If $f:\overline{U}\longrightarrow \overline{V}$ is a homeomorphism, then $f|_{Bd(U)}$ is a homemorphism on $Bd(V)$ Let $U,V\subset \mathbb R^n$ open and $f:\overline{U}\longrightarrow \overline{V}$ a homeomorphism. Let $U=int(\overline{U})$ and $V=int(\overline{V})$. Show that $f|_{Bd(U)}$ is a homeomorphism in $Bd(V)$. I recall that $Bd(A)=\overline{A}\backslash int(A)$ is the boundary.
Proof
By the invariance of domain, $f|_U$ is a homeomorphism in $f(U)$. How can I show that $f|_{Bd(U)}$ is a homeomorphism in $Bd(V)$ ? 
My idea is to show that $f(U)=V$ and thus $$f(Bd(U))=f(\overline{U}\backslash U)\underset{f\ bijection}{=}f(\overline{U})\backslash f(U)=\overline{V}\backslash V=Bd(V)$$
but I have difficulties to show that $f(U)=V$.
1) How can I do it ?
2) For the fact that $f|_{Bd(U)}$ is a homeomorphism in $Bd(V)$ is the a result that say that if $f:A\longrightarrow B$ is a homeomorphism then if $U\subset A$, then $f:A\longrightarrow f(A)$ is a homeomorphism ? If yes, then it's finish by the fact that $f(Bd(U))=Bd(V)$, no ?
 A: HINT:
If $A\subset X$, $B\subset Y$ and $f\colon A \to B$ is a homeomorphism then $f(\overset{\circ}{A})  \subset\overset{\circ}{B}$  and $f^{-1} (\overset{\circ}{B})  \subset\overset{\circ}{A}$, because $f$, $f^{-1}$ are open map ( no need for the invariance of domain result). Therefore, $f$ maps homeomorphically $\overset{\circ}{A}$ onto $\overset{\circ}{B}$, and so it also maps the complements  in, respectively, $A$ and $B$. But the complements are exactly $\partial A$ and $\partial B$, if $A$ and $B$ are closed in $X$, respectively $Y$. 
Now we also have to check that under the hypotheses $U = Int(\bar U)$ we have $\partial \bar U = \partial U$. Indeed, for a set $C$ we have $\partial C = \bar C \backslash \overset{\circ}{C}$. So $\partial \bar U  = \bar (\bar U)  \backslash \overset{\circ}{\bar U}= \bar U \backslash U = \partial U$.  
A: $f|_{Bd(U)}$ is obviously an homeomorphism in $f(Bd(U))$. As you have shown, $f(Bd(U))=Bd(V)$, therefore $f:Bd(U)\to Bd(V)$ is a homeomorphism. It's not more complicate than that.
For $f(U)=V$, you know that $U=\bigcup_{T\subset \bar U\ open}T$ and $V=\bigcup_{O\subset \bar V\ open}O$. Since $f$ is an homeomorphism, $$f(U)=\bigcup_{T\subset \bar U\ open}f(U)=\bigcup_{O\subset \bar V\ open}O=V$$
because $T\subset \bar U$ is open $\iff$ $f(T)\subset \bar V$ is open. 
