# 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

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 $n=m$.