Flat modules and K-flat resolutions

Written and edited by GPT-6.1 Sol (OpenAI), in Codex at Ultra, October 2026. Mathematically self-checked by the writing AI. Original contributions are CC0; the combined course is distributed under GFDL-1.2-or-later. Full authorship and source attribution appear in the course notice.

Tensoring a resolution should preserve the cohomology information it was meant to carry. Flat modules solve this problem for a single tensor factor in one degree. Unbounded complexes require a stronger condition, K-flatness. The two conditions are independent for unbounded complexes: a contractible complex can have nonflat terms, and an acyclic complex of free terms can fail to be K-flat.

The prerequisites are Sheaves of modules on a ringed space, Theorems 2.1, 3.1 and 4.4 and Lemmas 5.1–5.2, and Complexes, cones and localization, Lemmas 1.1–1.2, 2.4 and Proposition 5.2. We construct K-flat resolutions independently of the K-injective existence theorem. This avoids a dependency cycle when derived tensor products and Hom are developed.

Throughout this lesson $\mathcal O$ is a sheaf of commutative unital rings on an arbitrary topological space $X$. Tensor totalizations use direct sums, including for unbounded complexes. We impose no separation, compactness, Noetherian or finite homological-dimension assumption.

The construction sources are the Stacks project authors’ Modules on Ringed Spaces, “Flat modules”, and Cohomology of Sheaves, “Flat resolutions”, in the AI Integrated Stacks Project edition. We develop flat-module calculations, cycle attachments and compatible bounded resolutions. Spaltenstein’s Resolutions of unbounded complexes provides historical background. Source attribution and the licence for adapted passages appear in the course notice.

1. Flat sheaves and exact tensor sequences

A sheaf $M$ is flat if $-\otimes_{\mathcal O}M$ is exact. The tensor sheaf is the sheafification of the presheaf $U\mapsto F(U)\otimes_{\mathcal O(U)}M(U)$. Its balanced universal property follows by gluing the presheaf balanced maps, using structured sheafification from the first lesson. Tensor is right exact: a balanced map on a quotient is precisely a balanced map on the original module vanishing on the relations defining that quotient. Thus flatness amounts to preservation of injections.

Flat building blocks

Lemma 1.1 (local covers). Every module sheaf is a quotient of a direct sum of sheaves $B_U=j_{U!}\mathcal O_U$.

Proof. For an open $U\subset X$, Lemma 5.1 of the first lesson gives $\operatorname{Hom}_{\mathcal O}(B_U,M)=M(U). \tag{1.1}$ For every module sheaf $M$, form the set-indexed sum $L(M)=\bigoplus_{U\subset X\ {\rm open}}\ \bigoplus_{s\in M(U)}B_U . \tag{1.2}$ The map $\lambda_M:L(M)\to M$ sends the generator indexed by $s$ to $s$, using (1.1). Every germ is represented by such a local section, so $\lambda_M$ is an epimorphism. $\square$

This uses local sections on all opens. A sheaf need not be generated by its global sections.

Lemma 1.2 (flatness and stalks). A module sheaf $M$ is flat if and only if every $M_x$ is flat over $\mathcal O_x$. In particular $B_U$, $L(M)$, and direct sums of flat module sheaves are flat.

Proof. The tensor sheaf is the sheafification of the sectionwise tensor presheaf. Its stalk is $(F\otimes_{\mathcal O}M)_x=F_x\otimes_{\mathcal O_x}M_x . \tag{1.3}$ Here is the relevant finite-relation argument. A tensor is a finite sum of elementary tensors. Represent all its factors on one neighbourhood of $x$. An equality between tensors is generated by finitely many additive and balancing relations. Represent their scalar and module entries on a common smaller neighbourhood; each equality of germs in this finite certificate holds after a further shrinking. This proves both surjectivity and injectivity of (1.3), even though the section rings vary with the neighbourhood.

If every $M_x$ is flat, (1.3) and Theorem 2.1 of the first lesson show that tensoring with $M$ preserves exact sequences. Conversely, an injection $A\to B$ of $\mathcal O_x$-modules gives an injection $x_*A\to x_*B$ of module sheaves: these maps are injective on sections. If $M$ is flat, tensoring preserves this injection, and its stalk at $x$ is $A\otimes M_x\to B\otimes M_x$. Tensor products are right exact, by their presentation with generators and relations, so this proves that $M_x$ is flat.

