TITLE: Is the limit of a sequence of biholomorphisms surjective?
QUESTION [6 upvotes]: 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. =)

REPLY [5 votes]: 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 $n \rightarrow \infty$.