# Are all coproducts of 1 in a topos distinct ?

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.

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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

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 $n=m$.

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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