The assertion is true. The standard filtration ${\mathbb{CP}}^0\subset \cdots \subset {\mathbb{CP}}^n$ gives one cell in each dimension $0,2,\dots ,2n$. There are no odd-dimensional cells, so all cellular differentials vanish. The cellular homology theorem and the universal coefficient theorem for cohomology give one copy of $\mathbb{Z}$ in each even degree from zero to $2n$, and zero in every other degree.

Let $x$ be the Poincare dual of a projective hyperplane, with the complex orientation. It has degree two and evaluates to $+1$ on a complex projective line, so it is the positive generator of $H^2$. We use the intersection interpretation of the cup product: the product of duals of oriented submanifolds in transverse position is the dual of their oriented intersection. Distinct transverse complex hyperplanes intersect in ${\mathbb{CP}}^{n-j}$ after $j$ intersections, with positive complex orientation. Thus $x^j$ is the dual of that linear subspace.

Pairing $x^j$ with a transverse linear ${\mathbb{CP}}^j$ gives one positively oriented intersection point. Therefore $x^j$ is a primitive generator of $H^{2j}$, for every $0\le j\le n$. There is no cohomology above dimension $2n$, so $x^{n+1}=0$. These facts show that the surjective graded ring map from $\mathbb{Z}[x]$ has exactly the indicated kernel:This proves the cohomology ring of complex projective space, rather than only its additive groups. For $n=0$ it is $\mathbb{Z}$, with $x=0$.

Put $A={\Sigma }_h^{\partial }$. This surface with boundary deformation retracts onto a wedge of $2h$ circles, so $H^1(A;\mathbb{R})$ has dimension $2h$ and $H^2(A;\mathbb{R})=0$. If $r$ were a retraction, ${\iota }^{\ast }{r}^{\ast }=\mathrm{id}$ would make $r^{\ast }$ injective. Its image $W\subset H^1({\Sigma }_g;\mathbb{R})$ would have dimension $2h$.

Every pair of classes $u,v$ in $H^1(A;\mathbb{R})$ has $u⌣v=0$, since $H^2(A;\mathbb{R})=0$. Naturality givesThus $W$ is an isotropic subspace of a symplectic vector space for the nondegenerate skew Poincare duality pairing on the $2g$-dimensional space $H^1({\Sigma }_g;\mathbb{R})$. The stated linear-algebra bound gives $2h\le g$. Hence

The bound is sharp. Double $A$ along its boundary: $D(A)=A{\cup }_{\partial A}A$ is the closed oriented surface of genus $2h$. Identify each copy with $A$ and fold them onto one copy. The two maps agree on the joining circle, so they give a continuous retraction fixing the first copy pointwise. This includes $h=0$, where the double of a disc is a sphere.

More generally, for $g\ge 2h$, add $g-2h$ handles in the interior of the second copy. Pinch those extra handles onto their connecting point, keeping the boundary fixed, and then fold onto $A$. This is still a retraction. In fact the construction works for any embedding in the question: the connected complement has one boundary component and genus $g-h$ by Euler-characteristic additivity, and the classification theorem for surfaces identifies it, relative to that boundary, with a copy of $A$ having $g-2h$ additional handles. Thus the exact existence criterion is $2h\le g$, as expressed by retraction onto a punctured oriented surface.

Suppose $F:{\Sigma }_g\to {\Sigma }_h$ has degree one. For degree-one classes $u,v$ on the target, naturality of the cup product and the definition of the degree of a map between oriented manifolds giveIf a nonzero $u$ had $F^{\ast }u=0$, nondegeneracy of the Poincare duality pairing would supply $v$ with nonzero right side, a contradiction. Thus $F^{\ast }$ injects the $2h$-dimensional real degree-one cohomology into the $2g$-dimensional source. Hence $g\ge h$. This is the cohomological injectivity of a degree-one map in this setting.

