Inspired by the two solutions to Harry's question


[Can a topos ever be an abelian category?](https://mathoverflow.net/questions/10290/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.



Are all coproducts of 1 in a topos distinct ?
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 $n=m$.


