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?
Algebraic Geometry – When is the Presheaf of Image of Morphism of Sheaves a Sheaf?
algebraic-geometry
Best Answer
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.