Homeomorphism vs Homotopy Equivalence

Question

Is the following informal way of thinking about the difference of homeomorphisms and homotpy equivalence valid: Let $f\colon X \to Y$ be a homotopy equivalence, then

$$f\text{ is a homeomorphism} \Leftrightarrow f \text{ doesn't change the ''dimension'' of } X.$$

Where by dimension I mean for example take $S^1 \subset \mathbb{C}$ and the ''thick circle'' $A:=\lbrace z \in \mathbb{C} ; 1\leq |z| \leq 2 \rbrace \subset \mathbb{C}$. I can't name a specific homotopy equivalence from $S^1 \to A$ but it is obvious to me that there is one. The homotopy equivalence goes from a 1D object to a 2D object.

If this is true what I'm saying can I replace the informal meaning of dimension above with the formal Hausdorff dimension ?

$$f\text{ is a homeomorphism} \Leftrightarrow f \text{ doesn't change the the Hausdorff dimension of } X.$$

Answer 1

This isn't true. $(0,1)$ is homotopy equivalent to $[0,1]$ since both spaces are contractible (e.g.) but they are not homeomorphic spaces: one is compact, the other isn't. But, the "dimension" of both is the same, under any reasonable definition. They certainly have the same Hausdorff dimension (rigorously, we know this because the Lebesgue measure equals the Hausdorff measure up to a constant and these spaces have finite non-zero Lebesgue measure hence finite non-zero Hausdorff-$1$ measure therefore the dimension must be $1$). There are many examples of this form.

That said, $[0,1]$ is not a manifold and it is possible that if you restrict attention to (connected) manifolds and use the definition of dimension of a manifold then there is a theorem of the type you're looking for - I wouldn't know.

Answer 2

Here is an example of homotopy equivalent but non-homeomorphic manifolds with the same dimension (this answers the question FShrike asks in the last sentence of their post): https://mathoverflow.net/a/263065/123673

Answer 3

Let $f\colon X \to Y$ be a homotopy equivalence, then $$f\text{ is a homeomorphism} \Leftrightarrow f \text{ doesn't change the ''dimension'' of } X.$$

Any continuous function $\mathbb{R}^n\to\mathbb{R}^n$ is a homotopy equivalence (because $\mathbb{R}^n$ is contractible). But of course not all of them are homeomorphism. In fact, we can easily construct a surjective continuous map $\mathbb{R}^n\to\mathbb{R}^n$ that is not injective. It will be a homotopy equivalence that is not a homeomorphism. Its image is $\mathbb{R}^n$, so I guess the question is: does it change the dimension? Well, you did not define this term, but intuitively I would say "no".