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$\newcommand{\fs}{\mathscr{F}}\newcommand{\gs}{\mathscr{G}}$Let ${\bf C}$ be some algebraic category (e.g. commutative rings), ${\bf Psh}(X,{\bf C})$ the category of presheaves on $X$ taking values in ${\bf C}$, and ${\bf Sh}(X,{\bf C})$ the category of sheaves on $X$ taking values in ${\bf C}$. Let $\phi:\fs\to\gs$ be an epimorphism in ${\bf Sh}(X,{\bf C})$. Is it possible that $\phi$ is not an epimorphism in ${\bf Psh}(X,{\bf C})$ ?

(Sheaf monomorphism seems to imply presheaf monomorphism, if one assumes that every presheaf can be sheafified. The proof doesn't extend to epimorphisms.)

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At least in the category of (pre)sheaves of sets, epimorphisms of presheaves are the naive ones, i.e., they are surjective on sections over every object. Epimorphisms of sheaves can be characterized as the morphisms which induce surjections on all stalks. It's definitely possible to have an epimorphism of sheaves which is not an epimorphism of the underlying presheaves, although it is clear that if the underlying morphism of presheaves is an epimorphism, then it is an epimorphism of sheaves. This is discussed in relative detail in the Stacks Project sections ons sheaves on spaces and (more – Keenan Kidwell Jan 20 '13 at 3:44
generally) sheaves on sites, where other target categories are discussed (and the notion of an algebraic structure is made formal). – Keenan Kidwell Jan 20 '13 at 3:45
I should mention that the characterization I give of epimorphisms of sheaves in terms of stalks only works for sheaves on topological spaces. For sites, the condition is: for every object $U$ and $s\in\mathscr{G}(U)$, there is a covering $(U_i\rightarrow U)$ such that $s\vert_{U_i}$ is in the image of $\varphi_{U_i}:\mathscr{F}(U_i)\rightarrow\mathscr{G}(U_i)$ for all $i$. – Keenan Kidwell Jan 20 '13 at 5:32
up vote 3 down vote accepted

No. If sheaf epimorphisms were always presheaf epimorphisms then there would be no point to sheaf cohomology!

Here's an example. Let $X$ be $\mathbb{R}^2$ minus a point, and let $\Omega^*_X$ be the de Rham complex. $X$ is two dimensional, so we get an exact sequence $$\Omega^0_X \longrightarrow \Omega^1_X \longrightarrow \Omega^2_X \longrightarrow 0$$ in $\textbf{Sh}(X, \textbf{Vect}_\mathbb{R})$, because we can always locally integrate a closed $(n + 1)$-form to get an $n$-form using the Poincaré lemma. Let $B^1 (\Omega_X)$ be the image of $\Omega^0_X \to \Omega^1_X$ in $\textbf{Sh}(X, \textbf{Vect}_\mathbb{R})$. As usual, $\Omega^0_X \to B^1 (\Omega_X)$ is an epimorphism, but after taking global sections, $\Gamma (X, \Omega^0_X) \to \Gamma (X, B^1 (\Omega_X))$ is not. Indeed, we find that the $1$-form $$\mathrm{d} \theta = \frac{-y \, \mathrm{d} x + x \, \mathrm{d} y}{x^2 + y^2}$$ is closed, but $\mathrm{d} \theta$ cannot be exact because $$\int_S \mathrm{d} \theta = 2 \pi$$ where $S$ is the unit circle in $X$. Thus $\Gamma (X, \Omega^0_X) \to \Gamma (X, B^1 (\Omega_X))$ is not surjective, which is just as well, because de Rham's theorem tells us that $H^1(\Gamma(X, \Omega^*_X)) \cong H^1(X, \mathbb{R}) \cong \mathbb{R}$!

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