Lemma 21.43.10. Let $g : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ be as above. Then the functor $Lg^* : D(\mathcal{O}) \to D(\mathcal{O}')$ maps $\mathit{QC}(\mathcal{O})$ into $\mathit{QC}(\mathcal{O}')$.
Proof. Let $U' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$ with image $U = u(U')$ in $\mathcal{C}$. Let $pt$ denote the category with a single object and a single morphism. Denote $(\mathop{\mathit{Sh}}\nolimits (pt), \mathcal{O}'(U'))$ and $(\mathop{\mathit{Sh}}\nolimits (pt), \mathcal{O}(U))$ the ringed topoi as indicated. Of course we identify the derived category of modules on these ringed topoi with $D(\mathcal{O}'(U'))$ and $D(\mathcal{O}(U))$. Then we have a commutative diagram of ringed topoi
Pullback along the lower horizontal morphism sends $K$ in $D(\mathcal{O})$ to $R\Gamma (U, K)$. Pullback by the left vertical arrow sends $M$ to $M \otimes _{\mathcal{O}(U)}^\mathbf {L} \mathcal{O}'(U')$. Going around the diagram either direction produces the same result (Lemma 21.18.3) and hence we conclude
Finally, let $f' : U' \to V'$ be a morphism in $\mathcal{C}'$ and denote $f = u(f') : U = u(U') \to V = u(V')$ the image in $\mathcal{C}$. If $K$ is in $\mathit{QC}(\mathcal{O})$ then we have
as desired. Here we have used the observation above both for $U'$ and $V'$. $\square$
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