Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(if any?)

I understand that in the natural numbers, the sum of two numbers can be readily thought of as the disjoint union of two finite sets.

John Baez even spent a week talking about how you can extend this idea to thinking about the integers here: TWF 102. This led into a discussion of the homotopy groups of spheres.

But then you have to pass to a certain colimit to get the rationals, and take a certain completion to get the reals. It all gets very complicated.

One place where we rely on a correspondence between a sum of real numbers and a certain coproduct is in measure theory-- perhaps in analogy to the relation between finite sets and natural numbers, we should think of some measure space as the categorification of the real numbers. But this sounds unpromising-- what space would be in any sense a canonical categorification? Moreover, what I was really hoping for originally was a precise sense in which the $\sigma$-additivity of a measure states that it preserves coproducts or something, so I was hoping there might be more to it than sigma-algebras.

share|cite|improve this question
In Baez's categorification of the integers, sum isn't coproduct; it's just a monoidal operation. So you need to be asking a slightly different question. – Qiaochu Yuan Apr 5 '10 at 5:08
There's also something of a problem with thinking of sigma-additivity as preserving coproducts: the natural category structure to place on a measurable space is the one where objects are measurable subsets and morphisms are inclusions. Then coproducts are unions; I assume this is what you were talking about. However, sigma-additivity only applies to disjoint unions. – Qiaochu Yuan Apr 5 '10 at 5:28
Aren't coproducts of sets already disjoint unions? The union would be a pushout, right? – Tim Campion Apr 5 '10 at 21:19

2 Answers 2

up vote 11 down vote accepted

None (except trivially).

It's an elementary (though maybe not obvious) lemma that if $X$ and $Y$ are objects of a category and their coproduct $X + Y$ is initial, then $X$ and $Y$ are both initial.

Suppose there is some category whose objects are the real numbers, and such that finite coproducts of objects exist and are the same as finite sums of real numbers. In particular (taking the empty sum/coproduct), the real number $0$ is an initial object. Now for any real number $x$ we have $x + (-x) = 0$, so by the lemma, $x$ is initial. So every object is initial, so all objects of the category are uniquely isomorphic, so the category is equivalent to the terminal category 1.

If you just want non-negative real numbers then this argument doesn't work, and I don't immediately see an argument to take its place. But I don't think it's too likely that an interesting such category exists.

I wonder if it would be more fruitful to ask a slightly different question. Product and coproduct aren't the only interesting binary operations on a category. You can equip a category with binary operations (as in the concept of monoidal category). Sometimes this is a better thing to do.

For example, there is on the one hand the concept of distributive category, which is something like a rig (=semiring) in that it has finite products $\times$ and finite coproducts $+$, with one distributing over the other. On the other hand, there is the concept of rig category, which is a category equipped with binary operations $\otimes$ and $\oplus$, with one distributing over the other. Distributive categories are examples of rig categories. Any rig, seen as a category with no morphisms other than identities, is a rig category. Any ordered rig can be regarded as a rig category (just as any poset can be regarded as a category): e.g. $[0, \infty]$ is one, with its usual ordering, $\otimes = \times$, and $\oplus = +$.

share|cite|improve this answer
You didn't write this but it is easy to come up with examples of rig categories where 0 is not initial. In fact there are non-trivial bimonoidal categories (bipermutative in fact) which has $\mathbb{Z}$ as the underlying object-set with its usual ring structure induced from $\otimes$ and $\oplus$. – Thomas Kragh Apr 5 '10 at 10:11
Right; e.g. any nontrivial rig regarded as a discrete rig category gives an example where 0 is not initial. On the second point (re $\mathbb{Z}$): there's also a nontrivial rig category whose underlying rig of isomorphism classes of objects is $\mathbb{Z}[i]$. – Tom Leinster Apr 5 '10 at 15:06
I'm probably just slow here, but I don't see why the arrow $0 \rightarrow X$ should be unique without making some additional assumptions... That said, point taken. Any category like the one I asked for is not going to be nice, not in any nice sense going to be the reals, rationals, postive reals, ... Thanks. – Tim Campion Apr 5 '10 at 21:25
Don't worry about being slow... it's just something to wrap one's head around. For any finite family of numbers, you can take the sum. In particular, you can do it for the empty family; and the sum of the empty family is zero. (You might call that a "convention", but it's the only sensible one.) In the same way, for any finite family of objects of a category, you can (if the category allows) take the coproduct. In particular, the coproduct of the empty family is the initial object. More precisely, the empty family has a coproduct iff an initial object exists, and in that case they coincide. – Tom Leinster Apr 5 '10 at 23:09

Isomorphism classes of finitely generated right Hilbert modules over a II_1 factor are in a bijective correspondence with the nonnegative reals. The correspondence sends every module to its dimension. Moreover, the dimension function is additive with respect to the coproduct of modules. I believe you can obtain all reals using some form of supersymmetry (Quillen construction?), but then the addition of reals will no longer correspond to the coproduct of objects.

share|cite|improve this answer
+1. There were a couple of related questions by David Corfield: and – Tom Leinster Apr 5 '10 at 15:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.