Let $X$ be a Riemann surface and denote by $\mathcal{M}^*$ the sheaf of invertible (i.e. not constantly zero on any connected component) meromorphic functions.
I have seen claims that $H^1(X, \mathcal{M}^*) = 0$. How could one prove this?
Since we have an exact sequence $0 \to \mathcal{O}^* \to \mathcal{M}^* \to \mathrm{Div} \to 0$ ($\mathrm{Div}$ denotes the sheaf of divisors on $X$, which is soft and hence acyclic) this is equivalent to showing that $H^1(X, \mathcal{O}^*) \cong \mathrm{Cl}(X)$, i. e. that the map $\mathrm{Div} (X) \to \mathcal{O}^*(X)$ is surjective ($\mathrm{Cl}(X)$ is the divisor class group for $X$).