Equivalence of Continuous Monotonic Functions $[0, 1] \to [0, 1]$

Question

A couple of later clarifications to my original post ...

1. the monotonic non-decreasing functions mapping $[0, 1] \to [0, 1]$ are surjective - it appears from the comments that this was misunderstood.
2. Prof. Raussen has kindly assisted me in understanding the proofs in the paper, and I will at some point post an answer giving the proof. In outline, the result depends on three others......
   a) Given a countable set of points in [0, 1] - "stop values" - one can construct an element of $\mathscr M$ with non-trivial (closed) intervals - "stop intervals" - that map to these points.
   b) an element of $\mathscr M$ has at most countably many stop values.
   c) If the stop values of $\phi \subset $ stop values of $\eta$ then there is $\psi$ such that $\eta = \phi \circ \psi$ (all functions being elements of $\mathscr M$).

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Let $\mathscr M$ be the set of continuous monotonic non-decreasing functions mapping $[0, 1] \to [0, 1]$.
Then under the operation of composition $\mathscr M$ is a monoid - a group without inverse: it is easily seen that it is closed, associative and has an identity ($i: [0, 1] \to [0, 1], i(t) = t$).
One also sees that all such functions are surjective (e.g. intermediate value theorem).

In the paper "Reparametrizations of continuous paths - Ulrich Fahrenberg and Martin Raussen" https://arxiv.org/pdf/0706.3560.pdf it seems to be proven that given any $f, g \in \mathscr M$ there are $\mu, \nu \in \mathscr M$ such that $f \circ \mu = g \circ \nu$. I must admit the paper is a little beyond me, and I'm looking for a simple proof for this.

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Background.

The paper shows among other things that re-parameterisation of paths is an equivalence relation, which then formalizes the definition of a curve as an equivalence class of paths. Showing symmetry and reflexivity is easy - the proof above is needed to show transitivity.

(The paper also shows that every path is equivalent to a regular path - i.e. one which does not "stop" at any point. An alternate proof for this can be found here https://math.stackexchange.com/q/3317511.)

Answer 1

Here are a few lemmas that appear to provide an answer. Not so, on a second reading, Lemma 2 is not valid..so my answer is simple, but perhaps way too simple.

Lemma 1. Suppose that $\psi$ is a monotonic non-decreasing function mapping $[0,1]$ onto $[0,1]$. Then $\psi$ is continuous.

Lemma 2. Suppose that $f,h\in\mathscr M$ and let $\mu(x)=\sup f^{-1}([0,h(x)])$. Then $\mu$ is a monotonic non-decreasing function mapping $[0,1]$ onto $[0,1]$ (and hence $\mu$ is also continuous), and $h(x)=f(\mu(x))$, that is $h=f\circ\mu$.

Lemma 3. Suppose that $f,g\in\mathscr M$ and let $h=g\circ f$. Then take $\nu=f$, and $\mu$ as in Lemma 2. We have $f\circ\mu=h=g\circ\nu$.

Reason why Lemma 2 is not valid. Suppose that for some $x$ we have that $f^{-1}(h(x))=[a,b]$ for some $a<b$. Then $\mu(x)=b$ (as defined in Lemma 2) and $\mu^{-1}(t)=\emptyset$ if $a<t<b$, so $\mu$ is not onto, and is not continuous. This seems to suggest that my approach letting $\nu=f$ is way too simplistic and won't work, and that a better definition of $\nu$ should take into account non-degenerate intervals on which $f$ is constant. I may think about this later, will leave this answer as is for now.