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
Comments (0)
There are also: