I want to show that the functor category $A^J$ is an abelian category if $A$ is an abelian category. I know it's easy to define a null object, binary biproducts, kernels, and cokernels.

But I got stuck on the last condition that every monomorphism is a kernel, and every epimorphism is a cokernel.

Suppose that $\alpha \colon S \to T$ is a natural transformation, and it's a monomorphism in $A^J$, but I don't think we can conclude that $\alpha_j \colon S_j \to T_j$ is a monomorphism in $A$. Then it's difficult to verify the last condition by universal properties.


My solution I think we can conclude $\ker \alpha=0$ if $\alpha$ is monic. Since if $\alpha \circ \beta =0$, then $\beta = 0$ ($\alpha$ is monic) factors through $\ker \alpha =0$. So it's obvious that the last condition holds for $A^J$.

Could you tell me whether my solution is right? Thank you.

  • $\begingroup$ Can you explain what $A^J$ means? $\endgroup$ – user593746 Dec 17 '18 at 15:04
  • $\begingroup$ @Zvi It's a functor category whose objects are the functors from $J$ to $A$ and morphisms are natural transformation. $\endgroup$ – Kai Xing Dec 17 '18 at 15:54

In general, it is true that the components a natural transformation may not be monomorphisms even when the natural transformation is. Relatedly, a limit of functors can exist but fail to be computed pointwise. However, assuming all the limits of a particular shape exist in the target category, then limits of that shape will exist in the functor category and be computed pointwise.1

Now, $f:X\to Y$ is a monomorphism if and only if $$\require{AMScd}\begin{CD}X@>id>>X\\@VidVV@VVfV\\X@>>f> Y\end{CD}$$ is a pullback square.2 Therefore, if $\alpha:S\to T$ is a monomorphism, then $S$ is a pullback of $\alpha$ along itself that's witnessed by the identity natural transformation. If the underlying category has all (binary) pullbacks, which Abelian categories do1, then the pullback in the functor category is calculated by pullbacks of the same shape in the target category. The components of the identity natural transformation are identity arrows, so those pullbacks are also witnessed by identities. Thus all the components are monomorphisms.

Dualize this for epimorphisms.

1 You should prove this if you haven't before.

2 I would suggest proving this too, but there's nothing to prove. If you write out what the universal property of that pullback is, it just is the statement that $f$ is a monomorphism. You should see this for yourself though.

  • $\begingroup$ Thank for your patient answer! $\endgroup$ – Kai Xing Dec 17 '18 at 9:50

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