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That's a classical theorem: I last stumbled upon it in this blog post, for example.

But I don't know how to prove it. Can anyone provide a complete proof, or a reference? Here are some references that do not contain proofs.

  • Aguilar-Gitler-Prieto, Algebraic Topology from a Homotopical Viewpoint, theorem 6.4.15: if $G$ is path-connected, then it is equivalent to the "weak" product $\prod_{i\geq 0}' K(\pi_i G,i)$. He defines a "weak "product in page 222.

  • Strom, Modern Classical Homotopy Theory, problem 20.66: same thing, but he doesn't ask for $G$ to be path-connected. He defines a "weak" product to be the "colimit of the finite products", if I understand correctly.

  • McCord, in Classifying spaces and infinite symmetric products, page 295, attributes the theorem to Moore and says "every topological abelian group has the homotopy type of a product of Eilenberg-Mac Lane spaces".

So I can't quite make up what's the actual theorem, i.e. what are the correct hypotheses. It might seem that we need path-connectedness for a monoid, but for a group we can get away with it? What's the deal with this "weak product" business?

The repeated reference is Dold-Thom's original paper, "Quasifaserungen...", but alas, it is in German, and my knowledge of the language of Goethe is limited, to say the least.

Bruno Stonek
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  • Salut Bruno :) I am not very familiar with all of this, but if you need help translating some specific passages of the original paper I might probably be able to give you a hand. (Of course, Christian is probably a better choice to help you out...) – Daniel Robert-Nicoud Oct 12 '16 at 11:21
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    The connetivity hypothesis is certainly needed. For instance, consider the free commutative monoid on a space $X$, which has the form $G=\coprod X^n/\Sigma_n$. If $X=S^2$, then I think $X^n/\Sigma^n$ is supposed to be homeomorphic to $CP^n$. These are not Eilenberg-Mac Lane spaces. – Charles Rezk Oct 12 '16 at 12:48
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    I haven't looked at the references, but "weak product" vs. "product" is the same distinction as "direct sum" vs. "product" of abelian groups. In this particular case it makes no difference: the weak product is weakly equivalent to the product. (The homotopy groups of the weak product are the direct sum of the homotopy groups of the factors.) – Charles Rezk Oct 12 '16 at 12:54
  • @DanielRobert-Nicoud Salut et merci! But it looks like we might not need to go back to Dold-Thom, luckily. – Bruno Stonek Oct 12 '16 at 13:03
  • @CharlesRezk Thanks! Yeah, that's a counterexample. But I believe that if $G$ is a group then everything goes through (see my answer below). – Bruno Stonek Oct 12 '16 at 13:04

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Well, a reference was pointed out to me by Denis Nardin:

Theorem 4.K.7 (Hatcher): A path-connected, commutative, associative H-space with a strict identity element has the weak homotopy type of a product of Eilenberg-Mac Lane spaces.

In particular, a path-connected topological abelian monoid satisfies the hypotheses.

Charles Rezk points out in the comments below an example of why path-connectedness is needed. A reference for $SP_n(S^2)$ being $\mathbb{C}P^n$ is again Hatcher, example 4K.4.

Now, if $G$ is a topological abelian group, then I believe we don't need path-connectedness. I think one can prove $G$ to be weakly equivalent to the coproduct of its path-connected components. But again, as pointed out to me by Denis, the path-connected component of the identity is a path-connected topological abelian monoid, so we apply the theorem to it. All path components are homeomorphic, though, so that's how we put a group structure on this disjoint union. (Quite hand-wavy, I know).

Bruno Stonek
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