= Solution
Consider the singularity of the injective <entire function> $f$ at infinity. If removable, it is bounded outside a disk and therefore bounded globally; <Liouville theorem> makes it constant, a contradiction. If essential, the <Casorati-Weierstrass theorem> says its values outside every large disk are dense. Choose a small neighbourhood $U$ of a finite point; the <open mapping theorem> puts a nonempty open disk inside $f(U)$. A point outside a disk containing $U$ would map into that open disk, duplicating a value already attained in $U$. This contradicts injectivity. Thus infinity is a pole, so $f$ is a <polynomial>. If its degree were greater than one, its <derivative> would have a root; the local power-series form at that point has local degree at least two and cannot be injective. Hence
$$
\boxed{f(z)=az+b,\qquad a\ne0}.
$$
For a nonconstant <holomorphic map> of connected compact <Riemann surfaces> of degree $d$, the <Riemann-Hurwitz formula> is
$$
2g_R-2=d(2g_S-2)+\sum_{p\in R}(e_p-1),
$$
where $g_R,g_S$ are genera and $e_p$ is the local mapping degree, the multiplicity in the target fibre. There are finitely many points with $e_p>1$.
For a degree-two sphere map, the formula gives total ramification two. Since $e_p\le2$, there are exactly two distinct ramification points, each of local degree two. Their images are distinct, since a fibre has total multiplicity two. Choose a <Möbius transformation> $T$ taking $0,\infty$ to these domain points and $S_0$ taking their images to $0,\infty$. The resulting rational map has its only zero at zero of order two and its only pole at infinity of order two, hence is $cz^2$ with $c\ne0$. Replace $S_0$ by $S(z)=c^{-1}S_0(z)$, obtaining
$$
\boxed{S\circ f\circ T(z)=z^2}.
$$
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