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Let $M$ be a topological monoid. How does the homology-formulation of the group completion theorem, namely (see McDuff, Segal: Homology FIbrations and the "Group-Completion" Theorem)

If $\pi_0$ is in the centre of $H_\*(M)$ then $H_\*(M)[\pi_0^{-1}]\cong H_\*(\Omega BM)$

imply that $M\to \Omega BM$ is a weak homotopy equivalence if $\pi_0(M)$ is already a group? I don't see the connection to homology. Can one prove the latter (perhaps weaker) statement more easily than the whole group completion theorem?

A topological group completion $G(M)$ of $M$ should transform the monoid $\pi_0(M)$ into its (standard algebraic) group completion. But a space with this property is not unique. Why is $\Omega BM$ the "right" choice? Perhaps this is clear when I see the connection to the homology-formulation above.

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The statement that $M \to \Omega BM$ is a weak equivalence when $M$ is a group-like topological monoid is indeed easier: the map $EM = B(M \wr M) \to BM$ is then a quasi-fibration, has geometric fibre $M$ over the basepoint and homotopy fibre $\Omega BM$.

However the homological group-completion theorem also implies this: if $M$ is group-like then $\pi_0(M)$ already consists of units in $H_*(M)$, so it just says that $M \to \Omega BM$ is a homology equivalence. Each of these spaces has homotopy equivalent path components, so it is then enough to observe that the map of 0 components is a homology equivalence between simple spaces, so a weak homotopy equivalence.

However it is perverse to prove the "$M \simeq \Omega BM$" result this way.

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Thank you, Oscar. Where can I find statements about $EM$ an $BM$ with $M$ a topological monoid (and not a topological group)? The constructions are the same for monoids as for groups, I think, but how to prove that $EM\to BM$ is a quasi-fibration (is this a Serre-fibration?)? Is there something like a "$M$-principal bundle" where $M$ is a monoid? –  veit79 Mar 5 '10 at 12:55
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That $EM \to BM$ is a quasi-fibration when $M$ is group like is proved in e.g. J.P. May "Classifying Spaces and Fibrations" Memoirs AMS 155, Theorem 7.6. –  Oscar Randal-Williams Mar 5 '10 at 13:21
    
Do you know a free reference? I am not able to get this book without a lot of effort. –  veit79 Mar 5 '10 at 15:45
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It is available on May's homepage: math.uchicago.edu/~may/BOOKS/Classifying.pdf –  Oscar Randal-Williams Mar 5 '10 at 18:29
    
Tank you, Oscar. I suppose $M\wr M$ is a wreath product. Do you know where I can read about "topological" wreath products? –  veit79 Mar 14 '10 at 16:24
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Well, if $\pi_0=\pi_0(M)$ is already a group, then $H_\*(M)\approx H_\*(M)[\pi_0^{-1}]$. So $M$ and $\Omega B M$ have the same homology in this case. This isn't quite enough on its own, but if you can produce a map $M\to \Omega BM$ which induces this homology isomorphism, then the result follows using the Hurewicz theorem.

What McDuff-Segal actually do is show that if $M$ is a topological monoid which acts on a space $X$, in such a way that every $m\in M$ induces a homology equivalence $x\mapsto mx\colon X\to X$, then you can produce a "homology fibration" $f:X_M\to BM$ with fiber $X$. "Homology fibration" means that the fibers of $f$ are homology-equivalent to the homotopy fibers of $f$.

If $\pi_0M$ is an abelian group, you can find an $X$ such that $X_M$ is contractible, and the fiber of $f:X_M\to BM$ is $X$. This gives the homology equivalence you want, since the homotopy fibers of $f$ look like $\Omega BM$.

Take a look at McDuff and Segal's paper, it's nice. There is a also a treatment in terms of simplicial sets in Goerss-Jardine, *Simplicial Homotopy Theory".

Added: The functor $M\mapsto \Omega BM$ is the "total derived functor of group completion". The only convincing explanation of why this is so (that I'm aware of) is in Dwyer-Kan, Simplicial Localizations of Categories, JPAA (17) 267-283. Though they work simplicially, and work more generally (with categories in place of monoids), they show that $M$ is a cofibrant simplicial monoid, then the simplicial monoid $M[M^{-1}]$ is weakly equivalent to the space $\Omega |BM|$.

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Thank you, Charles. Why is a map inducing homology isomorphisms a homotopy equivalence if $M$ is not simply connected? –  veit79 Mar 5 '10 at 13:15
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This is not true in general, but is true when you have a map of grouplike H-spaces - in this case, the fundamental groups act trivially in the higher homotopy groups, and the relative Whitehead theorem tells you that the first relative homology group coincides with the first relative homotopy group. –  Tyler Lawson Mar 5 '10 at 14:51
    
Ah, I see. Thank you. –  veit79 Mar 5 '10 at 15:06
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To elaborate on Tyler's comment: One more step is needed to go from trivial action of the fundamental group on the higher homotopy groups of the two spaces to trivial action on the relative homotopy groups of the mapping cylinder. This can be done by an obstruction theory argument as in Proposition 4.74 of my book. –  Allen Hatcher Mar 5 '10 at 15:35
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