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Let $A$ and $B$ be topological spaces (or smooth manifolds with possibly empty boundary if it matters, I'm not sure either way).

Let $X$ be a fixed ambient topological space (or ambient manifold if it matters), i.e. $A, B \subset X$ (and $A, B$ have the subset topology, if that matters). Then I know that:

$A,B$ isotopic in $X$ $\implies$ $A,B$ homeomorphic $\implies$ $A,B$ homotopically equivalent.

Also, in case it matters, when I say "isotopy" I think that I mean "ambient isotopy", i.e. the sense used in knot theory, see for example Grumpy Parsnip's comment here: Isotopy and homeomorphism. Also maybe relevant: Isotopy and Homotopy.

My question:

Now assume that the ambient space is not fixed, i.e. we want to consider all possible ambient spaces for $A,B$ (is this the same as all spaces in which $A$ and $B$ can be embedded?). Then do we have

$A,B$ isotopic in some ambient space $\iff$ $A,B$ homeomorphic? (or $A,B$ diffeomorphic?)

Obviously one direction is true, but I am not sure about the other. I was watching Tadashi Todieka's lectures on algebraic/differential topology on YouTube and sort of got the impression that the other direction might be true, but he never said so either way.

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Chill2Macht
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    Sure just take $A \times I$ as the ambient space. – PVAL-inactive Aug 13 '16 at 17:47
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    @PVAL Doesn't your comment only show isotopic $\implies$ isotopic in some ambient space? To me it seems there is still way to go until we have homeomorphic $\implies$ isotopic. – Hagen von Eitzen Aug 13 '16 at 17:56
  • @HagenvonEitzen No, it shows homeomorphic implies (ambiently) isotopic in some topological space. Note it doesn't make much sense to talk about non-ambient isotopies between different spaces (where different means not identified by a specific homeomorphism). – PVAL-inactive Aug 13 '16 at 18:39
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    Both "isotopy" and "ambient isotopy" require $A$ and $B$ to be embedded in some ambient space (really, $A$ and $B$ have to already be homeomorphic images of some common domain, so homeomorphism is presupposed). The difference is that ambient isotopy presents a stronger condition on the way in which the embeddings are deformed, i.e. it demands that not only the images but also the ambient space be deformed through homeomorphisms. – silvascientist Feb 06 '19 at 16:41
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    The trefoil knot (or any tame knot really) is isotopic to the unknot as the "knotted" portion can be shrunk down injectively until it disapppears, but this cannot be accomplished ambient isotopically as this process does not extend to an isotopy of the ambient space (intuitively, the space around the knot gets "twisted up" too much to accomodate an ambient isotopy). – silvascientist Feb 06 '19 at 16:42
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    It's important to remember that knots are not just topological spaces, they are embeddings of the same space ($S^1$) into the ambient space $\mathbb{R}^3$. Two knots are automatically "homeomorphic", and they can only be distinguished by virtue of the way the circle is embedded in $\mathbb{R}^3$. – silvascientist Feb 06 '19 at 16:44

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