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
\[ \xymatrix{ \mathcal{F} \ar[r] \ar[d] & \mathcal{F}^\# \ar[r] \ar[d] & \Pi (\mathcal{F}) \ar[d] \\ \mathcal{G} \ar[r] & \mathcal{G}^\# \ar[r] & \Pi (\mathcal{G}) \\ } \]
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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