TITLE: Are all coproducts of 1 in a topos distinct ?
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: 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 $n=m$.