= Solution
For a <sheaf of abelian groups> $\mathcal F$ on a <topological space> $X$, put
$$
C(\mathcal F)(U)=\prod_{P\in U}\mathcal F_P.
$$
These arbitrary families of <germs> form a <sheaf>, and restrictions are projections. Extending a family by zero proves that $C(\mathcal F)$ is <flasque>. Sending a section to all its <germs> gives an injective <sheaf> map $\mathcal F\to C(\mathcal F)$, since a section with every <germ> zero vanishes locally and hence globally.
Set $Q^{-1}=\mathcal F$, $I^j=C(Q^{j-1})$ and $Q^j=\operatorname{coker}(Q^{j-1}\to I^j)$, using the <sheaf> <cokernel>. Compose $I^j\to Q^j\to I^{j+1}$ to obtain the <Godement resolution>
$$
0\longrightarrow\mathcal F\longrightarrow I^0\longrightarrow I^1\longrightarrow\cdots.
$$
It is exact by construction, with <flasque> terms. Define <sheaf cohomology> by the <cohomology> of its global-section complex:
$$
H^j(X,\mathcal F)=H^j\bigl(\Gamma(X,I^\bullet)\bigr),\qquad
H^0(X,\mathcal F)=\Gamma(X,\mathcal F).
$$
The <resolution principle for sheaf cohomology> identifies this with the computation using any <flasque resolution>.
The elementary input for acyclicity is the <flasque-kernel section-lifting lemma>: in a <short exact sequence> with <flasque> kernel, sections of the quotient lift over every <open subset>. Consequently a quotient of two <flasque sheaves> is <flasque>, because a quotient section lifts and that lift extends. If $\mathcal F$ itself is <flasque>, induction through $0\to Q^{j-1}\to I^j\to Q^j\to0$ makes every $Q^j$ <flasque>. The same lifting lemma makes these sequences exact on <global sections>. Hence the global-section complex has no positive-degree <cohomology>, proving
$$
\boxed{H^j(X,\mathcal F)=0\quad(j>0)\text{ for flasque }\mathcal F}.
$$
This deduction uses the lifting lemma, not the desired cohomological vanishing as an assumption.
For the rational-section construction, add two representative sections after restricting to the intersection of their dense domains. Multiply by a representative rational function in the same way. Finite intersections of dense <open sets> are dense, and further restricting both representatives does not change their resulting class. The <module> axioms hold on their common domain. Thus <rational sections> form a <module> over $\operatorname{Rat}(X)$; irreducibility is not needed for this <module> statement. When $X$ is irreducible, $\operatorname{Rat}(X)$ is the <function field> $K=k(X)$.
Suppose now that $\mathcal F$ is <locally free> and $X$ irreducible. A <germ> at $P$ is represented on a nonempty open neighbourhood, which is dense, so it defines a <rational section>. Two representatives of the <germ> agree on a neighbourhood of $P$ and therefore give the same class. If the class is zero, trivialize $\mathcal F$ on a neighbourhood of $P$. Its coefficient functions vanish on a dense <open subset>; <regular functions> on a reduced <irreducible variety> with this property vanish identically. The <germ> is therefore zero. This proves the <stalk inclusion into rational sections of a locally free sheaf>:
$$
\boxed{\mathcal F_P\hookrightarrow\operatorname{Rat}(\mathcal F)}.
$$
Local freeness matters: a nonzero torsion <germ> can vanish on a dense <open set>.
On an irreducible <algebraic curve>, every proper closed subset is finite, and every <open subset> is <quasi-compact> because the Zariski space is <Noetherian>. Let $R=\operatorname{Rat}(\mathcal F)$. The <constant sheaf> $\mathcal R(\mathcal F)$ has sections $R$ on every nonempty <open set>: such opens are irreducible and hence connected. Define
$$
\mathcal P(\mathcal F)(U)=\bigoplus_{P\in U}R/\mathcal F_P.
$$
Restriction drops the components outside the smaller <open set>. For a compatible family over an open cover of $U$, take a finite subcover by quasi-compactness. The union of the finite supports of these finitely many sections is finite. Their agreeing components therefore glue to a unique finite-support tuple on $U$. This verifies the <sheaf gluing axiom>, including arbitrary covers, rather than merely asserting that a presheaf direct sum is always a <sheaf>.
