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*)



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> If $\pi\_0$ is in the centre of $H\_\*(M)$ then $H\_\*(M)[\pi\_0^{-1}]\cong H\_\*(\Omega BM)$
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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. 



Group completion theorem
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.


