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My question was inspired by this question.

Let $\Omega\subset \mathbb C^n$ be an open bounded set, and $f\colon \overline \Omega \to\overline \Omega$ a continous map such that

$$\forall x\in \partial \Omega,\quad f(x)=x.$$

Is $f$ surjective?

We know that if $\Omega\subset \mathbb R^n$, this is the case (you can find a proof here or here).

But I don't think the result holds for $\mathbb C$ instead of $\mathbb R$, though I am not able to find a counterexample.

I also think that if $f$ is differentiable, then $f$ is holomorphic so for all $z\in \Omega$, $f(z)$ is uniquely determined by it's value on $\partial \Omega$ so $f=\mathrm{id}_\Omega$ and the result holds. So a counterexample would require $f$ continuous but not differentiable.

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    $\mathbb{C}^n =\mathbb{R}^{2n}$ – 2017-01-09
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    @Timkinsella Yes, but is it sufficient ? I thought the condition $f(x)=x$ for $x\in \partial \Omega$ was weaker in that case, but may be I was wrong. – 2017-01-09
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    @E.Joseph Why would you think this and what would be the meaning? – 2017-01-09
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    @MoisheCohen No, I was wrong, open sets of $\mathbb C$ are exactly the same as open sets of $\mathbb R^2$ so the question is trivial. – 2017-01-09

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