To what extent are homeomorphisms just deformations?

Question

Background. It is often said that two spaces are homeomorphic if, roughly speaking, one space can be continuously deformed into the other without any tearing and gluing. It is then emphasized that this is more of a guiding principle than a solid fact. Indeed, a well-known 'counterexample' to this heuristic would be double-twisted Möbius strip $M$, which is homeomorphic to the cylinder $C$, but when we visualize the two spaces in $\mathbb{R}^3$, we find that we cannot deform into each other without tearing.

Why does this happen? In a sense this is because $\mathbb{R}^3$ does not have enough room for the desired deformation; there simply aren't enough dimensions to 'untwist' the double twist without tearing things up. This leads us to the following:

Preliminary question. Suppose I were to embed $M$ and $C$ in $\mathbb{R}^n$ for some large $n$. Can I now deform one into the other without tearing or gluing?

The answer turns out to be yes, and in fact I believe $n = 4$ already suffices. If you care for the informal proof, read on; if not, scroll down to the main part. Let us start with a suitable embedding of a double twist.

This picture visualizes an embedding of the double twist within $\mathbb{R}^4$, where the first three dimensions are spatial, and the fourth dimension is represented by different colours. (The choice of colours may be unfortunate, but I did not have any other pens.) Now imagine moving the 'left strip' and the 'right strip' closer together, like so.

Notice that the 'double twist' starts to resemble a loop. Now, if we move the strips yet closer, they will start to overlap spatially. But notice that the colours will remain different on the overlap so that no actual tearing or gluing occurs. Now keep going until the two strips fully overlap each other. At this point the colors have aligned, so that the twist has turned into a 'loop'.

This loop is undesirable, but the solution to that issue is simple. During the proces in which we let the strips overlap, we simply let the size of the loop go to zero. At the very moment that the overlap is complete (and the colors have aligned), the size of the loop reaches, and as such the loop will vanish altogether. 'QED'.

To our main question. We see that our counterexample is no longer a counterexample. We might therefore wonder if this is part of a general phenomenon. Let us make this more precise.

Let $X$ be a topological space, and let $f : X\to \mathbb{R}^n$ and $g : X \to \mathbb{R}^n$ be two embeddings. Denote by $\varepsilon_{n,m} : \mathbb{R}^n \to \mathbb{R}^{n+m}$ the embedding which sends a point $x$ to $(x,0)$. Let's say $f$ can be deformed by embeddings into $g$ if there exists an $m$ and a homotopy from $H : X \times I \to \mathbb{R}^{n+m}$ from $\varepsilon_{n,m} \circ f$ to $\varepsilon_{n,m} \circ g$ such that for all $t\in [0,1]$, the map $H_t : X \to \mathbb{R}^{n+m}$ is a continuous embedding.

Question. Suppose $X$ is a reasonable space (say, a compact manifold with boundary). Let $f$ and $g$ be two embeddings of $X$ into some $\mathbb{R}^n$. Can $f$ always be deformed by embeddings into $g$?

This question can be generalized and specialized in many ways, so feel free to change some of its details.

Answer 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.