The Stacks project

Proof. To see this we use the general theory developed in Derived Categories, Section 13.14. Set $\mathcal{D} = K(\mathcal{O}_ Y)$ and $\mathcal{D}' = D(\mathcal{O}_ X)$. 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 $\mathcal{D} = K(\mathcal{O}_ Y)$. 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: (1) follows from Lemma 20.26.12 and to see (2) we have to show that for a quasi-isomorphism $\mathcal{K}_1^\bullet \to \mathcal{K}_2^\bullet $ between K-flat complexes of $\mathcal{O}_ Y$-modules 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. By Lemma 20.26.8 applied to the morphism $(X, f^{-1}\mathcal{O}_ Y) \to (Y, \mathcal{O}_ Y)$ the complexes $f^{-1}\mathcal{K}_1^\bullet $ and $f^{-1}\mathcal{K}_2^\bullet $ are K-flat complexes of $f^{-1}\mathcal{O}_ Y$-modules. Hence Lemma 20.26.13 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 K-flat, i.e., $Lf^* = LF$ is indeed computed in the way described above. $\square$

Proof. Let $E$ be an object of $D(\mathcal{O}_ Z)$. By construction $Lg^*E$ is computed by choosing a K-flat complex $\mathcal{K}^\bullet $ representing $E$ on $Z$ and setting $Lg^*E = g^*\mathcal{K}^\bullet $. By Lemma 20.26.8 we see that $g^*\mathcal{K}^\bullet $ is K-flat on $Y$. Then $Lf^*Lg^*E$ is given by $f^*g^*\mathcal{K}^\bullet = (g \circ f)^*\mathcal{K}^\bullet $ which also represents $L(g \circ f)^*E$. $\square$

Proof. We may assume that $\mathcal{F}^\bullet $ and $\mathcal{G}^\bullet $ are K-flat complexes. In this case $\mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y}^{\mathbf{L}} \mathcal{G}^\bullet $ is just the total complex associated to the double complex $\mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{G}^\bullet $. By Lemma 20.26.5 $\text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{G}^\bullet )$ is K-flat also. Hence the isomorphism of the lemma comes from the isomorphism

whose constituents are the isomorphisms $f^*\mathcal{F}^ p \otimes _{\mathcal{O}_ X} f^*\mathcal{G}^ q \to f^*(\mathcal{F}^ p \otimes _{\mathcal{O}_ Y} \mathcal{G}^ q)$ of Modules, Lemma 17.16.4. $\square$

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

Proof. We will use the existence of K-flat resolutions as in Lemma 20.26.8. 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 20.26.5. $\square$