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.

  • 2
    $\begingroup$ 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 $\endgroup$ – Denis Nardin Oct 1 '18 at 7:16
  • $\begingroup$ @DenisNardin It is just a question about the "state of the art", nothing else. $\endgroup$ – Philippe Gaucher Oct 1 '18 at 7:42

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.)

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

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