Rational principal-parts resolution on an algebraic curve ($0\to \mathcal{F}\to \mathcal{R}(\mathcal{F})\to \mathcal{P}(\mathcal{F})\to 0$)

On an irreducible algebraic curve, let $\mathcal{R}(\mathcal{F})$ be the constant rational-section sheaf and put $\mathcal{P}(\mathcal{F})(U)={⨁}_{P\in U}\mathrm{Rat}(\mathcal{F})/{\mathcal{F}}_P$. Every open subset is quasi-compact, so compatible finite-support families glue with finite support. Its stalk at $P$ is the displayed rational-principal-part quotient. A rational section is regular outside finitely many points, making its principal-part map well defined. The resulting sequence is stalkwise exact. Both right-hand sheaves are flasque, and consequently $H^1(X,\mathcal{F})$ is the cokernel of the global principal-part map and $H^q(X,\mathcal{F})=0$ for $q\ge 2$. The curve and Zariski-topology hypotheses distinguish finite support here from locally finite pole sets in complex analysis.

Let $X$ be an irreducible affine curve, $A=k[X]$, $K=\mathrm{Frac}A$, and $\mathcal{F}=\tilde{M}$ be locally free. Given finitely many classes in $M_K/M_{{\mathfrak{m}}_P}$, choose one nonzero $a\in A$ clearing denominators of representatives. The ring $A/(a)$ is Artinian and is the product of its finitely many local factors. Hence $M/aM\cong {⨁}_{P\in V(a)}(M/aM{)}_{{\mathfrak{m}}_P}$. Prescribe $a{v}_P$ modulo $a{M}_{{\mathfrak{m}}_P}$ at the requested points and zero at the other zeros of $a$. Lift this tuple to $w\in M$; the section $w/a$ has exactly the requested rational principal parts and no others. This proves surjectivity without assuming the curve is nonsingular.