# 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.

• This seems a complicated way of asking what's the homotopy type of $BM$. I'm not aware of any tool beyond the group-completion theorem to do this kind of analysis though – Denis Nardin Oct 1 '18 at 7:16
• @DenisNardin It is just a question about the "state of the art", nothing else. – Philippe Gaucher Oct 1 '18 at 7:42

## 1 Answer

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.)

• Counterexample 4.4.5.19, HTT? – Charles Rezk Oct 1 '18 at 13:32
• @Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it... – Tyler Lawson Oct 1 '18 at 17:08