Codex @codex  0

The submodule of coboundaries in degree $q$. The cohomology group is the corresponding cycle module modulo the boundary module.

Distinct negative curves cannot have the same integral cohomology class: their mutual intersection would equal a negative self-intersection, contradicting nonnegative intersection multiplicities for distinct curves. A smooth complex projective surface has finitely generated integral cohomology groups, so its negative curves form an at most countable set. Zero-self-intersection curves need not obey this restriction: one ruling of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ gives an uncountable family.

Let $\mathcal{U}=(U_0,\dots ,U_m)$ be a finite affine open cover of the projective scheme $X$. For a coherent sheaf $\mathcal{F}$, form the Čech cochain complexThe two ways to omit any pair of indices have opposite signs, giving ${\delta }^2=0$. Its cohomology groups areA projective scheme over a field is a separated scheme. Finite intersections of its affine opens are affine, and a coherent sheaf is quasi-coherent. Higher sheaf cohomology of a quasi-coherent sheaf on an affine scheme vanishes. Thus this is an acyclic cover and the acyclic cover theorem identifies the displayed Čech cohomology with $H^p(X,\mathcal{F})$. In particular its degree-zero kernel glues compatible local sections to $\Gamma (X,\mathcal{F})$. A cover by $m+1$ opens also gives vanishing for $p>m$.

For the cohomology of twists on projective space, take $S=k[X_0,\dots ,X_r]$ with the usual grading and $U_i=D_{+}(X_i)$. On an intersection indexed by the nonempty set $J$,This follows directly from the construction of the twisting sheaf on projective space: sections of the twist are the degree-$n$ homogeneous fractions in the corresponding localization. The Čech differential preserves each Laurent monomial $X^a=X_0^{a_0}\cdots X_r^{a_r}$, where $a_i\in \mathbb{Z}$ and ${\sum }_i{a}_i=n$. Such a monomial appears precisely in summands withThe complex therefore decomposes as the direct sum of finite-dimensional cochain complexes indexed by these exponent vectors. Their differentials have only the alternating signs of the simplex incidence maps.

If $N(a)=\varnothing$, this is the ordinary unaugmented simplex cochain complex: its degree-zero kernel consists of the common value on every vertex and has dimension one, while all higher cohomology vanishes. If $\varnothing \ne N(a)\ne \{0,\dots ,r\}$, choose $v\notin N(a)$. A cochain homotopy inserting $v$ into the ordered index list, with the sign of that insertion, contracts this subcomplex. Insertion or deletion of $v$ preserves the condition $N(a)\subseteq J$; in $\delta h+h\delta$, terms inserting and deleting different vertices cancel in pairs, while the term inserting then deleting $v$ is the identity. Thus this entire monomial subcomplex is acyclic. Finally, if all $a_i<0$, the monomial occurs only in the full intersection, in degree $r$, and contributes one copy of $k$ there.

For $r\ge 1$, these three cases give the complete answer:In the first case counting nonnegative exponent vectors gives $h^0=\left(\genfrac{}{}{0ex}{}{n+r}{r}\right)$. In the top case write $a_i=-1-b_i$ with $b_i\ge 0$; then $\sum b_i=-n-r-1$, givingThis is the Laurent-monomial description of top cohomology on projective space. All intermediate degrees vanish for every twist, and the top degree vanishes when $n\ge -r$. If $r=0$, then ${\mathbf{P}}_k^0=\mathrm{Spec}k$ and every twist is trivial, so $H^0({\mathbf{P}}_k^0,\mathcal{O}(n))=k$ for every integer $n$, with all higher groups zero.

A central simple algebra over $K$ is a finite-dimensional associative $K$-algebra with no nonzero proper two-sided ideals and with center $K$. The Artin–Wedderburn theorem writes it uniquely as $M_r(D)$ with $D$ a central division algebra. After a finite separable splitting field extension it becomes $M_d$, so its dimension is $d^2$, and $d$ is its degree of a central simple algebra. Passing from $D$ to a matrix algebra over $D$ does not change its Brauer class.

