# How cavalier can I be when demanding a category have direct sums?

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.

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Theo, I like your usage of "direct sum" rather than "biproduct". I do the same thing myself. – Tom Leinster Feb 14 '10 at 21:23
What is meant by taking the "quotient of a category by an ideal"? – Mike Shulman Feb 15 '10 at 2:03
@Mike: the version I am familiar with is: define an ideal as a class $I$ of morphisms closed under composition with morphisms, and consider the category with objects the original objets. and morphisms the classes of original morphisms modulo the elements of $I$ (in the relevant $\mathrm{hom}$-set) I have no idea what $ABGP$ mans, though :) – Mariano Suárez-Alvarez Feb 15 '10 at 2:18
@Mike: for example, the stable category ${}_A\underline{\mathrm{Mod}}$ of a module category ${}_A\mathrm{Mod}$, for a ring $A$, has objects the $A$-modules and as morphisms $f:M\to N$ the morphisms of $A$-modules modulo those that factor through a projective module. – Mariano Suárez-Alvarez Feb 15 '10 at 2:19
@Mike: Well, in an AbGp-enriched category, I think that an ideal should be (i) for each objects X,Y, an abelian subgroup of Hom(X,Y), such that (ii) the ideal is closed under left- and right- composition with arbitrary morphisms. I could similarly define "left ideal" and "right ideal". I think that this is the right notion for quotients of AbGp-enriched categories to be still AbGp-enriched. – Theo Johnson-Freyd Feb 15 '10 at 3:25

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.

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Awesome. Plus seven more characters. – Theo Johnson-Freyd Feb 15 '10 at 23:10

I don't think your definition of direct sum is quite right (even if you add the obviously necessary condition that the isomorphisms be natural). My understanding is that a direct sum / biproduct is an object that is both a product and a coproduct in a compatible way. This is usually phrased by saying that you have coproducts and products, and the unique morphisms $0\to 1$ and $X\sqcup Y \to X\times Y$ are isomorphisms. In terms of your definition, I think this would be equivalent to saying that you have a zero object, and the composite isomorphism $$\hom(Z,Z) \cong \hom(X,X)\times \hom(Y,X)\times \hom(X,Y)\times \hom(Y,Y)$$ relates $1_Z$ to $(1_X,0,0,1_Y)$ (where $0$ is the map factoring through the zero object). It's true, but not (I think) obvious, that if you have products and coproducts and an arbitrary natural family of isomorphisms $X\sqcup Y \cong X\times Y$, then you actually have biproducts. But I don't think this works as a definition for an individual biproduct.

As to your actual question, I don't have a complete answer, but one thing to note is that in the world of categories enriched over additive monoids (or groups), direct sums are absolute (co)limits, aka Cauchy (co)limits. That means that they are automatically preserved by any AbMon-enriched functor, and moreover the 2-category of AbMon-enriched categories with direct sums is reflective in the 2-category of all AbMon-enriched categories. Therefore, after performing any "free" or "quotient" or "colimit" construction on AbMon-enriched categories, you can always apply the reflector to add any direct sums that might be missing (and whatever direct sums you might already have had won't be changed). In particular, this provides a construction of an additive category "presented" by any notion of generators and relations: first generate the free AbGp-enriched category, then reflect into additive categories.

In general, it's not obvious to me that if you add some additional structure freely (like kernels or cokernels), then apply the above "Cauchy-completion" reflector, that the presence of the new thing you added is preserved by the reflector. But if it isn't, then perhaps some sort of sequential colimit of successive approximations could be performed. Note that of the other constructions you mentioned, splitting of idempotents is also an absolute (co)limit, so it behaves similarly to direct sums, whereas kernels and cokernels are not.

However, none of this really answers the question you actually asked, which is whether such "free" constructions in the world of AbGp-enrichment already preserve the presence of direct sums, without the need to Cauchy-complete. I would guess that in general they don't, but I don't have a counterexample.

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Oh, a question. I thought that I knew how to define a canonical morphism $X \sqcup Y \to X \sqcap Y$, but I realized that I do not. In particular, I see no way to do this in SET. E.g. if $X \neq \emptyset$ but $Y = \emptyset$, then there is no such map. Conversely, I see no canonical map the other way if $X = Y$ is a one-element set. – Theo Johnson-Freyd Feb 15 '10 at 23:05
You can only do it once you already have a zero object. – Mike Shulman Feb 16 '10 at 3:04

My understanding of your question is different to Mike's. I have not come across an example where Hom-sets are abelian monoids but not abelian groups. I'd be interested to see how this arises.

I am going to assume your Hom-sets are abelian groups (or modules over a commutative ring) and that composition is bilinear. Then you can add (and subtract) morphisms and you want to add objects. This can be done formally, just form a new category with objects finite lists of objects and morphism are written as matrices. That's it; there is no condition on the Hom-modules.

You may also want to do idempotent completion. Then you should first apply the above construction and then take the idempotent completion. This will be both additive and idempotent complete. If you do it the other way around it may not be idempotent complete; intuitively, if $A$ is an algebra $M_n(A)$ can have more idempotents than $n\times n$ diagonal matrices with entries in $A$.

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I don't know of any examples "in nature" in which Hom-sets are commutative monoids but not groups. The statement is that any category with direct sums is com-mon-enriched, but not necessarily ab-gp-enriched. I guess an example is monoids over $\mathbb N$, or another rig. I.e. commutative monoids with an $\mathbb N$-action (which is no extra data for $\mathbb N$, but is for other rigs). – Theo Johnson-Freyd Feb 15 '10 at 23:08
Enrichment over abelian monoids that aren't abelian groups often takes the form of enrichment over semilattices. – Mike Shulman Feb 16 '10 at 3:06