The Stacks project

Proof. The terms $f^*\mathcal{K}^ n$ are flat $\mathcal{O}'$-modules by Modules on Sites, Lemma 18.39.1. Choose a diagram

as in Lemma 21.17.10. We will use all of the properties stated in the lemma without further mention. Each $\mathcal{K}_ n^\bullet $ is a bounded above complex of flat modules, see Modules on Sites, Lemma 18.28.7. Consider the short exact sequence of complexes

defining $\mathcal{M}^\bullet $. By Lemmas 21.17.8 and 21.17.9 the complex $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet $ is K-flat and by Modules on Sites, Lemma 18.28.5 it has flat terms. By Modules on Sites, Lemma 18.28.10 $\mathcal{M}^\bullet $ has flat terms, by Lemma 21.17.7 $\mathcal{M}^\bullet $ is K-flat, and by the long exact cohomology sequence $\mathcal{M}^\bullet $ is acyclic (because the second arrow is a quasi-isomorphism). The pullback $f^*(\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet ) = \mathop{\mathrm{colim}}\nolimits f^*\mathcal{K}_ n^\bullet $ is a colimit of bounded below complexes of flat $\mathcal{O}'$-modules and hence is K-flat (by the same lemmas as above). The pullback of our short exact sequence

is a short exact sequence of complexes by Modules on Sites, Lemma 18.39.4. Hence by Lemma 21.17.7 it suffices to show that $f^*\mathcal{M}^\bullet $ is K-flat. This reduces us to the case discussed in the next paragraph.

Assume $\mathcal{K}^\bullet $ is acyclic as well as K-flat and with flat terms. Then Lemma 21.17.16 guarantees that all terms of $\tau _{\leq n}\mathcal{K}^\bullet $ are flat $\mathcal{O}$-modules. We choose a diagram as above and we will use all the properties proven above for this diagram. Denote $\mathcal{M}_ n^\bullet $ the kernel of the map of complexes $\mathcal{K}_ n^\bullet \to \tau _{\leq n}\mathcal{K}^\bullet $ so that we have short exact sequences of complexes

By Modules on Sites, Lemma 18.28.10 we see that the terms of the complex $\mathcal{M}_ n^\bullet $ are flat. Hence we see that $\mathcal{M} = \mathop{\mathrm{colim}}\nolimits \mathcal{M}_ n^\bullet $ is a filtered colimit of bounded below complexes of flat modules in this case. Thus $f^*\mathcal{M}^\bullet $ is K-flat (same argument as above) and we win. $\square$

Proof. To see this we use the general theory developed in Derived Categories, Section 13.14. Set $\mathcal{D} = K(\mathcal{O}')$ and $\mathcal{D}' = D(\mathcal{O})$. Let us write $F : \mathcal{D} \to \mathcal{D}'$ the exact functor of triangulated categories defined by the rule $F(\mathcal{G}^\bullet ) = f^*\mathcal{G}^\bullet $. We let $S$ be the set of quasi-isomorphisms in $\mathcal{D} = K(\mathcal{O}')$. This gives a situation as in Derived Categories, Situation 13.14.1 so that Derived Categories, Definition 13.14.2 applies. We claim that $LF$ is everywhere defined. This follows from Derived Categories, Lemma 13.14.15 with $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ the collection of K-flat complexes $\mathcal{K}^\bullet $ with flat terms. Namely, (1) follows from Lemma 21.17.11 and to see (2) we have to show that for a quasi-isomorphism $\mathcal{K}_1^\bullet \to \mathcal{K}_2^\bullet $ between elements of $\mathcal{P}$ the map $f^*\mathcal{K}_1^\bullet \to f^*\mathcal{K}_2^\bullet $ is a quasi-isomorphism. To see this write this as

The functor $f^{-1}$ is exact, hence the map $f^{-1}\mathcal{K}_1^\bullet \to f^{-1}\mathcal{K}_2^\bullet $ is a quasi-isomorphism. The complexes $f^{-1}\mathcal{K}_1^\bullet $ and $f^{-1}\mathcal{K}_2^\bullet $ are K-flat complexes of $f^{-1}\mathcal{O}'$-modules by Lemma 21.18.1 because we can consider the morphism of ringed topoi $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), f^{-1}\mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$. Hence Lemma 21.17.12 guarantees that the displayed map is a quasi-isomorphism. Thus we obtain a derived functor

see Derived Categories, Equation (13.14.9.1). Finally, Derived Categories, Lemma 13.14.15 also guarantees that $LF(\mathcal{K}^\bullet ) = F(\mathcal{K}^\bullet ) = f^*\mathcal{K}^\bullet $ when $\mathcal{K}^\bullet $ is in $\mathcal{P}$. The proof is finished by observing that bounded above complexes of flat modules are in $\mathcal{P}$ by Lemma 21.17.8. $\square$

Proof. Let $E$ be an object of $D(\mathcal{O}_\mathcal {E})$. We may represent $E$ by a K-flat complex $\mathcal{K}^\bullet $ with flat terms, see Lemma 21.17.11. By construction $Lg^*E$ is computed by $g^*\mathcal{K}^\bullet $, see Lemma 21.18.2. By Lemma 21.18.1 the complex $g^*\mathcal{K}^\bullet $ is K-flat with flat terms. Hence $Lf^*Lg^*E$ is represented by $f^*g^*\mathcal{K}^\bullet $. Since also $L(g \circ f)^*E$ is represented by $(g \circ f)^*\mathcal{K}^\bullet = f^*g^*\mathcal{K}^\bullet $ we conclude. $\square$

