Does a homeomorphism between two open subsets of $\mathbb{R}^n$ map boundary to boundary?

Question

Let $F:U\to V, U, V$ bounded open subsets of $\mathbb{R}^n,$ be a homeomorphism. We don't assume diffeomorphism. I think my question is related to this question, this question, this question, and this question.

Since $U,V$ are bounded, they have non-empty boundaries in $\mathbb{R}^n.$ Can we always say that the homeomorphism $F$ extends to a homeomorphism of the closure, again (by abusing notation) denoted by $F,$ so that $F(\partial U)=\partial V?$ I think the answer is no, but I'm having trouble finding a simple counterexample.

Here's how I'm thinking about it:

Let $u\in \partial U \implies u_n \in U, u_n \to u.$ Consider $L:=\{F(u_n)\} \subset V,$ which is bounded. So $L$ has a convergent subsequence $\{u_{n_k}\}$ converging to a limit point $v \in \bar{V}.$ But $v\notin V$, because if so, then the subsequence $\{u_{n_k}\}\to F^{-1}(v)\in U,$ contradicting the fact that $\{u_{n_k}\}\to u\in \partial U,$ assumed earlier. So $v \in \partial V.$

The above argument shows that any limit point or cluster/accumulation point of $L$ must lie in $\partial V.$ What it does not prove is that the limit point set of $L$ is a singleton set. So this seems like an obstacle in proving that $F$ maps boundary to boundary.

If limit point set of $L$ was indeed singleton, then we'd have: $F(u_n)\to v \in \partial V.$ I feel this also means $v$ must be equal to $F(u),$ but can't get my argument sorted (pretty sure I'm missing something obvious!). If $F$ is continuous up to the boundary, then of course $v=F(u).$ But is it still so if $F$ is not continuous up to the boundary? Maybe some help is useful here. I think that in a potential counterexample, the limit points of $L$ will not be unique...

Answer 1

Here is a simple counter-example. Let $\mathbb{D} = \{ z \in \mathbb{C} : |z| < 1 \}$ denote the unit disk, and let

$$ U = \mathbb{D} \setminus[-1, 0], \qquad V = \{ z \in \mathbb{D} : \Re(z) > 0\}. $$

Then define $F : U \to V$ by the restriction of the principal square root on $U$:

$$ f(z) = \sqrt{z}. $$

Clearly $F$ is continuous with the inverse $F^{-1}(w) = w^2$. However, for $0 < x < 1$,

$$ \lim_{\varepsilon \to 0^+} f(-x + i\varepsilon) = i\sqrt{x} \neq -i\sqrt{x} = \lim_{\varepsilon \to 0^+} f(-x - i\varepsilon) $$

and hence $f$ cannot extend to a continuous function on the closed unit disk $\overline{U} = \overline{\mathbb{D}}$.

Answer 2

It is not even true for biholomorphisms of open domains in $\Bbb C$. The Poincaré uniformisation theorem implies that two simply connected open domains of $\Bbb C$ are biholomorphic. Hence, the two following domains are biholomorphic, and hence homeomorphic.

However, it is clear that their boundaries are not homeomorphic.

Remark: in higher complex dimension, it is a result of Fefferman that a biholomorphism between bounded domains of $\Bbb C^n$, $n\geq 2$, always extends as a CR-diffeomorphism between the boundaries, and in particular, as a homeomorphism.

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Edit: here is a very easy example, pointed at by @PaulFrost.

Answer 3

If there is an extension of $F\colon U\to V$ to a map $\widehat F\colon \overline U\to\overline V$, you correctly proved that the boundary will map to the boundary. Since $\overline U$ will then be compact and hence $\widehat F$ will be uniformly continous, it follows that in order for $\widehat F$ to exist, $F$ needs to be uniformly continuous.

Suppose that $F$ is uniformly continuous. Let $x\in\partial U$. Let $x_n\longrightarrow x$ be a series in $U$ which converges to $x$ and put $a:= \lim_nf(x_n)$. By uniform continuity, for any $\epsilon>0$, you can pick a small enough neighborhood $x\in W\subset \overline U$ such that in $W\cap U$ there is no $y$ with $\left\|f(y)-a\right\|\ge\epsilon$, hence the limit $\lim_{x'\to x}f(x)=a$ is well defined. By a similar argument, we see that $\widehat F$ is continuous. (Again, uniform continuity prohibits the value from varying too much.) If the inverse is also uniformly continuous, it is a homeomorphism too, as you can do the same procedure again for the inverse map $F^{-1}\colon V\to U$, and $\widehat F\circ\widehat{F^{-1}}=\operatorname{id}_U$. No matter if $F^{-1}$ is unif. cont. or not, the map $\widehat F$ is surjective, as for any point $\xi\in\partial V$ together with a convergent sequence $\xi_n\longrightarrow \xi$ in $V$, $\widehat F(\lim_nF^{-1}(\xi_n))=\xi$. It is injective if and only if $F^{-1}$ is unif. cont., as then by compactness of $\overline F$ and bijectivity, there is a continuous inverse map $\overline V\to \overline U$, implying unif. cont. of $F^{-1}$.

Finally, for the fun part, I shall deliver two counterexamples. Let $U=V=\mathbb B_1(0)\subset \mathbb C\cong\mathbb R^2$ be the open unit disk, and \begin{align*}F\colon U\to V, &0\mapsto 0,\\ &x\mapsto x\mathrm e^{\frac{\mathrm i}{1-|x|}}. \end{align*}

Moreover, let $U=\mathbb D\setminus\{0\}$ and $V=\mathbb D\setminus\overline{\mathbb B_{1/2}(0)}$, $x\mapsto x\frac{1+|x|}2$. Then $F$ is not uniformly continuous, but the inverse $G\colon V\to U$ is. So the map $\widehat{G}\colon V\to U$ can be defined and it is surjective, but $\widehat F\colon U\to V$ cannot.

Summary: $F$ lifts to a map $\widehat F\colon\overline U\to \overline V$ with $\widehat F(\partial U)=\partial V$ iff $F$ is uniformly continuous. $\widehat F$ is a homeorphism iff both $F$ and its inverse are uniformly continuous.

Edit: I had forgotten the case of one direction being unif. cont. and the other direction not being that.

Answer 4

Here is a somewhat extreme counterexample. Let $\mathbb D$ be the open unit disc in the complex plane and define $f : \mathbb D \to \mathbb D$ by the formula $$f(z) = z \, \exp(i \, \theta(z)), \qquad \theta(z) = \tan(\pi \, |z| / 2 ) $$ As $|z|$ approaches $1$, $\theta(z)$ approaches $\infty$.

It follows that for any continuous path in $\mathbb D$ that limits on a point $z_0 \in \partial\mathbb D$ in the unit circle, the $f$ image of that path spirals infinitely around and around the inner boundary of the circle, getting closer and closer to the boundary, thus having every point of the unit circle as a limit point.