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.


  • 3
    $\begingroup$ Yes. This equivalence is true. It is exercise II.1.2(b) in Hartshorne. $\endgroup$ – Jared Apr 28 '13 at 3:10
  • $\begingroup$ Oh, you're right. Hartshorne does have this statement! Somehow, I skipped his statement. Thanks! $\endgroup$ – mr.bigproblem Apr 28 '13 at 4:04

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.


Note that surjectivity on stalks implies that $f$ is an epimorphism. The cokernel is then just the zero sheaf. The kernel of the cokernel (aka the image) is then clearly $\mathcal{G}$. So yes, $(2) \implies (1)$ is true.

  • $\begingroup$ I'm not sure if I understand your logic, or your expression. Your first sentence means $(2) \Rightarrow (1)$. Your second sentence is a direct consequence, which doesn't contain anything deep. Your third sentence is just restating the definition of an epimorphism, so this is equivalent to your first sentence. And the word "So" in your fourth sentence suggests to me that you are making a somewhat "merry-go-round" argument? I take your answer here, forgetting all the "round and round" logic, as a confirmation for the fact. Thanks! What you said in the fifth sentence is certainly true. $\endgroup$ – mr.bigproblem Apr 28 '13 at 4:01
  • $\begingroup$ I am not sure what you are referring to as round and round logic, but I guess it really depends on how you define things (in your case you seem to define an epimorphism as the image sheaf being $\mathcal{G}$ which is not really how I have seen epimorphisms being defined). Also, I did not claim that what I am saying is deep, because $(2) \implies (1)$ is not particularly difficult to prove. $\endgroup$ – Rankeya Apr 28 '13 at 4:10
  • $\begingroup$ What is your definition of an epimorphism of sheaves? I should correct my saying by excluding the word "deep". What I meant there is your second sentence follows immediately from your first sentence, in the way I understood it. $\endgroup$ – mr.bigproblem Apr 28 '13 at 4:14
  • $\begingroup$ It's how epimorphisms are defined in a category. $f$ is an epimorphism if for any morphisms $\eta_1, \eta_2: \mathcal{G} \rightarrow \mathcal{H}$ such that $\eta_1 \circ f = \eta_2 \circ f$, $\eta_1 = \eta_2$. $\endgroup$ – Rankeya Apr 28 '13 at 4:17
  • $\begingroup$ @mr.bigproblem: Dear mr.bigproblem, Rankeya is using the fact that two morphisms of sheaves coincide if they coincide on stalks. Using this, the property of being an epimorphism can be checked on stalks, and so one sees that being an epimorphism follows from surjectivity on stalks. This gives (2) $\implies$ (1). (I am just repeating Rankeya's argument, but in doing so, am hoping to convince you that the argument is not circular.) Regards, $\endgroup$ – Matt E Apr 28 '13 at 4:44

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