A continuous map that fixes the boundary of a domain pointwise is surjective

Solution 1:

The problem can be solved using degree theory.

Suppose otherwise, that $f(\overline{\Omega})\subset\overline{\Omega}$ is strict. Since $f(\partial\Omega)=\partial{\Omega}\subset f(\overline{\Omega})$, then there is some $p\in\Omega$ such that $p\notin f(\Omega)$.

Since $p\notin\partial\Omega=f(\partial\Omega)=\mathbb{I}(\partial\Omega)$, then $\text{deg}(f,\Omega,p)=\text{deg}(\mathbb{I},\Omega,p)=1$ by the Poincare-Bohl theorem.

But the basic properties of degree give that $\text{deg}(f,\Omega,p)\neq0\Longrightarrow\exists x\in\Omega$ such that $f(x)=p$, so we get a contradiction.

Solution 2:

Note that $\Omega$ can be written as a countable union of disjoint open intervals, i.e. $\Omega=\cup I_n$, where $I_n=(a_n,b_n)$. Then $f(a_n)=a_n$ and $f(b_n)=b_n$ and because $f$ is continous, $f$ satisfies the intermediate value property, hence, for each $u\in [a_n,b_n]$ with $f(a_n)\leq u\leq f(b_n)$, you can find $v\in [a_n,b_n]$ such that $f(v)=u$. This implies that $\overline{\Omega}\subset f(\overline{\Omega})$

Edit: As Ayman pointed out, I just proved that $\overline{\Omega}\subset f(\overline{\Omega})$ and because I think this might be helpful to someone, I will not delete the answer.

Remark: When I answered the question, $\Omega$ was a subset of $\mathbb{R}$.