Lemma 20.27.1. The construction above is independent of choices and defines an exact functor of triangulated categories $Lf^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$.

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

$f^{-1}\mathcal{K}_1^\bullet \otimes _{f^{-1}\mathcal{O}_ Y} \mathcal{O}_ X \longrightarrow f^{-1}\mathcal{K}_2^\bullet \otimes _{f^{-1}\mathcal{O}_ Y} \mathcal{O}_ X$

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

$LF : D(\mathcal{O}_ Y) = S^{-1}\mathcal{D} \longrightarrow \mathcal{D}' = D(\mathcal{O}_ X)$

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. $\square$

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