Homeomorphism preserves regular openness

Question

Let $f: X \to Y$ be a homeomorphism between two topological spaces. Let $U \subset X$ be a regular open connected set, i.e., $\text{int}(\bar{U}) = U$. Let $V = f(U)$. Then $f|_{U}: U \to V$ is a homeomorphism on the two subspaces. Is it true $V$ is also regular open? I know $f(\bar{U}) = \bar{V}$ and $f(\partial U) = \partial V$ if $f$ is homeomorphism. But how do we conclude $V = \bar{V} \setminus \partial V$?

Answer 1

Hint: Since $f$ is a bijection, $$f(U) = f(\bar{U}) \backslash f(\partial U).$$