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

$\xymatrix{ (\mathop{\mathit{Sh}}\nolimits (pt), \mathcal{O}'(U')) \ar[rr]_{U'} \ar[d] & & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \ar[d]^ g \\ (\mathop{\mathit{Sh}}\nolimits (pt), \mathcal{O}(U)) \ar[rr]^ U & & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) }$

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

$R\Gamma (U', Lg^*K) = R\Gamma (U, K) \otimes _{\mathcal{O}(U)}^\mathbf {L} \mathcal{O}'(U')$

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

\begin{align*} R\Gamma (V', Lg^*K) \otimes _{\mathcal{O}'(V')}^\mathbf {L} \mathcal{O}'(U') & = R\Gamma (V, K) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}'(V') \otimes _{\mathcal{O}'(V')}^\mathbf {L} \mathcal{O}'(U') \\ & = R\Gamma (V, K) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}'(U') \\ & = R\Gamma (V, K) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}(U) \otimes _{\mathcal{O}(U)}^\mathbf {L} \mathcal{O}'(U') \\ & = R\Gamma (U, K) \otimes _{\mathcal{O}(U)}^\mathbf {L} \mathcal{O}'(U') \\ & = R\Gamma (U', Lg^*K) \end{align*}

as desired. Here we have used the observation above both for $U'$ and $V'$. $\square$

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