Map $\mathbb{C}\setminus [-1,1]$ onto the open unit disk Let $G = \mathbb{C}\setminus [-1,1]$. I wan't to find an analytic function $f:G\rightarrow \mathbb{D}$ where $\mathbb{D}$ denotes the unit disk such that $f$ is onto, and preferably if possible one-to-one.
Now I've seen that $g(z) = \frac{1}{2}(z+1/z)$ maps the open unit disk to $\mathbb{C}\setminus [-1,1]$ in a one-to-one fashion and therefore the given candidate would be its inverse.
$$g(z) = w\Leftrightarrow z^2-2zw+1 = 0\Leftrightarrow (z-w)^2 = w^2-1.$$
Here is my problem: I want to show that $g$ is invertible and find a concrete formula for $g^{-1}$ however I am not sure how to do this using square roots. Say we want to use the square root defined from the branch of the logarithm which satisfies $\log re^{it} = \log r+it$ for $0<t<2\pi$. Then we want $w^2-1$ to stay away from $(-\infty,0]$. However writing $w = x+iy$ it is clear that we then need to restrict $w$ to lie in $\mathbb{C}\setminus \Big((-\infty,-1]\cup(1,\infty]\Big)$ which defeats the purpose since we want $w\in \mathbb{C}\setminus [-1,1]$. How should I go about choosing a branch in this case?

Clearly $g$ is undefined at $0$ so this only gives a bijective map from $G$ to $\mathbb{D}\setminus\{0\}$. Also by simple connectivity injectivity is not possible however can we find a map which is onto?

If $f:G\to \mathbb D$ is onto it cannot be one-to-one: else $G$ and $\mathbb D$ would be analytically isomorphic.
Indeed the inverse of a bijective analytic mapping between open subsets of $\mathbb C$ is automatically analytic.
Note that this is a non trivial result.
But this is absurd since these domains are not even homeomorphic:  $\mathbb D$ is simply-connected whereas $G$ is not.
