Given a germ, does there always exist a sequence of compatible germs restricting to it

This question comes from Vakil's Exercise 2.4.E (isomorphisms are determined by stalks) — my question has less to do with this fact, but instead a subtle nuance that I've run into chasing some of the diagrams. An isomorphism of sheaves $f :\mathscr{F} \to \mathscr{G}$ should mean we get an inverse $g : \mathscr{G} \to \mathscr{F}$ with $ g \circ f $ and $ f \circ g$ the identity morphism (this is what isomorphism means in every category).

By exercise 2.4.C and 2.4.A in Vakil, $\mathscr{F}(U)$ is isomorphic to its image under $\varrho$ (as sets) consisting of those sequences $ \prod_{p \in U} s_p $ with compatible germs. Now obviously $ \pi_p $ (the canonical projection) is surjective, but should it follow that when restricted to $ \varrho(\mathscr{F}(U)) $, the map is still surjective? In other words, shouldn't $f_p$ only be an isomorphism when restricted to germs with compatible stalks?

Solution 1:

No, there's no reason for this to be true. For instance, take $X=U=\Bbb P^1_k$, $\mathcal{F}=\mathcal{O}_X$, and $p=[0:1]$. Then $\mathcal{F}(U)\cong k$ while $\mathcal{F}_p\cong k[t]_{(t)}$ and the map $\pi_p\circ\varrho$ is $1\mapsto 1$ which is very far from surjective.

Instead, given an element $s$ in $\mathcal{F}_p$, there exists an open $U'$ so that $s$ is in the image of $res_{U',p}:\mathcal{F}(U')\to\mathcal{F}_p$. The previous counterexample shows that sometimes we have to pick a smaller $U$.