Analytic map from punctured disk to punctured plane If $\mathbb{D}$ is the open unit disk, we know from Liouville's theorem that there cannot exist a 1-1 and onto analytic map (i.e. biholomorphic) from $\mathbb{D}\to\mathbb{C}$? But could there exist one from $\mathbb{D}\setminus\{0\}\to\mathbb{C}\setminus\{0\}$?
 A: The answer is NO.
Assume it is possible. If $g=f^{-1}:\mathbb C\smallsetminus\{0\}\to D\smallsetminus\{0\}$ is holomorphic and 1-1, then $0$ cannot be an essential singularity of $g$, since near the essential singularity a holomorphic function is not 1-1. Thus, it is either removable or a pole.
But removable can not be, due to Liouville's Theorem.
If $0$ is a pole for $g$, then it should be a single pole, for otherwise $g$ will not be 1-1 near $0$ - to see this consider $h=1/g$, and if the pole is of order $k$, then $h$ would behave like $z^k$. So $g(z)=G(z)/z$, where $G$ is entire. But, the image $g[B(0,r)]$, for $r$ sufficiently small, is a neighborhood of infinity, which means that there exists an $R>0$, such that
$$
g[B(0,r)]\supset B(0,R).
$$ 
Hence $g$ is bounded far from $0$, i.e., there exists an $M>0$, with $\lvert g(z)\rvert\le M$, for $\lvert z\rvert\ge r$. Hence $\lvert G(z)\rvert\le M\lvert z\rvert$, for $\lvert z\rvert\ge r$, and as $G$ is entire, then $\lvert G(z)\rvert\le M\lvert z\rvert$, for all $z$. Thus $G(z)=az$, for some $a$ complex, and $g(z)=a$. Contradiction.
A: I think there is a faster way to prove that. There is an equivalent proposition for removable singularities.
Proposition: Let $f$ be a holomorphic function defined on an open set $G$ that has an isolated singularity $z_0$. Then $f$ is bounded on $ D(z_0,r) \backslash \{z_0\} \subset G$,  for some  $r \iff z_0$ is removable singularity of $f$.
So, let a holomorphic $f: \mathbb{C} \backslash \{0\} \to \mathbb{D} \backslash \{0\}$
Since the range is bounded, as a subset of $\mathbb{D} $, $f$ must have a removable singularity.
Thus $f$ can be analytically extended to $\mathbb{C}$. Being both Entire and Bounded, by Liouville's theorem must be constant.
So, $f$ can only be constant.
A: If there exists $g=f^{-1}:\mathbb C\smallsetminus\{0\}\to D\smallsetminus\{0\}$ with $g$ holomorphic, then we can conclude that $g$ is a constant map from the view of universal covering.
The universal covering of $\mathbb C\smallsetminus\{0\}$ is $\mathbb C$ and 
the universal covering of $\mathbb D\smallsetminus\{0\}$ is $\mathbb D$, hence by
Liouville's theorem the map between universal covering is constant so as $g$.
