Let $\phi: \mathscr{F} \to \mathscr{G}$ be a morphism of sheaves. Recall that $\phi$ is an epimorphism if given $\psi_1, \psi_2: \mathscr{G} \to \mathscr{H}$, then $\psi_1 \circ \phi = \psi_2 \circ \phi$ implies $\psi_1 = \psi_2$.

One direction of this is clear: let $p \in X$ (assuming $\mathscr{F}, \mathscr{G},$ and $\mathscr{H}$ are sheaves on $X$); if $\phi_p: \mathscr{F}_p \to \mathscr{G}$ is surjective for all $p$, then if $\psi_1\circ\phi = \psi_2\circ\phi$, they induce the same map on stalks. $\phi_p$ is surjective on stalks, so is an epimorphism in $\mathsf{Set}$ or $\mathsf{Grp}$ or whatever category. Thus $\psi_{1,p} = \psi_{2,p}$. Maps that are equal on stalks are equal, so $\psi_1 = \psi_2$.

It is the other direction that I'm failing to see clearly. I understand that we must exhibit a sheaf $\mathscr{H}$, along with maps $\psi_1, \psi_2: \mathscr{G} \to \mathscr{H}$ such that the following conditions hold:

  • $\psi_1 \circ \phi = \psi_2 \circ \phi$, and
  • $\psi_1 = \psi_2$ implies $\phi_p$ is surjective.

Any hints into what the right sheaf or sheaf morphisms to consider are would be greatly appreciated. For reference, this comes from Exercise 2.4.O in Vakil's Algebraic Geometry notes.


1 Answer 1


One way to get out of this may be as follows: Let $\phi_1, \phi_2$ be such that $\phi_1\not= \phi_2, \phi_1\circ \phi=\phi_2\circ \phi$. Since $\phi_1\not=\phi_2$, there exist some point $x\in X$ such that the morphisms on the stalks are different. But by definition after compose it with $\phi_{x}$ we get the same map on the stalks. So we reached a contradiction.

The forward proof you claimed seems to be fishy. For a concise proof, see:

An 'easy' way to prove that epimorphism of sheaves implies surjectivity on stalks

  • $\begingroup$ What's fishy about the forward direction? Looks fine to me. $\endgroup$
    – user5826
    Nov 23, 2020 at 20:49

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