$H: \mathbb{R^2 } \rightarrow \mathbb{R^2}$ be a homeomorphism such that $H(\partial B_1)= \partial B_2$. Show that $H(B_2)=B_1$.

Question

Let $B_1,\ B_2 \subset \mathbb{R^2}$ be disks, and let $H: \mathbb{R^2 } \rightarrow \mathbb{R^2}$ be a homeomorhism such that $H(\partial B_1)= \partial B_2$. Show that $H(B_2)=B_1$.

My attempt: Since $\partial B_1$, and $\partial B_2$ are 1-spheres, by Schönflies Theorem there exist homeomorphisms $H_1,H_2: \mathbb{R^2} \rightarrow \mathbb{R^2}$ such that $H_1(S^1)= \partial B_1$, and $H_2(S^1)= \partial B_2$.

So, $S^1 = H_i^{-1}(\partial B_i)$ for $i=1,2$.

$$H_1^{-1}(\partial B_1)=H_2^{-1}(\partial B_2)$$ $$H_2 \circ H_1^{-1}(\partial B_1) = \partial B_2$$

It is clear that $H_2 \circ H_1^{-1}$ is a homemorphism. Now, I wish to prove the following claim.

Claim: $H_2 \circ H_1^{-1}(B_1)=B_2$.

This is all I have done so far. Am I on the right track? I am open to any help.

Additionally, even though it seems a bit irrelevant I have another question. If we have a homeomorphism $H: \mathbb{R^n} \rightarrow \mathbb{R^n}$, and we have two homeomorphic subsets of $\mathbb{R^n}$, say $B$ and $D^2$,(unit disk), can we say directly that $H: B \rightarrow D^2$ is also a homeomorphism?

Answer 1

Hint: (a very simple special case of) the Jordan curve theorem tells you that both $\partial B_1$ and $\partial B_2$ separate the plane into an "inside" (bounded and connected) region (say $I_1$ for $\partial B_1$ and $I_2$ for $\partial B_2$) and an "outside" (unbounded and connected) region (say $O_1$ for $\partial B_1$ and $O_2$ for $\partial B_2$). Under your homeomorphism, a simple argument using connectivity says that you must have $H(I_1) = I_2$ or $H(I_1) = O_2$. But the latter case is impossible, because it implies that the compact set $H(I_1 \cup \partial B_1)$ is homeomorphic to the non-compact set $O_2 \cup \partial B_2$. So $H(B_1) = H(I_1 \cup \partial B_1) = I_2 \cup \partial B_2 = B_2$.

Answer 2

You are asked to prove something about a given homeomorphism $H$. But your proof shows nothing about $H$, it just shows something about two other homeomorphisms that you cooked up by two applications of the Schönflies Theorem.

Answer 3

HINT:

In lay terms, since it takes the boundary to the boundary, it should take interior (exterior) points to interior(exterior) points. So now we only have to define interior (exterior) point of a simple closed curve ( homeomorphic to $S^1$) in an topological way, which will be therefore invariant under homeomorphisms. It is the well known one: a point is inside (outside) a simple closed curve if and only the winding number of the point with respect to the curve is $\pm 1$ ($0$).

This approach is easily generalized to $\mathbb{R}^n$, with a corresponding definition of the index.

$\bf{Added:}$ There is a simpler way to define topologically the outside of a ball. A point $P$ is outside of a ball if and only if there exists a path $\gamma \colon [0, \infty) \to \mathbb{R}^n$, $\gamma(0) = P$, $\lim_{t\to \infty} \gamma(t) = \infty$ and $\gamma$ does not intersect the boundary (so it is in the connected component of $\infty$).