Lemma 91.14.4. 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 $FT$ is an equivalence of categories.
Proof. A solution of (91.13.0.1) for $\mathcal{F}$ in particular gives an extension of $f^{-1}\mathcal{O}_{B'}$-algebras
\[ 0 \to \mathcal{F} \to \mathcal{O}' \to \mathcal{O}_ X \to 0 \]
where $\mathcal{F}$ is an ideal of square zero. Similarly for $\mathcal{G}$. Moreover, given such an extension, we obtain a map $c_{\mathcal{O}'} : f^{-1}\mathcal{J} \to \mathcal{F}$. Thus we are looking at the full subcategory of such extensions of $f^{-1}\mathcal{O}_{B'}$-algebras with $c = c_{\mathcal{O}'}$. Clearly, if $\mathcal{O}'' = F(\mathcal{O}')$ where $F$ is the equivalence of Lemma 91.14.3 (applied to $X \to B'$ this time), then $c_{\mathcal{O}''}$ is the composition of $c_{\mathcal{O}'}$ and the map $\mathcal{F} \to \mathcal{G}$. This proves the lemma. $\square$
Comments (0)