Lemma 20.55.7. In Situation 20.55.2 there is an additive functor1 $L\eta _\mathcal {I} : D(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$ such that if $M$ in $D(\mathcal{O}_ X)$ is represented by a complex $\mathcal{F}^\bullet $ of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules, then $L\eta _\mathcal {I}M = \eta _\mathcal {I}\mathcal{F}^\bullet $. Similarly for morphisms.
Proof. Denote $\mathcal{T} \subset \textit{Mod}(\mathcal{O}_ X)$ the full subcategory of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. We have a corresponding inclusion
\[ K(\mathcal{T}) \quad \subset \quad K(\textit{Mod}(\mathcal{O}_ X)) = K(\mathcal{O}_ X) \]
of $K(\mathcal{T})$ as a full triangulated subcategory of $K(\mathcal{O}_ X)$. Let $S \subset \text{Arrows}(K(\mathcal{T}))$ be the quasi-isomorphisms. We will apply Derived Categories, Lemma 13.5.8 to show that the map
\[ S^{-1}K(\mathcal{T}) \longrightarrow D(\mathcal{O}_ X) \]
is an equivalence of triangulated categories. The lemma shows that it suffices to prove: given a complex $\mathcal{G}^\bullet $ of $\mathcal{O}_ X$-modules, there exists a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{G}^\bullet $ with $\mathcal{F}^\bullet $ a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. By Lemma 20.26.12 we can find a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{G}^\bullet $ such that the complex $\mathcal{F}^\bullet $ is K-flat (we won't use this) and consists of flat $\mathcal{O}_ X$-modules $\mathcal{F}^ i$. By the third characterization of Lemma 20.55.3 we see that a flat $\mathcal{O}_ X$-module is an $\mathcal{I}$-torsion free $\mathcal{O}_ X$-module and we are done.
With these preliminaries out of the way we can define $L\eta _ f$. Namely, by the discussion following Lemma 20.55.3 this section we have already a well defined functor
\[ K(\mathcal{T}) \xrightarrow {\eta _ f} K(\mathcal{T}) \to K(\mathcal{O}_ X) \to D(\mathcal{O}_ X) \]
which according to Lemma 20.55.6 sends quasi-isomorphisms to quasi-isomorphisms. Hence this functor factors over $S^{-1}K(\mathcal{T}) = D(\mathcal{O}_ X)$ by Categories, Lemma 4.27.8. $\square$
[1] Beware that this functor isn't exact, i.e., does not transform distinguished triangles into distinguished triangles.
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