Let $M$ be a topological monoid that is homotopy commutative. I've been trying to understand the proof of the group completion theorem from Hatcher's notes. Roughly speaking, this theorem says that the homology of $\Omega B M$ can be obtained from the graded ring $H_{\bullet}(M)$ by localizing at $\pi_0(M)$.
When $\pi_0(M)$ is finitely generated by elements $\{s_1,\ldots,s_m\}$, the model for this localization used in the above notes is the infinite telescope of the map $M \rightarrow M$ induced by multiplying by $t=s_1 \cdots s_m$.
This is not quite what I expected. The group completion of a discrete commutative monoid $N$ equals $N_{\text{gc}} = (N \times N) / N$, where $N$ acts diagonally. The completion map $N \rightarrow N_{\text{gc}}$ is the map taking $n \in N$ to the image of $(1,n)$ in $N_{\text{gc}}$. Since we're working in the homotopy category, it makes sense to replace the quotient by some kind of homotopy quotient and define $M_{\text{gc}}$ to be the homotopy quotient of $M \times M$ by $M$, which I suppose should be the simplicial space $(EM \times M \times M)/M$, where $M$ acts diagonally.
This brings me to my questions: first, is the homology of $M_{\text{gc}}$ the same as that of $\Omega B M$, and second, can we use $M_{\text{gc}}$ in place of the infinite telescope in the proof of the group completion theorem? The first question might have a positive answer without the giving a positive answer to the second question if you need the group completion theorem to calculate the homology of $M_{\text{gc}}$.
Assuming this actually works, it would have the advantage of not requiring the choice of a generating set (and also working when $\pi_0(M)$ is not finitely generated).