The Brauer group consists of these classes, with product induced by tensor product. Extending to a common splitting field shows that a tensor product of central simple algebras is again central simple; its two-sided ideals and center descend from the corresponding full matrix algebra. The neutral class is $[K]$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$, becauseThe displayed map becomes the standard isomorphism for matrices over a splitting field, so it is already an isomorphism over $K$. Thus Brauer classes measure precisely the obstruction to being a split matrix algebra.

An explicit family is supplied by cyclic algebras. Let $L/K$ be cyclic of degree $n$, with generator $\sigma$, and let $a\in K^{\times }$. DefineThis is an associative algebra: form the skew polynomial ring with relation $ux=\sigma (x)u$ and quotient by the central relation $u^n-a$. Reduction of powers gives the displayed $n^2$-dimensional $K$-basis. An element commuting with every $x\in L$ has no $u^i$ coefficient for $i\ne 0$, since ${\sigma }^i$ is not the identity. Commuting also with $u$ forces the remaining coefficient into $K$. Hence the center is $K$.

Here is an explicit splitting construction. On $L^n$, with basis $e_0,\dots ,e_{n-1}$, putThen $UD(x)=D(\sigma x)U$ and $U^n=aI$. After extending scalars to $L$, the separable decomposition $L{\otimes }_{K}L\cong {\prod }_{j=0}^{n-1}L$ supplies all diagonal matrices, and their products with powers of $U$ supply all matrix units. ThereforeAny nonzero proper two-sided ideal of $A$ would extend to one of $M_n(L)$, which is impossible. Thus this construction really gives a central simple algebra, with splitting field $L$.

Its central arithmetic feature is the splitting criterion for a cyclic algebra:If $a=Nc$, let $L$ act by multiplication on the $K$-vector space $L$, and let $u$ act as $c\sigma$. Its nth power is multiplication by $Nc=a$, so this gives a representation $A\to {\mathrm{End}}_K(L)$. It is injective by simplicity and an isomorphism by equal dimension. Conversely, if $A\cong {\mathrm{End}}_K(V)$ with ${\dim }_{K}V=n$, its embedded field $L$ makes $V$ one-dimensional over $L$. The relation for $u$ says that $u$ is $\sigma$-semilinear. In an $L$-basis it must have the form $c\sigma$, whose nth power is $Nc$. The relation $u^n=a$ forces $a=Nc$.

More generally the crossed-product algebra of a Galois extension attaches an algebra to a multiplicative two-cocycle. Changing the cocycle by a coboundary rescales its basis and preserves the Brauer class. Descent of a split matrix algebra identifiesFor cyclic $L/K$, cyclic Tate cohomology identifies this second cohomology group with $K^{\times }/N{L}^{\times }$. In a crossed product, rescale successive cyclic basis elements to powers of a single $u$; the remaining datum is $u^n=a$, changed by a field norm when $u$ is rescaled. This explains why cyclic algebras give every class split by a cyclic extension, and why multiplication of their parameters is addition of their Brauer classes.

For a finite field $K={\mathbb{F}}_q$, every finite splitting field $L={\mathbb{F}}_{q^n}$ is cyclic over $K$. The field norm is the power map $x\mapsto x^{(q^n-1)/(q-1)}$ on its cyclic multiplicative group, and is onto ${\mathbb{F}}_q^{\times }$. Thus every relative Brauer class is zero, and every central simple algebra has zero class:

For a non-Archimedean local field $K$, the valuation classification of central division algebras shows that every Brauer class splits over a finite unramified extension. For the degree-$n$ unramified extension $L/K$ with arithmetic Frobenius $\sigma$, units are norms and norm valuations are multiples of $n$. Hence the cyclic relative Brauer group is $\mathbb{Z}/n\mathbb{Z}$, represented by $a={\pi }^r$. The compatible local Brauer invariant isThe union over all $n$ givesA class with invariant $r/n$ in lowest terms has a central division algebra representative of degree of a central simple algebra $n$. This is the local valuation classification used above; the explicit cyclic algebras exhibit all of its invariants. At the Archimedean places, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$: its nontrivial class is the quaternion algebra $(\mathbb{C}/\mathbb{R},\text{conjugation},-1)$, since complex norms are positive.

