Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

Here are a few comments:

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)?

share|improve this question
    
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

1 Answer 1

up vote 3 down vote accepted

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 $\beta = \gamma$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.