Is the category of sheaves of objects from an abelian category abelian? Suppose we have an abelian category $\mathcal{A}$ and a topological space $X$. Let $\mathscr{F}^{\mathcal{A}}$ be the category of all sheaves of $\mathcal{A}$ objects over $X$. Is $\mathscr{F}^{\mathcal{A}}$ guaranteed to be an abelian category? I don't really have any evidence that this should be true other than that it happens to be true when $\mathcal{A}$ is Ab or $R$-Mod, and those are the principle abelian categories.
Of course, by the Freyd–Mitchell embedding theorem $\mathcal{A}$ is always equivalent to a full subcategory of $R$-Mod, but I'm not familiar enough with sheaves or the embedding theorem itself to know if the "subcategory" factor could play a problem here.
In the event that the answer to the posed question is "no", is there some extended criterion for $\mathscr{F}^{\mathcal{A}}$ to be abelian?
 A: In the first place, it is unclear to me what a sheaf of $\mathcal{A}$-objects on $X$ should be. The obvious candidate, if you are used to the functor definition of sheaf, is that a sheaf of $\mathcal{A}$-objects on $X$ is a functor $\textrm{Open} (X)^\textrm{op} \to \mathcal{A}$ that satisfies the usual sheaf condition. I think this is the wrong definition and will fail to give an abelian category in general – at least, it is not obvious to me where cokernels are going to come from. On the other hand, there is another candidate, more in line with the original definition of sheaf in terms of stalks, and this one yields an abelian category. In more detail:
Let $\mathcal{A}$ be an abelian category and suppose $\mathcal{A}$ is an exact full subcategory of an abelian category $\mathcal{B}$ that satisfies Grothendieck's axioms AB3, AB3*, and AB5. (For example, if $\mathcal{A}$ is a Grothendieck abelian category we may take $\mathcal{B} = \mathcal{A}$; or if $\mathcal{A}$ is an essentially small abelian category we may take $\mathcal{B} = \textbf{Ind} (\mathcal{A})$.)
Definition. A sheaf of $\mathcal{A}$-objects on $X$ is a functor $F : \textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ with the following properties:

*

*$F : \textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ satisfies the sheaf condition.

*For all $x \in X$, the stalk $F_x$ (computed in $\mathcal{B}$) is (isomorphic to) an object in $\mathcal{A}$.

Remark. It does not follow that $F (U)$ is (isomorphic to) an object in $\mathcal{A}$! For example, if $X$ is an infinite discrete set and $\mathcal{A}$ is the category of finite abelian groups (and $\mathcal{B}$ is the category of all abelian groups), then $F (X)$ can be infinite even if every $F_x$ is finite.
Proposition. The category of sheaves of $\mathcal{A}$-objects on $X$ is an abelian category.
Proof. The category of sheaves of $\mathcal{B}$-objects on $X$ is a full subcategory of the category of all functors $\textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ and has an exact reflector. The category of all functors $\textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ is an abelian category, so it follows that the category of sheaves of $\mathcal{B}$-objects on $X$ is also an abelian category. Taking stalks is also exact, so the category of sheaves of $\mathcal{A}$-objects is an exact full subcategory of the category of sheaves of $\mathcal{B}$-objects, hence is also an abelian category. ■
What is not clear to me is whether the category of sheaves of $\mathcal{A}$-objects on $X$ defined in this way is independent of the choice of embedding $\mathcal{A} \hookrightarrow \mathcal{B}$. If it is then I could be convinced this is the correct definition.
A: It seems that the answer to the posed question is "yes". 
It is noted, but without proof (there might be no difficulty with a direct proof without using the Freyd-Mitchell embedding theorem or whatever non-trivial result), in Weibel An introduction to homological algebra (1.6 "More on abelian categories").
The category Sheaves(X) of sheaves on X with values in any abelian category is an abelian category, however it is not an abelian subcategory of Presheaves(X). 
