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Let $f,g:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be smooth injective and let $n\leq m$. Let $k \in \mathbb{N}$, and let $\iota_m^{m+k}:\mathbb{R}^m\rightarrow \mathbb{R}^{m+k}$ be the canonical inclusion. Suppose also that $f(\mathbb{R}^n)\cong \mathbb{R}^n\cong g(\mathbb{R}^n)$ via some $C^{\infty}$-diffeomorphism.

Fix a compact subset $K\subseteq \mathbb{R}^n$. For what values of $k$, does there exist a homeomorphism $\phi:\mathbb{R}^{m+k}\rightarrow \mathbb{R}^{m+k}$ satisfying $$ \iota_m^{m+k}\circ f(x)= \phi\circ \iota_m^{m+k}\circ g(x) \qquad (\forall x \in K)? $$ It isn't difficult to see that $k\leq m+n$. However, what is the smallest such value of $k$ for which this holds? My intuition says 1...


Reduction to Extension Problem

I guess since $f$ and $g$ are homeomorphisms onto their image then, $h:=g\circ f^{-1}:f(K)\rightarrow g(K)$ is a homomorphism. So the problem reduces to finding an extension of $h$ to all of $\mathbb{R}^{m+k}$. But when does such an extension exist?

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  • $\begingroup$ No, not even if you ask that $f$ and $g$ be proper. For instance, any (long) knot in $\mathbb R^3$ gives an example. $\endgroup$ – Marco Golla Nov 21 at 15:22
  • $\begingroup$ @MarcoGolla I refined my question to bypass this pathology. So now, this comment implies that $k\geq 1$. Also, it is easy to see that $k\leq n+m$. $\endgroup$ – James_T Nov 21 at 15:26
  • $\begingroup$ I don't understand the question. It seems $f$ and $\phi \circ \iota_m^{m+k} \circ g$ don't have the same target. $\endgroup$ – skupers Nov 21 at 16:08
  • $\begingroup$ Ah, you mean it should read $\iota_m^{m+k}\circ f$? Sorry about the notation abuse. $\endgroup$ – James_T Nov 21 at 16:13
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It sounds like you're looking for something like the Klee trick. If $K,K' \subset \mathbb{R}^n$ are compact and homeomorphic, it gives a construction of a self-homeomorphism $\phi$ of $\mathbb{R}^{2n}$ such that $\iota_n^{2n}(K) = \phi(\iota_n^{2n}(K'))$.

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  • $\begingroup$ @skuppers, just one question. Will $\phi$ preserve orientation in general? $\endgroup$ – James_T Nov 21 at 16:55
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    $\begingroup$ You can assume it does by composing with a reflection that is the identity on the image of $\iota_n^{2n}$ if necessary. $\endgroup$ – skupers Nov 21 at 16:57
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    $\begingroup$ A nice reference for the Klee trick is Chapter 3 of sites.math.rutgers.edu/~sferry/ps/geotop.pdf. $\endgroup$ – skupers yesterday
  • $\begingroup$ Thanks a lot! I'll take a careful look :) $\endgroup$ – James_T 4 hours ago

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