By a domain I mean an open connected subset of ${\mathbb C}$. If $D$ is a domain, let $\operatorname{Aut}(D)$ denote the collection of holomorphic bijections $f:D\to D$. It is well-known that if $f$ is holomorphic, so is its inverse, so $\operatorname{Aut}(D)$ is actually a group.
We can give a topology to $H(D)$, the set of holomorphic maps with domain $D$, by setting $f_n\to f$ iff $f_n$ converges uniformly to $f$ on any compact subset of $D$. This actually makes $H(D)$ a complete metric space (Section 2.2 of Berenstein-Gay "Complex variables").
I have a few questions about $\operatorname{Aut}(D)$ with this topology. First,
Is $\operatorname{Aut}(D)$ a topological group?
It is easy to check that $f_n\circ g_n\to f\circ g$ if $f_n\to f$ and $g_n\to g$ in $H(D)$, so the question here is whether $f_n^{-1}\to f^{-1}$ if $f_n\to f$, and $f_n,f\in \operatorname{Aut}(D)$. (This may be trivial.) Second,
If $f_n\to f$ in $H(D)$ and $f_n\in \operatorname{Aut}(D)$ for all $n$, is $f\in \operatorname{Aut}(D)$ as well? Are there any reasonable assumptions on $D$ that make this true? (As pointed out in the comments below, if $D$ is simply connected, then $f$ needs not be injective.)
By Rouche, we have that $f$ is either constant or injective (Prop 2.6.19 in Berenstein-Gay "Complex Variables"). Perhaps, if it is injective, showing it is surjective should be easy?
Finally,
I've heard that $\operatorname{Aut}(D)$ is actually a Lie group. I suppose the topology is the one I mentioned, but I do not see where the manifold structure will come from if that's the case. Could you sketch why this is so, or point me in the right direction if I'm completely off the mark here?