# Extension of homeomorphisms

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?

• No, not even if you ask that $f$ and $g$ be proper. For instance, any (long) knot in $\mathbb R^3$ gives an example. – Marco Golla Nov 21 '20 at 15:22
• @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 $k\leq n+m$. – James_T Nov 21 '20 at 15:26
• I don't understand the question. It seems $f$ and $\phi \circ \iota_m^{m+k} \circ g$ don't have the same target. – skupers Nov 21 '20 at 16:08
• Ah, you mean it should read $\iota_m^{m+k}\circ f$? Sorry about the notation abuse. – James_T Nov 21 '20 at 16:13

## 1 Answer

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 $$\iota_n^{2n}(K) = \phi(\iota_n^{2n}(K'))$$.

• @skuppers, just one question. Will $\phi$ preserve orientation in general? – James_T Nov 21 '20 at 16:55
• You can assume it does by composing with a reflection that is the identity on the image of $\iota_n^{2n}$ if necessary. – skupers Nov 21 '20 at 16:57
• A nice reference for the Klee trick is Chapter 3 of sites.math.rutgers.edu/~sferry/ps/geotop.pdf. – skupers Nov 30 '20 at 15:08
• Thanks a lot! I'll take a careful look :) – James_T Dec 1 '20 at 14:38
• I was thinking some more about this. Does a dual result also hold? That is, if $f,g:\mathbb{R}^m\rightarrow \mathbb{R}^n$ are retracts; can we use Klee's trick to get the dual result to infer the existence of an auto-homeomorphism on $\mathbb{R}^{2m}$ taking $f(K)$ to $g(K)$? – James_T Dec 4 '20 at 15:29