Group completion theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:04:26Z http://mathoverflow.net/feeds/question/17128 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17128/group-completion-theorem Group completion theorem veit79 2010-03-04T20:07:24Z 2010-03-05T12:59:29Z <p>Let $M$ be a topological monoid. How does the homology-formulation of the group completion theorem, namely (see McDuff, Segal: <em>Homology FIbrations and the "Group-Completion" Theorem</em>)</p> <blockquote> <p>If $\pi_0$ is in the centre of $H_*(M)$ then $H_*(M)[\pi_0^{-1}]\cong H_*(\Omega BM)$</p> </blockquote> <p>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? </p> <p>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. </p> http://mathoverflow.net/questions/17128/group-completion-theorem/17134#17134 Answer by Oscar Randal-Williams for Group completion theorem Oscar Randal-Williams 2010-03-04T20:40:25Z 2010-03-04T20:40:25Z <p>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 $\Omega BM$.</p> <p>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.</p> <p>However it is perverse to prove the "$M \simeq \Omega BM$" result this way.</p> http://mathoverflow.net/questions/17128/group-completion-theorem/17135#17135 Answer by Charles Rezk for Group completion theorem Charles Rezk 2010-03-04T20:41:05Z 2010-03-04T20:46:58Z <p>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.</p> <p>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 $f$. </p> <p>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 $\Omega BM$. </p> <p>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". </p> <p><strong>Added:</strong> 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, <em>Simplicial Localizations of Categories</em>, JPAA (17) 267-283. Though they work simplicially, and work more generally (with categories in place of monoids), they show that $M$ is a <em>cofibrant simplicial monoid</em>, then the simplicial monoid $M[M^{-1}]$ is weakly equivalent to the space $\Omega |BM|$. </p>