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

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 ?

• Did you mean that $\Omega\subset\mathbb{R}$? Feb 23, 2013 at 14:44
• $\Omega \subset \mathbb{R}^n$ sorry Feb 23, 2013 at 14:57
• If we divide $\overline{\Omega}$ in it's connected components, I think we just have to show that if $U$ is an component, then $f(int(U))= U$ where $int$ is interior. Is this true? Feb 23, 2013 at 15:18
• @Tomás Yes, but you probably want to look at the components of $\Omega$. Otherwise $int(U)$ doesn't need to be connected. I think one can show fairly easily that it suffices to prove the result for a connected set. Feb 23, 2013 at 15:50
• For the sufficiency it is enough to prove that $f(int (U))\subset U$. Feb 23, 2013 at 15:59

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.

• What happens if $p\in f(\bar\Omega)\setminus\bar\Omega$ instead? Mar 2, 2013 at 2:41
• $f(\overline{\Omega})\backslash\overline{\Omega}=\emptyset$ since $f:\overline{\Omega}\longrightarrow\overline{\Omega}$.
– mdg
Mar 2, 2013 at 2:48
• D'oh, thanks! :-) Mar 2, 2013 at 2:51
• The various techniques which constitute the area of nonlinear (functional) analysis are all equally impressive and useful, and I believe deserve more attention and study than they get...!
– mdg
Mar 2, 2013 at 20:09

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 $\mathbb{R}$.

• The question says $f : \overline{\Omega} \to \overline{\Omega}$. So equality holds after all. Feb 23, 2013 at 15:00
• All you needed to prove was that $\overline{\Omega}\subseteq f\left(\overline{\Omega}\right),$ since we have $f:\overline{\Omega}\to\overline{\Omega}.$ However, the question has been edited so that $\Omega\subseteq\Bbb R^n$, so you'll need to take a different approach. Feb 23, 2013 at 15:02
• It is true, but now I have to change it. Feb 23, 2013 at 15:04
• so you say that we have only $\overline{\Omega}\subset f(\overline{\Omega})$ ? Feb 23, 2013 at 16:15
• @KarimaMht You have $f(\overline{\Omega})\subseteq \overline{\Omega}$ by assumption. Feb 23, 2013 at 16:21

## Wrong Solution

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.

• This is not correct because $K_i$ may be empty. Since $U$ is open and you are assuming $x\in U$, then eventually $\text{int}N_i\subset U$. But there is no guarantee that $F^{-1}(\text{int}N_i)\neq\emptyset$, which is exactly what we are trying to prove, i.e. there may be no points in $U$ that map to $\text{int}N_i$.
– mdg
Mar 2, 2013 at 7:36
• -blush- Well, I suppose I ought to leave this nonanswer up as a warning to those attempting a solution without using more heavy machinery. It's so tempting. Mar 6, 2013 at 20:17
• It would be nice to be able to prove such a simple statement directly. Using degree theory seems to be shooting an ant with an atom bomb...
– mdg
Mar 6, 2013 at 23:18