Assume $S$ is an $(n-1)$-topological sphere in $\mathbb{R}^n$ that is bicollared. Meaning, there exists an embedding $\phi:S\rightarrow \mathbb{R}^n$ that extends to an embedding $F:S\times [-1,1]\rightarrow \mathbb{R}^n$. Specifically, $F(s,0)=\phi(s)$ for all $s\in S$.
Why does there exist a homeomorphism $h:\mathbb{R}^n\rightarrow \mathbb{R}^n$ such that $h(S)=\mathbb{S}^{n-1}$? As the answer in the post suggests: Are any two $n$-balls in $\mathbb{R}^n$ isotopic?
The answer quotes the schoenflies theorem, but for me schoenflies is:
Theorem. Let $\phi:\mathbb{S}^{n-1}\rightarrow \mathbb{S}^n$ be a topological embedding, $n\geq 2$. $A$ is the closure of a component of $\mathbb{S}^n-\phi(\mathbb{S}^{n-1})$ and $A$ is a manifold with boundary, then $A$ is homeomorphic to $D^n$.
Another question: Why does it follow that any closed topological $B$ with collared boundary, there exists a homeomorphism on $\mathbb{R}^n$ taking it to the closed disk?
Approach:
Since we have a bicollaring $C$ of the boundary, assuming $C$ is open, we may identify $C$ with $\mathbb{R}^n-pt$. So we can find a homeomorphism $g:\mathbb{R}^n-pt\rightarrow \mathbb{R}^n-pt$ so that $g(S)=\mathbb{S}^{n-1}$. Now throw back in the point to get the homeomorphism $\mathbb{R}^n\rightarrow \mathbb{R}^n$.
Is this correct? Where was Schoenflies needed? as the answer in the linked post suggests?
Best Answer
Your approach does not work because although there may indeed exist a homeomorphism $g : \mathbb R^n - C \to \mathbb R^n - pt$, it is quite possible that the set $\mathbb R^n - C$ does not consist of a single point --- take, for example, $C = \{p \in \mathbb R^n \mid .5 < |p| < 1.5\}$ and $g(p) = \tan\bigl((|p|-.5) \pi / 2\bigr) \cdot p $. In this kind of situation I'm sure you can see that $g$ does not extend to a bijection $\mathbb R^n \mapsto \mathbb R^n$, let alone a homeomorphism.
Instead, one can prove in a reasonably straight forward manner that the following two versions of Schönflies Theorem are equivalent to each other:
To prove that Version 1 implies Version 2, suppose we are given $\phi$ and $A$ as in Version 2. Since $A$ is a manifold with boundary, its boundary $\partial A$ has a collar, consisting of an open neighborhood $V \subset A$ and a homeomorphism of pairs $$f : \bigl(\partial A \times [0,1), \partial A \times \{0\}\bigr) \to \bigl(V,V \cap \partial A\bigr) $$ Choose also a point $p \in \mathbb S^n - A$ and a homeomorphism $k : \mathbb S^n - \{p\} \to \mathbb R^n$. Note that $f(\partial A \times \{.5\}) \subset \mathbb S^n$ is bicollared; and we can restrict the bicollaring so that it is disjoint from $\{p\}$. It follows that $S = k \circ f(\partial A \times \{.5\})$ is bicollared. Version 1 therefore applies to $S$, we obtain a homeomorphism $h : \mathbb R^n \to \mathbb R^n$ such that $h(S) = \mathbb S^{n-1}$. Letting $A' \subset A - f(\partial A \times [0,.5))$, it follows that $h \circ k \circ f : A' \to D^n$ is a homeomorphism; it should now be very easy to piece this homeomorphism together with the homeomorphism $f : \partial A \times [0,.5] \to \text{closure}(A-A')$ to give a homeomorphism $A \to D^n$.
The proof that Version 2 implies Version 1 is easier. Given a bicollared topological $n-1$ sphere $S$ embedded in $\mathbb R^n$, we can use the 1-point compactification $\mathbb R^n \cup \{\infty\} \approx \mathbb S^{n}$, which turns $S$ into a bicollared embedding in $S^{n}$. The closures of both components of $\mathbb S^{n} - S$ are therefore manifolds with boundary, and Version 2 can be applied to each of those components, producing homeomorphisms from each to $D^n$. These two homeomorphisms can be easily pieced together, and the point at infinity then removed, to give a homeomorphism $\mathbb R^n \to \mathbb R^n$ taking $S$ to $\mathbb S^{n-1}$.