The stalks of $B_U$ are $\mathcal O_x$ or zero (Lemma 5.1 of the first lesson); both are flat. Stalks commute with direct sums. A direct sum of flat modules is flat because its tensor functor is the direct sum of their exact tensor functors, and direct sums of exact module sequences remain exact coordinatewise. This proves the final assertions. $\square$

Lemma 1.3 (flat quotients and flat exact sequences). If $0\to M'\to M\to M''\to0$ is exact and $M''$ is flat, then tensoring with any sheaf $N$ preserves that exact sequence. Consequently flat sheaves are closed under extensions and kernels of epimorphisms between flat sheaves. Filtered colimits of flat sheaves are flat, and a direct summand of a flat sheaf is flat.

Proof. We first prove the flat-quotient assertion over a ring $R$. Resolve a test module $N$ by a complex $F^\bullet$ of free modules in degrees at most zero. Choose a free module on the elements of $N$, map it onto $N$, choose a free module onto that kernel, and repeat. This construction gives $H^0(F)=N$ and zero negative cohomology. Flatness of $M''$ makes $F\otimes_R M''$ exact in negative degrees, by tensoring the short exact sequences of cycles. Free terms make

0\to F\otimes M'\to F\otimes M\to F\otimes M''\to0

termwise exact, because a free tensor is a direct sum of the original exact sequence. The long exact sequence of cohomology gives an injection $H^0(F\otimes M')\to H^0(F\otimes M)$, since the preceding $H^{-1}(F\otimes M'')$ is zero. Right exactness identifies these degree-zero groups with $N\otimes M'$ and $N\otimes M$. This proves the assertion without presupposing a Tor theorem. Apply it at each stalk, using Lemma 1.2, to obtain the sheaf assertion.

For an injection $N\to N_1$, compare the two exact tensor rows supplied by this assertion. If the outside sheaves $M',M''$ are flat, their vertical maps are monic and so is the middle one: an element in its kernel maps to zero in the right column and then is forced to be zero by the left column. This is the same kernel argument on every stalk. If $M,M''$ are flat, the injection $N\otimes M'\to N\otimes M$, followed by the monomorphism into $N_1\otimes M$, shows that the vertical map for $M'$ is monic as well.

Tensor commutes with colimits by its balanced universal property. Exactness of filtered colimits, proved in the first lesson, therefore proves flatness of a filtered colimit of flats. Finally $N\otimes(M_1\oplus M_2)=(N\otimes M_1)\oplus(N\otimes M_2)$; if this map preserves each injection, so does each summand. $\square$

The flat-quotient assertion is often phrased as vanishing of first Tor against a flat module. The proof here establishes the exact sequence needed before Tor is defined in the next lesson.

2. K-flat complexes and their closure properties

For cochain complexes define

(A\otimes P)^n=\bigoplus_{p+q=n}A^p\otimes P^q, \qquad d(a\otimes p)=d_Aa\otimes p+(-1)^{|a|}a\otimes d_Pp. \tag{2.1}

Two applications of $d$ have opposite mixed terms, while the remaining terms contain $d_A^2$ and $d_P^2$; hence $d^2=0$. A complex $P$ is K-flat if $A\otimes P$ is acyclic for every acyclic $A$. The graded flip $a\otimes p\mapsto(-1)^{|a||p|}p\otimes a$ is an isomorphism of complexes: on the two differential terms the sign exponents differ by exactly the degree increased in that term. It interchanges the tensor variables.

Lemma 2.1 (tensor respects homotopies and triangles). For any fixed complex $A$, tensoring with $A$ is an exact functor on the homotopy category. A K-flat $P$ therefore carries quasi-isomorphisms in the other variable to quasi-isomorphisms.

