Homotopy type of a specific discrete monoid Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is always onto).

What do we know about the homotopy type of this monoid (viewed as a
  one-object category) ? In particular, about its homotopy groups ?

My background on this subject is very small. By a paper from Dusa McDuff (On the classifying space of discrete monoids), every path-connected space has the same homotopy type as the classifying space of some monoid, and the fundamental group of $BM$ is the groupification of $M$.
EDIT: I am a bit confused between the English meaning and the French meaning of nondecreasing. The monoid I am talking about is not a group because a nondecreasing map preserving extremities is not necessarily one-to-one. I hope that this clarification will be helpful.
 A: This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
\begin{align*}
A(x) &=
\begin{cases} 2x &\text{if }x \leq 1/2\\1 &\text{if }x \geq 1/2\end{cases}\\
B(x) &=
\begin{cases} 0 &\text{if }x \leq 1/2\\2x-1 &\text{if }x \geq 1/2\end{cases}
\end{align*}
$$
Define three monoid homomorphisms $Id, U, V: M \to M$ by:
$$
\begin{align*}
(Id(f))(x) &= f(x)\\
(Uf)(x) &=
\begin{cases} \tfrac{1}{2}f(2x) &\text{if }x \leq 1/2\\x &\text{if }x \geq 1/2\end{cases}\\
(Vf)(x) &= x
\end{align*}
$$
For any $f \in M$, we have the following identities:
$$
\begin{align*}
A \circ (Uf) &= f \circ A\\
B \circ (Uf) &= (Vf) \circ B
\end{align*}
$$
As a result, we can reinterpret this in terms of the one-object category $M$: we get three functors $Id,U,V: M \to M$ and natural transformations $A: U \to I$ and $B: U \to V$.
Upon taking geometric realization, we get a space $BM$, these functors turn into continuous maps $Id, U, V:BM \to BM$ and homotopies from $U$ to $Id$ and from $U$ to $V$. However, $V$ is a constant map, and so this says that the homotopy type of $BM$ is contractible.
(I believe that I learned this from somewhere in a paper of Lurie's, but I can't find it currently.)
