Lemma 17.19.1. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$. Let $\mathcal{F}$ be a sheaf of sets on $X$. There exists a set $I$ and for each $i \in I$ an element $U_ i \in \mathcal{B}$ and a finite set $S_ i$ such that there exists a surjection $\coprod _{i \in I} j_{U_ i!}\underline{S_ i} \to \mathcal{F}$.

Proof. Let $S$ be a singleton set. We will prove the result with $S_ i = S$. For every $x \in X$ and element $s \in \mathcal{F}_ x$ we can choose a $U(x, s) \in \mathcal{B}$ and $s(x, s) \in \mathcal{F}(U(x, s))$ which maps to $s$ in $\mathcal{F}_ x$. By Sheaves, Lemma 6.31.4 the section $s(x, s)$ corresponds to a map of sheaves $j_{U(x, s)!}\underline{S} \to \mathcal{F}$. Then

$\coprod \nolimits _{(x, s)} j_{U(x, s)!}\underline{S} \to \mathcal{F}$

is surjective on stalks and hence surjective. $\square$

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