Proof. If $f-g=dh+hd$, then $H(a\otimes p)=(-1)^{|a|}a\otimes h(p)$ gives $1_A\otimes(f-g)=dH+Hd$; expanding the mixed terms cancels them. The shift comparison $A\otimes P[1]\to(A\otimes P)[1]$ multiplies a component in $A^p$ by $(-1)^p$. For the cone convention of the common reading, the same factor on the shifted second summand gives

A\otimes\operatorname{Cone}(f) \cong\operatorname{Cone}(1_A\otimes f). \tag{2.2}

Indeed, the off-diagonal term on the left is $(-1)^pf$; the factor $(-1)^p$ makes it the off-diagonal term on the right. On the second diagonal block, the resulting differential is the negative of the total differential on $A\otimes P$. The comparison respects the cone inclusion and the negative projection used in the distinguished triangle. Thus shifts and cone triangles are preserved. The graded flip gives the assertion in the other variable. A quasi-isomorphism has acyclic cone. If $P$ is K-flat, its tensor with that cone is acyclic, and (2.2) proves the last assertion. $\square$

Lemma 2.2 (stalks and pullbacks). A complex $P$ of module sheaves is K-flat if and only if every stalk $P_x$ is K-flat over $\mathcal O_x$. Pullback along any morphism of commutative ringed spaces preserves flat sheaves and K-flat complexes.

Proof. Stalks commute with the tensor sheaf by Lemma 1.2 and with direct sums by the first lesson. Hence $(A\otimes P)_x=A_x\otimes P_x$. If the stalks of $P$ are K-flat, an acyclic $A$ has acyclic tensor on all stalks, and the tensor sheaf complex is acyclic. Conversely, for an acyclic complex $C$ over $\mathcal O_x$, the complex of skyscrapers $x_*C$ is acyclic: the skyscraper functor is exact on sections, as established in Lemma 5.2 of the first lesson. Its stalk at $x$ is $C$. Applying K-flatness of $P$ to this complex and taking that stalk proves K-flatness of $P_x$.

For a ring map $R\to S$ and K-flat $P$ over $R$, any acyclic complex $C$ of $S$-modules remains acyclic after restriction of scalars, and

C\otimes_S(S\otimes_R P)=C\otimes_R P

is acyclic. The isomorphism sends $c\otimes(s\otimes p)$ to $sc\otimes p$; its inverse sends $c\otimes p$ to $c\otimes(1\otimes p)$, and both commute with the total differentials. This proves base change of K-flatness. The identical one-degree argument proves base change of flat modules. The stalk formula for pullback from the first lesson, with $R=\mathcal O_{Y,f(x)}$ and $S=\mathcal O_{X,x}$, proves the sheaf statements. No flatness of $S$ over $R$ is needed. $\square$

Lemma 2.3 (operations preserving K-flatness). A flat sheaf in a single degree is K-flat. Direct sums and filtered colimits of K-flat complexes are K-flat. If $0\longrightarrow P\longrightarrow Q\longrightarrow R\longrightarrow0 \tag{2.3}$ is a degreewise split exact sequence of complexes and $P,R$ are K-flat, then $Q$ is K-flat. Every bounded-above complex of flat sheaves is K-flat.

Proof. If $A$ is acyclic and $M$ is flat, tensor the exact sequences $0\to Z^nA\to A^n\to Z^{n+1}A\to0$ with $M$. They identify the kernels and images in $A\otimes M$, proving its acyclicity. Shifting a complex does not change this conclusion.

Tensor totalization commutes with direct sums and filtered colimits. These operations are exact for module sheaves by Theorem 3.1 of the first lesson (an arbitrary sum is the filtered colimit of its finite partial sums). They therefore commute with cohomology: kernels and images in an exact filtered colimit are the colimits of the corresponding kernels and images. Apply this to $A\otimes P_i$ to obtain the assertions about sums and colimits.

After tensoring (2.3) with any complex $A$, the sequence remains degreewise split exact, including after totalization. Its long exact cohomology sequence proves the extension assertion. The splitting is only a splitting of the underlying graded sheaves; it need not respect the differentials.

A bounded complex of flat sheaves is K-flat by induction on its length, using its bottom-degree term as the quotient and the remaining tail as the subcomplex in (2.3). If $P$ is merely bounded above, its brutal lower truncations $P^{\geq m}$ are bounded complexes and are subcomplexes of $P$. As $m$ decreases their union is $P$. The filtered-colimit assertion proves the last statement. $\square$

Lemma 2.4 (tensor products and two out of three). The tensor product of two K-flat complexes is K-flat. Shifts, direct summands, and homotopy equivalent complexes preserve K-flatness. Two out of three objects in a distinguished triangle of the homotopy category being K-flat forces the third to be K-flat. The same two-out-of-three statement holds in a short exact sequence of complexes whose quotient has flat terms.

