Show that there is no analytic bijection from the unit disc to $\mathbb{C}$.
I am quite unsure how to proceed here. I know for a fact that there is no analytic function from $\mathbb{C}$ to the open disc by Louvilles theorem.
Suppose that $f$ is ideed an analytic bijection from the unit disc to $\mathbb{C}$. Then consider the map $g(z)=f(\frac{1}{z})$. Now $g$ is an analytic bijective function from $\mathbb{C}/\{0\}$ to $\mathbb{C}$. Can this be extended to a proof?
Best Answer
Denote unit disc as $\mathbb{D}$.
If there exists an analytical $\space f: \mathbb{C}\to\mathbb{D}$, then $f$ is a bounded entire function, and therefore by Liouville's theorem, $\space f$ is constant, so it cannot be a bijection.