group completion theorem by using homology fibrations In the paper Homology fibrations and group completion theorem, McDuff-Segal (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 (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 (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!
 A: 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.
