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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.

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    $\begingroup$ I understand what "injective on $\Omega$" means. When you write "bijective on $\Omega$", are you trying to convey information other than injectivity? If so, I'm not getting the message. $\endgroup$ – user31373 Jun 25 '12 at 16:51
  • $\begingroup$ This is false for trivial reasons: take the unit disk as $\Omega$ and $f(z)=(1/2) z$. But perhaps you are ruling this example out by saying "bijective on $\Omega$"? $\endgroup$ – Giuseppe Negro Jun 25 '12 at 16:52
  • $\begingroup$ I take bijective to mean that for each point $f(a)\in\Omega \iff a\in\Omega$, I will add this detail to the question. $\endgroup$ – EuYu Jun 25 '12 at 16:55
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Yes. If $f$ is continuous on $\overline{\Omega}$ and $f\colon \Omega\to\Omega$ is a bijection, then $f(\partial\Omega)\subset \partial \Omega$.

Indeed, $f(\partial\Omega)\subset \overline{\Omega}$ by continuity. Suppose $f(x)=y\in \Omega $ for some $x\in\partial \Omega$. Pick a sequence $(x_n)\subset \Omega$ such that $x_n\to x$. By the bijectivity assumption $y=f(x')$ for some $x'\in \Omega$. By bijectivity & continuity, there exists a neighborhood $U$ of $y$ such that $f^{-1}(U)\cap \Omega$ is compactly contained in $\Omega$. Yet $f(x_n)\in U$ for all large $n$, a contradiction.

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    $\begingroup$ Bijectivity is a strong assumption. I think the idea here is that $f$ will map the boundary of $\Omega$ into itself provided $f:\Omega \rightarrow \Omega$ is a proper map, i.e. the preimage by $f$ of every compact set is compact. $\endgroup$ – Malik Younsi Jun 25 '12 at 23:08
  • $\begingroup$ @MalikYounsi Sure thing; I was just answering the question as it was asked. $\endgroup$ – user31373 Jun 25 '12 at 23:11
  • $\begingroup$ Kovaled : I should have posted this as a comment to the OP's question. Anyway, nice answer, +1! $\endgroup$ – Malik Younsi Jun 25 '12 at 23:17
  • $\begingroup$ "By bijectivity & continuity, there exists a neighborhood UU of yy such that f−1(U)∩Ωf−1(U)∩Ω is compactly contained in ΩΩ. " Can someone further explain this part? I don't understand this step. $\endgroup$ – Ameet Sharma Sep 15 '16 at 17:16

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