Comment: A meromorphic function on a Riemann surface \( X \) is a holomorphic map from \( X \) to the Riemann Sphere \( \hat{\mathbb{C}} \) which is not identically equal to \( \infty \).
Let \( f \) be the given meromorphic function. Then \( f \ \colon S \longrightarrow \hat{\mathbb{C}} \) is a nonconstant holomorphic map on the compact Riemann surface \( S \), so it is proper. This means that the degree \[ \displaystyle d = d(y) = \sum_{x \in f^{1}(y)} v_{f}(x) \] of \( f \) is welldefined, i.e. independent of the choice of \( y \in \hat{\mathbb{C}} \). Let \( x \) be the unique pole of \( f \). Then we can compute the degree of \( f \) using \( y = \infty \), and since \( x \) is a simple pole of \( f \), i.e. the ramification index of \( f \) at \(x\) is \( v_{f}(x) = 1\), we conclude that \( d = 1 \), as \( f \) has a unique pole. By the definition of the degree, it follows that \( f \) is bijective (as it assumes every value \( y \in \hat{\mathbb{C}} \)). Hence, \( f \) is biholomorphic, as a holomorphic bijection. This means that \( S \) is biholomorphic to the Riemann Sphere.
