$\mathbf{coim}(f) = \mathbf{im}(f)$ for sheave morphisms

Question

I'm trying to prove that the category of sheaves on the category of $R$-modules is an Abelian, which requires one last step: $\mathbf{coim}(f) = \mathbf{im}(f)$ for arbitrary morphism $f$. The only reference I found that uses this definition of Abelian categories is the stack project. However, I don't really understand the statement that "constructions of kernels and cokernels above these stalks are the coimage and image" of $R$-modules (probably because of my lack of experience on sheaves). Can anyone decompose the argument here? Thanks.

Answer 1

There is a canonical morphism $\operatorname{coim} f \to \operatorname{im} f$. You want to show it is an isomorphism. To show a sheaf morphism is an isomorphism it is sufficient to check it on stalks. Taking stalks preserves cokernels and kernels, so it also preserves coimages and images, so the stalk at $x$ of this morphism is also the canonical morphism $\operatorname{coim} f_x \to \operatorname{im} f_x$. But now you are in a category which you already know is abelian.