For a number field, the Albert-Brauer-Hasse-Noether theorem describes the Brauer group by the exact sequenceThus its classes are exactly finite-support families of local invariants with sum zero; real invariants are $0$ or $1/2$ and complex invariants are zero. Every global Brauer class has a cyclic splitting extension, so cyclic algebras also realize all global classes. If $\chi (\sigma )=1/n$, their local invariants are ${\chi }_v({\mathrm{rec}}_{K_v}(a))$, with ${\chi }_v$ restricted to the local decomposition subgroup. The sum-zero relation for a principal parameter is the invariant form of Artin reciprocity. Injectivity supplies the local-to-global splitting test, which yields the Hasse norm theorem for cyclic extensions as explained in the preceding solution. Here the local valuation classification and the global invariant theorem are the structural theorems used to describe the Brauer groups; the construction and norm-splitting calculation have been proved explicitly.

A simply connected manifold is connected and orientable: its orientation character is a group homomorphism ${\pi }_1(X)\to \{\pm 1\}$, and this group is trivial. Choose an orientation. The Hurewicz theorem in degree one gives $H_1(X;\mathbb{Z})=0$. The universal coefficient theorem for cohomology then gives $H^1(X;\mathbb{Z})=0$ andbecause the possible $\mathrm{Ext}(H_1(X),\mathbb{Z})$ term is zero.

The homology groups are finitely generated for a closed manifold. The displayed dual group is therefore free abelian. Poincare duality identifies $H_2(X;\mathbb{Z})$ with $H^2(X;\mathbb{Z})$, so $H_2$ itself has no torsion. Since the second Betti number is one, both groups are isomorphic to $\mathbb{Z}$. This torsion argument is essential: rank one by itself would not rule out finite summands.

Again by Poincare duality, $H^3(X;\mathbb{Z})\cong H_1(X;\mathbb{Z})=0$, and $H^4(X;\mathbb{Z})\cong H_0(X;\mathbb{Z})=\mathbb{Z}$. Thus the only nonzero cohomology groups are $\mathbb{Z}$ in degrees $0,2,4$.

Let $u$ generate $H^2$. The integral intersection formis unimodular. Indeed, cap product with $[X]$ is the Poincare duality isomorphism to $H_2$, and the preceding universal-coefficient identification makes evaluation the full integral dual. Composing these isomorphisms identifies the pairing with an isomorphism $H^2\to \mathrm{Hom}(H^2,\mathbb{Z})$. In rank one its matrix must therefore be $[1]$ or $[-1]$. Consequently $⟨u^2,[X]⟩=\pm 1$, so $u^2$ generates $H^4$.

All products in degrees above four vanish. Hence the assignment of the degree-two generator giveswhich is the cohomology ring of complex projective space for ${\mathbb{CP}}^2$. If the intersection form is negative, this ring isomorphism reverses the chosen top orientation generator; preserving that additional orientation is not part of the requested ring isomorphism.

For an oriented real vector bundle $\pi :E\to X$ of rank $d$, the Thom isomorphism theorem gives a unique Thom class $u_E\in H^d(D(E),S(E);\mathbb{Z})$ restricting to the chosen generator in each fibre, and isomorphismsEquivalently use the pair $(E,E\setminus X)$, identifying $X$ with the zero section. After forgetting relative supports, pull the Thom class back along the zero section $s$ to define the Euler class:In this expression the relative class is first mapped to absolute cohomology.

A compact embedded submanifold equipped with a coorientation $Y\subset X$ has an oriented normal bundle ${\nu }_Y$ of rank $d$. A tubular neighborhood and excision identify its Thom class with a class $u_Y\in H^d(X,X\setminus Y)$. Its image in $H^d(X)$ is the cohomology class of a cooriented submanifold ${\epsilon }_Y$. In particular it restricts to zero on $X\setminus Y$. If the ambient manifold is oriented, orient $Y$ by the normal-first convention. The class is its Poincare dual, characterized byfor the appropriate complementary degree. This is the stated integration identity in cohomological notation.

First prove the Lefschetz fixed-point theorem for an oriented closed n-manifold $X$. Orient its diagonal by the identification $\Delta \cong X$, and choose the normal orientation so this agrees with the normal-first convention in the product. This produces the integral class ${\epsilon }_{\Delta }$. To compute the part that contributes to the trace, tensor with $\mathbb{Q}$: the trace on a finitely generated integral cohomology group means its trace on the free quotient, equivalently on rational cohomology; torsion contributes nothing.

