The Stacks project

Lemma 65.18.12. Let $S$ be a scheme. Let

\[ \xymatrix{ X' \ar[r] \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

be a cartesian square of algebraic spaces over $S$. Let $\mathcal{F}$ be a sheaf on $X_{\acute{e}tale}$. If $g$ is ├ętale, then

  1. $f'_{small, *}(\mathcal{F}|_{X'}) = (f_{small, *}\mathcal{F})|_{Y'}$ in $\mathop{\mathit{Sh}}\nolimits (Y'_{\acute{e}tale})$1, and

  2. if $\mathcal{F}$ is an abelian sheaf, then $R^ if'_{small, *}(\mathcal{F}|_{X'}) = (R^ if_{small, *}\mathcal{F})|_{Y'}$.

Proof. Consider the following diagram of functors

\[ \xymatrix{ X'_{spaces, {\acute{e}tale}} \ar[r]_ j & X_{spaces, {\acute{e}tale}} \\ Y'_{spaces, {\acute{e}tale}} \ar[r]^ j \ar[u]^{V' \mapsto V' \times _{Y'} X'} & Y_{spaces, {\acute{e}tale}} \ar[u]_{V \mapsto V \times _ Y X} } \]

The horizontal arrows are localizations and the vertical arrows induce morphisms of sites. Hence the last statement of Sites, Lemma 7.28.1 gives (1). To see (2) apply (1) to an injective resolution of $\mathcal{F}$ and use that restriction is exact and preserves injectives (see Cohomology on Sites, Lemma 21.7.1). $\square$

[1] Also $(f')_{small}^{-1}(\mathcal{G}|_{Y'}) = (f_{small}^{-1}\mathcal{G})|_{X'}$ because of commutativity of the diagram and (65.18.11.1)

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Your email address will not be published.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 03LR. Beware of the difference between the letter 'O' and the digit '0'.