The Stacks project

Lemma 7.41.4. Let $f : \mathcal{D} \to \mathcal{C}$ be a morphism of sites given by the functor $u : \mathcal{C} \to \mathcal{D}$. Assume that for every object $V$ of $\mathcal{D}$ there exist objects $U_ i$ of $\mathcal{C}$ and morphisms $u(U_ i) \to V$ such that $\{ u(U_ i) \to V\} $ is a covering of $\mathcal{D}$. In this case the functor $f_* : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ reflects injections and surjections.

Proof. Let $\alpha : \mathcal{F} \to \mathcal{G}$ be maps of sheaves on $\mathcal{D}$. By assumption for every object $V$ of $\mathcal{D}$ we get $\mathcal{F}(V) \subset \prod \mathcal{F}(u(U_ i)) = \prod f_*\mathcal{F}(U_ i)$ by the sheaf condition for some $U_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and similarly for $\mathcal{G}$. Hence it is clear that if $f_*\alpha $ is injective, then $\alpha $ is injective. In other words $f_*$ reflects injections.

Suppose that $f_*\alpha $ is surjective. Then for $V, U_ i, u(U_ i) \to V$ as above and a section $s \in \mathcal{G}(V)$, there exist coverings $\{ U_{ij} \to U_ i\} $ such that $s|_{u(U_{ij})}$ is in the image of $\mathcal{F}(u(U_{ij}))$. Since $\{ u(U_{ij}) \to V\} $ is a covering (as $u$ is continuous and by the axioms of a site) we conclude that $s$ is locally in the image. Thus $\alpha $ is surjective. In other words $f_*$ reflects surjections. $\square$


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Your email address will not be published.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 04D9. Beware of the difference between the letter 'O' and the digit '0'.