Conversely, for $g\ge h$, express ${\Sigma }_g$ as ${\Sigma }_h\#{\Sigma }_{g-h}$. Collapse the second punctured summand and the joining circle to a point. The quotient of the retained punctured summand by its boundary is homeomorphic to ${\Sigma }_h$, giving a continuous map to that surface. Its restriction to a small oriented disc away from the collapsing region is an orientation-preserving homeomorphism, and a point in this disc has exactly one preimage. The induced map on local top homology, and hence on the fundamental class, has coefficient $+1$. The map therefore has degree one. For $h=0$, the same construction is the familiar collapse of the complement of a disc to obtain $S^2$.

ConsequentlyThis proves both directions of the degree-one maps between closed oriented surfaces criterion. The case $g=h$ also admits the identity map.

Give the closed orientable surface its standard CW complex structure: one zero-cell, $2g$ one-cells, and one two-cell attached by the product of $g$ commutators. The cellular boundary of the two-cell is zero, since every edge occurs once with each orientation in that word; the one-cell boundaries are also zero. HenceHere we use the cellular homology theorem, identifying cellular and singular homology, and the universal coefficient theorem for cohomology: its exact sequence has terms $\mathrm{Ext}(H_{q-1},\mathbb{Z})$ and $\mathrm{Hom}(H_q,\mathbb{Z})$. All the homology groups here are free, so the Ext terms vanish.

The ring structure comes from Poincare duality and algebraic intersection number of curves on an oriented surface. For a closed oriented surface, cap product with its fundamental class identifies degree-one cohomology with degree-one homology; evaluating the cup product of two such classes equals the signed intersection number of their dual one-cycles. Choose the usual $g$ pairs of handle curves, each pair meeting positively once, and distinct pairs disjoint. Their dual classes can accordingly be named $a_1,b_1,\dots ,a_g,b_g$ so that, for the positive orientation class $\omega$,These formulas include squares. More generally, graded commutativity of the cup product kills every degree-one square here because $H^2$ is torsion-free. The unit generates $H^0$, and products involving $\omega$ and any positive-degree class vanish for dimensional reasons. These additive groups and multiplication rules completely describe the cohomology ring of a closed oriented surface, including $g=0$, when there are no degree-one generators. The intersection pairing is integral and unimodular, rather than merely nondegenerate over a field.

For any finite-dimensional bounded cochain complex $(V^{\ast },d)$ over a field, let $r_q=\mathrm{rank}(d:V^q\to V^{q+1})$, with zero ranks outside its degree range. ThenTaking the alternating sum cancels the two rank sums. Thus taking cohomology preserves the Euler characteristic of a finite graded complex. Apply this to $V^q=H^q(M;{\mathbb{F}}_p)$ and $d=\beta$, using part (b).

A closed three-dimensional manifold has finite-dimensional cohomology, vanishing above degree three. Its Euler characteristic is zero, even when it is not orientable: Poincare duality with ${\mathbb{F}}_2$ coefficients gives $b_q=b_{3-q}$, so the alternating sum vanishes. A finite triangulation, or finite CW complex model, shows that the alternating sum is the same integer with coefficients in any field, since it equals the alternating count of cells. ThereforeThe primeness assumption makes $\mathbb{Z}/p$ a field; no orientation assumption on $M$ is needed.

The standard CW complex structure on Real projective space has one cell in each dimension from zero to three. Its integral cellular boundary is multiplication by $2$ in even positive degrees and zero in odd degrees. Thus the integral cellular cochain complex for ${\mathbb{RP}}^3$ isin degrees $0,1,2,3$. With coefficients ${\mathbb{F}}_2$, all its differentials vanish, so every one of these four cohomology groups is one-dimensional.

