Embedding of half open half closed $n$-set in $n$-space

Question

Let $n\geq 2$. Set $\Sigma= \{x\in \mathbb{R}^n: 1\leq |x|<2\}$. Assume $h:\Sigma \rightarrow \mathbb{R}^n$ is continuous and injective.

Question: Must $h$ also be an embedding?

Some thoughts:

1. $h|_{\operatorname{int}\Sigma}$ is an embedding.

2. $h(\operatorname{int}\Sigma)$ is open.

3. If $h$ defined on $\overline{\Sigma}$ then YES.

4. If $\dim \Sigma <n$ then NO. Consider $\alpha:[0,1)\rightarrow \mathbb{R}^2$, $\alpha(t)=(\cos(2\pi t),\sin(2\pi t))$.

5. The map $h|_{\mathbb{S}^{n-1}}$ is an embedding.

Observe $h(\Sigma)=h(\operatorname{int} \Sigma)\coprod h(\mathbb{S}^{n-1})$.

Answer 1

The answer is yes. It suffices to show that $h$ restricted to each of the sets $1\leq |x|\leq 3/2$ and $3/2\leq |x|<2$ is an embedding, because the image of points near the boundary $|x|=2$ will be away from the image of $1\leq |x|\leq 3/2$. Indeed, by the generalized Jordan-Brouwer separation theorem $1\leq |x|\leq 3/2$ and $3/2\leq |x|<2$ will be in different components of $\mathbb{R}^n\setminus f(\{|x|=3/2\})$.

Thus the only problem is to verify that $h$ is an embedding near $|x|=2$. However, according to The generalized Schoenflies theorem of Mazur and Brown the map $h$ restricted to $3/2\leq |x|<2$ extends to an embedding of $|x|<2$ and hence $h$ is an embedding of $3/2\leq |x|<2$.

Here is a second (even more elementary) argument:

According to Brouwer's invariance of domain theorem $h$ restricted to $1<|x|<2$ is an embedding. It is also an embedding of $1\leq |x|\leq 3/2$ (by compactness of the domain). Since the image of $|x|=1$ is at a positive distance to the image of $3/2\leq |x|<2$ (by the Jordan-Brouwer separation theorem as explained above), it follows that $h$ is an embedding of $1\leq |x|<2$.