Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a continuous function bijective on an open and bounded set $\Omega$. If $f$ maps $\Omega$ to itself, that is $f(\Omega)\subseteq \Omega$ then is it necessarily true that $f$ maps the boundary of $\Omega$ to itself? The condition that $f$ is bijective is clearly necessary, since we can take $f$ to be a constant function defined on $\Omega$ otherwise. Given that $f$ is bijective and continuous, does it necessarily have to map the boundary to itself? If not, are there sufficient smoothness conditions (holomorphic?) we can impose to make it true? I realize that we may need the boundary of $\Omega$ to be smooth also, feel free to impose any kind of restrictions to make the question well defined.
Edit: By bijection on $\Omega$, I mean that the function satifies $f(a)\in\Omega \iff a\in\Omega$ Hopefully, this will help to rule out some trivial counterexamples.