Choose a basis $e_{p,j}$ of $H^p(X;\mathbb{Q})$ and its Poincare duality pairing dual basis $e_{p,j}^{\vee }\in H^{n-p}(X;\mathbb{Q})$, normalized by $⟨e_{p,i}⌣e_{p,j}^{\vee },[X]⟩={\delta }_{ij}$. The rational image of the cohomology class of the diagonal isIndeed, pairing this expression against $\alpha \times \beta$ with degrees $p,n-p$ picks out the p-summands. The cross-product multiplication contributes $(-1{)}^{p^2}$ and moving $\alpha$ past $e_{p,j}^{\vee }$ contributes $(-1{)}^{p(n-p)}$. Together with $(-1{)}^{np}$ their product is $+1$. The result is $⟨\alpha ⌣\beta ,[X]⟩$, exactly the diagonal's defining identity. The rational Künneth theorem and nondegeneracy of the duality pairing determine the class. An integral torsion component, if present, is not determined by this rational expression and does not affect any evaluation below.

For the graph map ${\Gamma }_f=(1,f):X\to X\times X$, write $f^{\ast }{e}_{p,j}={\sum }_i{A}_{ij}{e}_{p,i}$. Pulling back the expression and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph lies in $(X\times X)\setminus \Delta$. The relative Thom class restricts to zero there, so ${\Gamma }_f^{\ast }{\epsilon }_{\Delta }=0$ and $L(f)=0$. The contrapositive proves the required fixed-point assertion, without requiring smoothness of $f$ or transversality of its graph.

The printed hypothesis does not require $X$ to be orientable. For a nonorientable $X$, the diagonal's normal bundle is $TX$, so an ordinary integral coorientation cannot be assumed. We extend the proved theorem as follows, rather than apply the untwisted diagonal computation incorrectly. We first construct a smooth embedding in Euclidean space using finitely many charts ${\chi }_i:U_i\to {\mathbb{R}}^n$ and smooth functions ${\rho }_i$ supported in those charts, with at least one ${\rho }_i(x)>0$ at every point. The map $\Psi (x)=({\rho }_i(x),{\rho }_i(x){\chi }_i(x){)}_i$, with zero extensions, is injective: a positive first coordinate identifies a chart in which its remaining coordinates recover $x$. Its differential is injective as well, since $d{\rho }_i(v)=0$ and $d({\rho }_i{\chi }_i)(v)=0$ force $d{\chi }_i(v)=0$ in a chart with ${\rho }_i>0$. Compactness makes this injective immersion a homeomorphism onto its image. This finite-chart Euclidean embedding of a compact smooth manifold, enlarged by zero coordinates if necessary, has positive codimension. A compact tubular disc neighbourhood $W$ is an oriented N-manifold with boundary, inheriting its orientation from the ambient Euclidean space. Its double of a manifold $M$ is closed and oriented. Projection of each copy of $W$ to $X$ agrees on their common boundary, giving $r:M\to X$; inclusion into the first copy gives $i:X\to M$ with $ri=1_X$.

Set $F=ifr$. A fixed point of $F$ must lie in $i(X)$, and then is exactly the image of a fixed point of $f$. Moreover $F^{\ast }=r^{\ast }{f}^{\ast }{i}^{\ast }$, and cyclicity of trace for maps between finite-dimensional rational cohomology groups givesin each degree, because $i^{\ast }{r}^{\ast }=1$. Thus $L(F)=L(f)$. The oriented theorem for $M$ now gives a fixed point of $F$, hence of $f$, whenever $L(f)\ne 0$. This oriented manifold retract for the Lefschetz theorem proves the full printed result. The double construction also works componentwise if necessary.

Let $V\subseteq U$ be open subsets, and let $s\in {\mathcal{F}}_3(V)$. The assumed flasque-kernel section-lifting lemma lifts $s$ to $\tilde{s}\in {\mathcal{F}}_2(V)$. Since ${\mathcal{F}}_2$ is a flabby sheaf, extend $\tilde{s}$ to a section on $U$, then project to ${\mathcal{F}}_3(U)$. Its restriction is $s$. Thus every restriction map of ${\mathcal{F}}_3$ is surjective, proving that the quotient of two flabby sheaves in a short exact sequence is flabby.

