The Stacks project

Proof. It is probably better for the reader to find their own explanation of this than to read the proof here. In fact the lemma is true for the same reason as why the presheaf of continuous function is a sheaf, see Example 6.7.3 (and this analogy can be made precise using the “espace étalé”).

Anyway, let $U = \bigcup U_ i$ be an open covering. Suppose that $s_ i = (s_{i, u})_{u \in U_ i} \in \mathcal{F}^{\# }(U_ i)$ such that $s_ i$ and $s_ j$ agree over $U_ i \cap U_ j$. Because $\Pi (\mathcal{F})$ is a sheaf, we find an element $s = (s_ u)_{u\in U}$ in $\prod _{u\in U} \mathcal{F}_ u$ restricting to $s_ i$ on $U_ i$. We have to check property $(*)$. Pick $u \in U$. Then $u \in U_ i$ for some $i$. Hence by $(*)$ for $s_ i$, there exists a $V$ open, $u \in V \subset U_ i$ and a $\sigma \in \mathcal{F}(V)$ such that $s_{i, v} = (V, \sigma )$ in $\mathcal{F}_ v$ for all $v \in V$. Since $s_{i, v} = s_ v$ we get $(*)$ for $s$. $\square$


Comments (0)

There are also:

  • 4 comment(s) on Section 6.17: Sheafification

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Your email address will not be published.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 007Y. Beware of the difference between the letter 'O' and the digit '0'.