Lemma 91.14.3. In the situation above assume that $X$ is quasi-compact and quasi-separated and that $DQ_ X(\mathcal{F}) \to DQ_ X(\mathcal{G})$ (Derived Categories of Spaces, Section 75.19) is an isomorphism. Then the functor $F$ is an equivalence of categories.
Proof. Recall that $\mathop{N\! L}\nolimits _{X/B}$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$, see More on Morphisms of Spaces, Lemma 76.21.4. Hence our assumption implies the maps
\[ \mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathop{N\! L}\nolimits _{X/B}, \mathcal{F}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathop{N\! L}\nolimits _{X/B}, \mathcal{G}) \]
are isomorphisms for all $i$. This implies our functor is fully faithful by Lemma 91.13.1. On the other hand, the functor is essentially surjective by Lemma 91.13.3 because we have the solutions $\mathcal{O}_ X \oplus \mathcal{F}$ and $\mathcal{O}_ X \oplus \mathcal{G}$ in both categories. $\square$
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