I am trying to understand a statement from http://arxiv.org/abs/1312.1562, saying "... $H^1(\Sigma^x,T\Sigma) = 0$ since $H^1(\Sigma^x,\mathcal{O}) = 0$ and the Mittag-Leffler problem is solvable on the non-compact Riemann surface $\Sigma^x = \Sigma\backslash\{X\}$." (p.16). Here $\Sigma$ is a compact Riemann surface, $X\in\Sigma$ and $T\Sigma$ is the holomorphic tangent sheaf.
To my understanding, the second and third statement of the quote are equivalent, but how does the first one ($H^1(\Sigma^x,T\Sigma) = 0$) follow? Can the holomorphic tangent sheaf of a non-compact Riemann surface be identified with the sheaf of holomorphic functions?
Yes, the tangent sheaf of a non-compact Riemann surface $X$ can be identified with the sheaf of holomorphic functions:
Any holomorphic locally free sheaf of any rank on a non-compact Riemann surface is holomorphically trivial !
Amazing, eh?
You can find a proof on page 229 of Forster's masterful book.
Edit
At Jason's request let me translate the displayed statement into an equivalent one not involving sheaves:
Any holomorphic vector bundle of any rank on a non-compact Riemann surface is holomorphically trivial !