Let $X$ be a compact Riemann surface, $\mathcal M_X$ the sheaf of meromorphic functions, $\mathcal M_X^\ast$ the sheaf of invertible meromorphic function (a sheaf in abelian groups for the multiplication of meromorphic functions).
How to prove that $H^1(X,\mathcal M_X^\ast)=0$?
The result is true. Here is a proof: from Forster, Lectures on Riemann Surfaces, Exercise 16.4, we have an exact sequence $$\text{Div} X \rightarrow H^1(X,\mathcal O_X^\ast) \rightarrow H^1(X,\mathcal M_X^\ast) \rightarrow 0,$$ where $\mathcal O_X^\ast$ is the sheaf of invertible holomorphic function. Since it is known that every invertible sheaf (an element of $H^1(X,\mathcal O_X^\ast)$) comes from a divisor, the first arrow is surjective, hence $H^1(X,\mathcal M_X^\ast)=0$. But is there a more direct proof, which does not use the interpretation of $H^1(X,\mathcal O_X^\ast)$ as the group of invertible sheaves?