The Stacks project

Lemma 81.10.1. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $Z \subset X$ closed subspace such that $f^{-1}Z \to Z$ is integral and universally injective. Let $\overline{y}$ be a geometric point of $Y$ and $\overline{x} = f(\overline{y})$. We have

\[ (Rf_*Q)_{\overline{x}} = Q_{\overline{y}} \]

in $D(\textit{Ab})$ for any object $Q$ of $D(Y_{\acute{e}tale})$ supported on $|f^{-1}Z|$.

Proof. Consider the commutative diagram of algebraic spaces

\[ \xymatrix{ f^{-1}Z \ar[r]_{i'} \ar[d]_{f'} & Y \ar[d]_ f \\ Z \ar[r]^ i & X } \]

By Cohomology of Spaces, Lemma 69.9.4 we can write $Q = Ri'_*K'$ for some object $K'$ of $D(f^{-1}Z_{\acute{e}tale})$. By Morphisms of Spaces, Lemma 67.53.7 we have $K' = (f')^{-1}K$ with $K = Rf'_*K'$. Then we have $Rf_*Q = Rf_*Ri'_*K' = Ri_*Rf'_*K' = Ri_*K$. Let $\overline{z}$ be the geometric point of $Z$ corresponding to $\overline{x}$ and let $\overline{z}'$ be the geometric point of $f^{-1}Z$ corresponding to $\overline{y}$. We obtain the result of the lemma as follows

\[ Q_{\overline{y}} = (Ri'_*K')_{\overline{y}} = K'_{\overline{z}'} = (f')^{-1}K_{\overline{z}'} = K_{\overline{z}} = Ri_*K_{\overline{x}} = Rf_*Q_{\overline{x}} \]

The middle equality holds because of the description of the stalk of a pullback given in Properties of Spaces, Lemma 66.19.9. $\square$


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Your email address will not be published.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