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I have been seing a lot of videos about topology where they describe homeomorphism as the action of "squishing and moving without cutting" or things along this line.

I thought of a way to formalize it and I came up with this:

Let $A,B \subseteq \mathbb R^n$ thus they are homeomorphic if there exist a continous function $H:A×I \to \mathbb R^n$ such that:

  1. For all $a \in A$: $H(a,0)=a$

  2. For all $t \in [0,1]$ the function $H|_{A×{t}}$ is injective(because when we squish and move in the real world no two points of can be in the same place)

  3. $H(A×\{1\})=B$

I haven't been able to prove or disprove this in general(the injective assumption ruined my attempts to disprove it).

I was able to prove it under the assumption that $A$ is open. Indeed the function $f:A \to \mathbb R^n$ defined by $f(a)=H(a,1)$ is an injective continous map from an open subset of $\mathbb R^n$ to $\mathbb R^n$ and thus from the invarience of domain theorem the sets $A,g(A)=B$ are homeomorphic as we want. But this doesn't realy use most of the assumtion of $H$.

Can we prove this at least for closed sets or simply connected sets?

If this is not true, Is there a way to make this interpetation formal such that it will work at least for open and closed sets?

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    $\begingroup$ I presume you are familiar with the notion of a homotopy (since your definition seems to be based on the idea). You should look up regular homotopy, isotopy, and ambient isotopy. They are subsequent strengthenings of homotopy along the lines you wrote that are strictly stronger than homotopy. $\endgroup$ Oct 4, 2022 at 7:08
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    $\begingroup$ Since the pushforward can be taken to be the identity map on $\mathbb R^n$ (by identifying the tangent spaces with the space itself), the concept of an immersion is exactly an injective map. In that sense, what you have defined above is essentially exactly a regular homotopy. If you instead demand injectivity of the images (so the manifolds don't appear to "pass through" themself extrinsically during the homotopy) then you will essentially end up with an isotopy. $\endgroup$ Oct 4, 2022 at 7:13
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    $\begingroup$ It's definitely a common confusion to mix up isotopy and homeomorphism since people usually show the former when trying to give an example of the latter (which can even yield some false intuition). For some discussion on the distinction, see here: math.stackexchange.com/q/1571166/269764 $\endgroup$ Oct 4, 2022 at 7:22
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    $\begingroup$ It's also worth noting a homeomorphism requires an inverse, and that homotopies really talk about maps and not spaces. If we do the natural thing and equate a space with its natural embedding map (inclusion into the fixed ambient space), then your question boils down to asking if isotopic maps induce homeomorphisms of their images and vice-versa $\endgroup$ Oct 4, 2022 at 7:27
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    $\begingroup$ Final note: it's true that homeomorphic spaces have the same isotopy type, but the converse is false (which I think answers your question): for example, consider a sphere with a line segment coming off it and let your homotopy slowly shrink the line segment until it "disappears". At no point is this not injective nor continuous, but the sphere is clearly not homeomorphic to itself with a line segment sticking out (for more on isotopy see here:encyclopediaofmath.org/wiki/…) $\endgroup$ Oct 4, 2022 at 7:50

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Your approach does not work properly. Of course you can restrict to subspaces $A, B \subset \mathbb R^n$. You try to decribe a deformation of $A$ into $B$ inside $\mathbb R^n$. Here are some problems.

  1. You only require the $H_t$ to be injective.

This allows to deform $[0,1) \cup [2,3] \subset \mathbb R$ into $[0,2] \subset \mathbb R$. These sets certainly do not deserve to be considered as homeomorphic.

  1. You could strenghten your assumptions by requiring that the $H_t$ are embeddings.

This would be circular because the concept of embedding is based on the concept of homeomorphism.

Anyway, if you work with this strengthened variant, you get the concept of isotopy in the ambient space $\mathbb R^n$. This is stronger than homeomorphism, and it depends on the ambient space $\mathbb R^n$.

  1. The depedency on the ambient $\mathbb R^n$ is inadequate for the concept of homeomorphism.

For $n = 2$ let $A$ be the union of two concentric circles and $B$ be the union of two "distant" circles not enclosing a common point. Then $A, B$ are not isotopic. However, if you consider the same situation for $n =3$ (where $A, B$ are contained in $\mathbb R^2 \times \{0\} \subset \mathbb R^2$), they are isotopic.

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  • $\begingroup$ Knots are classified by smooth or ambient isotopy. All (tame) knots are isotopic in the topological category (simply let the loops shrink over time... This ruins the smooth structure, but gives a topological isotopy). You mention ambient space in your post so it seems you meant to write this, but it's worth noting to the OP that isotopies generally aren't ambient. $\endgroup$ Oct 4, 2022 at 13:09
  • $\begingroup$ @BrevanEllefsen I did not speak about ambient isotopy. For a definition see en.wikipedia.org/wiki/Ambient_isotopy. But also "ordinary" isotopies require an ambient space in which the isotopy takes place. See section "Isotopy" in en.wikipedia.org/wiki/Homotopy#Isotopy. And it is not true that all (tame) knots are isotopic in the topological category. You cannot shrink to a point because the final map is not an embedding. Also see en.wikipedia.org/wiki/Knot_theory. $\endgroup$
    – Paul Frost
    Oct 4, 2022 at 13:31
  • $\begingroup$ I'm well aware of the definitions. All knots are absolutely isotopic in the continuous category, e.g. math.stackexchange.com/q/1311865/269764 $\endgroup$ Oct 4, 2022 at 14:41
  • $\begingroup$ The final map is an embedding trivially, as it's just the identity map $S^1$ to $S^1$ $\endgroup$ Oct 4, 2022 at 15:00
  • $\begingroup$ @BrevanEllefsen You are right concerning tame knots. I shall delete the knot part because a correct presentation would be too technical here. $\endgroup$
    – Paul Frost
    Oct 4, 2022 at 15:46

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