Proof. Direct-sum reindexing identifies $A\otimes(P\otimes Q)$ with $(A\otimes P)\otimes Q$. On a term of degrees $a,p,q$, both differentials have signs $1,(-1)^a,(-1)^{a+p}$. If $A$ is acyclic, successively applying K-flatness of $P$ and $Q$ gives acyclicity. Shifts and homotopy equivalences follow from Lemma 2.1. A direct summand of an acyclic tensor complex is acyclic, giving the summand assertion. Tensoring a distinguished triangle gives a distinguished triangle by that lemma, and its long exact cohomology sequence gives two out of three. For a short exact sequence with flat quotient terms, Lemma 1.3 makes its tensor with each term of any complex exact. Direct sums preserve exactness, so totalization gives a short exact sequence of tensor complexes. Its long exact sequence supplies the last claim. $\square$

Only filtered colimits occur in Lemma 2.3. An arbitrary cokernel of K-flat complexes need not be K-flat; Section 5 supplies the smallest counterexample.

3. A surjective resolution by attaching cycles

Theorem 3.1 (K-flat resolutions). Every complex $G$ of $\mathcal O$-modules has a quasi-isomorphism $P\to G$, surjective in each degree, where $P$ is K-flat and every $P^n$ is flat.

Proof. For a sheaf $M$, let $D_n(M)$ have $M$ in degrees $n,n+1$, identity differential between them, and zero elsewhere. A map $M\to G^n$ extends uniquely to a chain map $D_n(M)\to G$, with upper component its composite with $d_G$. Start with $\begin{gathered} P_0=\bigoplus_{n\in\mathbb Z}D_n(L(G^n)),\\ \epsilon_0:P_0\longrightarrow G. \end{gathered} \tag{3.1}$ Use $\lambda_{G^n}$ on the lower term of the $n$-th summand. The map is onto in every degree. Each disk is contractible. Tensoring a contraction $h$ on the second factor with a complex $A$ gives the contraction $a\otimes p\mapsto(-1)^{|a|}a\otimes h(p)$. Thus $P_0$ is K-flat; its terms are flat by Lemma 1.2.

Suppose $P_s\to G$ has been built, and put $K_s=\ker(\epsilon_s)$, a subcomplex. For every integer $q$, open $U$, and section $z\in Z^q(K_s)(U),$ add a copy of $B_U$ in degree $q-1$. Write its generator locally as $e_z$, and prescribe $d e_z=z,\qquad \epsilon_{s+1}(e_z)=0. \tag{3.2}$ These are morphisms of sheaves by (1.1). The two conditions needed for them to define a complex over $G$ hold: $dz=0$ and $\epsilon_s(z)=0$. Explicitly, if $V_s^n$ is the sum of the newly added $B_U$'s in degree $n$, the underlying graded sheaf is $P_s\oplus V_s$, with differential $d(p,v)=(d_{P_s}p+\alpha_s(v),0),$ where $\alpha_s:V_s^n\to P_s^{n+1}$ sends each generator to its chosen cycle. Since $d_{P_s}\alpha_s=0$, its square is zero.

The inclusion $P_s\to P_{s+1}$ is degreewise split, and the quotient is $V_s$ with zero differential. It is a direct sum of shifts of the flat sheaves $B_U$, hence K-flat. Lemma 2.3 proves by induction that every $P_s$ is K-flat. Its terms remain direct sums of the flat building blocks.

