Let $F:X\to Y$ be a nonconstant <holomorphic map> of degree $d$ between compact connected Riemann surfaces. At $p\in X$, let $e_p$ be the <ramification index of a holomorphic map>. Choose a triangulation of $Y$ containing every branch value among its vertices, and lift it to $X$.

If the target triangulation has $V$ vertices, $E$ edges, and $T$ triangles, then the lifted triangulation has $dE$ edges and $dT$ triangles, because no branch value lies in an edge or face interior. For a target vertex $v$, local degree counting gives
$$
\sum_{p\in F^{-1}(v)}e_p=d.
$$
Consequently the number of vertices above $v$ is
$$
|F^{-1}(v)|
=d-\sum_{p\in F^{-1}(v)}(e_p-1).
$$
Writing $R=\sum_{p\in X}(e_p-1)$, the <Euler characteristic> is therefore
$$
\chi(X)=(dV-R)-dE+dT=d\chi(Y)-R.
$$
Since a compact orientable surface of genus $g$ has Euler characteristic $2-2g$, this <Triangulation proof of the Riemann-Hurwitz formula> gives
$$
2-2g_X=d(2-2g_Y)-\sum_{p\in X}(e_p-1),
$$
or equivalently
$$
\boxed{2g_X-2=d(2g_Y-2)+\sum_{p\in X}(e_p-1)}.
$$

Now extend a cubic polynomial $f:\mathbb C\to\mathbb C$ to a degree-three holomorphic map
$$
F:\mathbb P^1(\mathbb C)\to\mathbb P^1(\mathbb C)
$$
of the <Riemann sphere>. Both genera are zero, so Riemann--Hurwitz gives
$$
\sum_p(e_p-1)=2\cdot3-2=4.
$$
The point at infinity is totally ramified with $e_\infty=3$, contributing two. Thus the finite points contribute exactly two. There are consequently two cases.

If there is one finite ramification point $\alpha$, it has index three. Choose an affine source map $h$ sending zero to $\alpha$. Then
$$
(f\circ h)'(z)=cz^2
$$
for some $c\ne0$, and integration gives
$$
f\circ h(z)=\frac c3z^3+d.
$$
An affine target map subtracts $d$ and rescales by $3/c$, yielding $z^3$.

Otherwise there are two distinct finite ramification points, each of index two. Choose an affine source map $h$ sending $-1$ and $1$ to them. The derivative of $f\circ h$ is then a nonzero multiple of $z^2-1$:
$$
(f\circ h)'(z)=c(z^2-1).
$$
Hence
$$
f\circ h(z)=c\left(\frac{z^3}{3}-z\right)+d.
$$
The affine target map $g(w)=(w-d)/c$ gives
$$
g\circ f\circ h(z)=\frac{z^3}{3}-z
=z\left(\frac{z^2}{3}-1\right).
$$
Therefore the <affine normal forms of a complex cubic polynomial> are precisely
$$
\boxed{z^3\quad\text{and}\quad z(z^2/3-1)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 24F / Solution

Up to invertible affine changes in source and target, every complex cubic polynomial is exactly one of
$$
z^3,
\qquad
z\left(\frac{z^2}{3}-1\right).
$$
The extension to the <Riemann sphere> has two finite units of ramification. A single finite critical point has ramification index three and gives the first form; two distinct simple critical points can be moved to $\pm1$, after which integration gives the second.

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