# Is the limit of a sequence of biholomorphisms surjective?

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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Did you mean to add the hypothesis that $\Omega$ is simply connected? Otherwise the claim that you can map biholomorphically into the disk is incorrect in general. Also, $\frac{z}{n}$ gives a counterexample on $\mathbb C\setminus\{0\}$. (But Malik's example is both more interesting and has simply connected domain.) – Jonas Meyer Jan 9 '12 at 3:10
@JonasMeyer: Oh my... I somehow thought the first part of the proof 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! – Sam Jan 9 '12 at 13:46

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 $n \rightarrow \infty$.
Ah, I see. So the last of my 'observations' was wrong. Since the derivatives can go to $\infty$ near the boundary, I cannot be so careless with taking the limes outside of the integral. Thank you! – Sam Jan 8 '12 at 22:58
Better yet, $\psi$ is neither surjective nor injective. – Jesse Madnick Jan 9 '12 at 1:01
Yes indeed. Actually, injectivity woulf follow in the case where $\psi$ is non-constant : If $\psi_n$ are injective and holomorphic in a domain $\Omega$ and $\psi_n \rightarrow \psi$ locally uniformly in $\Omega$, then either $\psi$ is constant or $\psi$ is injective. This follows from a theorem of Hurwitz. – Kalim Jan 9 '12 at 1:23
What if we require that the limit $\psi$ is non-constant? Do we obtain surjectivity? – Dimitrios Nt Jun 7 '14 at 19:30