= Holomorphic surjection from the unit disc onto the complex plane
{title2=$F(z)=z^2/(1-z)^2$}
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 $\operatorname{Re}r\geq0$ and put $z=r/(1+r)$. The identity $|1+r|^2-|r|^2=1+2\operatorname{Re}r>0$ 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.
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