In the paper *[Homology fibrations and group completion theorem, McDuff-Segal](http://www.maths.ed.ac.uk/~aar/papers/mcdsegal.pdf)* (www.maths.ed.ac.uk/~aar/papers/mcdsegal.pdf), page 281:


Let $M$ be a topological monoid such that $\pi\_0M$ is generated by $s\_1,s\_2, \cdots,s\_k$. Let $s=s\_1s\_2\cdots s\_k$. Then 
$$
H\_\*(M)[\pi\_0M^{-1}]=H\_\*(M)[s^{-1}].
$$
However, $\pi\_0M$ may not be able to be generated by $s$. 
Let $m\_1\in M$ be a point lying in $s$ ($s$ corresponds to a path-connected component of $M$). Let $M\_\infty$ be the mapping telescope formed by the following sequence of maps
$$
M\to ^{\cdot m\_1}M\to ^{\cdot m\_1}M\to ^{\cdot m\_1}M\to ^{\cdot m\_1}\cdots.
$$
It is claimed in *[Homology fibrations and group completion theorem, McDuff-Segal](http://www.maths.ed.ac.uk/~aar/papers/mcdsegal.pdf)* (www.maths.ed.ac.uk/~aar/papers/mcdsegal.pdf), page 281, line 22- line 23 that "the preceding argument applies" for the proof of group completion theorem for $M$. 


In "the preceding argument" of *[Homology fibrations and group completion theorem, McDuff-Segal](http://www.maths.ed.ac.uk/~aar/papers/mcdsegal.pdf)* (www.maths.ed.ac.uk/~aar/papers/mcdsegal.pdf), $\pi\_0M$ is assumed to be $(\mathbb{Z}\_{\geq 0},+)$, i.e., $$
\pi\_0(M)= [M\_1]^{\mathbb{Z}\_{\geq 0}}=\{1, s=[M\_1], [M\_1]^2, [M\_1]^3,\cdots\} 
$$
where $1, [M\_1],[M\_1]^2,\cdots,[M\_1]^j,\cdots$ are the path-connected components of $M$. The following claim is essential for the proof in "the preceding argument".


***Claim.***[line 14-line 15, page 281].
*For any $m\in M$, the left action of $m$ on $M\_\infty$ given by  

$$m(x\mapsto xm\_1\mapsto xm\_1^2\mapsto\cdots)= (mx\mapsto mxm\_1\mapsto mxm\_1^2\mapsto \cdots)$$ is a homology equivalence.*


***Question:*** **For the case that $\pi\_0M$ is generated by $s\_1,s\_2, \cdots,s\_k$, since the proof of the above claim does not hold any longer, how can "the preceding argument applies"?**


Apologize: I apologize sincerely that I originally posted the question on math.stackexchange and then deleted the post and shifted it here. Sorry for disturbing!



group completion theorem by using homology fibrations
The submonoid (isomorphic to $\mathbb{Z}\_{\geq 0}$) generated by the product $s = s\_1 \cdots s\_k$, while not equal to $\pi\_0(M)$, is cofinal in it, and so has the same localization. More directly: the mapping telescope for $s$ and the iterated mapping telescope for each of the $s\_i$ have isomorphic homologies. 


This is perhaps easiest to see if $k=2$. For an element $x \in \pi\_0 M$, write $M[x^{-1}]$ for the telescope for multiplication by $x$. There is a comparison map $\alpha: M[s\_1^{-1}] \to M[s^{-1}]$ induced by the comparison of direct systems defining each. I don't know how to do diagrams in mathoverflow, so I can't draw what would be very helpful here. But the comparison of the direct systems comes from the commutativity of $s\_1$ and $s\_2$.


From $\alpha$, we may build a second comparison of direct limits to get a map $\beta: M[s\_1^{-1}][s\_2^{-1}] \to M[s^{-1}]$. The domain is $M[s\_1^{-1}][s\_2^{-1}] = M[s\_1^{-1}, s\_2^{-1}] = M[\pi\_0^{-1}]$. There's also a natural map in the opposite direction, since $s \in \pi\_0$. I claim that these are inverse homology equivalences, as can be seen by noting that in


$$H\_\*(M[s^{-1}]) = H\_\*(M)[s^{-1}]$$


an inverse to the element $s\_1$ is given by $s\_2 s^{-1}$, and an inverse for $s\_2$ is $s\_1 s^{-1}$. Thus $H\_\*(M)[s^{-1}] = H\_\*(M)[s\_1^{-1}, s\_2^{-1}]$.


For $k>2$, iterate.