Set $P=\varinjlim_{s\geq0}P_s$. The maps to $G$ agree, and the map $\epsilon:P\to G$ is termwise onto already from $P_0$. Its terms are flat, and $P$ is K-flat by Lemma 2.3. Exactness of filtered colimits gives $\ker(\epsilon)=\varinjlim_s K_s,\qquad H^q(\ker\epsilon)=\varinjlim_s H^q(K_s). \tag{3.3}$ Each transition $H^q(K_s)\to H^q(K_{s+1})$ is zero. Indeed, a stalk class is locally represented by a cycle section $z$; the next stage supplies $e_z$ in the kernel with $de_z=z$. Thus every term in the colimit on the right of (3.3) maps to zero at the next stage, and the colimit vanishes. The kernel of $\epsilon$ is acyclic. The long exact sequence for $0\to\ker\epsilon\to P\to G\to0$ proves that $\epsilon$ is a quasi-isomorphism. $\square$

The stages can create new cycles; the next stage kills them. A germ involves a finite stage, so a countable sequence of these stages suffices. Nothing in the construction asks sections to lift through an arbitrary sheaf epimorphism: generators are added for sections that actually exist on each open.

Lemma 3.2 (making another resolution termwise surjective). If $Q\to G$ is a quasi-isomorphism with $Q$ K-flat, then $Q\oplus P_0\to G$, the sum of that map and (3.1), is a termwise-surjective K-flat resolution. If $Q$ also has flat terms, so does this enlargement.

Proof. The new map is onto because $\epsilon_0$ is. The complex $P_0$ is contractible, and the inclusion $Q\to Q\oplus P_0$ is a quasi-isomorphism commuting with the maps to $G$. The new map is therefore a quasi-isomorphism. The direct sum is K-flat by Lemma 2.3; this uses no termwise-flatness assumption on $Q$. If its terms are flat, Lemma 1.2 gives flatness of the enlarged terms as well. $\square$

Remark 3.3 (what the construction does not prove). Extensions by zero of structure sheaves need not be projective objects in the category of module sheaves. The construction gives a resolution suitable for tensor products, not a K-projective resolution for computing arbitrary Hom functors. Neither an inverse limit of injective truncations nor exactness of products of sheaves was used.

4. Compatible bounded resolutions and factorization

The cycle construction proves the unbounded existence theorem directly. We also prove the stronger compatible bounded-above form used in the Stacks treatment, with every term a direct sum of the flat building blocks $B_U$.

Lemma 4.1 (bounded-above resolutions). If $C$ vanishes in degrees greater than $b$, there is a termwise-surjective quasi-isomorphism $E\to C$, with $E$ vanishing above $b$ and its terms direct sums of $B_U$.

Proof. At degree $n$, replace the current term $C^n$ by its surjective cover $u:L(C^n)\to C^n$. Replace degree $n-1$ by the pullback $C^{n-1}\times_{C^n}L(C^n)$, with differential into degree $n$ its second projection. The next differential is $d_Cu$; the preceding differential sends $c$ to $(d_Cc,0)$. These prescriptions have square zero because the old differential had square zero. They give a termwise epimorphic chain map to the old complex; the pullback projection is epic because it is a pullback of $u$. Its kernel is $\ker u$ in degrees $n-1,n$, with identity differential, so the replacement is a quasi-isomorphism.

Perform these replacements for $n=b,b-1,b-2,\ldots$. A fixed term changes at most twice and is finally a chosen cover $L(-)$. Define $E$, its differential and its map to the original complex by these stabilized components. The finite compositions into any fixed original term are epic. For cohomology in degree $r$, the three terms and two differentials in degrees $r-1,r,r+1$ stabilize after finitely many replacements. At that point the finite composition is a quasi-isomorphism, so the stabilized map is an isomorphism on $H^r$. This proves the result without assuming exact inverse limits. $\square$

Lemma 4.2 (an explicit extension over a cone). Given $a:P\to H$ and a termwise-surjective quasi-isomorphism $u:Q\to C(a)$, write $u(q)=(u_Hq,u_Pq)$, where $u_P:Q^n\to P^{n+1}$. Define

\begin{gathered} N^n=P^n\oplus Q^n,\\ d_N(p,q)=(d_Pp-u_Pq,d_Qq),\\ f(p,q)=a(p)+u_Hq. \end{gathered} \tag{4.1}

Then $P\to N$ is a degreewise split inclusion, its quotient is $Q$, $f:N\to H$ is a termwise-surjective quasi-isomorphism, and $f|_P=a$.

