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Let $F$ and $G$ be sheaves of groups $\mathcal{S}^{op}\to Groups$ and $f:F\to G$ an epimorphism (of sheaves of sets).

If $F$ is a sheaf of commutative groups, is $G$ also a sheaf of commutative groups?

If $f$ would be a surjection on every section then for $x,y\in G(V)$ there would be $x', y'\in F(V)$ with $f(V)(x')=x$ and $f(V)(y')=y$ and $$ xy=f(V)(x')f(V)(y')=f(V)(x'y')=f(V)(y'x')=f(V)(y')f(V)(x')=yx. $$ but $f$ is only surjective on stalks.

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up vote 2 down vote accepted

Yes. You may find the following fact helpful: a sheaf of groups on a topological space is a sheaf of abelian groups if and only if all its stalks are abelian groups.

The claim is also true for a general site, but then it makes no sense to talk about stalks. Instead the best approach is to treat the question internally in the topos of sheaves of sets and observe that being an abelian group is a condition that can be expressed in terms of the commutativity of certain diagrams of sheaves of sets; thus, the defining property of epimorphisms is enough to imply that quotients of sheaves of groups are sheaves of abelian groups. This amounts to internalising the usual proof that any group quotient of an abelian group is again an abelian group. (I reiterate my view that sheaves of algebraic structures are best thought of as internal models in a topos rather than sheaves taking values in some category.)

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