I work with weak Hausdorff $k$-spaces (so all spaces are $T_1$). The internal hom is denoted by $\mathbf{TOP}(-,-)$. Let $\mathcal{G}$ be the topological group of nondecreasing homeomorphisms from $[0,1]$ to itself (the composition law being the composition of maps). I denote by $\mu_\ell:[0,\ell]\to [0,1]$ the homothetie $x\mapsto x/\ell$ with $\ell>0$.

I consider a triple $(X,P_{a,b},P_{b,c})$ where $X$ is a topological space, $a,b,c$ are three distinct points of $X$, $P_{a,b}$ is an arbitrary subspace of $\mathbf{TOP}([0,1],X)$ such that for all $\gamma\in P_{a,b}$, $\gamma(0)=a$ and $\gamma(1)=b$, $P_{b,c}$ is an arbitrary subspace of $\mathbf{TOP}([0,1],X)$ such that for all $\gamma\in P_{b,c}$, $\gamma(0)=b$ and $\gamma(1)=c$. I suppose $P_{a,b}$ and $ P_{b,c}$ closed under the reparametrization by $\mathcal{G}$.

I consider the space of paths $P_{a,b}*P_{b,c} \subset \mathbf{TOP}([0,1],X)$ defined as follows: every path $\gamma$ of $P_{a,b}*P_{b,c}$ is of the form :

  1. We choose $(\ell_1,\ell_2)$ such that $0<\ell_1,\ell_2<1$, $\ell_1+\ell_2=1$
  2. $\gamma = (\gamma_1.\mu_{\ell_1}) * (\gamma_2.\mu_{\ell_2})$ with $\gamma_1\in P_{a,b}$ and $\gamma_2\in P_{b,c}$ where $*$ is the Moore composition.

I consider another triple $(X',P_{a',b'},P_{b',c'})$ as above such that there exists a continuous map $f:X\to X'$ such that $f(a)=a'$, $f(b)=b'$, $f(c)=c'$ inducing weak homotopy equivalences $P_{a,b} \to P_{a',b'}$ and $P_{b,c} \to P_{b',c'}$.

Is it true that the continuous map $g:P_{a,b}*P_{b,c} \to > P_{a',b'}*P_{b',c'}$ induced by $f$ is a weak homotopy equivalence ?

The difficulty of this question is to understand $P_{a,b}*P_{b,c}$. There is a continuous map $$\mathrm{Int}(\Delta^1) \times P_{a,b}\times P_{b,c} \to P_{a,b}*P_{b,c} \ \ (1)$$ where $\mathrm{Int}(\Delta^1)$ is the interior of the $1$-simplex which takes $((\ell_1,\ell_2),\gamma_1,\gamma_2)$ to $(\gamma_1.\mu_{\ell_1}) * (\gamma_2.\mu_{\ell_2})$ which is surjective. The map (1) is actually a quotient map. It is not necessarily one-to-one because the paths of $P_{a,b}*P_{b,c}$ may stop at $b$. If none of the paths of $P_{a,b}*P_{b,c}$ stops at $b$ (i.e. the inverse image is one point of $]0,1[$, then the map (1) above is a homeomorphism. And if none of the paths of $P_{a',b'}*P_{b',c'}$ stops at $b'$ as well, then the map $g$ is a weak homotopy equivalence.

2) MOTIVATION (EDIT 02/13/2020)

I add the motivation in the hope that someone could have a suggestion by reading it. The proof of the left properness of the q-model category of multipointed $d$-spaces given in Left properness of multipointed d-spaces is incomplete. I forgot to treat the case of the generating cofibration $R:\{0,1\}\to \{0\}$ identifying two states, i.e. that the pushout of a weak equivalence along $R:\{0,1\}\to \{0\}$ is a weak equivalence. The problem above is a particular case. The similar fact for the category of flows is trivial. The generating cofibration $R:\{0,1\}\to \{0\}$ is not necessary to build the cellular objects so it would not be a real problem if the category of multipointed $d$-spaces was only "almost" left proper.


If the spaces are not $T_1$, then I run into several complicated point-set topology obstacles because there are three non-homeomorphic topologies on two points: the discrete one, the indiscrete one and the Sierpinski topology. To avoid pointless complicated mathematical arguments, it is sufficient to work with the correct separability condition for $\Delta$-generated spaces as explained in the model category structure and its left determinedness (Section 3). The question above is written down in the framework of weak Hausdorff $k$-spaces because I do not think that the local presentability condition has any role in this problem.

  • $\begingroup$ I don't understand how $\mathcal{G}$ is a group? Shouldn't the inverse of a nondecreasing function be nonincreasing? $\endgroup$ – Jeff Strom Jan 22 at 22:18
  • $\begingroup$ @JeffStrom $\mathcal{G}$ is a group for the composition of maps. It contains only homeomorphisms. $\endgroup$ – Philippe Gaucher Jan 22 at 22:34
  • $\begingroup$ Of course. Too silly. $\endgroup$ – Jeff Strom Jan 23 at 3:22

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