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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?

insipidintegrator
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RT1
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  • You can think about the function H as the movement of the points in A along the time – RT1 Oct 04 '22 at 07:01
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    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. – Brevan Ellefsen Oct 04 '22 at 07:08
  • Thanks, I will look it up – RT1 Oct 04 '22 at 07:10
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    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. – Brevan Ellefsen Oct 04 '22 at 07:13
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    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: https://math.stackexchange.com/q/1571166/269764 – Brevan Ellefsen Oct 04 '22 at 07:22
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    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 – Brevan Ellefsen Oct 04 '22 at 07:27
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    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:https://encyclopediaofmath.org/wiki/Isotopy_(in_topology)#:~:text=Two%20spaces%20%2C%20are%20said%20to,analogy%20with%20the%20homotopy%20type).) – Brevan Ellefsen Oct 04 '22 at 07:50
  • Thank you very much, if you want you can write it as an answer and I will accept it – RT1 Oct 04 '22 at 07:52
  • But when you shrink the line, in the end the points of the line will not go(via H) to points of the sphere which will make it not injective? – RT1 Oct 04 '22 at 07:54
  • That's the issue: it's injective at each point in time, and the deformation is continuous, but intuitively we are "collapsing" structure. It thus satisfies the conditions you gave (its an isotopy of embeddings) but we don't get a homeomorphism. This is a good example of why isotopy is actually a rather boring invariant much of the time (we usually exclude some weirdness by requiring smooth isotopy or ambient isotopy, e.g. in knot theory where all knots are isotopic) – Brevan Ellefsen Oct 04 '22 at 08:04
  • But I want that $H|_{A×{1}}$ will be injective too, otherwise we can also take H to be H(x,t)=(1-t)x betwewn $\mathbb R^n$ to $ {0}$ it is injective for all $[0,1)$ and of course they aren't homeomorphic. I'm sorry if I mosunderstood you but I don't understend why the isotopy you have created is injective on $A×{1}$ – RT1 Oct 04 '22 at 08:12
  • I was assuming that was a minor typo on your part, since otherwise your construction doesn't do much: let $G$ be the restriction of $H$ at $t = 1$ with the codomain corestricted to be $B$ (we replace the codomain with something smaller containing the image, in this case the image itself). Then $G$ is an injective and continuous map (by hypothesis) and surjective (since corestricted) from $A \times {1}$ to $B$). Since it's bijective it has an inverse, and since $A \times {1} \cong A$ your claim follows iff the inverse is continuous, which is not generally implied by your other conditions. – Brevan Ellefsen Oct 05 '22 at 06:03

1 Answers1

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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.

Paul Frost
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  • 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. – Brevan Ellefsen Oct 04 '22 at 13:09
  • @BrevanEllefsen I did not speak about ambient isotopy. For a definition see https://en.wikipedia.org/wiki/Ambient_isotopy. But also "ordinary" isotopies require an ambient space in which the isotopy takes place. See section "Isotopy" in https://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 https://en.wikipedia.org/wiki/Knot_theory. – Paul Frost Oct 04 '22 at 13:31
  • I'm well aware of the definitions. All knots are absolutely isotopic in the continuous category, e.g. https://math.stackexchange.com/q/1311865/269764 – Brevan Ellefsen Oct 04 '22 at 14:41
  • The final map is an embedding trivially, as it's just the identity map $S^1$ to $S^1$ – Brevan Ellefsen Oct 04 '22 at 15:00
  • @BrevanEllefsen You are right concerning tame knots. I shall delete the knot part because a correct presentation would be too technical here. – Paul Frost Oct 04 '22 at 15:46