2
$\begingroup$

EDIT:

  1. First edit after an interesting answer.
  2. $(S,\mathcal{T}_1)$ and $(S,\mathcal{T}_2)$ are homotopy equivalent to the same Quillen cofibrant space.

Let $S$ be a set with two topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ such that the identity map induces a continuous map $(S,\mathcal{T}_1)\to (S,\mathcal{T}_2)$ which is also a homotopy equivalence. I suppose these two topological spaces $\Delta$-generated.

Is it enough to conclude that $\mathcal{T}_1=\mathcal{T}_2$ ?

Motivation: I need to prove that some continuous bijection is a homeomorphism. Since the spaces are $\Delta$-generated, it suffices to prove that every continuous map from $[0,1]$ to $(S,\mathcal{T}_2)$ is a continuous map from $[0,1]$ to $(S,\mathcal{T}_1)$. I would like to use in some way the homotopy equivalence.

$\endgroup$
7
  • $\begingroup$ Is the given bijection part of a homotopy equivalence or are they unrelated? $\endgroup$
    – David Roberts
    Commented Nov 14, 2020 at 10:20
  • $\begingroup$ @DavidRoberts In fact in my situation, the continuous bijection is also a homotopy equivalence. I realized that by reading Gabriel's answer. $\endgroup$ Commented Nov 14, 2020 at 11:08
  • 1
    $\begingroup$ @DavidRoberts You are absolutely right ! I had not realized that either. I'm going to modify my question. $\endgroup$ Commented Nov 14, 2020 at 12:36
  • 1
    $\begingroup$ It's easy to construct a counterexample which is a weak homotopy equivalence, where the target space is a Warsaw circle. I don't know if that would be of interest for you $\endgroup$ Commented Nov 14, 2020 at 14:56
  • 1
    $\begingroup$ @DenisNardin Is the Warsaw circle $\Delta$-generated? It seems to me that maps in from an interval shouldn't be able to "see" the interesting part of the topology. $\endgroup$ Commented Nov 14, 2020 at 15:57

2 Answers 2

6
$\begingroup$

Let $X = S^1$ and $X^\delta$ its discretisation. Then the identity map $\iota: X^\delta \to X$ is a continuous bijection of $\Delta$-generated spaces, although not a homotopy equivalence. But the map of unreduced cones $$C(\iota): C(X^\delta) \to C(X)$$ is a homotopy equivalence and a continuous bijection, and both spaces are $\Delta$-generated.

$\endgroup$
1
$\begingroup$

[answer is to the original question before edit which posited only the existence of a homotopy equivalence rather than the identity map being a homotopy equivalence]

No. Let $T_2$ be the wedge of countably many circles and $T_1$ be the wedge of countably many circles and an interval.

$\endgroup$
2
  • $\begingroup$ A half-open interval? $\endgroup$
    – David Roberts
    Commented Nov 14, 2020 at 10:19
  • 1
    $\begingroup$ @DavidRoberts Yes a half-open interval. Morally speaking, the idea is to dig a hole by adding open subsets. Anyway, that cannot happen in my situation so this answer is very helpful for me. $\endgroup$ Commented Nov 14, 2020 at 11:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .