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For given a morphism of sheaves, in general, I know that the presheaf of image(or the presheaf of cokernel) is not a sheaf.

is there when the the presheaf of image(or the presheaf of cokernel) of morphism of sheaves is a sheaf?

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    $\begingroup$ It deserves to be mentioned that the presheaf image of an injective morphism of sheaves is automatically a sheaf. $\endgroup$
    – Tomo
    Oct 28, 2015 at 22:10

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Short version: The presheaf cokernel of a morphism of sheaves $\mathcal{F'} \rightarrow \mathcal{F}$ is a sheaf if $F'$ is acyclic. A common type of acyclic sheaf are flasque/flabby sheaves, sheaves where all restriction maps are surjective.

Assume we have an exact sequence $0 \rightarrow\mathcal{F'} \rightarrow \mathcal{F} \rightarrow \mathcal{F''} \rightarrow 0.$ Consider the forgetful functor from the category of sheaves to the category of presheaves. This functor is left exact, so we now have a left-exact sequence $0 \rightarrow F' \rightarrow F \rightarrow F''.$ ($F$ is the presheaf corresponding to $\mathcal{F'}$; I'm not aware of the convention regarding this, so this is probably not good notation.)

The derived functor of the forgetful functor gives us a long exact sequence $0\rightarrow F' \rightarrow F \rightarrow F'' \rightarrow H^1F' \rightarrow H^1F\rightarrow H^1F''\rightarrow \cdots$ (once again, I do not know the standard notation for what I call $H^1F$; I'm more used to working with the derived functor of the global sections functor than the forgetful functor.)

So the left-exact sequence of presheaves will be exact (in which case $F''$, a sheaf, corresponds to the cokernel) if $H^1F'$ is $0$. Then the sheaf $F'$ is called acyclic; an important class of acyclic sheaves are flasque (also called flabby) sheaves, which are sheaves where all restriction maps are surjective.

Image is just kernel of cokernel, so if the cokernel is a sheaf, so is the image.

EDIT: Sorry, some of my if and only ifs were incorrect. They should be fixed now.

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    $\begingroup$ Let $i$ denote the inclusion functor from sheaves into presheaves. Let $\mathcal{H}^p(F)$ denote the presheaf $U \mapsto H^p(U, F)$ (for $F$ a sheaf). Here $H^p(U, .)$ is the derived functor of $\Gamma(U, .)$. I claim that $R^p i = \mathcal{H}^p$. Indeed, it is a cohomological delta-functor (pretty much because $H^p$ is), and it is effaceable (because the restriction of flasque sheaves to open subsets is still flasque). $\endgroup$ Oct 9, 2012 at 20:54
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    $\begingroup$ Actually, I think a sheaf is called acyclic if $H^p(F) = 0$ for all $p > 0$, where here $H^p$ denotes derived functor of global sections. So my above argument shows that we need the restriction of $F$ to any open subset to be acyclic. This is true for example for flasque sheaves. If I recall correctly injective implies flasque, so it is true for injective sheaves as well. $\endgroup$ Oct 9, 2012 at 20:56
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    $\begingroup$ In fact, thinking about it, we can reach your conclusion without talking about the derived functor of $i$. Just note that the presheaf cokernel of $0 \to F' \to F$ is equal to the sheaf cokernel iff $0 \to F'(U) \to F(U) \to F''(0) \to 0$ is exact for all open subsets $U$ (because the presheaf cokernel is just $U \mapsto F(U)/F'(U)$). This is the case iff $H^1(U, F') = 0$ for all $U$. $\endgroup$ Oct 9, 2012 at 21:01

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