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For an affine morphism $f:X\to Y$ and a quasi-coherent sheaf $\mathcal{F}$ on $X$, the direct image sheaf has the same sheaf cohomology on $Y$ as $\mathcal{F}$ on $X$. To prove this, take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Its direct images are flasque sheaves. On each affine open $V\subset Y$, the complex of sections of the direct images is $\Gamma (f^{-1}V,{\mathcal{I}}^{\bullet })$. By vanishing of quasi-coherent cohomology on an affine scheme, this augmented complex is exact. Checking on the affine basis proves that $f_{\ast }{\mathcal{I}}^{\bullet }$ resolves $f_{\ast }\mathcal{F}$. Its global sections are identical to those of the original resolution.

For a short exact sequence of sheaves with flasque sheaf kernel, every section of the quotient on any open set lifts to the middle sheaf on that set. Use the Zorn lemma to choose a maximal partial lift, with chains glued by the sheaf gluing axiom. A new local lift differs from the old one on their overlap by a kernel section. Extend that difference by flasqueness, adjust the new lift, and glue, contradicting maximality unless its domain is the entire open set. Consequently a quotient of two flasque sheaves is flasque. With an injective resolution of sheaves, the long exact sequence in sheaf cohomology and repeated dimension shifts prove that flasque sheaves have zero positive-degree cohomology.

A flasque resolution is an exact sequence of sheaves of abelian groups whose terms ${\mathcal{I}}^j$ are flasque sheaves. Every such sheaf embeds by its section germs into the sheaf $U\mapsto {\prod }_{P\in U}{\mathcal{F}}_P$. Its restriction maps are projections, so it is flasque. Apply this construction successively to the cokernels to obtain a resolution. The sheaf cohomology of $\mathcal{F}$ is the cohomology of the cochain complex $\Gamma (X,{\mathcal{I}}^{\bullet })$. A quotient of two flasque sheaves in a short exact sequence of sheaves is flasque: the flasque kernel allows local lifts to be glued into a section of the middle sheaf over the smaller open set; extend that lift using flasqueness of the middle sheaf, and project. Consequently the resolution's successive cokernels are flasque when $\mathcal{F}$ is flasque, and exactness of sections shows that all its positive-degree cohomology vanishes.

Write the affine variety as $X=\mathrm{Spec}A$. Its coordinate ring is Noetherian, and a coherent sheaf is $\mathcal{F}=\tilde{M}$ for a finitely generated $A$-module $M$. We first justify the acyclic resolution used to compute its sheaf cohomology.

If $I$ is an injective module, then $\tilde{I}$ is a flasque sheaf. Indeed every open $U$ in this Noetherian scheme has the form $X\setminus V(\mathfrak{a})$ for a finitely generated ideal. Clearing denominators givesHere a homomorphism $h:{\mathfrak{a}}^n\to I$ gives, on a principal open $D(f)$ with $f\in \mathfrak{a}$, the fraction $h(f^n)/f^n$; the fractions agree on overlaps. Conversely choose fractions for a section on a finite principal-open cover. A common power clears their denominators and the finitely many compatibility relations. Multiplication by a further power of $\mathfrak{a}$ kills any elements supported on $V(\mathfrak{a})$, producing a well-defined map on some ${\mathfrak{a}}^n$. Changing that power accounts for the direct limit. This is the sheafification of an injective module is flasque construction. Since $I$ is injective, each map ${\mathfrak{a}}^n\to I$ extends to $A\to I$. Thus every section on every open extends to $X$. Extending globally and restricting again also proves surjectivity for restrictions between arbitrary opens.

Choose an injective module resolution $0\to M\to I^0\to I^1\to \cdots$. Localization is exact, so sheafifying gives a flasque resolution of $\mathcal{F}$. Its global-section complex is exactly $I^{\bullet }$, since $\Gamma (\mathrm{Spec}A,\tilde{I})=I$. This complex is exact in positive degrees, and a flasque resolution computes sheaf cohomology. ConsequentlyThe argument actually proves vanishing of quasi-coherent cohomology on an affine scheme, without finite generation of $M$.

