In my meaning, a **direct sum** in a category should really be called a "biproduct". If $X,Y$ are objects, then a direct sum $X \oplus Y$ is an object $Z$ along with isomorphisms $\hom(Z,A) = \hom(X,A) \times \hom(Y,A)$ and $\hom(A,Z) = \hom(A,X) \times \hom(A,Y)$ for all objects $A$. A direct sum is unique up to canonical isomorphism if it exists, of course. A category **has (finite) direct sums** if it has a zero object (an object that is both initial and terminal; i.e. "the direct sum of zero things") and if $X\oplus Y$ exists for any objects $X,Y$. If a category has direct sums, then it is naturally enriched in abelian monoids. I believe that an **additive category** is a category with direct sums in which all the hom-sets (which are already abelian monoids) are actually abelian *groups*.

There are many times when people say "include all direct sums". For example:

**Example:**

Let $\mathcal C$ be any category (enriched over $\rm SET$). Then I can make it enriched over $\rm ABGP$ by applying the $\rm Free: SET \to ABGP$ functor to each hom-set. So now I have a new catefory ${\rm Free}(\mathcal C)$ in which I can add morphisms. But often I want to add objects, too, so I do something like "take the matrix category" ${\rm Mat}(\mathcal C)$, whose objects are finite sequences of objects in $\mathcal C$ and whose morphism are matrices of morphism in ${\rm Free}(\mathcal C)$. Then it's more or less obvious that ${\rm Mat}(\mathcal C)$ is an additive category. If $\mathcal C$ is freely generated by some (objects and) morphisms, then ${\rm Mat}(\mathcal C)$ is presumably "the free additive category generated by those morphisms".

But often I'm not content with free additive categories. For example, I might want to present a category by generators and relations.

Question:Is it clear that when I take the quotient of an additive category by some ideal (as an $\rm ABGP$-enriched category), that it still has direct products?

Or perhaps I really want the *abelian* category presented by generators and relations. Or maybe I just want every idempotent to split, in which case I might take the Karoubi envelope.

Question:If I extend my category to split all idempotents, or to include kernels and cokernels, or ..., is it clear that it still has direct products?

A very explicit application contained in these constructions is the formation of the exterior tensor product of categories: if $\mathcal B,\mathcal C$ are additive categories, then $\mathcal B \boxtimes \mathcal C$ is the free additive category generated by $\mathcal B \times \mathcal C$ with a bunch of relations.

thinkthat this is the right notion for quotients of AbGp-enriched categories to be still AbGp-enriched. $\endgroup$ – Theo Johnson-Freyd Feb 15 '10 at 3:25