Embedding of half open half closed set in n-space

Question

Edited: (After comments)

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

Is this a true statement?

Remarks:

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

2. $h(int\Sigma)$ 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.

The reason $h|_{int \Sigma}$ is an embedding is from invariance of domain, which states:

If $f:U\rightarrow \mathbb{R}^n$ is continuous injective map and $U$ open in $\mathbb{R}^n$ then $f(U)$ is open and $f$ is homeomorphism onto image .

The reason $h|_{\mathbb{S}^{n-1}}$ is an embedding is because $h$ is continuous and $K=\mathbb{S}^{n-1}$ is compact. So, $h|_{K}$ must be an embedding.

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