if $\mathbb{D}$ is simply connected,the same must be true of $Ω$?

Question

If $F : Ω → \mathbb{D}$ is conformal map (bijective and holomorphic ) and if $\mathbb{D}$ is simply connected,the same must be true of $Ω$.

$\mathbb{D}$ is the unit disc. Can anyone give a hint how to prove this statement?

Answer 1

In fact more is true. Such a conformal map has a non-vanishing derivative, since points where the derivative are $0$ fail to be conformal, and so we can use the fact that holomorphic implies real differentiable, and the nowhere-vanishing property of the derivative to conclude that $f$ is actually a diffeomorphism, since it is already assumed to be bijective, and you can use the inverse function theorem. This means that all the topological properties are preserved, so, im particular, $\Omega$ is simply connected.