Sheaf morphisms are determined by stalks

In "THE RISING SEA: Foundations of Algebraic Geometry" Exercise 2.4.E says the following:

A morphism of sheaves of sets $\phi: \mathcal{F} \to \mathcal{G}$ is an isomorphism iff. $\forall p \in X$ the induced morphism on the stalk $\psi_p : \mathcal{F}_p \to \mathcal{G}_p$ is an isomorphism.

Could somebody show me a proof of this? I'm stuck showing surjectivity of $\phi_U : \mathcal{F}(U) \to \mathcal{G}(U)$.

• Maybe you're not making this mistake, but a warning: to say that a morphism of sheaves is surjective does not mean that $\phi_U: \mathcal{F}(U) \to \mathcal{G}(U)$ for all $U$. This is a strictly stronger condition. – André 3000 Jun 22 '17 at 16:00
• Thank you very much I really made that mistake. Although it's still true in this case. – lush Jun 25 '17 at 9:03

Surjectivity is interesting because you are using injectivity in the proof ! Indeed the equivalence "for all $U$ open, $\psi_U : F(U) \to G(U)$ is surjective iff for all $p \in X$, $\psi_p : F_p \to G_p$ is surjective" is wrong, this is more or less what measure cohomology. On the other hand it's true if "surjective" is replaced by "injective".
Let's go for the proof of the surjectivity of $\psi_U$ : if $s \in G(U)$, we want to find $t \in F(U)$ with $\psi_U(t) = s$. For all $p$, by hypothesis there are some $t_p$ with $\psi_p(t_p) = s_p$, and this equality means by definition that this is also true on a little neighborhood, i.e we have a covering of $U$ by open $U_i$ and section $s_i \in G(U_i), t_i \in F(U_i)$ with $\phi_{U_i}(t_i) = s_i$. By injectivity of $\psi_{U_i}$ for all $i$, these sections agree on the various intersection $U_i \cap U_j$ so they glue together to a unique section $t$ and by construction $\psi_U(t) = s$.
Edit : In fact, a good way of seeing what is going on is to consider the morphism of sheaf $\exp : \mathcal O \to \mathcal O^*$ (where $X = \Bbb C, \mathcal O$ is the sheaf of holomorphic functions, and $\mathcal O^*$ the sheaf of nowhere zero holomorphic functions). Is this morphism surjective ? Is this true that for any open $\exp_U$ is surjective ? Why can't we apply the same proof ?