Consider sheaves of sets on a topological space $X$. A standard fact (and exercise) is the following equivalence for a morphism $\phi\colon \mathscr{F}\rightarrow \mathscr{G}$ of such sheaves:

(a) $\phi$ is an epimorphism in this category of sheaves

(b) The induced morphisms $\phi_x$ on stalks are surjective for all $x\in X$

I found it surprisingly hard to come up with an idea for a proof of (a)=>(b), though I think I managed to do it using an appropriate skyscraper sheaf and thus showing that the $\phi_x$ are epimorphisms too.

Since I found shorter proofs for the analogous statements for monomorphisms and injectivity, I was wondering if there was some elegant and at the same time 'elementary' way to do it. That is, I'd like to see where the surjectivity comes from.

I know there are short proofs using the fact that colimits commute with left adjoints, but I don't want to use that. Examples for what I would like are usage of the product of all stalks, the fact that morphisms are equal if they agree on all stalks or something alike. I didn't have any idea up to now, though.

Thanks for any insight.

TL;DR: Is there an elegant proof for the equivalence not using adjoints?

  • 3
    $\begingroup$ Using adjoints is the elegant proof, in my view! The fact that the family of all stalk functors is jointly conservative plus the fact that taking stalks preserves colimits gives this result very easily. I doubt there is an elementary proof, because being an epimorphism is a colimit condition, and as you know, colimits in the category of sheaves are a bit difficult to understand. $\endgroup$
    – Zhen Lin
    Oct 2, 2012 at 6:58
  • $\begingroup$ @ZhenLin: Thanks for your comment. I guess the general nonsense will have to do it for an elegant solution. It is not so difficult as I thought at first. If I want to have it more explicit, the longer solutions will still make visible where the surjectivity comes from. $\endgroup$ Oct 19, 2012 at 20:32
  • $\begingroup$ @ZhenLin: why do you need that the family of all stalk functors is conservative? Just apply the stalk functor to the exact sequence $\mathcal F \rightarrow \mathcal G \rightarrow 0$ to get $\mathcal F_p \rightarrow \mathcal G_p \rightarrow 0_p = 0$ $\endgroup$
    – Rodrigo
    Oct 26, 2013 at 17:34
  • $\begingroup$ @Rodrigo Conservativity is needed for (b) => (a). $\endgroup$
    – Zhen Lin
    Oct 26, 2013 at 17:37
  • $\begingroup$ @ZhenLin: being conservative means that an induced iso on the stalks is induced by an iso $\mathcal F \xrightarrow{\sim} \mathcal G$. Is there a reason why showing this result is easier than showing surjectivity, as in (b) => (a)? $\endgroup$
    – Rodrigo
    Oct 26, 2013 at 17:39

2 Answers 2


Eric's answer is very nice, but addresses the case when $\mathcal{F}$ and $\mathcal{G}$ are sheaves of abelian groups, whereas the original question asks about sheaves of sets. I thought I might show how to modify things in this case. The idea is quite similar; one again wants to use a skyscraper sheaf (as noted in the OP) to make an ''indicator function''-like argument. I'll use the notation of Vakil's ''Foundations of Algebraic Geometry''.

Let $\phi \colon \cal{F} \to \cal{G}$ be our epimorphism of sheaves. Define $\rho \colon \cal{G} \to i_{p, \ast}\{0, 1\}$ by

$$\rho(U)(g) = \begin{cases} 0 \text{ if } (g, U) \in Im(\phi_{p}) \\ 1 \text{ otherwise } \end{cases}$$

for open sets $U$ containing $p$, and trivial maps otherwise. It is straightforward to see that $\rho$ is a morphism of sheaves. Now define $\alpha \colon \cal{G} \to i_{p, \ast}\{0, 1\}$ by

$$\alpha(U)(g) = 0$$

for open sets $U$ containing $p$, and trivial maps otherwise. Likewise, it is obvious that $\alpha$ is a morphism of sheaves. Furthermore, it is clear that $\rho \circ \phi = \alpha \circ \phi$, so $\rho = \alpha$, since $\phi$ is an epimorphism. Surjectivity of the stalk maps is is now easily deduced.


Let $\mathcal{F}$ and $\mathcal{G}$ be sheaves on a space $X$, and let $\mathcal{F} \overset{\Phi}{\to} \mathcal{G}$ be an epimorphism of sheaves.

Claim: $\Phi_x$ is surjective on each stalk. (Note that the converse is easily seen to hold.)

Proof: Let $\mathcal{H}$ be the skyscraper sheaf at $x$ with value $\mathcal{G}_x/\operatorname{Im}\Phi_x$. Define a map $\mathcal{G} \overset{\Psi}{\to} \mathcal{H}$ by sending $f \mapsto \bar{f_x}$, if $f$ is a section over a set containing $x$. This is clearly a homomorphism of abelian groups. Now define two maps $\mathcal{G} \to \mathcal{H} \oplus \mathcal{H}$ by $(\Psi,0)$ and $(0,\Psi)$. If $\Phi_x$ is not surjective, then these maps differ, but they are the same when precomposed with $\Psi$, so we have established the contrapositive.

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    $\begingroup$ Dear Eric: nice argument. I think one could rephrase this also by defining $\Gamma \colon \cal{G} \to \cal{H}$ to be trivial over all open sets, and observing that the compositions $\Gamma \circ \Phi$ and $\Psi \circ \Phi$ agree. $\endgroup$ Nov 14, 2016 at 5:41

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