Let $\Omega$ be an open, bounded from $\mathbb{R}^n$ and $f: \overline{\Omega} \rightarrow \overline{\Omega}$ a contiuous function such that $f(x)=x, \forall x \in \partial \Omega$.
Prove that $f(\overline{\Omega})=\overline{\Omega}$.
how to solve it ? any idea please ?
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.
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 $\mathbb{R}$.
If $\Omega = \emptyset$, then we are done.
The plan for nonempty $\Omega$ (as hinted at by JSchlather and Tomás) is to show that $f(\overline{U}) = \overline{U}$ for any connected component $U$ of $\Omega$. The result follows from this, since the connected components exhaust $\Omega$.
We already know $f(\overline{U}) \subset \overline{\Omega}$.
Now, $\overline{U}$ is connected (since $U$ is connected), so $f(\overline{U})$ is connected, since the continuous image of a connected set is connected.
Consequently, $f(\overline{U})$ lies entirely within one connected component of $\Omega$.
Since $f(x) = x$ on $\partial U \subset \overline{U}$, it must be that $f(\overline{U}) \subseteq \overline{U}$.
Therefore, all we need to show now is that $f(\overline{U}) \supseteq \overline{U}$.
Let $B \supset \overline{U}$ be a closed ball neighborhood of $\overline{U}$. Extend $f$ to a function $F: B \to B$ by the identity map on $B\smallsetminus\overline{U}$. Then $F$ fixes the boundary of $B$ (in addition to lots of other elements of $B$), and is continuous. Also $F$ takes $B\smallsetminus\overline{U}$ to itself, and $\overline{U}$ to itself. So it will suffice to show that $F$ is surjective for any map $F: B \to B$ from a ball to itself fixing the boundary.
Let $N_1 \supseteq N_2 \supseteq \cdots$ be a decreasing sequence of closed balls whose intersection is $x$. Let $K_i = F^{-1}(N_i)$. Then $K_1 \supseteq K_2 \supseteq \cdots$ is a decreasing sequence of closed sets such that $F^{-1}(\mbox{int }N_i) \subset K_i$; in particular, $K_i$ is nonempty. (The statement in bold is wrong, as pointed out by @G. Smith.)
Since $B$ is compact, the intersection of all the $K_i$ is nonempty; call it $\mathcal{K}$. Then $f(\mathcal{K}) \subseteq N_i$ for all $N_i$. Therefore $f(\mathcal{K}) = x$.
Now, $x$ was arbitrary, so $f$ is surjective, and we are done.
