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We prove the required duality pairing explicitly. This also identifies the word “canonical” in the final conclusion: the isomorphism will be the intrinsic residue trace on a smooth projective curve, not a choice of basis for a one-dimensional vector space.

Begin on the projective line. With $s=t^{-1}$, the equation $ds=-t^{-2}dt$ identifies its sheaf of Kähler differentials over a field with $\mathcal{O}(-2)$. The Laurent description in Question 3 pairs $H^1(\mathcal{O}(a))$ with $H^0(\mathcal{O}(-a-2))$ by multiplication and extraction of the coefficient of $t^{-1}dt$. For $a\le -2$ the respective bases areand their pairing matrix has entriesThis is an invertible antidiagonal matrix. For $a\ge -1$ both spaces vanish. The coefficient extraction is well defined on the Čech cohomology quotient: a differential regular on the first chart has no $t^{-1}dt$ term, and one regular on the second chart is a sum of terms $t^{-2-j}dt$ with $j\ge 0$, also with no such term.

To extend this calculation to every vector bundle $E$ on the projective line, we give the splitting argument. Every line bundle on ${\mathbb{P}}^1$ is $\mathcal{O}(a)$: choose a nonzero rational section to obtain a divisor, and use the divisors of $t-c$ to move each finite point to infinity. The denominator-clearing argument in Question 4 makes both $E(n)$ and $E^{\vee }(n)$ globally generated for large $n$. The first assertion ensures $H^0(E(-a))\ne 0$ for some integer $a$; the second gives an embedding $E\hookrightarrow \mathcal{O}(n{)}^{\oplus r}$, so this is impossible for $a>n$. Choose the largest such integer $a$.

A nonzero section gives $\mathcal{O}(a)\to E$. Saturating its image gives a line subbundle; if it were strictly larger, it would be $\mathcal{O}(a+d)$ for a positive integer $d$, recording the section's zeros, contrary to maximality. Thus the quotient $F$ is a vector bundle. Induct on rank and write $F={⨁}_j\mathcal{O}(b_j)$. Twist the sequence by $-a-1$. Since $H^0(E(-a-1))=0$ and $H^1(\mathcal{O}(-1))=0$, exactness forces $H^0(F(-a-1))=0$, whence every $b_j\le a$. Local lifts of a quotient summand differ on overlaps by a Čech cocycle in $\mathcal{O}(a-b_j)$. But $H^1(\mathcal{O}(a-b_j))=0$ by Question 3. Adjusting the lifts by a coboundary makes them glue, splitting the sequence. This proves the Birkhoff–Grothendieck theorem here. The direct-sum Laurent calculation consequently proves the perfect pairingAlthough a splitting proves nondegeneracy, the pairing itself uses only evaluation followed by the coefficient trace on $H^1({\Omega }_{{\mathbb{P}}^1}^1)$, so does not depend on that splitting. This is residue duality on the projective line.

Now choose a rational function $t$ on $C$ with $dt\ne 0$. Such a choice is available over the perfect field $k$: a local parameter at a smooth point has nonzero differential in the rank-one sheaf of Kähler differentials over a field. Since $dt$ spans the differentials of $k(C)/k$, the extension $k(C)/k(t)$ has zero relative differentials and is separable. Question 1 gives a finite generically separable morphism $\pi :C\to {\mathbb{P}}^1$. It is flat: locally its finite module over a discrete valuation ring is torsion-free, hence free. For the invertible sheaf $\mathcal{M}$, its pushforward $E={\pi }_{\ast }\mathcal{M}$ is therefore a vector bundle.

We next prove the differential-valued dual identification needed to transfer the pairing. At a point of the base, complete the local rings. Each branch is of the formIf $e$ is the order of $f$, formal division writes $B=A[u]$ with basis $1,u,\dots ,u^{e-1}$ and a monic degree-$e$ equation $P(u)=0$. In particular this argument does not assume tame ramification or the special equation $s=u^e$. For a polynomial $g$ the elementary interpolation identity isTo see it, pass to a field containing the distinct roots $u_i$ of $P$. Lagrange interpolation writes the remainder as ${\sum }_{i}g(u_i)P(T)/((T-u_i)P^{\prime }(u_i))$. Comparison of its leading coefficient gives the identity, and descent returns it to the original field. The top-coefficient pairing on $A[T]/(P)$ is perfect: in the power basis its matrix has zero entries for $i+j<e-1$ and ones for $i+j=e-1$, so its determinant is a unit. It follows that the trace-dual fractional module is exactly $P^{\prime }(u{)}^{-1}B$.

