The Stacks project

Lemma 20.34.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $Z \subset X$ be a closed subset. Let $j : U \to X$ be the inclusion of an open subset with $U \cap Z = \emptyset $. Then $R\mathcal{H}_ Z(Rj_*K) = 0$ for all $K$ in $D(\mathcal{O}_ U)$.

Proof. Choose a K-injective complex $\mathcal{I}^\bullet $ of $\mathcal{O}_ U$-modules representing $K$. Then $j_*\mathcal{I}^\bullet $ represents $Rj_*K$. By Lemma 20.32.9 the complex $j_*\mathcal{I}^\bullet $ is a K-injective complex of $\mathcal{O}_ X$-modules. Hence $\mathcal{H}_ Z(j_*\mathcal{I}^\bullet )$ represents $R\mathcal{H}_ Z(Rj_*K)$. Thus it suffices to show that $\mathcal{H}_ Z(j_*\mathcal{G}) = 0$ for any abelian sheaf $\mathcal{G}$ on $U$. Thus we have to show that a section $s$ of $j_*\mathcal{G}$ over some open $W$ which is supported on $W \cap Z$ is zero. The support condition means that $s|_{W \setminus W \cap Z} = 0$. Since $j_*\mathcal{G}(W) = \mathcal{G}(U \cap W) = j_*\mathcal{G}(W \setminus W \cap Z)$ this implies that $s$ is zero as desired. $\square$

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Your email address will not be published.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 0G73. Beware of the difference between the letter 'O' and the digit '0'.