# Epimorphisms of sheaves of sets

Let $X$ be a topological space, and $F$ and $G$ be two sheaves of sets on $X$. Let $\eta : F \rightarrow G$ be a morphism of sheaves.

Then how would you show the following:

$\eta$ is an epimorphism in the category of sheaves of sets on $X$ if and only if for all $x \in X$, the induced maps on stalks $\eta_{x}: F_x \rightarrow G_x$ are surjective. (1)

In Algebraic Geometry Chapter 2, Proposition 1.1, Hartshorne proves that in the setup as above $\eta$ is an isomorphism if and only if for all $x \in X$ the induced maps on stalks $\eta_x : F_x \rightarrow G_x$ are bijective (he considers sheaves of abelian groups, so in his statement $\eta_x$ are isomorphisms, but the proof remains valid if we consider sheaves of sets). But, in this proof he relies on the following nice fact: $\eta$ is an isomorphism if and only if for all open $U \subset X$, the component maps $\eta(U) : F(U) \rightarrow G(U)$ are bijective.

I have never really seen a proof of the following analogous fact (which could be helpful in solving the problem): $\eta$ is an epimorphism if and only if for all open $U \subset X$, $\eta(U): F(U) \rightarrow G(U)$ is surjective (2).

((I think (2) is true, but hopefully there is a solution to my problem without using this fact.))

Even if we assume (2) to be a true statement, then to prove the backward implication of (1) we still cannot extend Hartshorne's method, for when he shows that for all open $U \subset X$, $\eta(U): F(U) \rightarrow G(U)$ is surjective, he needs the assumption that $\eta(U)$ are injective, which he proves beforehand.

With these comments, which probably show that I am thinking about the problem incorrectly, can someone give me an indication of how to prove (1)?

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I apologize that this question might actually require one to look at the proof of Proposition 1.1 in Hartshorne... – Rankeya Nov 20 '11 at 7:32
Your claim (2) is false. See this question. – Zhen Lin Nov 20 '11 at 8:39
I have to get this point clarified. Isn't a morphism of sheaves of sets a natural transformation in the categorical sense? Since, the category $Sets$ of sets has arbitrary products, then according to the answer provided to this question: math.stackexchange.com/questions/82568/…, any morphism of sheaves of sets that is epic, should in fact have epic (surjective) components, so that claim (2) seems to be true. What am I arguing wrong? – Rankeya Nov 20 '11 at 12:05
If it's the full category of presheaves, yes. But otherwise, no. More generally, subcategories may have different monomorphisms and epimorphisms compared to the original category. – Zhen Lin Nov 20 '11 at 15:13
You are right. I was being silly in regarding morphism of sheaves as being the same as morphism of presheaves. – Rankeya Nov 20 '11 at 15:29

Let $X$ be a topological space, and let $\alpha : \mathscr{F} \to \mathscr{G}$ be a morphism of sheaves on $X$. Let $j : \{ x \} \hookrightarrow X$ be the inclusion of a point into $X$. Then, by definition, the stalk of $\mathscr{F}$ at $x$ is just the inverse image sheaf $j^* \mathscr{F}$. But we know $j^* \dashv j_*$ (i.e. the inverse image functor $j^*$ is left adjoint to the direct image functor $j_*$), and left adjoints preserve colimits, hence if $\alpha$ is epic, so is $\alpha_x : \mathscr{F}_x \to \mathscr{G}_x$.
Conversely, suppose $\alpha_x : \mathscr{F}_x \to \mathscr{G}_x$ is surjective for every $x$ in $X$. Let $\beta, \gamma : \mathscr{G} \to \mathscr{H}$ be two further morphisms of sheaves, and suppose $\beta \circ \alpha = \gamma \circ \alpha$. We need to show $\beta = \gamma$. Certainly, we have $\beta_x \circ \alpha_x = \gamma_x \circ \alpha_x$, and $\alpha_x$ is epic by hypothesis, so we have $\beta_x = \gamma_x$ for all $x$ in $X$. Let $s \in \mathscr{G}(U)$, and consider $\beta_U(s)$ and $\gamma_U(s)$. Since $\beta$ and $\gamma$ agree on germs, at each $x$ in $U$ there is an open neighbourhood $V$ on which $\beta_V(s|_V) = \gamma_V(s|_V)$, and so by the unique collation property, we must have $\beta_U(s) = \gamma_U(s)$, and therefore we indeed have $\beta = \gamma$.