The Stacks project

Theorem 7.10.10. With $\mathcal{F}$ as above

  1. The presheaf $\mathcal{F}^+$ is separated.

  2. If $\mathcal{F}$ is separated, then $\mathcal{F}^+$ is a sheaf and the map of presheaves $\mathcal{F} \to \mathcal{F}^+$ is injective.

  3. If $\mathcal{F}$ is a sheaf, then $\mathcal{F} \to \mathcal{F}^+$ is an isomorphism.

  4. The presheaf $\mathcal{F}^{++}$ is always a sheaf.

Proof. Proof of (1). Suppose that $s, s' \in \mathcal{F}^+(U)$ and suppose that there exists some covering $\{ U_ i \to U\} $ such that $s|_{U_ i} = s'|_{U_ i}$ for all $i$. We now have three coverings of $U$: the covering $\{ U_ i \to U\} $ above, a covering $\mathcal{U}$ for $s$ as in Lemma 7.10.8, and a similar covering $\mathcal{U}'$ for $s'$. By Lemma 7.10.5, we can find a common refinement, say $\{ W_ j \to U\} $. This means we have $s_ j, s'_ j \in \mathcal{F}(W_ j)$ such that $s|_{W_ j} = \theta (s_ j)$, similarly for $s'|_{W_ j}$, and such that $\theta (s_ j) = \theta (s'_ j)$. This last equality means that there exists some covering $\{ W_{jk} \to W_ j\} $ such that $s_ j|_{W_{jk}} = s'_ j|_{W_{jk}}$. Then since $\{ W_{jk} \to U\} $ is a covering we see that $s, s'$ map to the same element of $H^0(\{ W_{jk} \to U\} , \mathcal{F})$ as desired.

Proof of (2). It is clear that $\mathcal{F} \to \mathcal{F}^+$ is injective because all the maps $\mathcal{F}(U) \to H^0(\mathcal{U}, \mathcal{F})$ are injective. It is also clear that, if $\mathcal{U} \to \mathcal{U}'$ is a refinement, then $H^0(\mathcal{U}', \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F})$ is injective. Now, suppose that $\{ U_ i \to U\} $ is a covering, and let $(s_ i)$ be a family of elements of $\mathcal{F}^+(U_ i)$ satisfying the sheaf condition $s_ i|_{U_ i \times _ U U_ j} = s_ j|_{U_ i \times _ U U_ j}$ for all $i, j \in I$. Choose coverings (as in Lemma 7.10.8) $\{ U_{ij} \to U_ i\} $ such that $s_ i|_{U_{ij}}$ is the image of the (unique) element $s_{ij} \in \mathcal{F}(U_{ij})$. The sheaf condition implies that $s_{ij}$ and $s_{i'j'}$ agree over $U_{ij} \times _ U U_{i'j'}$ because it maps to $U_ i \times _ U U_{i'}$ and we have the equality there. Hence $(s_{ij}) \in H^0(\{ U_{ij} \to U\} , \mathcal{F})$ gives rise to an element $s \in \mathcal{F}^+(U)$. We leave it to the reader to verify that $s|_{U_ i} = s_ i$.

Proof of (3). This is immediate from the definitions because the sheaf property says exactly that every map $\mathcal{F} \to H^0(\mathcal{U}, \mathcal{F})$ is bijective (for every covering $\mathcal{U}$ of $U$).

Statement (4) is now obvious. $\square$

Comments (0)

There are also:

  • 6 comment(s) on Section 7.10: 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 00WB. Beware of the difference between the letter 'O' and the digit '0'.