Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Inspired by the two solutions to Harry's question

Can a topos ever be an abelian category?

I was wondering whether all coproducts of 1 in a topos are distinct up to isomorphism? That is $1 + 1 + \dots + 1 \cong 1 + 1 + \dots + 1$ iff there are an equal number of 1s on each side?

Edit: In order to make the question (possibly) non-trivial, let's assume that the topos is not equivalent to the terminal category.

share|improve this question
Ah, a followup to this question. Since this is true, according to Mike, doesn't that mean that we can construct a "natural numbers object" in any topos? –  Harry Gindi Jan 2 '10 at 3:38
No, why would it mean that? –  Mike Shulman Jan 2 '10 at 16:13
For example, the category of finite sets is a topos but has no natural number object. –  Steven Gubkin Jan 3 '10 at 3:07

1 Answer 1

up vote 7 down vote accepted

At least if you're talking about finite coproducts, then the answer is yes. If $n\le m$, then we have a canonical inclusion $\sum_{i=1}^n 1 \hookrightarrow \sum_{j=1}^m 1$, which is in fact a complemented subobject with complement $\sum_{k=1}^{m-n} 1$. If this inclusion is an isomorphism, then its complement is initial, and hence (assuming the topos is nontrivial) $n=m$. Now if we have an arbitrary isomorphism $\sum_{i=1}^n 1 \cong \sum_{j=1}^m 1$, then composing with the above inclusion we get a monic $\sum_{i=1}^m 1 \hookrightarrow \sum_{j=1}^m 1$. However, one can show by induction that any finite coproduct of copies of $1$ in a topos is Dedekind-finite, i.e. any monic from it to itself is an isomorphism. (See D5.2.9 in "Sketches of an Elephant" vol 2.) Thus, the standard inclusion is also an isomorphism, so again $n=m$.

share|improve this answer
It's a more interesting result than I gave it credit for at first. Consider the "Kennison topos", in which an object is a set X together with a bijection between X and X+X. The terminal object is the Cantor set 2^N with the obvious bijection. Coproducts are as you'd guess. You might at first think (as I did) that in this case we'd have 1 isomorphic to 1+1, but by the result you prove, that must be false. What this says is that although there are loads of bijections between 2^N and 2^N+2^N, there's none that commutes with the structure maps. –  Tom Leinster Jan 2 '10 at 2:02
What do you mean by "structure maps"? Is there another name for these, because I'd look them up if I could, but I didn't see them on nlab. –  Harry Gindi Jan 2 '10 at 3:28
Harry, this is what I mean. An object of the Kennison topos is a pair (X, xi) where X is a set and xi: X --> X + X is a bijection. By "structure map" I mean xi. (This usage is informal, with the same kind of linguistic status as "forgetful functor".) Write (2^N, gamma) for the terminal object of the topos. The coproduct of two copies of the terminal object is of the form (2^N+2^N, delta). The result that Mike proved implies that there can be no isomorphism between these two objects of the topos, i.e. no bijection f: 2^N --> 2^N+2^N such that delta f = (f+f) gamma. –  Tom Leinster Jan 2 '10 at 6:12

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.