TITLE: How cavalier can I be when demanding a category have direct sums?
QUESTION [13 upvotes]: 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.

REPLY [14 votes]: 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.