Proof. By our construction of derived pullback in Lemma 21.18.2. and the existence of resolutions in Lemma 21.17.11 we may replace $\mathcal{F}^\bullet $ and $\mathcal{G}^\bullet $ by complexes of $\mathcal{O}'$-modules which are K-flat and have flat terms. In this case $\mathcal{F}^\bullet \otimes _{\mathcal{O}'}^{\mathbf{L}} \mathcal{G}^\bullet $ is just the total complex associated to the double complex $\mathcal{F}^\bullet \otimes _{\mathcal{O}'} \mathcal{G}^\bullet $. The complex $\text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}'} \mathcal{G}^\bullet )$ is K-flat with flat terms by Lemma 21.17.5 and Modules on Sites, Lemma 18.28.12. Hence the isomorphism of the lemma comes from the isomorphism

whose constituents are the isomorphisms $f^*\mathcal{F}^ p \otimes _{\mathcal{O}} f^*\mathcal{G}^ q \to f^*(\mathcal{F}^ p \otimes _{\mathcal{O}'} \mathcal{G}^ q)$ of Modules on Sites, Lemma 18.26.2. $\square$

Proof. Let $\mathcal{F}$ be an $\mathcal{O}$-module and let $\mathcal{G}$ be an $\mathcal{O}'$-module. Then $\mathcal{F} \otimes _{\mathcal{O}} f^*\mathcal{G} = \mathcal{F} \otimes _{f^{-1}\mathcal{O}'} f^{-1}\mathcal{G}$ because $f^*\mathcal{G} = \mathcal{O} \otimes _{f^{-1}\mathcal{O}'} f^{-1}\mathcal{G}$. The lemma follows from this and the definitions. $\square$

Proof. Proof of (2). If $\mathcal{C}$ has enough points and $\mathcal{K}_ p^\bullet $ is K-flat for all points $p$ of $\mathcal{C}$ then we see that $\mathcal{K}^\bullet $ is K-flat because $\otimes $ and direct sums commute with taking stalks and because we can check exactness at stalks, see Modules on Sites, Lemma 18.14.4.

Proof of (1). Assume $\mathcal{K}^\bullet $ is K-flat. Choose a quasi-isomorphism $a : \mathcal{L}^\bullet \to \mathcal{K}^\bullet $ such that $\mathcal{L}^\bullet $ is K-flat with flat terms, see Lemma 21.17.11. Any pullback of $\mathcal{L}^\bullet $ is K-flat, see Lemma 21.18.1. In particular the stalk $\mathcal{L}_ p^\bullet $ is a K-flat complex of $\mathcal{O}_ p$-modules. Thus the cone $C(a)$ on $a$ is a K-flat (Lemma 21.17.6) acyclic complex of $\mathcal{O}$-modules and it suffuces to show the stalk of $C(a)$ is K-flat (by More on Algebra, Lemma 15.59.5). Thus we may assume that $\mathcal{K}^\bullet $ is K-flat and acyclic.

Assume $\mathcal{K}^\bullet $ is acyclic and K-flat. Before continuing we replace the site $\mathcal{C}$ by another one as in Sites, Lemma 7.29.5 to insure that $\mathcal{C}$ has all finite limits. This implies the category of neighbourhoods of $p$ is filtered (Sites, Lemma 7.33.2) and the colimit defining the stalk of a sheaf is filtered. Let $M$ be a finitely presented $\mathcal{O}_ p$-module. It suffices to show that $\mathcal{K}^\bullet \otimes _{\mathcal{O}_ p} M$ is acyclic, see More on Algebra, Lemma 15.59.9. Since $\mathcal{O}_ p$ is the filtered colimit of $\mathcal{O}(U)$ where $U$ runs over the neighbourhoods of $p$, we can find a neighbourhood $(U, x)$ of $p$ and a finitely presented $\mathcal{O}(U)$-module $M'$ whose base change to $\mathcal{O}_ p$ is $M$, see Algebra, Lemma 10.127.6. By Lemma 21.17.4 we may replace $\mathcal{C}, \mathcal{O}, \mathcal{K}^\bullet $ by $\mathcal{C}/U, \mathcal{O}_ U, \mathcal{K}^\bullet |_ U$. We conclude that we may assume there exists an $\mathcal{O}$-module $\mathcal{F}$ such that $M \cong \mathcal{F}_ p$. Since $\mathcal{K}^\bullet $ is K-flat and acyclic, we see that $\mathcal{K}^\bullet \otimes _\mathcal {O} \mathcal{F}$ is acyclic (as it computes the derived tensor product by definition). Taking stalks is an exact functor, hence we get that $\mathcal{K}^\bullet \otimes _{\mathcal{O}_ p} M$ is acyclic as desired. $\square$

Proof. This follows from Lemma 21.18.6, Modules on Sites, Lemma 18.36.4, and More on Algebra, Lemma 15.59.3. $\square$

Proof. We will use the existence of K-flat resolutions with flat terms (Lemma 21.17.11), we will use that derived pullback is computed by such complexes (Lemma 21.18.2), and that pullbacks preserve these properties (Lemma 21.18.1). If we choose such resolutions $\mathcal{P}^\bullet \to \mathcal{K}^\bullet $ and $\mathcal{Q}^\bullet \to \mathcal{M}^\bullet $, then we see that

commutes. However, now the left hand side of the diagram is the left hand side of the diagram by our choice of $\mathcal{P}^\bullet $ and $\mathcal{Q}^\bullet $ and Lemma 21.17.5. $\square$