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The following is an exercise in Bloch's Intro to Geometric Topology

Let $B \subseteq \Bbb R^2$ be a set homeomorphic to the closed unit disk and $h :\partial B \to \partial B$, a homeomorphism. By Schonflies we can find a homeomorphism $F$ of $\Bbb R^2$ that is $F(D^2)=B$ and $F$ is the identity outside a disk. Then we can expand $F^{-1}\circ h\circ F$ to homeomorphism $g$ of the unit disk. Then $F \circ g \circ F^{-1}$ will give us a homeomorphism of $B$ that is $h$ on the boundary.

My question is if there is a way to expand $F \circ g \circ F^{-1}$ in all $\Bbb R^2$?

Amontillado
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2 Answers2

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Assuming that you do actually know how to extend $F^{-1} \circ h \circ F$ to a homeomorphism of the unit disc, and assuming you know how to do this so that the origin $\mathcal O$ is fixed, then this is possible.

Simply work in the one-point compactification $\mathbb R^2 \cup \{\infty\}$, and use the "inversion" homeomorphism $$g : \mathbb R^2 \cup \{\infty\} \to \mathbb R^2 \cup \{\infty\}, \qquad g(x) = \begin{cases} \frac{x}{|x|^2} & \quad\text{if $x \not\in \{0,\infty\}$} \\ \mathcal O &\quad \text{if $x = \infty$} \\ \infty &\quad \text{if $x=\mathcal O$} \end{cases} $$ You can then restrict $g^{-1} \circ (F^{-1} \circ h \circ F) \circ g$ to $\partial B$, next you can extend that to a homeomorphism $k : B \to B$ which fixes $\mathcal O$, and then the map $g \circ k \circ g^{-1}$, suitably restricted, is the extension that you want.

Lee Mosher
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Any continuous map $f : S^1 \to S^1$ extends to a continuous map $e(f) : \mathbb{R^2} \to \mathbb{R^2}$ by defining $$e(f)(x) = \begin{cases} 0 & x = 0 \\ \lVert x \rVert f(\frac{x}{\lVert x \rVert }) & x \ne 0 \end{cases} $$ Note that $e(g \circ f) = e(g) \circ e(f)$. Thus, if $h : S^1 \to S^1$ is a homeomorphism, then $e(h)$ is a homeomorphism. In fact, $e(h^{-1})$ is the inverse homeomorphism.

Paul Frost
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