Processing math: 100%

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)


Your email address will not be published.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.