The Stacks project

Lemma 96.22.2. Let $S$ be a scheme. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks over $S$. Assume $\mathcal{X}$, $\mathcal{Y}$ are representable by algebraic spaces $F$, $G$. Denote $f : F \to G$ the induced morphism of algebraic spaces.

  1. For any $\mathcal{F} \in \textit{Ab}(\mathcal{X}_{\acute{e}tale})$ we have

    \[ (Rf_*\mathcal{F})|_{G_{\acute{e}tale}} = Rf_{small, *}(\mathcal{F}|_{F_{\acute{e}tale}}) \]

    in $D(G_{\acute{e}tale})$.

  2. For any object $\mathcal{F}$ of $\textit{Mod}(\mathcal{X}_{\acute{e}tale}, \mathcal{O}_\mathcal {X})$ we have

    \[ (Rf_*\mathcal{F})|_{G_{\acute{e}tale}} = Rf_{small, *}(\mathcal{F}|_{F_{\acute{e}tale}}) \]

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

Proof. Part (1) follows immediately from Lemma 96.22.1 and (96.10.3.1) on choosing an injective resolution of $\mathcal{F}$.

Part (2) can be proved as follows. In Lemma 96.10.3 we have seen that $\pi _ G \circ f = f_{small} \circ \pi _ F$ as morphisms of ringed sites. Hence we obtain $R\pi _{G, *} \circ Rf_* = Rf_{small, *} \circ R\pi _{F, *}$ by Cohomology on Sites, Lemma 21.19.2. Since the restriction functors $\pi _{F, *}$ and $\pi _{G, *}$ are exact, we conclude. $\square$

Comments (0)

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Your email address will not be published.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 075N. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 96.22.2 as pdf