Let $X$ be riemann surface (not supposed compact) and $\mathcal D$ be sheaf of divisors on $X$.Remind that this means for $U\subset X$ open then $\mathcal D(U)$ is group of divisors on $U$. How to prove that $H^1(X,\mathcal D)=0$ . This is exercise in §16 of Forster book Riemann Surfaces (but is not homework for me) and he says hint is to use discontinuous partitions of unit but I don't understand. Any other method for proving is for me also satisfying . I suppose $H^2(X,\mathcal D)$ is also zero, no?
Addedd: Problem is in analytic category, not algebraic ( where is trivial : $\mathcal D$ is flabby). So I don't understand Matt E's proof (but thank you very much for answer Matt) because I don't know if cohomology commutes with co-limits in non noetherian case.