A <rational section> is regular on some dense <open set>, whose complement on a curve is finite. Its classes $[s]_P\in R/\mathcal F_P$ therefore have finite support, defining a <sheaf> map $\mathcal R(\mathcal F)\to\mathcal P(\mathcal F)$. At $P$, the first <sheaf> has <stalk> $R$. The second has <stalk> $R/\mathcal F_P$: any finite tuple can be restricted to a neighbourhood omitting all its other support points, and its $P$-component is unaffected by such restriction. The resulting sequence on <stalks> is
$$
0\to\mathcal F_P\to R\to R/\mathcal F_P\to0.
$$
Exactness of <sheaves> can be tested on <stalks>, so this proves the <rational principal-parts resolution on an algebraic curve>
$$
\boxed{0\to\mathcal F\to\mathcal R(\mathcal F)\to\mathcal P(\mathcal F)\to0}.
$$
Both right-hand <sheaves> are <flasque>: restrictions for the constant rational-section <sheaf> are identities between nonempty opens, and restrictions for the rational principal-parts <sheaf> are projections, with extension by zero. Consequently
$$
H^1(X,\mathcal F)\cong
\operatorname{coker}\left(R\to\bigoplus_{P\in X}R/\mathcal F_P\right),\qquad
H^j(X,\mathcal F)=0\quad(j\ge2).
$$
Surjectivity on <stalks> has not been confused with surjectivity on global sections.
For an explicit failure of global surjectivity take $X=\mathbb P^1$ and $\mathcal F=\mathcal O(-2)$. Use coordinate $z=X_1/X_0$ and frame $e_0=X_0^{-2}$ on the finite chart. At infinity use $w=1/z$ and $e_\infty=X_1^{-2}=z^{-2}e_0$. Prescribe the <rational principal part> of $z^{-1}e_0$ at zero and zero at every other point. If a <rational section> $r(z)e_0$ realized this tuple, then $r-z^{-1}$ would be regular at zero and $r$ would be regular at every other finite point. Thus $r=z^{-1}+p(z)$ with $p\in k[z]$. Regularity at infinity would require
$$
z^2r(z)=w^{-1}+w^{-2}p(w^{-1})
$$
to be regular at $w=0$. Its $w^{-1}$ term cannot cancel with any term of the <polynomial> contribution, whose powers are at most $-2$. This is impossible. Hence \b[the global rational-principal-part map need not be surjective].
For an affine curve it is always surjective, and here is a direct construction. Write $X=\operatorname{Spec}A$, $K=\operatorname{Frac}A$ and $\mathcal F=\widetilde M$. By the <affine module sheaf> construction, $R=M\otimes_AK=:M_K$ and $\mathcal F_P=M_{\mathfrak m_P}$. Choose representatives $v_P\in M_K$ for a prescribed finite tuple, and choose one nonzero $a\in A$ with $av_P\in M$ for all of them. The zero set $T=V(a)$ is finite. The ring $A/(a)$ is <Artinian>, since it is Noetherian of dimension zero, and its <Artinian decomposition into local factors> gives
$$
M/aM\cong\bigoplus_{P\in T}(M/aM)_{\mathfrak m_P}.
$$
Prescribe the class of $av_P$ in the indicated component for each requested point in $T$, and zero in the other components. A requested point outside $T$ already has zero principal part because $a$ is a unit there. Lift the tuple to $w\in M$. Then $s=w/a\in M_K$ satisfies $s-v_P\in M_{\mathfrak m_P}$ at every requested point in $T$, and is regular at every other point of $T$. Outside $T$ it is regular because $a$ is invertible. It therefore realizes exactly the prescribed tuple. This proves <affine principal-parts interpolation for a locally free sheaf> and gives
$$
\boxed{R\longrightarrow\bigoplus_{P\in X}R/\mathcal F_P
\text{ is surjective when }X\text{ is affine};\quad H^1(X,\mathcal F)=0}.
$$
No nonsingularity assumption was used; the construction applies to singular irreducible affine curves as well.
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