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.



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> **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?
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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](http://ncatlab.org/nlab/show/Karoubi+envelope).



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> **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?
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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.



How cavalier can I be when demanding a category have direct sums?
The answer to the first question is yes. If A and B have a direct sum A ⊕ B in C, then there are inclusions iA : A → A ⊕ B, iB : B → A ⊕ B and projections pA : A ⊕ B → A, pB : A ⊕ B → B such that pAiA = 1, pBiB = 1, and iApA + iBpB = 1. Conversely, the existence of such maps in an Ab-enriched category make A ⊕ B a direct sum of A and B, even if we do not asssume a priori that A and B have a direct sum. Now if we form the quotient C/I by an ideal I, and two objects A and B with a direct sum A ⊕ B in C, the image of this system of maps presents the image of A ⊕ B as the direct sum of the images of A and B. In short, direct sums are absolute colimits, and as the quotient functor C → C/I is essentially surjective (indeed, bijective on objects), every pair of objects of C/I inherits a direct sum from C.


