Holomorphic surjection from the unit disc onto the complex plane ($F(z)=z^2/(1-z{)}^2$)

The rational function $F(z)=z^2/(1-z{)}^2$ is holomorphic on the unit disc and maps it onto the entire complex plane. For any $w\in \mathbb{C}$, choose a square root $r$ with $\mathrm{Re}r\ge 0$ and put $z=r/(1+r)$. The identity $|1+r{|}^2-|r{|}^2=1+2\mathrm{Re}r>0$ gives $|z|<1$, while $F(z)=r^2=w$. It is not injective: the two choices $r=\pm i$ give distinct points in the disc with image $-1$.

A holomorphic bijection in this direction is impossible: its inverse would be a bounded entire function, contradicting the Liouville theorem. Thus a surjection onto the complex plane can exist even when a conformal equivalence cannot.