 Solution

View infinity as the puncture of the Riemann sphere and consider $g(1/z)$ near $z=0$. If this map had an essential singularity, the Great Picard theorem in a coordinate chart would make almost every nearby value occur infinitely often, contradicting the injectivity of $g$. The singularity is therefore removable as a map into the compact Riemann surface $R$, and $g$ extends to a nonconstant holomorphic mapThe extension is injective. A collision between $\tilde{g}(\infty )$ and $g(z_0)$, or local degree greater than one at infinity, would give nearby values two distinct preimages by the local normal form of a nonconstant holomorphic map, again contradicting injectivity on $\mathbb{C}$. Thus $\tilde{g}$ is a holomorphic bijection between compact Riemann surfaces and hence a biholomorphism. Therefore $R\cong {\mathbb{P}}^1$, whose genus is zero.