action of a monoid on a mapping telescope In the paper Homology fibrations and group completion theorem, McDuff-Segal, page 281, line 14-line 15:
For a topological monoid $M$, if $\pi_0(M)=\{0,1,2,3,......\}$, then the action of $M$ on $M_\infty$ on the left is by homology equivalences. 
Notations: 
(1). The space $M_\infty$ is constructed as follows: Let $M=\bigsqcup_{j=0}^\infty M_j$ where $M_j$'s are the path-connected components of $M$ such that $M_j$ is the component corresponding to $j\in\pi_0M$. We choose $m_1\in M_1$ and consider the sequence 
\begin{eqnarray}
M\overset{\cdot m_1} \longrightarrow M\overset{\cdot m_1} \longrightarrow M\overset{\cdot m_1} \longrightarrow \cdots
\end{eqnarray}
From this sequence we can form a mapping telescope
$$
M_\infty=(\bigsqcup_{i=1}^\infty [i-1,i]\times M)/\sim
$$
where $\sim$ is generated by the relations
$
(i,x)\sim (i, x m_1)
$
for any $x\in M$ and $i\geq 1$.
(2). "the action of $M$ on $M_\infty$ on the left is by homology equivalences" means:
For any $m\in M$, the left action of $m$ on $M_\infty$ given by
$$m(x\mapsto xm_1\mapsto xm_1^2\mapsto\cdots)= (mx\mapsto mxm_1\mapsto mxm_1^2\mapsto \cdots)$$
induces an isomorphism on homology.  
Question: 
Why the action of $M$ on $M_\infty$ on the left is by homology equivalences?

You are missing a hypothesis they are assuming, which is that $H_*(M)[\pi^{-1}]$ (which is just $H_*(M)[m_1^{-1}]$ in this case) can be constructed by right fractions.  This implies that $H_*(M_\infty)=H_*(M)[m_1^{-1}]$, as the colimit that computes $H_*(M_\infty)$ is exactly the right fractions for $H_*(M)[m_1^{-1}]$.  Since every element of $M$ is homotopic to some power of $m_1$ and $m_1$ acts invertibly on $H_*(M)[m_1^{-1}]$ (on either side) by definition, the result follows.
