EDIT: The original version regarding coker is false.

Consider a morphism of sheaves of abelian groups $\Phi:\mathscr{F}\to\mathscr{G}$.

It is not in general true that Im$_{pre}\Phi$ is a sheaf. However, if $\Phi$ is a monomorphism, then $\Phi_U$ is a monomorphism on open sets and Im$_{pre}\Phi$ seems to be a sheaf. Consider the explicit construction of Im$_{pre}\Phi$ where $$\text{Im}_{pre} \Phi(U) := \text{Im}\; \Phi_U \subset \mathscr{G}(U)$$ since we are dealing with sheaves of abelian groups, this seems to be a sheaf.

Is this true? What if we are dealing with a monomorphism $\Phi:\mathscr{F}\to\mathscr{G}$ of sheaves of some unknown abelian category, is it still true that Im$_{pre} \Phi$ is a sheaf? And if so what is a purely abstract argument?

  • $\begingroup$ One observation, and really I wouldn't dignify it by calling it anything more, is that the presheaf image is already a sheaf in this case. That's useful. $\endgroup$ – Hoot Jan 25 '15 at 20:45
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    $\begingroup$ To explain: the presheaf image here is isomorphic as a presheaf to $\mathscr{F}$, so it's a sheaf. $\endgroup$ – Hoot Jan 25 '15 at 22:54

Perhaps I've misunderstood what you wrote, but it seems to me that your statement about the presheaf cokernel being a sheaf is false.

E.g. take $X := \mathbb C^{\times}$ with its usual topology, let $\mathscr G := \mathscr O_X$ denote the sheaf of holomorphic functions on $X$, and let $\mathscr F$ denote the sheaf of locally constant functions on $X$ which take values in $2\pi i \mathbb Z.$

Then $\mathscr F \hookrightarrow \mathscr G$ (any locally constant function is holomorphic).

Let $\mathscr O_X^{\times}$ denote the sheaf of nowhere-zero holomorphic functions on $X$ (which is a sheaf of abelian groups via multiplication).

There is a natural isomorphism $\mathscr G/\mathscr F \buildrel \sim \over\longrightarrow \mathscr O_X^{\times}$, given via the morphism $f \mapsto e^f.$

This map is not surjective on global sections: there is no global section of $\mathscr G$ which maps to the global section $z$ of $\mathscr O_X^{\times}.$ (Concretely, there is no single valued branch of $\log z$ defined on $X$.)

Thus $\mathscr G(U) \to (\mathscr G/\mathscr F)(U)$ is not surjective, and in particular, the natural injection $$\mathscr G(U)/\mathscr F(U) \to (\mathscr G/\mathscr F)(U)$$ is not surjective.

(This failure of surjectivity is the source of sheaf cohomology. In the particular case I've described here, the related cohomological fact is that the $1$st cohomology of $X$ with integral coefficients is non-zero.)

  • $\begingroup$ This answer is indeed perfectly correct, contra mez's assertion. $\endgroup$ – Georges Elencwajg Jan 25 '15 at 20:07
  • $\begingroup$ You are right, I was always thinking about image and naively just reduced to cokernel. What about current version regarding image? $\endgroup$ – mez Jan 25 '15 at 22:49
  • $\begingroup$ @mez: If $\Phi$ is a monomorphism then it induces an isomorphism of $\mathscr F$ onto its image, and so yes, the presheaf and sheaf images coincide. $\endgroup$ – tracing Jan 26 '15 at 0:48

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