Is it true, that $H^1(X,\mathcal{K}_{x_1,x_2})=0$? - The cohomology of the complex curve with a coefficient of the shaeaf of meromorphic functions... Let X be complex curve (complex manifold and $\dim X=1$).
For $x_1,x_2\in X$ we define the sheaf $\mathcal{K}_{x_1,x_2}$(in complex topology) of meromorphic functions vanish at the points $x_1$ and $x_2$.
Is it true, that $H^1(X,\mathcal{K}_{x_1,x_2})=0$?
In general, what are sufficient conditions for the $\mathcal{F}$ to $H^1(X,\mathcal{F})=0$ if X is curve?
 A: The answer is yes for a non-compact Riemann surface   $H^1(X, \mathcal K_{x_1,x_2})=0$ .  
The key is the exact sequence of sheaves on $X$:$$0\to \mathcal K_{x_1,x_2} \to \mathcal K  \xrightarrow {truncate  }   \mathcal Q_1\oplus \mathcal Q_2\to 0$$ where $\mathcal Q_i$ is the sky-scraper sheaf at $x_i$ with fiber the Laurent tails (locally of the form $\sum_{j=0}^Na_jz^{-j}$).
Taking cohomology we get a long exact sequence $$\cdots  \mathcal K(X)  \xrightarrow {\text {truncate}}  \mathcal Q_1(X) \oplus \mathcal Q_2(X)\to H^1(X, \mathcal K_{x_1,x_2})\to   H^1(X, \mathcal K)  \to   \cdots   $$
The vanishing  of the cohomology group $H^1(X, \mathcal K_{x_1,x_2})$ then follows from the two facts:
1) $H^1(X, \mathcal K)=0$
2) The morphism $ \mathcal K(X)  \xrightarrow {\text {truncate}} \mathcal Q_1(X) \oplus \mathcal Q_2(X)$ is surjective because of the solvability of the Mittag-Leffler problem on a non-compact Riemann surface.
For  a compact Riemann surface of genus $\geq1$ the relevant Mittag-Leffler problem is not always solvable, so that we have $H^1(X, \mathcal K_{x_1,x_2})\neq 0$ (however  for the Riemann sphere $H^1(\mathbb P^1, \mathcal K_{x_1,x_2})=0$)
