The Stacks project

Lemma 20.11.7. Let $X$ be a ringed space. Let

\[ 0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0 \]

be a short exact sequence of $\mathcal{O}_ X$-modules. Let $U \subset X$ be an open subset. If there exists a cofinal system of open coverings $\mathcal{U}$ of $U$ such that $\check{H}^1(\mathcal{U}, \mathcal{F}) = 0$, then the map $\mathcal{G}(U) \to \mathcal{H}(U)$ is surjective.

Proof. Take an element $s \in \mathcal{H}(U)$. Choose an open covering $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ such that (a) $\check{H}^1(\mathcal{U}, \mathcal{F}) = 0$ and (b) $s|_{U_ i}$ is the image of a section $s_ i \in \mathcal{G}(U_ i)$. Since we can certainly find $\mathcal{U}$ such that (b) holds it follows from the assumptions of the lemma that we can find $\mathcal{U}$ such that (a) and (b) both hold. Consider the sections

\[ s_{i_0i_1} = s_{i_1}|_{U_{i_0i_1}} - s_{i_0}|_{U_{i_0i_1}}. \]

Since $s_ i$ lifts $s$ we see that $s_{i_0i_1} \in \mathcal{F}(U_{i_0i_1})$. By the vanishing of $\check{H}^1(\mathcal{U}, \mathcal{F})$ we can find sections $t_ i \in \mathcal{F}(U_ i)$ such that

\[ s_{i_0i_1} = t_{i_1}|_{U_{i_0i_1}} - t_{i_0}|_{U_{i_0i_1}}. \]

Then clearly the sections $s_ i - t_ i$ satisfy the sheaf condition and glue to a section of $\mathcal{G}$ over $U$ which maps to $s$. Hence we win. $\square$


Comments (3)

Comment #8354 by Et on

Can you not deduce this immediately from the exact sequence of Cech cohomology and the fact that the 0'th cohomology is just the global sections for sheaves?

Comment #8355 by Et on

Edit: I see now... it is an exact sequence of sheaves and the cech functor is only a delta functor on the pre-sheaf category... Perhap it's worth starting the proof with "note that lemma 20.10.2 does not apply here" to avoid confusion for future readers.

Comment #8961 by on

This is a reasonable concern to have, but since the proof is fine I am going to leave this as is.


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Your email address will not be published.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