Suppose that $f: \mathcal{F} \rightarrow \mathcal{G}$ is a morphism of sheaves on a topological space $X$. Consider the following statements.
1) $f$ is surjective, i.e. $\text{Im } f = \mathcal{G}$.
2) $f_{p}: \mathcal{F}_p \rightarrow \mathcal{G}_p$ is surjective for every $p\in X$.
$(1) \Rightarrow (2)$ is always true. I was wondering if $(2) \Rightarrow (1)$ is also true and I found Germ and sheaves problem of injectivity and surjectivity, which claims it positively. I double check all the details of the arguments made in that thread and found no mistake. I just want a confirmation that $(1) \Leftrightarrow (2)$ is right, so that we have a criterion for surjectivity of morphisms of sheaves.
In case you asked why this fact, which is already established in a previous thread, is repeated here, followings are my reasons:
a) I'm always skeptical, even with myself;
b) I have not seen this statement in popular texts. Maybe it's in EGA but I can't read French (if it is, would be nice if someone points it out please!);
c) In the proof for the fact that $f$ is isomorphic iff $f_p$ is isomorphic for all $p \in X$ (Prop 1.1, p.g. 63, Hartshorne's), to prove that $\mathcal{F}(U) \rightarrow \mathcal{G}(U)$ is surjective for all $U$ open, the proof requires injectivity of $f_p$ for all $p \in X$. This is not directly related to our situation but it provides one a caution that surjectivity of $f$ is a little bit subtle.
Thanks!
This can be found in every complete introduction to sheaves or algebraic geometry and comes down to the fact that the functor $F \mapsto (F_x)_{x \in X}$ is faithful and exact.