$\newcommand{\Fr}{\text{Bd}}$ I denote $\text{Bd}(A)$ the boundary of the et $A$. Let $U,V\subset \mathbb R^n$ open and $f:\overline{U}\longrightarrow \overline{V}$ an homeomorphism. Therefore $$x\in\text{Bd}(\overline{U})\iff f(x)\in \text{Bd}(\overline{V}).$$
I have a theorem that say:
If $U\subset \mathbb R^n$ open and $f:U\longrightarrow \mathbb R^n$ continue and injective, therefore $f$ is open. In particular $f(U)\subset \mathbb R^n.$
1) Does the theorem still hold for $U=\mathbb R^n$ ? I would say yes, but it's just to be sure.
2) My proof is: $$\text{Fr}(U)=\overline{U}\setminus U\quad\text{and}\quad\text{Fr}(V)=\overline{V}\setminus V.$$ Then $$f(\Fr(U))=f(\overline{U}\setminus U)\underset{injectivity}{=}f(\overline{U})\setminus f(U)\underset{surjectivity}{\subset }\overline{V}\setminus f(U).$$ Now, I think that the fact that $f$ is an homeomorphism, then $f(U)=V$, but I'm not really sure about that. And if it is, I don't see how to proove it, so any help is welcome. For the reciprocal, it's the same proof with $f^{-1}:\overline{V}\longrightarrow \overline{U}.$
Let $x\in \text{Bd}(U)$. Since $x\in \text{Bd}(U)$, for all open neighborhood $W$ of $x$ we have $W\cap int(\bar{U})\neq \emptyset$ and $W\cap \bar{U}^c\neq \emptyset$.
If $y=f(x)\notin \text{Bd}(V)\Rightarrow y\in V$, which implies there is an open neighborhood $V_y\subset V$ such that $y\in V_y$. By the theorem, we can show this:
Since $f$ is bijective, $x=f^{-1}(y)\in f^{-1}(V_y)\subset \bar{U}$. But this contradicts $x\in \text{Bd}(U)$ since $f^{-1}(V_y)\cap \bar{U}^c=\emptyset$