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Here are two supposedly equivalent definitions of a smooth isotopy (M and N are smooth manifolds):

  1. A smooth level preserving imbedding $M \times I \rightarrow N \times I$

  2. A smooth map $ F: M\times I \rightarrow N$ such that each $F_t$ is an imbedding

For the life of me I cannot show that an $F$ satisfying 2 will induce an imbedding as in 1. Clearly the map will be an injective immersion but why is it a homeomorphism onto its image? I tried showing its proper but I can't see why a sequence of points escaping from compact sets in M couldn't be mapped into a single compact set in N at different times, even though the times converge to some time in the interval. Please help! I'm probably just being very stupid. Or perhaps there's a counter example?

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This seems to be a counterexample. please let me know if it sounds fishy. I'd appreciate any feedback at all:

Choose an isotopy,$H'$, between the identity map on $\mathbb{R}$ ($:=H'_1$) and some diffeomorphism

$H'_0: \mathbb{R} \rightarrow (-\infty , -1)$

which is stable on $(-\infty , -2]$. Also Choose this isotopy so that it is stable on the set $(-\infty , -2]$. Also ensure that for all $t \neq 0$, we have that $H'_t$ is a diffeomorphism of $\mathbb{R}$ with itself (we can do this right?). Now define an isotopy

$H'' : \mathbb{R} \times I \rightarrow \mathbb{R}^2, (x,t) \mapsto (H'(x,t)^2 +t, H'(x,t))$

Intuition: we are lowering the graph of a parabola so that the vertex approaches the origin but so that the points that map to the vertex at different times diverge to positive infinity.

Finally define an isotopy (in the sense of 2)

$H : (\mathbb{R} \sqcup \mathbb{R}) \times I \rightarrow \mathbb{R}^2$ as $H := H'' \sqcup \{(x,t) \mapsto (x,0)\}$.

Then this will be an isotopy in the sense of 2, but the level-preserving map it induces is not proper: there is a sequence of points on the y-axis of $\mathbb{R}^2$ (the vertex of each parabola) which converge to the origin but whose preimages stretch out to positive infinity in the first copy of $\mathbb{R}$. If we include the origin with this sequence then we get a compact set whose preimage is not compact.

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Yes you can create $H'$ of this kind in the form $f(x)+t*(x-f(x))$ where $f$ is bounded above and agrees with identity for small $t$. However, I only see a contradiction with the map being proper. Definition 1 did not require a proper embedding, just an embedding. – user53153 Jan 11 '13 at 6:22
Thanks Pavel. But surely if a map is to be a homeomorphism onto its image then the preimage of a compact set contained in the image must be compact, no? – Pliny the Ill Jan 11 '13 at 6:29

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