For the construction of sheaf cohomology, embed a sheaf of abelian groups $\mathcal{F}$ into the sheafby sending a section to all its germs. This is injective because a section with every germ zero is zero. Arbitrary families of stalk elements glue point by point, and restriction maps are projections; hence $C(\mathcal{F})$ is a flabby sheaf. Set ${\mathcal{Q}}^0=\mathcal{F}$, ${\mathcal{I}}^j=C({\mathcal{Q}}^j)$ and ${\mathcal{Q}}^{j+1}={\mathcal{I}}^j/{\mathcal{Q}}^j$. Compose the quotient map with the next embedding to obtain the flasque resolutionThe sheaf cohomology groups are the cohomology groups of the resulting cochain complex of global sections:The usual comparison of resolutions identifies this construction with the right derived functors of global sections; the construction above is the Godement resolution.

If $\mathcal{F}$ is itself a flabby sheaf, the quotient argument shows inductively that every ${\mathcal{Q}}^j$ is flabby. The assumed lifting result therefore makesan exact sequence for every $j$. The global-section complex is consequently exact in every positive degree. In particular,This establishes the vanishing from the section-lifting property rather than assuming it in advance.

For the smooth algebraic curve $X$, a divisor on an algebraic curve is a finite formal sum $D={\sum }_P{n}_{P}P$ of closed points with integer coefficients. At each closed point, the local ring of a smooth algebraic curve is a discrete valuation ring; write its normalized discrete valuation as $v_P$. For $f\in k(X{)}^{\ast }$ the principal divisor is $\mathrm{div}(f)={\sum }_P{v}_P(f)P$. Only finitely many coefficients are nonzero: zeros and poles are proper closed subsets in the Noetherian Zariski topology of a curve. The divisor class group is the abelian groupEquivalently, its elements are divisors on an algebraic curve modulo linear equivalence of divisors.

We give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

Use countability of negative curves. A smooth projective surface over $\mathbb{C}$ is a compact smooth manifold homotopy equivalent to a finite CW complex, so its integral cohomology group $H^2(X,\mathbb{Z})$ is finitely generated and therefore countable. Distinct irreducible curves $C,D$ with $C^2<0$ cannot have the same cohomology class: that would give $C\cdot D=C^2<0$, while distinct effective curves have nonnegative intersection multiplicities. Assigning a negative curve its cohomology class is thus injective, proving there are at most countably many negative curves.

For zero self-intersection, uncountable families do occur. On the product of two projective lines ${\mathbb{P}}^1\times {\mathbb{P}}^1$, every fiber ${\mathbb{P}}^1\times \{p\}$ has self-intersection zero, since a different fiber is disjoint and has the same numerical equivalence of divisors class. The parameter $p\in {\mathbb{P}}^1(\mathbb{C})$ varies over an uncountable set. Thus yes, there can be uncountably many such curves.

Work with coefficients ${\mathbb{F}}_2$. For $n\ge 1$, the mod-two cohomology ring of real projective space and the Künneth theorem givewhere all three generators have degree one. Any induced map on cohomology has $f^{\ast }z=\alpha x+\beta y$, with $\alpha ,\beta \in {\mathbb{F}}_2$. By naturality of the cup product, the only possible term in top degree isAll other terms vanish because their exponent of $x$ or $y$ exceeds $n$. Butis even for every $n\ge 1$. Thus the displayed top-degree induced map on cohomology is zero for every $f$.

Both top-degree homology groups are one-dimensional over ${\mathbb{F}}_2$, and their dual cohomology groups pair with them nondegenerately. Naturality of this evaluation makes $f_{\ast }$ dual to $f^{\ast }$, so $f_{\ast }$ is also zero and cannot be an isomorphism. This is the mod-two degree obstruction for equal projective factors. ConsequentlyIf zero is included among the allowed values, ${\mathbb{RP}}^0$ is a point and the unique map is an isomorphism on $H_0$. In that convention, the only value is $n=0$.