Proof. The chain-map equation for $u$ gives $d_Pu_P+u_Pd_Q=0$ and $d_Hu_H+a u_P=u_Hd_Q$. These identities give $d_N^2=0$ and $d_Hf=f d_N$ by substitution. Surjectivity of $u$ followed by the graded projection $C(a)^n\to H^n$ proves surjectivity of $u_H$, and hence of $f$.

Compare the triangle from $0\to P\to N\to Q\to0$ with the cone triangle of $a$. The maps on $P,N,Q$ are respectively $1,f,u$. The first square commutes strictly. In the second square, $u\pi-i f$ is the homotopy boundary of $h(p,q)=(0,-p)$ into $C(a)^{n-1}$, because

d_Ch+h d_N=(-a(p),u_Pq)=u\pi-i f.

The boundary of the split extension is $-u_P:Q\to P[1]$, since that is its off-diagonal differential, and it equals the negative cone projection composed with $u$. Thus all squares commute on cohomology. The outside vertical maps are an identity and a quasi-isomorphism; the long exact cohomology sequences and the Five Lemma prove that $f$ is a quasi-isomorphism. $\square$

Proposition 4.3 (compatible bounded-above system). For any complex $G$, there are termwise-surjective quasi-isomorphisms $E_n\to\tau_{\le n}G$, for $n=1,2,\ldots$, with $E_n$ bounded above and all its terms sums of $B_U$. They form a commutative direct system whose transitions are termwise split injections with quotients termwise sums of $B_U$. The induced map $\varinjlim E_n\to G$ is a termwise-surjective quasi-isomorphism from a K-flat complex of flat terms.

Proof. Recall that $\tau_{\le n}G$ agrees with $G$ below $n$, has $\ker d_G^n$ at $n$, and vanishes above $n$. Its inclusions into the next truncation are chain maps, and their union is $G$. Lemma 4.1 supplies $E_1$. Suppose $E_n\to\tau_{\le n}G$ has been constructed. Compose it into $H=\tau_{\le n+1}G$, and take the cone $C(a)$. This cone is bounded above. Resolve it by a termwise-surjective $Q\to C(a)$ as in Lemma 4.1 and apply Lemma 4.2. Set $E_{n+1}=N$ and its map to $H$ equal to $f$. Its terms are sums of the desired blocks, it is bounded above, and the displayed formula makes the transition square strictly commute. Its graded quotient by $E_n$ is $Q$, with the required terms.

Exact filtered colimits commute with cohomology, so the union map is a quasi-isomorphism. It is onto in each degree because the truncations exhaust $G$. Each $E_n$ is K-flat by Lemma 2.3, so their filtered colimit is K-flat. Its terms are flat by Lemma 1.3, or directly from the chosen graded splittings as sums of the flat blocks. $\square$

Corollary 4.4 (factoring a map from a K-flat complex). If $a:P\to H$ and $P$ is K-flat, then $a$ factors through a K-flat $N\to H$ that is a termwise-surjective quasi-isomorphism. If $P$ has flat terms, $N$ can have flat terms too.

Proof. Resolve $C(a)$ by Theorem 3.1 and apply Lemma 4.2. The split exact sequence $0\to P\to N\to Q\to0$ makes $N$ K-flat by Lemma 2.3. Its terms $P^n\oplus Q^n$ are flat when the terms of $P$ are. The factorization is strict in this construction; the corresponding homotopy-category statement therefore follows. $\square$

Lemma 4.5 (comparison after tensoring). A quasi-isomorphism between K-flat complexes remains a quasi-isomorphism after tensoring with any complex. An acyclic K-flat complex consequently has acyclic tensor with every complex.

Proof. For a complex $A$, choose a K-flat resolution $F\to A$ by Theorem 3.1. If $P\to Q$ is a quasi-isomorphism between K-flats, in the tensor square with rows $F\otimes P\to F\otimes Q$ and $A\otimes P\to A\otimes Q$, the vertical maps are quasi-isomorphisms by K-flatness of $P,Q$, and the top map is one by K-flatness of $F$. Two out of three forces the bottom map to be one. Apply this with $P\to0$ for the last assertion. $\square$

5. Examples and failed shortcuts

K-flat with nonflat terms. The disk $D_0(\mathbb Z/2)$ is contractible. The contraction used for the disks in Theorem 3.1 makes its tensor with every complex contractible, so it is K-flat. Its nonzero terms are not flat over $\mathbb Z$: tensoring the injection $2:\mathbb Z\to\mathbb Z$ gives zero on the nonzero group $\mathbb Z/2$.

