The Stacks project

Lemma 20.11.10. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{I}$ be an injective $\mathcal{O}_ X$-module. Then

  1. $\check{H}^ p(\mathcal{V}, f_*\mathcal{I}) = 0$ for all $p > 0$ and any open covering $\mathcal{V} : V = \bigcup _{j \in J} V_ j$ of $Y$.

  2. $H^ p(V, f_*\mathcal{I}) = 0$ for all $p > 0$ and every open $V \subset Y$.

In other words, $f_*\mathcal{I}$ is right acyclic for $\Gamma (V, -)$ (see Derived Categories, Definition 13.15.3) for any $V \subset Y$ open.

Proof. Set $\mathcal{U} : f^{-1}(V) = \bigcup _{j \in J} f^{-1}(V_ j)$. It is an open covering of $X$ and

\[ \check{\mathcal{C}}^\bullet (\mathcal{V}, f_*\mathcal{I}) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}). \]

This is true because

\[ f_*\mathcal{I}(V_{j_0 \ldots j_ p}) = \mathcal{I}(f^{-1}(V_{j_0 \ldots j_ p})) = \mathcal{I}(f^{-1}(V_{j_0}) \cap \ldots \cap f^{-1}(V_{j_ p})) = \mathcal{I}(U_{j_0 \ldots j_ p}). \]

Thus the first statement of the lemma follows from Lemma 20.11.1. The second statement follows from the first and Lemma 20.11.8. $\square$

Comments (2)

Comment #2335 by Keenan Kidwell on

In the final sentence of the statement of the lemma, shouldn't for open be for open?

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Your email address will not be published.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 01EX. Beware of the difference between the letter 'O' and the digit '0'.