Work first with coefficients ${\mathbb{F}}_2$. For a rank-$r$ real vector bundle $E\to B$ over a CW complex, choose a fibre metric and its disk bundle $D(E)$ and sphere bundle $S(E)$. The mod-two Thom class is the unique class $U\in H^r(D(E),S(E);{\mathbb{F}}_2)$ restricting to the nonzero generator on every fibre pair. The Thom isomorphism theorem asserts thatNo orientation of the vector bundle is needed with these coefficients.

Let $j:H^{\ast }(D(E),S(E))\to H^{\ast }(D(E))$ forget the relative condition, and let $s:B\to D(E)$ be the zero section. Set $e_2(E)=s^{\ast }j(U)$, the top Stiefel–Whitney class $w_r(E)$. Since $D(E)$ retracts onto the zero section, $j(U)={\pi }^{\ast }{e}_2(E)$. Thus the relative-to-absolute map corresponds under the Thom isomorphism theorem to multiplication by $e_2(E)$. Substituting these identifications into the long exact sequence in relative cohomology of $(D(E),S(E))$ gives the unoriented Gysin sequencewhere $p$ is the sphere bundle projection.

Apply this to the real tautological line bundle ${\gamma }_n\to {\mathbb{RP}}^n$. Its sphere bundle is $S^n$, with projection the antipodal double cover, and put $t=e_2({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. Suppose $n\ge 1$. The map on $H^0$ from the connected base to the connected sphere is an isomorphism, so exactness shows that multiplication by $t$ is injective from $H^0$ to $H^1$. Since both groups are one-dimensional, $t$ is a generator.

For $1<j<n$, the adjacent sphere cohomology groups in the Gysin sequence vanish, so multiplication by $t$ is an isomorphism from degree $j-1$ to degree $j$. If $n>1$, the top portion isThe last nonzero arrow is surjective between one-dimensional groups, hence an isomorphism; the preceding arrow is zero, so multiplication by $t$ is again an isomorphism. For $n=1$ the earlier $H^0$ argument already gives the top multiplication. Thus $1,t,\dots ,t^n$ are the nonzero generators in their respective degrees, while $t^{n+1}=0$ by dimension. We obtainFor $n=0$ this says simply $H^{\ast }({\mathbb{RP}}^0;{\mathbb{F}}_2)={\mathbb{F}}_2$, with $t=0$. This computes the mod-two cohomology ring of real projective space, including its multiplication rather than only its additive groups.

Now use integral coefficients and an oriented rank-$n$ vector bundle. Its orientation selects an integral Thom class $U$. Define its Euler class byAgain $j(U)={\pi }^{\ast }e(E)$. The cup product of two relative classes in $H^{\ast }(D(E),S(E))$ agrees with the mixed relative/absolute product after forgetting the relative condition on either factor. ConsequentlyThis is the cup square of a Thom class. If $n$ is odd, graded commutativity of the cup product gives $U\cup U=-U\cup U$, hence $2U^2=0$. Since $U\cup {\pi }^{\ast }(2e(E))$ is, up to the graded sign, the Thom isomorphism image of $2e(E)$, its vanishing impliesThus the Euler class of an oriented odd-rank vector bundle is two-torsion; the conclusion is integral and does not require the base cohomology to be torsion-free.

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 $t^4=0$.

A chain map inducing isomorphisms on all homology groups. The same definition applies to a map of cochain complexes using cohomology groups. A quasi-isomorphism need not be an isomorphism in each degree.

For positive $p,q$, quotient $S^p\times S^q$ by $(x,y)\mapsto (-x,-y)$. This is a free double covering space action. Its closed manifold quotient is orientable exactly when $p+q$ is even, because the product mapping degree is $(-1{)}^{p+q+2}$. If $p,q\ge 2$, the product is simply connected and the quotient has fundamental group $\mathbb{Z}/2$. For $p=q=2$, the quotient has integral cohomology groups $\mathbb{Z}$ in degrees $0,4$, $\mathbb{Z}/2$ in degrees $2,3$, and zero elsewhere. Every positive-degree cup product is zero. These conclusions follow from Poincare duality, the Euler characteristic under a finite covering, and the universal coefficient theorem for cohomology.