Free terms without K-flatness. Over $R=k[\epsilon]/(\epsilon^2)$, let $P$ have $R$ in every integer degree and differential $\epsilon$. It is acyclic since both kernel and image are $\epsilon R$, and its terms are free. Let $F$ be its left half in degrees at most zero, with augmentation $F\to k=R/(\epsilon)$. Then $F$ resolves $k[0]$: the negative degrees are exact and the degree-zero cokernel is $k$. It is K-flat by the bounded-above lemma. Its tensor with the acyclic $P$ is therefore acyclic. But $k\otimes P$ has one copy of $k$ in every degree and zero differential. If $P$ were K-flat, it would carry $F\to k[0]$ to a quasi-isomorphism, which is impossible. More explicitly, the acyclic test $C(F\to k[0])$ has tensor with $P$ of cohomology $k$ in every degree, by the cone exact sequence. This displays the test that fails.

Why a general colimit is different. The cokernel of $2:\mathbb Z[0]\to\mathbb Z[0]$ is $(\mathbb Z/2)[0]$. Both source and target are K-flat, but the cokernel is not: tensoring the acyclic complex $\mathbb Z\xrightarrow{2}\mathbb Z\to\mathbb Z/2$ in degrees $0,1,2$ gives a complex with zero first differential and nonzero degree-zero cohomology. Thus the word “filtered” in the colimit theorem cannot be dropped.

Local generators need not be projective. On $[0,1]$ with the constant ring sheaf $\mathbb Z$, let $j:(0,1)\hookrightarrow[0,1]$. The sequence $0\to j_!\mathbb Z\to\mathbb Z_{[0,1]}\to i_*\mathbb Z_{\{0,1\}}\to0$ is exact on stalks. Its global-section map in the last two terms is the diagonal $\mathbb Z\to\mathbb Z^2$, so it is not onto. Therefore $\operatorname{Hom}(B_X,-)=\Gamma(X,-)$ is not exact and the flat building block $B_X=\mathcal O$ is not projective. The resolution is suited to tensor products; it supplies no K-projectivity theorem for Hom.

6. Exercises with checked solutions

Exercise 1 (easy: summands). Prove that a direct summand of a flat sheaf is flat, and that a direct summand of a K-flat complex is K-flat.

Solution. Tensor distributes over the finite direct sum defining the splitting. For an injection of test sheaves, injectivity of the direct sum of the two tensor maps forces injectivity of each component. Right exactness gives flatness of each sheaf summand. For complexes, cohomology of a finite direct sum is the direct sum of cohomologies. If tensoring with an acyclic test has acyclic sum, each tensor summand has zero cohomology in every degree. That is K-flatness.

Exercise 2 (medium: the converse stalk test). Why can the stalk K-flatness theorem be proved using a skyscraper complex even when the point is not closed?

Solution. The skyscraper functor in the first lesson is defined on every point: sections are the given stalk module on opens containing the point and zero otherwise. An exact sequence of these modules gives an exact section sequence on every open, hence an exact sheaf sequence. Its stalk at the chosen point is the original module, because every neighbourhood there contributes that module with identity restrictions. No assertion about stalks at other points is needed. For an acyclic stalk complex $C$, use $x_*C$ as the sheaf test, then take its tensor stalk $C\otimes P_x$. Thus the proof applies without a closed-point assumption.

Exercise 3 (medium: the bound direction). Prove that a bounded-above complex of flat sheaves is K-flat and explain why the same proof does not establish the assertion for a bounded-below complex of flat sheaves.

Solution. A bounded flat complex is built from its individual terms by graded split exact sequences: remove its lowest-degree term as quotient, leaving the higher-degree tail as subcomplex. Lemma 2.3 gives induction on length. If $P$ is bounded above, the brutal lower truncations $P^{\ge m}$ are bounded subcomplexes, because the differential raises degree. Their inclusions as $m$ decreases form a filtered system with union $P$, so filtered closure proves the assertion. Brutal upper truncations do not form subcomplexes when the outgoing differential at their top degree is nonzero. Thus the proposed bounded-below analogue lacks the system of inclusions used in this proof; no such analogue is inferred.

