4 deleted 3 characters in body

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?) 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.

3 fixed TeX, maybe

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 $H_(M)$ H_*(M)$then$H_(M)[\pi_0^{-1}]\cong H_*(M)[\pi_0^{-1}]\cong H_*(\Omega BM) $BM)$

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.

2 added 4 characters in body

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 $H_H_(M)$ M) $then$ H_H_(M)[\pi_0^{-1}]\cong H_*(\Omega BM)$BM)$

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.

1