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I wondered about the following some time ago:

Let $\Omega \subsetneq \mathbb C$ be a domain. Let $\psi_n: \Omega \to \Omega$ be a sequence of biholomorphisms converging to some $\psi$ locally uniformly on $\Omega$.

Is $\psi$ necessarily surjective?

Some observations:

  • We have $\psi_n(z) = \frac{z}{n}$ as a counterexample on $\Omega = \mathbb C$.

  • There is a biholomorphism $\phi$, which maps $\Omega$ into the unit disc $\mathbb D$. So considering $(\phi\circ \psi_n\circ \phi^{-1}): \phi(\Omega) \to \phi(\Omega)$ we can reduce the general case to the case of $\Omega \subset \mathbb D$ being bounded.

  • Assuming $\Omega$ to be bounded: The derivatives \psi'_n of $\psi_n$ also converge locally uniformly to \psi', so $ \begin{align} \mu(\psi(\Omega)) &= \iint_{\Omega} |J_{\psi}(z)| \; \mathrm dx\,\mathrm dy \\ &\ne \lim_{n\to \infty} \iint_{\Omega} |J_{\psi_n}(z)| \; \mathrm dx\,\mathrm dy \\ &= \lim_{n\to \infty} \iint_{\psi_n(\Omega)} \; \mathrm dx\,\mathrm dy \\ &= \mu(\Omega) \end{align} $ i.e. $\psi$ is 'almost surjective'.

I don't know how one might proceed from here (I hope I haven't made a mistake in my obeservations).

I'd be interested to see an answer to this question. =)

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    @JonasMeye$r$: Oh my... I somehow thought the first $p$a$r$t of the $p$roof of the Riemann mapping theorem would go through; but simple connectedness is crucial there "to make some room" for a disc in the complement. It seems I really haven't thought this through properly... :( Thanks for pointing it out!2012-01-09

1 Answers 1

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No. Consider $\psi_n(z)=\frac{z-(1-1/n)}{1-(1-1/n)z}.$ Then each $\psi_n$ is a biholomorphism of the unit disk $\mathbb{D}$ onto itself, but $\psi_n$ converges locally uniformly on $\mathbb{D}$ to the function $\psi(z)\equiv -1$. Indeed, if $|z|\leq r$, then

$\left|\frac{z-(1-1/n)}{1-(1-1/n)z} - (-1) \right| = \left|\frac{(1/n)(1+z)}{1-(1-1/n)z}\right| $

$\leq (1/n)\left|\frac{2}{1-(1-1/n)r} \right|$ which tends to zero as $n \rightarrow \infty$.

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    What if we require that the limit $\psi$ is non-constant? Do we obtain surjectivity?2014-06-07