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On ${\mathbb{P}}_k^n$ with $n>0$, let $\mathcal{F}=i_{\ast }k$ at a rational point. It has one-dimensional global sections, but its dual of a sheaf is zero: a local map from the residue field into the local polynomial ring must have image annihilated by a nonzero parameter, which is impossible in an integral domain. Thus $H^n({\mathcal{F}}^{\vee }(-n-1))=0$. Serre duality for arbitrary coherent sheaves instead uses an Ext functor; replacing this by the ordinary dual can lose torsion information.

Write ${\mathbb{P}}^r={\mathbb{P}}_k^r$ over its ground field; the argument also works over any Noetherian affine base. Since $\iota$ is a closed immersion, $X$ is a noetherian scheme and a quasi-compact scheme. LetHere the pullback of a sheaf of modules makes $L$ an invertible sheaf, the restriction of the twisting sheaf on projective space. It is an ample line bundle.

By global generation by sufficiently positive twists, choose $n\ge 0$ such that $F(n)$ is a globally generated sheaf. Finitely many global sections suffice: at each point, finite generation of the stalk of the coherent sheaf $F(n)$ lets us select finitely many generating germs of global sections. Those sections generate on a neighbourhood, because their cokernel sheaf is a coherent sheaf with zero stalk at that point. A finite subcover, using that $X$ is a quasi-compact scheme, supplies finitely many global sections generating everywhere. This is finite global generation on a quasi-compact scheme, and gives a surjective sheaf morphism

Tensor with $(L^{\vee }{)}^{\otimes n}$, where $L^{\vee }$ is the dual of a sheaf. Here tensoring with an invertible sheaf is exact: locally the tensor product of sheaves is tensoring with a free module of rank one. Using the permitted identity $L\otimes L^{\vee }\cong {\mathcal{O}}_X$, we obtainA finite direct sum of sheaves of invertible sheaves is a locally free sheaf of finite rank, and it is a coherent sheaf on the noetherian scheme $X$. This constructs the required $G$. No regularity or smoothness assumption on $X$ is needed.

Use the quotient convention for projective space. To a scheme $T$, associate the set of pairswhere $\mathcal{L}$ is a line bundle, modulo isomorphisms $\mathcal{L}\to {\mathcal{L}}^{\prime }$ carrying $q$ to $q^{\prime }$. Pullback defines the contravariant functor. On $T=\mathrm{Spec}A$, this means rank-one projective module quotients of $A^{n+1}$; the quotient need not be free. Thus merely listing generating tuples modulo multiplication by a unit would omit points over rings with nontrivial Picard group.

This functor is represented by ${\mathbb{P}}_{\mathbb{Z}}^n$. For each $i$, the locus where $q(e_i)$ generates $\mathcal{L}$ is open. These loci cover $T$, since a surjection onto a rank-one free stalk must have some coefficient a unit in its local ring. On this locus, trivialize by $q(e_i)$ and writeThese functions specify a map to the standard chart $U_i=\mathrm{Spec}\mathbb{Z}[t_j:j\ne i]$. Their transition rules are exactly those of projective space, so the chart maps glue. Conversely the universal quotient on projective space pulls back to the given pair, and the construction is natural and unique. This proves that projective space represents invertible quotients.

Write its universal exact sequence asA first-order change of the quotient, with target trivialized locally, is a map $\delta q:V\otimes \mathcal{O}\to \mathcal{O}(1)$. Changing the target trivialization by $1+\epsilon h$ changes $\delta q$ by $hq$. Restriction to $\mathcal{K}$ removes this ambiguity, and every local map $\mathcal{K}\to \mathcal{O}(1)$ extends because the displayed sequence splits locally. Consequently the tangent sheaf of a scheme isThis identification can also be seen on $U_i$: the first-order change of $t_j$ is $\delta q(e_j)-t_j\delta q(e_i)$ after setting $q(e_i)=1$. Thus these maps are precisely the coordinate tangent directions.

Apply $\mathcal{H}om(-,\mathcal{O}(1))$ to the universal sequence. Since its terms are locally free and it splits locally, this gives the Euler sequenceIt remains exact after any base change. Taking its dual of a sheaf over a field $k$ givesFor $n=1$, the first term has rank one; taking determinants yieldsExplicitly, for $u=t^{-1}$ on the overlap of the two affine charts, $du=-t^{-2}dt$, the same transition as $\mathcal{O}(-2)$ after changing one local frame's sign.

Choose the rational point $x=[1:0:\cdots :0]$ and the coherent sheaf $\mathcal{F}=i_{\ast }k$ supported there. By skyscraper sheaf cohomology, $H^0(X,\mathcal{F})=k$.

Its dual of a sheaf ${\mathcal{F}}^{\vee }=\mathcal{H}om(\mathcal{F},{\mathcal{O}}_X)$ is zero. Away from $x$ its source is zero. At $x$, a homomorphism $k\to {\mathcal{O}}_{X,x}$ must send one to an element annihilated by the maximal ideal. Since $n>0$, that ideal contains a nonzero coordinate parameter; the local ring is an integral domain, so the image must vanish. This proves that every local homomorphism vanishes. ConsequentlyThere is no contradiction with Serre duality: the ordinary sheaf dual suffices for locally free sheaves, but general coherent sheaves require an Ext functor. This example illustrates failure of ordinary sheaf-dual Serre duality for a skyscraper sheaf.

The relative tangent sheaf is the dual of a sheaf of Kähler differentials. Its sections are derivations of the structure sheaf that kill base scalars. For a smooth morphism it is locally free and describes first-order deformations of sections into the scheme.

For an invertible sheaf $L$, tensoring sheaves of modules with $L$ is an exact functor: on an open cover trivializing $L$, it is naturally the identity functor. Equivalently, its stalks are free modules of rank one and hence flat modules. Its inverse functor is tensoring with the dual of a sheaf $L^{\vee }$, because $L\otimes L^{\vee }\cong {\mathcal{O}}_X$. This justifies untwisting a surjection onto a positively twisted coherent sheaf.