Exercise 4 (hard: a two-term Sierpiński example). Let $X=\{\eta,s\}$ have opens $\varnothing,\{\eta\},X$ and constant structure ring $\mathbb Z$. Set $B=B_X$, $E=B_{\{\eta\}}$, and let $i:E\to B$ be the canonical inclusion. Let $M$ have section diagram $M(X)=\mathbb Z/2\to M(\{\eta\})=0$, and $T=j_{\eta!}(\mathbb Z/2)$. Construct a bounded-above resolution by sums of $B,E$ of the two-term complex $G=(M\xrightarrow{0}T)$ in degrees $0,1$, and the compatible system of Proposition 4.3 for this $G$.

Solution. On this space a module sheaf is exactly its restriction map from $X$ to $\{\eta\}$; the only other sheaf condition is on the empty set. Here $B$ has diagram $\mathbb Z\xrightarrow{1}\mathbb Z$, and $E$ has $0\to\mathbb Z$. Define

P^{-2}=E,\quad P^{-1}=E\oplus B,\quad P^0=B\oplus E,\quad P^1=E,

with zero other terms and differentials

t\longmapsto(-2t,i(t)),\qquad (a,b)\longmapsto(i(a)+2b,0),\qquad (b,c)\longmapsto2c.

Both adjacent composites vanish. The augmentation reduces the $B$ coordinate in degree zero onto $M$ and reduces the degree-one $E$ onto $T$; it is epic on every stalk. At $s$, the $E$ stalks vanish and the complex is $\mathbb Z\xrightarrow{2}\mathbb Z$ in degrees $-1,0$, resolving $M_s$. At $\eta$, the first three components contributed by $M$ are $\mathbb Z\to\mathbb Z^2\to\mathbb Z$, with maps $(-2,1)$ and $(1,2)$. The first map is injective, its image is the kernel of the second, and the second is onto. The remaining $E$ coordinates give $\mathbb Z\xrightarrow{2}\mathbb Z$ in degrees $0,1$, resolving $T_\eta$ in degree one. Thus the augmentation is a quasi-isomorphism. Each term is a sum of the required flat blocks, and bounded-above flatness proves K-flatness. Since $\tau_{\le n}G=G$ for all $n\ge1$, take every $E_n=P$ and every transition the identity. These are split injections with zero quotient, which is the empty sum of flat blocks, and meet all properties of Proposition 4.3.

Exercise 5 (hard: the extension signs). In Lemma 4.2, verify both the chain-map equation for $f$ and the homotopy in the second triangle square. What breaks if only the sign of $u_P$ in $d_N$ is reversed?

Solution. The first component of the equation $d_Cu=u d_Q$ is $d_Hu_H+a u_P=u_Hd_Q$. Therefore

d_Hf(p,q)=a d_Pp+u_Hd_Qq-a u_Pq=f d_N(p,q).

For $h(p,q)=(0,-p)$, the two homotopy terms are $(-a p,d_Pp)$ and $(0,-d_Pp+u_Pq)$, whose sum is $(-a p,u_Pq)=u\pi-i f$. Replacing just the off-diagonal sign gives $f d_N=a d_Pp+a u_Pq+u_Hd_Qq$, which differs from $d_Hf$ by $2a u_Pq$ in general. It also reverses the extension boundary without reversing the cone projection. The coherent negative signs in the displayed construction are essential.

The next lesson uses these resolutions to define the derived tensor product and Tor sheaves, proving choice independence from Lemma 4.5 and the localization universal property.

Sources and licensing: Stacks project authors, Modules on Ringed Spaces, flat modules; Cohomology of Sheaves, flat resolutions; and Derived Categories, compatible direct systems. The complete used proofs were read and their short module/cone arguments expanded here. The cycle construction is an expository alternative to the source's compatible bounded resolutions, not a new existence theorem. N. Spaltenstein, Resolutions of unbounded complexes, Compositio Mathematica 65 (1988), 121–154, is available from Numdam. See the combined licence notice and GNU FDL text.