TITLE: Topology of the space of univalent functions
QUESTION [5 upvotes]: Let $D\subset CP^1$ be a domain (a nonempty open connected subset) and let $S_D$ denote the space of conformal embeddings $D\to CP^1$ equipped with topology of uniform convergence on compacts. Is it true that this space is connected? Has homotopy type of $PSL(2,C)$? 
The answer is easy and positive if $D$ is simply-connected, but I already do not know what happens if $D$ is an annulus.

REPLY [3 votes]: Here is a proof of connectivity for an annulus, 
but it generalizes to any domain of finite connectivity.
WLOG, $D$ is an annulus $a<|z|<1$. Let $f:D\to P^1$ be the embedding. First we approximate
it by an embedding which sends the boundary to quasicircles. 
For this we first restrict $f$ on
$D'=\{ z: r_1<|z|<r_2\}$, where $a<r_1<r_2<1$. Let $\phi$ be a quasiconformal map
$D\to D'$. Then there exists a quasiconformal map $\psi:P^1\to P^1$ such that
$g=\psi\circ f\circ\phi$ is conformal on $D$. If $r_1$ and $r_2$ are close to $a$ and $1$,
then $\phi$ can be chosen uniformly close to the identity, and $\psi$ can be chosen uniformly close to the
identity.
On the second stage, we extend $g$ to the homeomorphism of the Riemann shpere which
is quasiconformal on $|z|<a$ and $|z|>1$. This is possible to do because the images
of these circles are quasicircles. Let $\mu$ be the Beltrami coefficient of $g$.
Then we can deform $g$ to identity by constructing a family of normalized quasiconformal
homeomorphisms $g_t$ with Beltrami coefficients $\mu_t=t\mu$, where $0\leq t\leq 1.$
This gives a deformation of your homeomorphism to the identity.
Reference on quasiconformal maps, quasicircles and Beltrami equation: 
L. Ahlfors, Lectures on quasiconformal mappings.