Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 printed variant / 4 / b / Solution

If a holomorphic bijection existed, its inverse would be an entire function taking values in the unit disc. Liouville theorem would make that inverse constant, contradicting bijectivity. Thus no holomorphic bijection exists.

A continuous bijection does exist, even a homeomorphism:Both maps preserve the argument and change the radius by the mutually inverse strictly increasing functions $r\mapsto r/(1-r)$ and $s\mapsto s/(1+s)$. The displayed formulas also prove continuity at zero and everywhere else.