/6\mathbb Z\longrightarrow\mathbb Z/3\mathbb Z. \tag{6.1} \]

It is surjective because the class of two generates the group of order three. Its kernel before taking the quotient is \(2\mathbb Z\cap3\mathbb Z=6\mathbb Z\). Explicitly the classes of \(0,2,4\) map to \(0,2,1\), giving a bijection and a group isomorphism with the required structural maps.

Exercise 2 (intermediate: every premise has a job). Over any field \(k\), give five counterexamples to the monic part of Theorem 4.1, omitting in turn epicity of \(f_0\), monicity of \(f_1\), monicity of \(f_3\), upper exactness at \(X_2\), and lower exactness at \(Y_1\). Retain all the other premises in each example.

Solution. Describe each row by its four vector spaces; all arrows are zero unless an identity is explicitly indicated. Every resulting row is a complex.

To omit epicity of \(f_0\), take the upper row \((0,k,k,0)\) with \(a_1=1_k\), and the lower row \((k,k,0,0)\) with \(b_0=1_k\). Set \(f_1=1_k\); all other comparison maps are zero. Upper exactness at \(X_2\) and lower exactness at \(Y_1\) both identify the image of an identity with the kernel of a zero map. The maps \(f_1,f_3\) are monic, while \(f_0:0\to k\) is not epic and \(f_2:k\to0\) is not monic. The two identity arrows land in different row positions, so all three comparison squares commute.

To omit monicity of \(f_1\), keep the upper row \((0,k,k,0)\) with \(a_1=1_k\), and take the lower row to be zero. All comparison maps are zero. The two exact positions hold; \(f_0\) is epic and \(f_3\) monic because both are identities of zero. But \(f_1\) and \(f_2\) have nonzero source and zero target.

To omit monicity of \(f_3\), take the upper row \((0,0,k,k)\) with \(a_2=1_k\), and the lower row zero. All comparisons are zero. Upper exactness at \(X_2\) reads \(0=\ker1_k\), and lower exactness holds in zero. The maps \(f_0,f_1\) satisfy their premises, whereas \(f_3\) and \(f_2\) are not monic.

To omit upper exactness, use upper row \((0,0,k,0)\), lower row zero, and all maps zero. All three comparison premises hold and lower exactness holds. At \(X_2\), however, the incoming image is zero and the outgoing kernel is \(k\). Again \(f_2\) is not monic.

To omit lower exactness, take upper row \((0,k,k,0)\) with \(a_1=1_k\), lower row \((0,k,0,0)\), and \(f_1=1_k\), with all other maps zero. The squares commute, upper exactness holds, and the three comparison premises hold. Lower exactness fails because the incoming image at \(Y_1\) is zero while its outgoing kernel is \(k\). The map \(f_2:k\to0\) is not monic. All five constructions work in every characteristic.

Exercise 3 (hard: convert square signs into a differential). Let \(M^{p,q}\) be a diagram in an additive category with horizontal maps \(h\) of bidegree \((1,0)\) and vertical maps \(v\) of bidegree \((0,1)\). Assume \(h^2=v^2=0\) and that every diagonal \(p+q=n\) has only finitely many nonzero objects. If every square anticommutes, prove that \(h+v\) defines a differential on \(T^n=\bigoplus_{p+q=n}M^{p,q}\). If squares commute instead, find the sign in the vertical differential that gives a complex. Exhibit a commuting square of abelian groups where omitting the sign fails.

Solution. A map from a finite biproduct is determined by its restrictions to the summands. On \(M^{p,q}\), the square of \(h+v\) is \(h^2+vh+hv+v^2\). The first and last terms vanish, and anticommutativity says \(vh+hv=0\). Thus its square is zero on every summand and hence on \(T^n\).

For commuting squares, use

\[ D|_{M^{p,q}}=h+(-1)^p v. \tag{6.2} \]

On that summand, the two cross terms in \(D^2\) are \((-1)^{p+1}vh\) and \((-1)^p hv\). They cancel because \(vh=hv\). The two same-direction terms vanish by the row and column complex conditions. This proves \(D^2=0\) in an arbitrary additive category.

Put \(\mathbb Z\) at each corner of the square with bidegrees \((0,0),(1,0),(0,1),(1,1)\), and identity on all four edges; put zero outside it. Rows and columns are complexes, and its square commutes. Without signs, the square of the total map from \(T^0=\mathbb Z\) to \(T^2=\mathbb Z\) is multiplication by two, which is nonzero. Formula (6.2) makes the two paths have opposite signs. The finite-diagonal assumption is exactly what supplies the displayed biproducts without assuming any infinite coproducts.

Exercise 4 (advanced: pointwise sections need not be compatible). Consider the category of linear maps \(V\to W\) over a field \(k\), with morphisms the commuting pairs of linear maps. Let \(E=(k\xrightarrow1 k)\), \(B=(k\to0)\), and \(A=(0\to k)\). Define the evident componentwise sequence \(0\to A\to E\to B\to0\). Prove that it is short exact in this category and that its two component sequences split, while it has no section as a sequence of diagrams. Explain why this does not contradict the lifting results.

Solution. Commuting pairs form an abelian group under componentwise addition, and composition is bilinear. For a morphism of arrows \((\alpha,\beta):(V\xrightarrow d W)\to(V'\xrightarrow{d'}W')\), the equation \(\beta d=d'\alpha\) makes \(d\) restrict to \(\ker\alpha\to\ker\beta\). This arrow is its categorical kernel: a commuting pair killed by \((\alpha,\beta)\) factors uniquely through the two vector-space kernels, and their restrictions still commute. Similarly the equation induces a map \(\operatorname{coker}\alpha\to\operatorname{coker}\beta\). A pair killing the original morphism descends uniquely to these quotients, and the descended pair commutes, proving the cokernel universal property. Zero and biproducts are componentwise. Componentwise coimage-to-image isomorphisms give a commuting pair whose inverse also commutes, so this arrow category is abelian.

The inclusion \(A\to E\) has components \(0\to k\) and \(1_k\); the quotient \(E\to B\) has components \(1_k\) and \(k\to0\). Their kernel and cokernel are exactly the displayed arrows by the preceding constructions. Thus the sequence is short exact. At the first component it is \(0\to0\to k\xrightarrow1 k\to0\); at the second it is \(0\to k\xrightarrow1 k\to0\to0\). Both split.

A section \(B\to E\) would have first component \(1_k\), since the first component of the quotient is the identity, and second component the zero map \(0\to k\). Compatibility with the arrows would require \(1_k\circ1_k=0\), which is false. So there is no section of diagrams. The lifting theorem applies inside this abelian arrow category: for the identity test of \(B\), the epimorphism \(E\to B\) itself is a cover, with lift \(1_E\). Neither it nor componentwise splitting promises a compatible section on the original test object.

7. References