The lift-and-divide construction of the Bockstein homomorphism turns the integral differential $2$ into $1$ modulo $2$. Therefore $\beta :H^1\to H^2$ is an isomorphism, while the maps from degrees $0,2,3$ are zero. The Bockstein cohomology is consequentlyFor comparison, in the mod-two cohomology ring of real projective space ${\mathbb{F}}_2[t]/(t^4)$, $|t|=1$, this says $\beta (t)=t^2$, $\beta (t^2)=0$ and $\beta (t^3)=0$. The last two formulas also follow from the Bockstein derivation rule and the truncation $t^4=0$.

Write $\rho :H^{\ast }(X;\mathbb{Z})\to H^{\ast }(X;\mathbb{Z}/n)$ for coefficient reduction. Compare the two coefficient sequences in part (a): the maps from the integral sequence to the finite sequence are reduction modulo $n$ on the left, reduction modulo $n^2$ in the middle, and the identity on the right. The square involving the injections commutes because $[na{]}_{n^2}=j([a{]}_n)$.

Naturality of the connecting homomorphism gives the Bockstein factorization through integral cohomologyOne can see this directly without a diagram: lift a modulo-$n$ cocycle to an integral cochain $a$. Its coboundary has the form $\delta a=nb$. Then $\hat{\beta }[a]=[b]$, whereas $\beta [a]=[b\mathrm{mod}n]$.

Exactness of the integral coefficient sequence gives $\hat{\beta }\rho =0$: a class obtained by reducing an integral cocycle has zero integral connecting class. ConsequentlyThis Bockstein square-zero identity holds without requiring $n$ to be prime. At the cochain level, ${\delta }^{2}a=0$ and the torsion-free integral cochain groups imply $\delta b=0$, which also makes the second connecting class vanish.

First account for the unlabelled coefficient-sequence construction. The two short exact sequences of abelian groups areThe singular chain groups of $X$ are free abelian. Applying $\mathrm{Hom}(C_q(X),-)$ therefore preserves these exact sequences, degree by degree, giving short exact sequences of cochain complexes. The associated long exact sequence from a coefficient sequence gives the displayed maps in cohomology; the connecting maps are the integral Bockstein homomorphism $\hat{\beta }$ and the modulo-$n$ Bockstein homomorphism $\beta$. The first omitted map is multiplication by $n$, and the second is induced by $j$.

For the requested example, attach an $(i+1)$-cell to $S^i$ using a map of degree $n$. The resulting Moore space $X=M(\mathbb{Z}/n,i)$ has positive-degree cellular chain complexin degrees $i+1,i$. This construction also works for $i=1$, using the degree-$n$ map of the circle. In cellular cohomology with coefficients $\mathbb{Z}/n$, the differential is zero, so both $H^i$ and $H^{i+1}$ are $\mathbb{Z}/n$.

Lift the cochain taking value $1$ on the $i$-cell to a cochain with coefficients $\mathbb{Z}/n^2$. Its coboundary takes value $n$ on the $(i+1)$-cell, which is $j(1)$. The definition of the connecting homomorphism therefore sends the degree-$i$ generator to the degree-$(i+1)$ generator. HenceIt is nonzero for every $i\ge 1$ and $n\ge 2$, including composite $n$. This is the Bockstein on a cyclic Moore space.

Eliminate $t_3$ using the linear equation. The homogeneous coordinate ring becomesThe cubic is a nonzero element of a polynomial integral domain, hence a non-zero-divisor. It gives the exact sequence of graded modulesThus $\dim B_m=\left(\genfrac{}{}{0ex}{}{m+2}{2}\right)-\left(\genfrac{}{}{0ex}{}{m-1}{2}\right)$ for $m\ge 3$. Equivalently, use the Hilbert series $(1-z^3)/(1-z{)}^3$, or the sheafified exact sequence on ${\mathbb{P}}^2$. The Hilbert polynomial of the plane cubic isIts degree is three and its arithmetic genus is one, although the curve is singular: its affine equation near $[1:0:0]$ is the cusp $t_2^2+t_1^3=0$. The Hilbert polynomial records the arithmetic genus, rather than the genus of the normalization.