Formal division of $f(T)-s$ gives $f(T)-s=W(T,s)P(T,s)$ with $W$ a unit. Differentiate in $T$ and evaluate at $u$. This shows that $f^{\prime }(u)$ and $P^{\prime }(u)$ differ by a unit. Since $ds=f^{\prime }(u)du$, we obtainThis is the trace-dual module of a finite curve map identity. Its map sends a differential $\eta$ to the functional $b\mapsto \mathrm{Tr}(b\eta )$, where the trace of $hds$ is $\mathrm{Tr}(h)ds$. For several branches take the direct sum of these maps. Completion is faithfully flat, so their being isomorphisms proves the corresponding isomorphism of the original local modules. A local trivialization of $\mathcal{M}$ now gives, compatibly on overlaps,

The inverse images of the two standard affine charts and their intersection are affine. As in Question 4, the Čech cochain complexes identify $H^1(C,\mathcal{M})$ with $H^1({\mathbb{P}}^1,E)$. Global sections in the last display identify the other factor with $H^0(C,{\mathcal{M}}^{\vee }\otimes {\Omega }_C^1)$. Thus the perfect pairing already calculated on the projective line givesIn particular the stated vanishing of global sections forces $H^1(C,\mathcal{M})=0$. The proof has established the relevant case of Serre duality, rather than invoked it as the requested argument.

To identify its intrinsic trace, use the rational principal-parts resolution on an algebraic curve. The constant sheaf of rational sections and the sheaf of their finite-support principal parts are both flasque sheaves. Consequently a class of $H^1(C,\mathcal{M})$ is represented by finitely many rational principal parts $(m_P)$, modulo the family coming from one global rational section. For a global section $\eta$ of ${\mathcal{M}}^{\vee }\otimes {\Omega }_C^1$, evaluation gives the pairingThe algebraic residue of a rational differential is its $u^{-1}du$ coefficient, independent of the local parameter. Adding a regular representative changes no residue. The local trace calculation above also givesOne can check this directly in the monogenic description: expand a differential in the basis $1,u,\dots ,u^{e-1}$, use the displayed interpolation formula for its trace, and extract the coefficient of $s^{-1}ds$. Formal division of $f(u)-s$ makes that coefficient exactly the coefficient of $u^{-1}du$; for a product of branch fields the trace is the sum. Thus the identity remains valid in positive characteristic, including wild ramification. On ${\mathbb{P}}^1$, partial fractions show that a rational differential's residues at finite points sum to the negative of its residue at infinity. Tracing to ${\mathbb{P}}^1$ therefore proves that the sum of residues of every global rational differential on $C$ is zero. Adding a global rational section to $(m_P)$ consequently changes the pairing by zero. This defines the same pairing as the finite-pushforward calculation, intrinsically, without a chosen map to ${\mathbb{P}}^1$.

Finally take $\mathcal{M}={\Omega }_C^1$. Its differential-valued dual is ${\mathcal{O}}_C$, and Question 4 proved $H^0(C,{\mathcal{O}}_C)=k$. Perfectness identifies $H^1(C,{\Omega }_C^1)$ with the dual of this space. Evaluation at the distinguished constant section $1$ givesThis residue trace on a smooth projective curve uses neither a basis choice nor an auxiliary finite map. It is the canonical isomorphism.

For a sheaf of abelian groups $\mathcal{F}$ on a topological space $X$, putThese 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 $Q^j=\mathrm{coker}(Q^{j-1}\to I^j)$, using the sheaf cokernel. Compose $I^j\to Q^j\to I^{j+1}$ to obtain the Godement resolutionIt is exact by construction, with flasque terms. Define sheaf cohomology by the cohomology of its global-section complex: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 $0\to Q^{j-1}\to I^j\to Q^j\to 0$ 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, provingThis 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 $\mathrm{Rat}(X)$; irreducibility is not needed for this module statement. When $X$ is irreducible, $\mathrm{Rat}(X)$ 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: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 $R=\mathrm{Rat}(\mathcal{F})$. The constant sheaf $\mathcal{R}(\mathcal{F})$ has sections $R$ on every nonempty open set: such opens are irreducible and hence connected. DefineRestriction 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 isExactness of sheaves can be tested on stalks, so this proves the rational principal-parts resolution on an algebraic curveBoth 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. ConsequentlySurjectivity 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 requireto 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 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 $X=\mathrm{Spec}A$, $K=\mathrm{Frac}A$ and $\mathcal{F}=\tilde{M}$. By the affine module sheaf construction, $R=M{\otimes }_{A}K=:M_K$ 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 $a{v}_P\in M$ 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 givesPrescribe the class of $a{v}_P$ 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 givesNo nonsingularity assumption was used; the construction applies to singular irreducible affine curves as well.

Use the rational principal-parts resolution on an algebraic curve to represent a class by finitely many rational differential principal parts. Sum their algebraic residues of a rational differential. The sum of residues of a global rational differential is zero: trace to the projective line, where partial fractions prove the assertion. Hence the sum descends to cohomology. Pairing an invertible sheaf with its differential-valued dual gives the perfect Serre duality pairing. Since the global regular functions on an integral projective curve over an algebraically closed field are $k$, evaluation at the constant function $1$ identifies this trace canonically with an isomorphism to $k$.