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In looking for a rigorous proof of the fact that Alexander's Horned Sphere along with the volume that it bounds is homeomorphic to a 3-ball, I have found a number of posts on the web that attempt to explain this fact by showing that we can continuously deform a 3-ball into Alexander's Horned Sphere and the volume that it bounds without moving two distinct points into the same place at any time. The thing is, that's not a homeomorphism, that's an isotopy. In general, when I would like to find a proof of the fact that two spaces are homeomorphic, I often just find numerous links to some video or picture that illustrates an isotopy between the two spaces. However, as far as I can tell, the fact that two topological spaces are isotopic is in no way a sufficient condition for those spaces being homeomorphic, and I never see any justification for the fact that these isotopies yield homeomorphisms. I find this very frustrating because I can generally deduce that two spaces are isotopic with relative ease, but find it much trickier to see that two spaces are homeomorphic. Thus, I end up learning nothing from the time that I spend researching these questions. I guess what I'm asking then, is why do I see these two concepts equated so frequently in specific examples? Moreover, under what conditions is isotopic equivalent to homeomorphic? My guess is that there must be some well-known fact relating the two concepts that I am unaware of, but if this is the case, then I can't seem to find this fact anywhere. Thank you for your time, any answers will be greatly appreciated.

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    $\begingroup$ Alexander horned sphere is not homeomorphic to the 3-ball. It's homeomorphic to $S^2$. $\endgroup$ Dec 11, 2015 at 18:03
  • $\begingroup$ What do you mean by a "space" and an isotopy here? I guess you're thinking specifically of subsets of $\Bbb R^3$, but I can't quite figure out your definition of isotopy. $\endgroup$
    – user98602
    Dec 11, 2015 at 18:06
  • $\begingroup$ Oh yes, I meant the sphere and the volume it bounds. $\endgroup$ Dec 11, 2015 at 18:07
  • $\begingroup$ It even goes so far tha tthe popular explanation of what topology is about is that objects are considered "the same" (i.e., homeomorphic) iff they can be transformed "without tearing and glueing" into another (i.e., have homotopic embeddings into $\Bbb R^3$) $\endgroup$ Dec 11, 2015 at 18:10
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    $\begingroup$ @MichaelFerguson: Then your domains are compact, your $f_0$ is by definition a homeomorphism, and your $f_1$ is a continuous bijection onto your mystery space. Continuous bijections from compact spaces are homeomorphisms. (Normally when one talks about isotopy they just demand that each $f_t$ is an embedding - a homeomorphism onto its image. For noncompact spaces this is stronger than one-to-one.) $\endgroup$
    – user98602
    Dec 11, 2015 at 18:15

1 Answer 1

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1) Isotopy is an equivalence relation on subsets of some space $Y$. (Better yet, on embeddings $X \to Y$.) It usually says that $f: X \to Y$ and $g: X \to Y$ are isotopic if there is a homotopy through embeddings between them. In particular, it's not subsets that are usually called isotopic, it's maps. (One can pass to subsets if they so desired by saying that there is an embedding from $X$ whose image is each subset that the embeddings are homotopic, but... blah.) But in particular, even if you do this, it's automatic that the subsets are homeomorphic. One isn't really interested in the homeomorphism type of the subset, but rather how they're embedded. (See 3).)

2) As mentioned in the comment, this is no longer the same thing if your map is just injective: there are lots of injective maps that are not embeddings. One is to draw the letter 8; that's an injective map from $\Bbb R$ whose image is not homeomorphic to $\Bbb R$. But if your domain $X$ is compact (and codomain Hausdorff, but I have never heard anybody care about isotopy when the codomain isn't, like, a manifold) then this is the same thing as an embedding.

3) Isotopy, as stated, is a dumb equivalence relation. Take knots in $S^3$. Unless we modify the definition, every "tame" knot is isotopic (this includes every knot you've ever seen). You really want to either work with i) ambient isotopy or ii) some smooth/locally flat/PL condition on your embeddings and your isotopies. (i and ii are basically equivalent.) Then, say, the trefoil and circle are no longer isotopic.

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