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 direct image of a coherent sheaf under a closed immersion puts $\mathcal{G}=f_{\ast }\mathcal{F}$ in the category of coherent sheaves on ${\mathbb{P}}_k^n$. By sheaf cohomology under a closed inclusion and the supplied compatibility of twisting with direct image,Here is a proof of the required Serre vanishing on projective space. A coherent sheaf on ${\mathbb{P}}_k^n$ is the sheaf associated with a graded module for a finite graded $k[t_0,\dots ,t_n]$-module. Equivalently, it has a presentation by finite sums of twisting sheaves. Use a finite twisting resolution of a coherent sheaf on projective space: resolve the graded module by a finite graded free resolution, using the Hilbert syzygy theorem, and sheafify; exactness of localization preserves the resolution. Its terms are finite sums of $\mathcal{O}(a)$.

Choose $d$ sufficiently large that all twists $a+d$ occurring in these finitely many terms are nonnegative. The cohomology of twisting sheaves on projective space then vanishes in every positive degree for every resolution term. In a short exact sequence $0\to {\mathcal{K}}_{j+1}\to {\mathcal{E}}_j\to {\mathcal{K}}_j\to 0$, with ${\mathcal{K}}_0=\mathcal{G}$, the long exact sequence in sheaf cohomology identifies $H^p({\mathcal{K}}_j(d))$ with $H^{p+1}({\mathcal{K}}_{j+1}(d))$ for $p>0$. Iterating to the final acyclic term provesThe standard graded-module description used here is given in Stacks Project, Section 30.15; the vanishing follows from the displayed finite-resolution argument.

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 $x$.

For the closed inclusion $f$, the direct image sheaf has stalk ${\mathcal{F}}_x$ at $x\in X$ and zero stalk at points outside $X$: outside the closed set there is an open neighbourhood disjoint from $X$. Hence $f_{\ast }$ is exact, since exact sequences of sheaves are detected on stalks.

Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Direct image preserves flasqueness, because restrictions are the restrictions on inverse-image opens. Exactness makes $f_{\ast }{\mathcal{I}}^{\bullet }$ a flasque resolution of $f_{\ast }\mathcal{F}$. MoreoverThe complexes computing sheaf cohomology are identical. ThereforeThis is sheaf cohomology under a closed inclusion; the closedness hypothesis is used in the exactness of direct image.

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.

The first implication is false. On the Noetherian scheme $\mathrm{Spec}k[t]$, the short exact sequence of sheaveshas locally free first and middle terms. The quotient is not locally free: at $(t)$ it is a nonzero module annihilated by $t$, whereas a free module over the integral domain $k[t{]}_{(t)}$ has no such element.

The second implication is true. Near any point trivialize $\mathcal{F}$ and $\mathcal{H}$. Choose a finite local basis of $\mathcal{H}$. Surjectivity as a sheaf gives local lifts of its basis sections; shrink to the intersection of their finitely many neighbourhoods. The lifts define a splitting $\mathcal{H}\to \mathcal{G}$ on that neighbourhood. Thus $\mathcal{G}\cong \mathcal{F}\oplus \mathcal{H}$ there and is locally free. This is local splitting when a quotient sheaf is locally free; no global splitting is asserted.

Take the hyperplane divisor $D=(t_0=0)$. On $U_i=D_{+}(t_i)$ its local equation is $g_i=t_0/t_i$, a non-zero-divisor; on $U_0$ it is the unit one. Thus it is an effective Cartier divisor. On overlaps, $g_i/g_j=t_j/t_i$ is a unit.

The divisor line bundle ${\mathcal{O}}_X(D)$ is locally generated by $1/g_i=t_i/t_0$. Map this generator to the local generator $t_i$ of the twisting sheaf on projective space ${\mathcal{O}}_X(1)$. On overlaps both generators transform by $t_i/t_j$, so the maps glue and giveFor $n=0$ the hyperplane is empty and this is the zero Cartier divisor; the conclusion still holds.

