To what extent are homeomorphisms just deformations?

Solution 1:

Here's a theorem. Let $X$ be a closed subset of $\newcommand{\R}{\Bbb R}\R^m$ and $Y$ be a closed subset of $\R^n$. Let $X$ and $Y$ be homeomorphic. Embed $\R^m$ and $\R^n$ into $\R^{m+n}$ by $(x_1,\ldots,x_n)\mapsto (x_1,\ldots,x_m,0,\ldots,0)$ and $(y_1,\ldots,y_n)\mapsto (0,\ldots,0,y_1,\ldots,y_n)$. So we can think of $X$ and $Y$ as closed subsets of $\R^{m+n}$. Then the pairs $(\R^{m+n},X)$ and $(\R^{m+n},Y)$ are homeomorphic.

Let $f:X\to Y\subseteq\R^n$ be a homeomorphism. By Tietze's theorem, this extends to a continuous map $F:\R^m\to\R^n$. Now $(x,y)\mapsto(x,y+F(x))$ is a homeomorphism $\R^{m+n}\to\R^{m+n}$ taking $X$ to the graph $\Gamma$ of $f$. Therefore the pair $(\R^{m+n},X)$ is homeomorphic to $(\R^{m+n},\Gamma)$. Considering $f^{-1}:Y\to X$ instead gives $(\R^{m+n},Y)$ homeomorphic to $(\R^{m+n},\Gamma)$.

This homeomorphism $(\R^{m+n},X)\to(\R^{m+n},\Gamma)$ is part of a continuous family of homeomorphisms $(x,y)\mapsto(x,y+tF(x))$.

So the answer to your question is yes, for closed subsets of $\R^n$ and expanding the ambient Euclidean space to enough dimensions.