Lemma 34.3.12. The category of sheaves on $S_{Zar}$ is equivalent to the category of sheaves on the underlying topological space of $S$.

Proof. We will use repeatedly that for any object $U/S$ of $S_{Zar}$ the morphism $U \to S$ is an isomorphism onto an open subscheme. Let $\mathcal{F}$ be a sheaf on $S$. Then we define a sheaf on $S_{Zar}$ by the rule $\mathcal{F}'(U/S) = \mathcal{F}(\mathop{\mathrm{Im}}(U \to S))$. For the converse, we choose for every open subscheme $U \subset S$ an object $U'/S \in \mathop{\mathrm{Ob}}\nolimits (S_{Zar})$ with $\mathop{\mathrm{Im}}(U' \to S) = U$ (here you have to use Sets, Lemma 3.9.9). Given a sheaf $\mathcal{G}$ on $S_{Zar}$ we define a sheaf on $S$ by setting $\mathcal{G}'(U) = \mathcal{G}(U'/S)$. To see that $\mathcal{G}'$ is a sheaf we use that for any open covering $U = \bigcup _{i \in I} U_ i$ the covering $\{ U_ i \to U\} _{i \in I}$ is combinatorially equivalent to a covering $\{ U_ j' \to U'\} _{j \in J}$ in $S_{Zar}$ by Sets, Lemma 3.11.1, and we use Sites, Lemma 7.8.4. Details omitted. $\square$

Comment #3068 by Herman Rohrbach on

I think there is a small typo in the definition of $\mathcal{G}'$: $\mathcal{G}(U) = \mathcal{G}(U'/S)$ should be $\mathcal{G}'(U) = \mathcal{G}(U'/S)$.

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