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. $\square$

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