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Let $X$ and $Y$ be topological spaces. Suppose we have an isotopy between maps $f, g: X\to Y$.

The question is that is there a homeomorphism $h: Y\to Y$ such that $h\circ f =g$?

I am especially interested in the case when $Y$ is a surface and $f, g$ are embeddings.

Is this true or do we need more conditions? Is there any analogous result?

Thank you in advance.

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What exactly do you mean by isotopy if $f$ and $g$ aren't embeddings? –  Jason DeVito Jan 12 '13 at 1:10
    
(Assume f and g are embeddings) In the smooth manifold setting, If $X$ is compact and $G$ is the isotopy youve assumed, there is an "ambient isotopy" of the identity map on Y. I.e. an isotopy $H: Y\times I \rightarrow Y$ such that $H(f,t)= G$ and $H_0 = 1$. Then $H_1$ is your homeomorphism. See Kosinski's "Differential Manifolds" theorem 5.2 in chapter 2. I'm not sure about the non compact case. Weirdly I was just discussing this in the "differential-topology" section. –  Tim kinsella Jan 12 '13 at 1:18
    
Be aware that often when people say "isotopy" they mean "ambient isotopy," and ambient isotopy does give a homeomorphism like the one you ask for. –  Grumpy Parsnip Jan 12 '13 at 12:19

1 Answer 1

As pointed out here, this is not true in general in the topological case (although it is true in the differential case, as observed by Tim above). The interesting link in that answer is this.

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