Abstractly (and ignoring topology), the groups $\mathbb{R}$ and $\mathbb{R}^n$ are isomorphic: both are $\mathbb{Q}$-vector spaces of dimension $2^{\aleph_0}$. A group is isomorphic to a subgroup of such iff it is torsionfree and of rank bounded by $2^{\aleph_0}$. But given what you are looking for, KConrad's suggestion makes a lot of sense.

Thanks, Nicholas: you did indeed explain the connection; not sure why I missed that. I'll add the a-t tag back in, and I think you should feel free to rollback to the previous version, but maybe it doesn't matter too much after Derek's answer.

Well, yes, toposes are cartesian closed by definition. As for string-like diagrams in symmetric monoidal closed categories, you can read what Baez and Stay say in their Rosetta Stone paper. But I'm not sure how truly useful this would be for studying the effective topos.

Alex, could you expand on what you mean by "polynomial rings are rigid (unlike fields)"?

Since I've heard Chomsky speak on this theme before, I think the real point he's leading up to is that despite its conceptual power, mathematics isn't, and can't be, used to solve scientific problems the way a Laplacean demon could. We can't understand in detail what is going on in a cup of coffee, much less "how a dog works", etc. etc.

I don't have a ready explanation, but in terms of LaTeX output, it should render the same way with or without those spaces, so I wouldn't worry about that. Thanks for explaining.

Dear Naveen: you might not be aware, but each edit bumps your question to the top of the queue and thus bumps another question vying for attention off the top page. Thus very many small edits in a short time span might be resented by other members of the community.

Thanks! The colimit in the second case is also known as a Prüfer group: en.wikipedia.org/wiki/Pr%C3%BCfer_group. The only thing I'd add to your proof is that the inclusion $Grp \to sSet$ is fully faithful, so that if the colimits on the simplicial set side were isomorphic, they'd have to be isomorphic on the group side, leading to a contradiction as you explained.

@Andrea, why don't you write that up? You could take for example $\mathbb{Z} \to \mathbb{Z} \to \ldots$ where in one case it's all identity maps, and in the other it's all multiplication by 2 (in groups).

@PeterLeFanuLumsdaine Learn something new every day. Thanks! (And now that I look again, the tag abelian-categories seems inappropriate. I think I'll remove it. The semiabelian-categories tag has a description which matches the one OP meant.)

@MarcoFarinati He said semi-abelian. So your comment doesn't apply. ncatlab.org/nlab/show/semi-abelian+category

@aginensky Please feel free to flag others that you see.

David, sorry I wasn't more clear: in view of your question "is this true?", it was meant partly to deflect from the idea that it was Russell asserting it, to the idea that maybe he was just quoting Wikipedia. I agree it would be good to get an expert opinion.

I imagine Russell was going by the Wikipedia article: en.wikipedia.org/w/…. Maybe one of the references there has details.