Let $A$ be an additive category. We have that for all $X, Y$ objects in $A, Hom_A(X, Y)$ is an abelian group. However, what is the abelian group structure? If $f, g: X \rightarrow Y$ are two morphisms in $Hom_A(X, Y)$ then it doesn't quite make sense to compose them. So would the operation be the addition of morphisms?
If this is true, does this tell us some underlying structure of $Y?$ That is, we can add and subtract things in $Y,$ because $-f,$ im assuming, is defined as the inverse of $x$ in $Y.$ Furthermore, the identity map would map everything to $0(?)$ in $Y.$ So does it follow that the image of $X$ in $Y$ has an abelian group structure? Associativity is something we would have to consider, but I guess the general question is if the structure of $Hom_A(X, Y)$ does say something about the structure of $X, Y…$
Best Answer
A priori, the abelian group structure on $\text{Hom}(X, Y)$ is just some arbitrary abelian group structure, given to you as part of the data of $A$ being an additive category. So a priori there's nothing more to say than "it's whatever abelian group structure was given to you, by virtue of however it is you know that $A$ is an additive category."
However, in fact the abelian group structure turns out to be determined by the structure of $A$ as a bare category. (This should be quite surprising.) More explicitly, if $f, g : X \to Y$ are parallel morphisms, their sum is the composite
$$X \xrightarrow{\Delta} X \oplus X \xrightarrow{f \oplus g} Y \oplus Y \xrightarrow{\nabla} Y$$
where $\Delta$ is the diagonal map (which exists in any category with finite produts) and $\nabla$ is the codiagonal map (which exists in any category with finite coproducts). See this blog post where I describe how this works in excruciating detail.
What this tells us is that, by the Yoneda lemma, every object $Y$ in an additive category has a canonical structure of an abelian group object.