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. $\square$

There are also:

• 4 comment(s) on Section 6.17: Sheafification

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).