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 ( 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

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

Comment #7292 by on

Yes, we use this all the time: a map in 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.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0944. Beware of the difference between the letter 'O' and the digit '0'.