If $X\subset\Bbb R^n$ is a topological ball, is $\partial X$ a topological sphere?

Question

Given a set $X\subset\Bbb R^n$ (I am most interested in $n=4$) that is homeomorphic to the $n$-ball $B^n:=\{x\in\Bbb R^n\mid \|x\|\le 1\}$. Is it true that the boundary $\partial X$ is homeomorphic to the $(n-1)$-sphere $S^{n-1}:=\partial B^n$?

This sounds like an inverse of the generalized Schönflies theorem, but I haven't found anything on this.

Answer 1

This follows from invariance of domain. Let $f:B^n\to X$ be a homeomorphism and let $U$ be the interior of $B^n$. Then $f(U)$ is a subset of $\mathbb{R}^n$ homeomorphic to $\mathbb{R}^n$, so it is open in $\mathbb{R}^n$ by invariance of domain. Thus $\partial X\subseteq f(\partial B^n)$. For the reverse inclusion, if $x$ is in the interior of $X$, then $x$ has a neighborhood in $X$ which is homeomorphic to $\mathbb{R}^n$, so $f^{-1}(x)$ has a neighborhood in $B^n$ which is homeomorphic to $\mathbb{R}^n$, which implies $f^{-1}(x)\in U$ (for instance, by invariance of domain considering $B^n$ as embedded in $\mathbb{R}^n$). So $\partial X=f(\partial B^n)$ and is homeomorphic to $\partial B^n$ via $f$.

Answer 2

It is true. As Randall suggests in his comment, we can apply the invariance of domain theorem.

We have an embedding $f : B^n \to \mathbb R^n$ such that $f(B^n) = X$. The restriction of $f$ to the open unit ball $\mathring B^n$ is a continuous injection, thus $X' = f(\mathring B^n)$ is open in $\mathbb R^n$. Since $X' \subset X$, we have $X' \subset \operatorname{int} X =$ topological interior of $X$ in $\mathbb R^n$. Let $f^{-1} : X \to B^n$ be the inverse of $f$. The map $\phi : \operatorname{int} X \to \mathbb R^n, \phi(x) = f^{-1}(x)$, is a continuous injection, thus $\phi(\operatorname{int} X)$ is open in $\mathbb R^n$. Since $\phi(\operatorname{int} X) \subset B^n$, we see that $\phi(\operatorname{int} X) = f^{-1}(\operatorname{int} X) \subset \operatorname{int} B^n = \mathring B^n$. Thus $\operatorname{int} X \subset X'$.

Hence $\operatorname{int} X = X'$ which implies $\partial X = X \setminus \operatorname{int} X = X \setminus X' = f(S^{n-1})$, i.e. $\partial X \approx S^{n-1}$.