Lemma 6.17.3. Let $\mathcal{F}$ be a presheaf of sets on $X$. Any map $\mathcal{F} \to \mathcal{G}$ into a sheaf of sets factors uniquely as $\mathcal{F} \to \mathcal{F}^\# \to \mathcal{G}$.

**Proof.**
Clearly, there is a commutative diagram

So it suffices to prove that $\mathcal{G} = \mathcal{G}^\# $. To see this it suffices to prove, for every point $x \in X$ the map $\mathcal{G}_ x \to \mathcal{G}^\# _ x$ is bijective, by Lemma 6.16.1. And this is Lemma 6.17.2 above. $\square$

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