Sheaf epimorphisms

Question

$\newcommand{\fs}{\mathscr{F}}\newcommand{\gs}{\mathscr{G}}$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.)

Answer 1

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 $H^1(\Gamma(X, \Omega^*_X)) \cong H^1(X, \mathbb{R}) \cong \mathbb{R}$!