Lemma 21.50.1 (0944)—The Stacks project

Lemma 21.50.1. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi. Let $E \in D(\mathcal{O}_\mathcal {C})$ and $K \in D(\mathcal{O}_\mathcal {D})$ . If $K$ is perfect, then

$Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {D}} K = Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {C}} Lf^*K)$

in $D(\mathcal{O}_\mathcal {D})$ .

Proof. To check (21.50.0.1) is an isomorphism we may work locally on $\mathcal{D}$ , i.e., for any object $V$ of $\mathcal{D}$ we have to find a covering $\{ V_ j \to V\}$ such that the map restricts to an isomorphism on $V_ j$ . By definition of perfect objects, this means we may assume $K$ is represented by a strictly perfect complex of $\mathcal{O}_\mathcal {D}$ -modules. Note that, completely generally, the statement is true for $K = K_1 \oplus K_2$ , if and only if the statement is true for $K_1$ and $K_2$ . Hence we may assume $K$ is a finite complex of finite free $\mathcal{O}_\mathcal {D}$ -modules. In this case a simple argument involving stupid truncations reduces the statement to the case where $K$ is represented by a finite free $\mathcal{O}_\mathcal {D}$ -module. Since the statement is invariant under finite direct summands in the $K$ variable, we conclude it suffices to prove it for $K = \mathcal{O}_\mathcal {D}[n]$ in which case it is trivial. $\square$

Comments (2)

Comment #7139 by Hao Peng on March 25, 2022 at 02:34

it took me a while to figure why the first sentence is true. I first thought this is because $U\mapsto D(\mathcal O_U)$ is a prestack, but I cannot see this. Then I realized it suffices to show it induces isomorhisms on $H^i$ , and then reduce to the fact $U\mapsto Mod(\mathcal O_U)$ is a prestack(even a stack). But now I still wonder if $U\mapsto D(\mathcal O_U)$ is a prestack?

Comment #7292 by Johan on May 03, 2022 at 21:17

Yes, we use this all the time: a map $K \to L$ in $D(\mathcal{O})$ is an isomorphism if and only if it is an isomorphism on cohomology sheaves which may be checked locally in the topology on the site, i.e., one object at a time.

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