The degree-one generation condition for Proj ensures that the opens $D_{+}(f)$ for $f\in S_1$ cover $X$. Put $B=S_f$ and $R=B_0$. Multiplication by $f^n$ is an $R$-linear bijection $R\to B_n$ for every integer $n$, because $f$ is an invertible degree-one element. The convention $S(n{)}_d=S_{n+d}$ therefore identifies $\widetilde{S(n)}$ on this chart with the free rank-one $R$-module $B_n$, generated by $f^n$. Hence every twisting sheaf ${\mathcal{O}}_X(n)$ is invertible, including negative twists.

For the tensor product of sheaves, consider the natural mapIt is an isomorphism by degree-one localization of a graded module. Explicitly, a homogeneous tensor $m_a\otimes n_{-a}$ of total degree zero is represented on the left by $f^{-a}{m}_a\otimes f^a{n}_{-a}$. Multiplying a tensor factor by a homogeneous element of $B$ gives the same result after using the tensor relation, since that element is a degree-zero coefficient times a power of $f$. This defines the inverse. Since localization commutes with tensor products, the target is $((M{\otimes }_{S}N{)}_f{)}_0$. These natural chart maps agree on overlaps, soThe degree-one hypothesis matters: on general Proj constructions, twisting sheaves need not be line bundles. The local trivializations are also recorded in Stacks Project, Section 27.10.

Work first over an affine neighbourhood $\mathrm{Spec}A$ of $y=\mathfrak{p}$, with an affine chart $\mathrm{Spec}B$ in its inverse image. The scheme-theoretic fibre on this chart has ringwhere $\kappa (\mathfrak{p})$ is the residue field. The prime ideal correspondence for localization and the correspondence for a quotient identify its points with primes $\mathfrak{q}\subset B$ containing $\mathfrak{p}B$ and disjoint from $S$. These two conditions are exactly $\mathfrak{q}\cap A=\mathfrak{p}$.

Under this bijection, each principal open $D(b\otimes 1)$ corresponds to $D(b)\cap f^{-1}\{y\}$. These are bases for the two topologies: every element of the fibre ring is represented by $b/s$, with the image of $s$ invertible. Thus the bijection is a homeomorphism. The identifications agree on overlapping charts and glue, provingThe scheme-theoretic fibre may nevertheless carry nilpotents, so this topological statement does not identify its structure sheaf with a naive restriction of ${\mathcal{O}}_X$.

A fibre product of schemes $Z=X{\times }_{S}Y$ has projection morphisms to $X$ and $Y$ with equal composites to $S$, and for every scheme $T$ the induced map is a bijectionThis universal property determines $Z$ uniquely up to unique isomorphism. On affine charts over $\mathrm{Spec}R$, takeThe tensor product of commutative algebras has the required universal property for ring maps. Localization identifies these constructions on principal-open overlaps. Cover the base and its two inverse images by affine charts and glue the resulting affine products along those overlaps; their local universal properties give the global one.

For a counterexample to preservation of irreducibility, take $X=Y=\mathrm{Spec}\mathbb{C}$ and $S=\mathrm{Spec}\mathbb{R}$ with their usual structure morphisms. All three are integral schemes, but the Chinese remainder theorem givesThe product scheme is the disjoint union of two points, so it is not irreducible.

An isomorphism of schemes over $\mathbb{R}$ induces a bijection on rational points. A real point of this projective plane curve would be represented by a nonzero real triple with $t_0^2+t_1^2+t_2^2=0$. Each summand is nonnegative, so every coordinate would vanish. Thus $X(\mathbb{R})=\varnothing$, whereas $[1:0]\in {\mathbb{P}}^1(\mathbb{R})$. The two real schemes are not isomorphic. The obstruction is the real conic without real rational points; nonsingularity alone does not make a real conic a projective line over its ground field.