Prove that $H^1(\mathcal{M}^*)=0$. Let $X$ be a compact Riemann surface. For an open set $U$, let $\mathcal{M}^*(U)$ be the multiplicative group of nonzero meromorphic functions on $U$ ("nonzero" meaning "not identically zero"). This makes a sheaf $\mathcal{M}^*$.

Proposition: $H^1(\mathcal{M}^*)=0$.

Miranda remarks this, claiming that $\mathcal{M}^*$ is a constant sheaf and the proposition follows.


*

*A sheaf $\mathcal{F}$ is constant iff there is a group $A$ and an open cover $\{U_i\}$ such that $\mathcal{F}(U_i) = A$ for all $i$, right? Is it equivalent to say that $\mathcal{F}_p=A$ for all $p\in X$?

*If $\mathcal{F}$ is constant and you want to prove $H^1(\mathcal{F})=0$, it seems like you need to use the fact that $X$ is compact, and that any two Zariski-open sets intersect. If I use the Zariski topology, don't I have to appeal to GAGA? Or something like that?

*I'm not certain what Miranda means when he says "$\mathcal{M}^*$ is a constant sheaf". Is he referring to the fact that every open set is dense, so a meromorphic function will be determined by its values on any open subset? Why would this imply $H^1(\mathcal{M}^*)=0$?
 A: You could also see it as a consequence of the fact that every holomorphic line bundle on a compact Riemann surface is the line bundle associated to some divisor. Indeed, this statement is equivalent to the mapping $\iota_*$ being the zero map in the long exact sequence
$$
H^0(X,\mathcal{M}^*/\mathcal{O}^{\times}) \overset{\delta}\to H^1(X,\mathcal{O}^{\times}) \overset{\iota_*}\to H^1(X,\mathcal{M}^*) \overset{\pi_*}\to H^1(X,\mathcal{M}^*/\mathcal{O}^{\times}) \to 0
$$
associated to the short exact sequence
$$
0 \to \mathcal{O}^{\times} \overset{\iota}\to \mathcal{M}^* \overset{\pi}\to \mathcal{M}^*/\mathcal{O}^{\times} \to 0.
$$
Now if $\iota_* \equiv 0$, then $\pi_*$ is injective so it suffices to see that $H^1(X,\mathcal{M}^*/\mathcal{O}^{\times}) = 0$. 
Here's one way we can see this directly: Consider an open cover $(U_i)$ of open sets and take a representative $f_{ij} \in \mathcal{M}^{*}(U_{ij})$ for a Cech 1-cocycle $f_{ij} \in (\mathcal{M}^*/\mathcal{O}^{\times})(U_{ij})$. To construct sections $f_k \in (\mathcal{M}^*/\mathcal{O}^{\times})(U_{k})$ for $k=i,j$ restricting to $f_{ij}$ on $U_{ij}$, we only need to find meromorphic functions $f_k \in \mathcal{M}^*(U_k)$ with exactly the same poles as $f_{ij}$ on $U_{ij}$. Indeed, the quotient $f_i/f_j$ would then be holomorphic and nowhere-zero on $U_{ij}$, i.e. $f_i|_{U_j}$ would be equal to $f_j|_{U_i}$ as elements of $(\mathcal{M}^*/\mathcal{O}^{\times})(U_{ij})$. To do this, suppose $\{p_k\}_{k=1}^N$ is the set of poles of our representative $f_{ij}$ on $U_{ij}$ (say $p_k$ is of order $r_k$). Now we take coordinate charts $(V_l,z_l)$ for $l=1, \ldots, N$ with $V_l$ centered on $p_l$ and $V_l \cap V_{l'} = \varnothing$ if $l \neq l'$, and we extend this to an open covering $(V_l)_{l=1}^{N'}$ of $U_i \cup U_j$. Let 
$$
f_k(z_k) := f_{ij}(z_k) = \sum_{m=-r_k}^0 a_mz_k^m + h_k(z_k)
$$
where $h_k \in \mathcal{O}^{\times}(V_k)$ for $k=1, \ldots, N$ and for the other $V_k$'s we can just let $f_k(z_k)$ be any non-vanishing holomorphic function $f_k \in \mathcal{O}^{\times}(V_k)$ (for example a constant). Now since these functions $(f_k)_{k=1}^{N'}$ always have the same poles with the same orders when they intersect on some $V_k$, the $f_k|_{U_i}$'s can be glued together to form a well-defined section $f_i$ of $(\mathcal{M}^*/\mathcal{O}^{\times})(U_{i})$, and similarly the $f_k|_{U_j}$'s give a section $f_j$ of $(\mathcal{M}^*/\mathcal{O}^{\times})(U_{j})$. Again, on the intersection $U_i \cap U_j$, $f_i$ and $f_j$ have the same poles with the same orders so they both restrict to $f_{ij}$ as a section of $(\mathcal{M}^*/\mathcal{O}^{\times})(U_{ij})$, showing that $f_{ij}$ is indeed a coboundary.
