# Is a functor category of an $\mathbf{Ab}$-category an $\mathbf{Ab}$-category itself?

In Weibel's An introduction to homological algebra, exercise 2.6.4 reads

Show that $\operatorname{colim}$ is left adjoint to $\Delta$. Conclude that $\operatorname{colim}$ is a right exact functor when $\mathcal{A}$ is abelian (and $\operatorname{colim}$ exists).

in particular, $\mathcal{A}$ is required to be abelian because from what we learned in section 1.6 this implies that, for any category $C$, the functor category $\mathcal{A}^C$ is abelian. Then we have an adjoint pair of additive functors and we can conclude (he talks about exactness only for additive functors).

But, and this is my question, couldn't we use the same argument assuming less? Or, is it true that if $\mathcal{A}$ is an $\mathbf{Ab}$-category, then $\mathcal{A}^C$ is such, too?

## 1 Answer

Yes. If $X,Y \in A^C$ and $f,g : X \to Y$, then $f+g$ is defined by $(f+g)(c)=f(c)+g(c)$ for all $c \in C$. This is a morphism $X \to Y$, since for all $i : c \to c'$ we have

$$Y(i) (f+g)(c)=Y(i) (f(c)+g(c))=Y(i) f(c) + Y(i) g(c)=f(c') X(i) + g(c') X(i) = (f(c')+g(c')) X(i)=(f+g)(c') X(i).$$

Similarily $-f$ and $0$ are defined. It is obvious that $A^C$ becomes an $\mathsf{Ab}$-category.