First sheaf cohomology $H^1(\mathscr{O}_D, \mathbb{D})=0$ Can I get a hint on this problem?
Given a finite divisor $D=p_1+\dots +p_m -q_1 -\dots -q_n$ on the unit disk $\mathbb{D}$, how do I show that the first sheaf cohomology group $H^1(\mathscr{O}_D, \mathbb{D})=0$?
I was first only considering effective divisors, which would make $\mathscr{O}_D$ a subset of the holomorphic functions. We can also pick a covering $(U_i)$ such that each $p$ is only contained in a single open set $U$. Then I could split any cochain $(f_{ij})$ into $g_j-g_i$ of holomorphic functions. Then I want to somehow get the functions $g_i$ to be zero at the necessary points, maybe by multiplying $g_i$ by $(z-p_i)$ locally in $U_i$. But this doesn't seem to be likely to work.
My motivation for doing this is to show that on any compact riemann surface X, a covering $(U_i)$ where all the $U_i$ are isomorphic to disks is a leray covering relative to any sheaf of a divisor $\mathscr{O}_D$.
 A: Let's do a simple example. Namely, let's show that $\mathcal{O}(-p)$ (for some point $p$) has vanishing cohomology. 
Begin by considering that we have the following short exact sequence
$$0\to\mathcal{O}(-p)\to \mathcal{O}_{\mathbb{D}}\to i_\ast\mathcal{O}_p\to 0$$
where $i$ is the inclusion of the point $p$ into $\mathbb{D}$ (and $\mathcal{O}_p$ is the structure sheaf on the point). Then, we get the LES
$$0\to \mathcal{O}(-p)(\mathbb{D})\to \mathcal{O}_{\mathbb{D}}(\mathbb{D})\to \mathbb{C}\to H^1(\mathbb{D},\mathcal{O}(-p))\to H^1(\mathbb{D},\mathcal{O}_{\mathbb{D}})$$
But, note that using the exponential sequence
$$0\to\mathbb{Z}\to \mathcal{O}_{\mathbb{D}}\to\mathcal{O}_{\mathbb{D}}^\times\to 0$$
we have the long exact sequence
$$0\to \mathbb{Z}\to \mathcal{O}_{\mathbb{D}}(\mathbb{D})\to \mathcal{O}_{\mathbb{D}}(\mathbb{D})^\times\to H^1(\mathbb{D},\underline{\mathbb{Z}})\to H^1(\mathbb{D},\mathcal{O}_{\mathbb{D}})\to H^1(\mathbb{D},\mathcal{O}_{\mathbb{D}}^\times)\to H^2(\mathbb{D},\underline{\mathbb{Z}})$$
But, since $\mathbb{D}$ is contractible
$$H^1(\mathbb{D},\underline{\mathbb{Z}})=H^2(\mathbb{D},\underline{\mathbb{Z}})=0$$
and so
$$H^1(\mathbb{D},\mathcal{O}_{\mathbb{D}})=H^1(\mathbb{D},\mathcal{O}_{\mathbb{D}}^\times)=\mathrm{Pic}(\mathbb{D})$$
But, $\mathrm{Pic}(\mathbb{D})=\mathrm{Cl}(\mathbb{D})=0$. The last equality follows by the Mittag-Leffler theorem.
Thus, from this side computation we see that we have the exact sequence
$$0\to\mathcal{O}(-p)(\mathbb{D})\to\mathcal{O}_\mathbb{D}(\mathbb{D})\to\mathbb{C}\to H^1(\mathbb{D},\mathcal{O}(-p))\to 0$$
and thus proving that $H^1(\mathbb{D},\mathcal{O}(-p))=0$ is equivalent to showing that 
$$0\to \mathcal{O}(-p)(\mathbb{D})\to \mathcal{O}_{\mathbb{D}}(\mathbb{D})\to \mathbb{C}\to 0$$
is exact.
Do you see how to do this? Do you see how to generalize this?