For the intersection assertion, varieties are taken to be separated, as usual. The diagonal morphism $X\to X{\times }_{k}X$ is a closed immersion. Its inverse image over $U{\times }_{k}V$ identifies $U\cap V$ with a closed subvariety of the affine variety $U{\times }_{k}V$. A closed subvariety of an affine variety is affine: its coordinate ring is the ambient coordinate ring modulo its defining ideal. Thus $U\cap V$ is affine. Repeated application gives affine finite intersections. Without separatedness this statement need not hold.

Choose a finite affine open cover $(U_{\alpha })$ of the projective variety $X$. For ordered indices write $U_{{\alpha }_0\cdots {\alpha }_p}=U_{{\alpha }_0}\cap \cdots \cap U_{{\alpha }_p}$. The answer is the cohomology of the Čech cochain complexIn particular $H^0=\ker (C^0\to C^1)$ and, in positive degree,The chart modules $\Gamma (U_{\alpha },\mathcal{F})$, together with their restriction and gluing maps, supply the modules on overlaps by localization. The gluing maps are essential; abstract chart modules alone do not specify a sheaf.

For completeness, take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$ and its double complex of Čech cochain complexes on this cover. In the resolution direction, its cohomology is $H^q(U_{{\alpha }_0\cdots {\alpha }_p},\mathcal{F})$, which vanishes for $q>0$ by the affine result. In the cover direction, the augmented Čech cochain complex of a flasque sheaf is exact: sections can be extended from overlaps, successively adjusting a cocycle to eliminate one cover member at a time. Equivalently the augmented sheaf complex is stalkwise contracted by inserting a cover index containing the point. Thus computing the total cohomology in the two directions identifies Čech cohomology with sheaf cohomology, proving the displayed description.

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.

The affine space $X={\mathbb{A}}_k^n$ is an integral scheme, hence its underlying topological space is irreducible. Every nonempty open subset of an irreducible topological space is irreducible and in particular connected. A section of the constant sheaf with value the function field $K=k(t_1,\dots ,t_n)$ is a locally constant function, so on any such open subset it is constant. Thus its sections are $K$ on each nonempty open subset and zero on the empty set. All restriction maps are either identities on $K$ or maps onto zero.

The constant sheaf is therefore a flasque sheaf. It has its natural ${\mathcal{O}}_X$-module structure because a regular function on a nonempty open subset embeds in the function field and acts by multiplication. Part (i) now givesThis is an instance of constant sheaves on irreducible spaces are flasque; no restriction on $n$ is needed.

Recall that a flasque sheaf has surjective restriction maps on every inclusion of open subsets. We first prove the flasque-kernel section-lifting lemma: in an exact sequence $0\to \mathcal{A}\to \mathcal{B}\to \mathcal{C}\to 0$ of sheaves, with $\mathcal{A}$ a flasque sheaf, every section of $\mathcal{C}$ on any open subset $V$ lifts to a section of $\mathcal{B}$ on $V$.

Fix such a section $c$. Order its lifts on open subsets $W\subseteq V$ by extension. A chain has an upper bound by the sheaf gluing axiom, so Zorn's lemma gives a maximal lift $b_W$. If $W\ne V$, choose a point outside $W$ and a neighborhood $T\subseteq V$ on which $c$ has a local lift $b_T$. On $W\cap T$, the difference $b_T-b_W$ is a section of $\mathcal{A}$. Extend it to $T$ by flasqueness and subtract it from $b_T$. The corrected lift now agrees with $b_W$ and glues on $W\cup T$, contradicting maximality. Thus $W=V$, proving the lemma.

It follows also that if both $\mathcal{A}$ and $\mathcal{B}$ are flasque sheaves, then $\mathcal{C}$ is a flasque sheaf. Given a section of $\mathcal{C}$ on $U\subseteq V$, first lift it to $\mathcal{B}$ on $U$, extend that lift to $V$, and project to $\mathcal{C}$.

Next, an injective sheaf of modules $\mathcal{I}$ is a flasque sheaf. For $U\subseteq V$, let $j_U$ and $j_V$ denote the open inclusions into $X$. Extension by zero gives a monomorphism $j_{U!}{\mathcal{O}}_U\to j_{V!}{\mathcal{O}}_V$: on every stalk it is either $0\to {\mathcal{O}}_{X,x}$, the identity, or $0\to 0$. By the adjunction for extension by zero,Injectivity therefore makes $\Gamma (V,\mathcal{I})\to \Gamma (U,\mathcal{I})$ surjective. This proves injective module sheaves are flasque.

