Cokernel of a sheaf morphism not being a sheaf

Question

Does anyone know a good example where the cokernel of a sheaf morphism is not a sheaf?

Answer 1

Here's a simple example: consider shaves on the unit circle, which I will write as $\mathbf{R} / 1$. Let $F$ be the sheaf of sections of the bundle $\mathbf{R} \to \mathbf{R}/1$ given by the projection.

The cokernel of the "add 1" map is the sheaf $G$ of sections of the bundle $\mathbf{R}/1 \to \mathbf{R}/1$ given by the identity.

$F$ has no global sections but $G$ does: therefore $G$ is not the cokernel of the "add 1" map when computed as a map of presheaves.

EDIT: This is actually an example for the coequalizer of sheaves of sets, rather than a cokernel of abelian sheaves. (I'm talking about the coequalizer of the identity and the add 1 maps)

For another example in the same spirit, consider the abelian sheaves $F$ of continuously differentiable functions of period $1$ and $G$ of continuous functions of period $1$. What is the cokernel of the derivative map? On any proper open subset, $F(U) \to G(U)$ is surjective, so the cokernel sheaf is zero. However, the cokernel of $F(\mathbf{R}/1) \to G(\mathbf{R}/1)$ is $\mathbb{R}$, so the cokernel presheaf is nonzero.

The projection to the cokernel, incidentally, is

$$ f \mapsto \int_{0}^1 f(t) \, dt $$