The category $\operatorname{Fun}(\sf{C},\sf{A})$ is abelian is $\sf{A}$ is abelian and $\sf{C}$ is small. Why the set-theoretic condition? Proposition 1.4.4 in F. Borceux Handbook of Categorical Algebra II says that if $\sf{A}$ is an abelian category and $\sf{C}$ is a small category, then the category of functors $\operatorname{Fun}(\sf{C},\sf{A})$ is abelian.
I have two questions:

*

*Is this true without demanding $\sf{C}$ to be small? The only place where Borceux proof seem to demand this condition is where he uses that in this case limits and colimits are computed pointwise. But I think I know how to prove that products, kernels and cokernels are computed pointwise without demanding $\sf{C}$ to be small.

*If the theorem is true without this condition, then why aren't limits and colimits computed pointwise in functor categories without this assumption?

 A: If $\mathcal{C}$ is not small, the functor category $\operatorname{Fun}(\mathcal{C},\mathcal{A})$ will not be locally small in general. The morphisms in this category are natural transformations, whose components are indexed in $\operatorname{Ob}\mathcal{C}$, which is a proper class/large set if $\mathcal{C}$ is not small. Thus $\operatorname{Fun}(\mathcal{C},\mathcal{A})$ cannot be $\mathsf{Ab}$-enriched and hence cannot be abelian.
A: It is "merely" a problem of formalities.
Borceux mentions both the universe axiom and Gödel-Bernays (NBG) set theory as solutions to the problem of working with large categories, but they are not equally powerful.
Borceux's definition of "category" requires that the collection of objects is a class and that the collection of all morphisms between any two objects is a set – so he is implicitly working in NBG.
Thus, in this definition, there are two obstructions to forming the category of all functors $\mathcal{A} \to \mathcal{B}$ when $\mathcal{A}$ is not small:

*

*The collection of all functors $\mathcal{A} \to \mathcal{B}$ is so big that it is not even a class!
If it were a class, then a functor $\mathcal{A} \to \mathcal{B}$ would be a set, since a set is defined to be a collection that is a member of some class.
But then we could use the axiom of replacement to deduce that the class of objects of $\mathcal{A}$ is a set – contradiction.


*The collection of natural transformations between two functors $\mathcal{A} \to \mathcal{B}$ is also too big to be a class, let alone a set, by the same argument.
That said, these are not serious problems.
If one uses the universe axiom instead of NBG then there is a more appropriate definition of "category" that permits the formation of functor categories between any two categories.
All the basic theorems hold, so if $\mathcal{B}$ has limits of a certain shape then limits of that shape exist and are computed componentwise in the category of all functors $\mathcal{A} \to \mathcal{B}$, etc.
In particular, if $\mathcal{B}$ is an abelian category then the category of all functors $\mathcal{A} \to \mathcal{B}$ is also abelian.
