= Solution
Choose coordinates $z$ at $p$ and $w$ at $f(p)$ vanishing at those points. The local expansion is $w\circ f\circ z^{-1}(\zeta)=a\zeta^v+O(\zeta^{v+1})$, $a\ne0$. Define the branching order $v_f(p)=v$. A change of either coordinate has nonzero linear coefficient, so it preserves the leading exponent, proving well-definedness.
Ramification means $v>1$, equivalently vanishing of the derivative in local coordinates. These zeros are isolated unless the local derivative is identically zero, which by analytic continuation would make $f$ constant. An infinite set of <ramification points> on a compact surface would have an accumulation point, contradicting isolation. Thus that set is finite.
A nonconstant <holomorphic map> of compact connected Riemann surfaces is onto, since its image is open by the local form and closed by <compactness>. It has a finite degree $d\geq1$. The <Riemann-Hurwitz formula> is
$$
2g_R-2=d(2g_S-2)+\sum_{p\in R}(v_f(p)-1).
$$
If both genera equal $g>1$, positivity of $2g-2$ forces $d=1$ and every ramification term zero. The map is bijective and has nonzero derivative everywhere. The <holomorphic inverse function theorem> says that a <holomorphic function> with nonzero derivative has a local holomorphic inverse; these inverses patch, so \b[$f$ is biholomorphic]. For $g=1$, multiplication by two on $\mathbb C/\Lambda$ is a degree-four <holomorphic map> of an <elliptic curve> to itself and is not biholomorphic.
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