# Abelian group structure of the additive category

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 $$X, Y...$$

• To get your head straight, start considering the category of groups versus abelian groups. Commented Mar 10, 2019 at 1:22
• Why would it tell you anything about $Y$? The morphisms of a category have absolutely nothing to do with the objects, in general. Commented Mar 10, 2019 at 2:46
• @Eric: it tells you that every object of an additive category is canonically an abelian group object, by the Yoneda lemma. Commented Mar 10, 2019 at 2:55
• @QiaochuYuan: I don't think that's the sort of thing OP is talking about--they are asking whether $Y$ has to literally be a group. Of course, saying $Y$ is an abelian group object is actually a statement about certain morphisms in the category involving $Y$, not a statement about $Y$ itself as a mathematical object outside of the category. Commented Mar 10, 2019 at 3:13

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.

• Dear Qiaochu, I am very grateful for that blog post. Commented Sep 25, 2022 at 9:07
• That said, I have some queries. I am really very stuck on why the addition thus defined should be expected to be commutative: $$\nabla(f\oplus g)\Delta\overset{?}{=}\nabla(g\oplus f)\Delta$$You prove this by proving that the (co)diagonals are (co)commutative, and you prove *that* in a way which is mysterious to me. Eg you say more or less that: “composing $\nabla$ in all ways to get arrows $M\sqcup M\cdots\sqcup M\to M$ will always give you the same composite arrow”. I don’t know why this might mean $\nabla(f\oplus g)=\nabla(g\oplus f)$, that is, I don’t know in what sense this is commutative Commented Sep 25, 2022 at 11:04
• @FShrike: can you post a separate question about this? This is a little too much to get into in the comments. Commented Sep 25, 2022 at 15:31

Qiaochu's answer is great but there is an additional very important point which you seem to be misunderstanding which I wanted to address. In a category, the objects and morphisms can be anything. The morphisms don't have to be functions between the objects, or even have any actual relation to the objects at all. All that matters is that we have defined what it means for a morphism to have a certain object as a domain or a codomain, and we have defined composition and identity morphisms, and these satisfy the axioms for a category. The same thing applies to the addition operation in an additive category: it is just some abstract operation which need not have anything to do with what the objects and morphisms actually are, beyond satisfying the axioms. (As Qiaochu mentions, it is actually uniquely determined by the category structure, but the category structure itself still need not have anything to do with what the objects and morphisms are.)

As a result, most of the questions you are asking do not even have any meaning: there is no such thing as "the image of $$X$$ in $$Y$$", for instance, because our morphisms may not be functions or anything remotely resembling functions. More generally, nothing in category theory can ever give us information about "some underlying structure" of an object, except insofar as that structure is defined using the category itself.

• Thank you for this clarification - I did not realize this gap in my knowledge! Much appreciated Commented Mar 10, 2019 at 5:48