Lemma 52.6.7. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ be a morphism of ringed topoi. Let $\mathcal{I} \subset \mathcal{O}$ and $\mathcal{I}' \subset \mathcal{O}'$ be sheaves of ideals such that $f^\sharp $ sends $f^{-1}\mathcal{I}$ into $\mathcal{I}'$. Then $Rf_*$ sends $D_{comp}(\mathcal{O}', \mathcal{I}')$ into $D_{comp}(\mathcal{O}, \mathcal{I})$.

**Proof.**
We may assume $f$ is given by a morphism of ringed sites corresponding to a continuous functor $\mathcal{C} \to \mathcal{D}$ (Modules on Sites, Lemma 18.7.2 ). Let $U$ be an object of $\mathcal{C}$ and let $g$ be a section of $\mathcal{I}$ over $U$. We have to show that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(\mathcal{O}_{U, g}, Rf_*K|_ U) = 0$ whenever $K$ is derived complete with respect to $\mathcal{I}'$. Namely, by Cohomology on Sites, Lemma 21.35.1 this, applied to all objects over $U$ and all shifts of $K$, will imply that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, g}, Rf_*K|_ U)$ is zero, which implies that $T(Rf_*K|_ U, g)$ is zero (Lemma 52.6.2) which is what we have to show (Definition 52.6.4). Let $V$ in $\mathcal{D}$ be the image of $U$. Then

where $g' = f^\sharp (g) \in \mathcal{I}'(V)$. The second equality because $K$ is derived complete and the first equality because the derived pullback of $\mathcal{O}_{U, g}$ is $\mathcal{O}'_{V, g'}$ and Cohomology on Sites, Lemma 21.19.1. $\square$

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