I'm beginning to study abelian categories. The definition I'm using is this:

A category is said to be abelian if:

$1)$ it has a zero object

$2)$ it has all binary products and binary coproducts

$3)$ it has all kernels and cokernels

$4)$ all monomorphisms and epimorphisms are normal.

I'm trying to prove that if a category $C$ is abelian, than for every $A,B\in\text{Ob}(C)$, the set $\hom_C(A,B)$ has an abelian group structure with neutral element $0:A\to B$ (the zero morphism), and that the composition of morphism is bilinear with respect to this group operation.

In simple examples like the category of $R$-modules or $K$-vector spaces, there is an obvious additive group structure, but in the general context, I have no idea how to define this group operation.

What is the idea?

• I'm not sure this is right; I think you need to also specify that products and coproducts agree (in a particular way). See qchu.wordpress.com/2012/09/14/… for a discussion. – Qiaochu Yuan Mar 26 '17 at 18:29
• @QiaochuYuan, that was a great article, thanks! – rmdmc89 Mar 26 '17 at 19:04
• @QiaochuYuan, I'm looking at the following theorem from the article you linked: "A semiadditive category $C$ is canonically enriched over $\text{CMon}$. The identity in each commutative monoid $\text{Hom}(a, b)$ is the zero morphism $0_{a, b}$". In the beginning of the proof, you say that $f\oplus 0: a\oplus a\to b\oplus b$ factors through $a\oplus 0\to b\oplus 0$, and using $a\oplus 0\simeq a$, $b\oplus 0\simeq b$, you conclude that $f+0=f$. How do you formalize this conclusion? – rmdmc89 Mar 26 '17 at 22:49
• You don't need to specify that products and coproducts agree ; it is always the case by the short five lemma, but there's a lot of work needed to do that. It is done in Chapter 2 of Peter Freyd's "Abelian Categories", which is available online at tac.mta.ca/tac/reprints/articles/3/tr3abs.html – Arnaud D. Mar 28 '17 at 13:47
• – Arnaud D. Mar 28 '17 at 13:49