Let $U,V\subset\mathbb{R}^n$ two bounded open sets and let $F:\overline{U}\to\mathbb{R}^n$ be a continuous map. Assume that $F$ maps $\partial U$ homeomorphically onto $\partial V$.
The question: is it always the case that $V\subseteq F(U)$?
Using the multiplicative property of Brouwer degree, I am able to show that this is true if $\mathbb{R}^n\setminus\overline{U}$ is connected.
In general it is false: let $n=1$ and choose a bi-infinite sequence $(a_n)_{n\in\mathbb{Z}}\subset\mathbb{R}$ which is bounded and increasing and put $U:=\cup_n (a_{2n},a_{2n+1})$ and $V:=\cup_n (a_{2n-1},a_{2n})$. Here $\partial U=\partial V$ and $F:=id_\mathbb{R}$ provides a counterexample.
I'd like to know what can be said under some reasonable hypothesis: what happens if $U$ and $V$ are regular open sets? What if we only require that $\mathbb{R^n}\setminus\partial U$ and $\mathbb{R^n}\setminus\partial V$ have finitely many connected components?