Take an injective resolution of sheaves of $\mathcal{F}$. Write ${\mathcal{C}}^0=\mathcal{F}$ and, successively,The initial ${\mathcal{C}}^0$ is a flasque sheaf; each ${\mathcal{I}}^r$ is a flasque sheaf; and the quotient argument proves inductively that all ${\mathcal{C}}^r$ are flasque sheaves. The section-lifting lemma makes each displayed sequence exact after global sections. Consequently the complex of global sections of the injective resolution of sheaves has zero positive cohomology. Since that complex defines sheaf cohomology,

Take the affine scheme $X={\mathbb{A}}_k^1$, its function field $K=k(t)$, and the constant sheaf $\mathcal{K}$ with value $K$. By Question 4(ii), $\mathcal{K}$ is a flasque sheaf. At the two distinct closed rational points $p_0=0$ and $p_1=1$, let $S_0,S_1$ be skyscraper sheaves with value $K$: on an open subset $V$, $S_i(V)=K$ if $p_i\in V$ and is zero otherwise. Both are flasque sheaves.

There is a natural morphism of sheaveswhich sends a constant rational function to the same element in every summand present on the open subset. At $p_i$, its stalk map is the identity $K\to K$, and at any other point the target stalk is zero. Thus $\rho$ is a surjection of sheaves, although it is not surjective on global sections. Put $\mathcal{F}=\ker \rho$.

This can even be an example of sheaves of ${\mathcal{O}}_X$-modules: let regular functions act on each copy of $K$ by multiplication after embedding them in $K$. At the closed points this action is multiplication by a rational function, not evaluation at the point. With these actions, $\rho$ is ${\mathcal{O}}_X$-linear.

The exact sequence $0\to \mathcal{F}\to \mathcal{K}\to S_0\oplus S_1\to 0$, together with flasque acyclicity, givesHence $H^0(X,\mathcal{F})=0$ and $H^1(X,\mathcal{F})=K^2/\Delta K\simeq K\ne 0$. In contrast, use the one-member affine open cover $\mathcal{U}=\{X\}$. Its alternating Čech cochain complex has no positive-degree terms, so ${\check{H}}^1(\mathcal{U},\mathcal{F})=0$. We have the explicit discrepancyThe sheaf $\mathcal{F}$ is not a quasi-coherent sheaf, as also follows from its nonzero higher sheaf cohomology on an affine scheme. Thus this example does not contradict the acyclic cover theorem for quasi-coherent sheaves on affine covers.

For a point $x\in X$ and an abelian group $A$, the skyscraper sheaf $i_{\ast }A$, with $i:\{x\}\hookrightarrow X$, isRestrictions are identities or the map to zero, so it is a flasque sheaf. The global sections equal $A$ and higher cohomology vanishes:No closed-point assumption is needed for this calculation of skyscraper sheaf cohomology. On an arbitrary topological space, its nonzero stalks are at points of $\overline{\{x\}}$ when $A\ne 0$; only for a closed point is the nonzero stalk confined to $x$.

A flasque sheaf has surjective restriction maps $\mathcal{F}(V)\to \mathcal{F}(U)$ for every pair of opens $U\subseteq V$. To prove the claim for an injective sheaf of modules, let $j_U:U\hookrightarrow X$ and $j_V:V\hookrightarrow X$. The natural mapis a monomorphism: its stalks are either the identity on ${\mathcal{O}}_{X,x}$, the map from zero to that stalk, or the zero-to-zero map. Here $j_{!}$ is extension by zero for module sheaves.

The extension-by-zero adjunction identifiesThe injective object property extends every morphism from $(j_U{)}_{!}{\mathcal{O}}_U$ to one from $(j_V{)}_{!}{\mathcal{O}}_V$. Under the displayed identification this is exactly surjectivity of $I(V)\to I(U)$. Injective module sheaves are therefore flasque. This argument works on an arbitrary ringed space, without Noetherian or separation assumptions.