Example of epimorphisms such that the product is not an epimorphism in the category of sheaves I've heard that in the category of sheaves over a topological space $X$, products of epimorphisms are not epimorphisms. I think that it's equivalent to saying that $\mathbf{Sh}(X)$ does not satisfy AB4. However I could not find an example.
 A: The product of finitely many epimorphisms is an epimorphism; the problem is infinite products. Let $X$ be a topological space with lots of open subsets, e.g. $[0, 1]$. Choose a sequence of open covers $\mathfrak{U}_0, \mathfrak{U}_1, \mathfrak{U}_2, \ldots$ of $X$ with the following properties:

*

*Each $\mathfrak{U}_{n+1}$ refines $\mathfrak{U}_n$.

*If $U$ is an open subset of $X$ such that $U$ is contained in some member of $\mathfrak{U}_n$ for all sufficiently large $n$, then $U = \emptyset$.

For example, if $X = [0, 1]$, we could take $\mathfrak{U}_n$ to be the collection of open balls of radius $2^{-n}$ centred on points in $2^{-n} \mathbb{Z}$.
Now, let $\mathscr{F}_n$ be the abelian sheaf freely generated by $\mathfrak{U}_n$. (This is the sheaf associated with the presheaf $U \mapsto \bigoplus_{U' \in \mathfrak{U}_n, U \subseteq U'} \mathbb{Z}$.) It is not hard to see that, for any given open $U \subseteq X$, we have $\Gamma (U, \mathscr{F}_n) = 0$ for all sufficiently large $n$. Nonetheless, there is an epimorphism $\mathscr{F}_n \to \mathbb{Z}$ defined by sending each generator to $1$. However, consider $\prod_{n \ge 0} \mathscr{F}_n \to \prod_{n \ge 0} \mathbb{Z}$. We have $\Gamma (U, \prod_{n \ge 0} \mathscr{F}_n) \cong \prod_{n \ge 0} \Gamma (U, \mathscr{F}_n)$ for all $U$, so $(1, 1, 1, \ldots)$ is not even locally in the image. Thus  $\prod_{n \ge 0} \mathscr{F}_n \to \prod_{n \ge 0} \mathbb{Z}$ is not an epimorphism.
That said, there are topological spaces where products of epimorphisms are still epimorphisms: for instance, the spectrum of a local Dedekind domain.
