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An element killed by the next differential of a cochain complex. Its class modulo coboundaries is a cohomology class.

The largest number of positive-degree cohomology classes whose cup product is nonzero, with a specified coefficient ring $R$; it can be infinite. It is zero if there is no nonzero positive-degree class. The vanishing cup product from an open cover argument implies that Lusternik–Schnirelmann category is at least cup length plus one. For Complex projective space ${\mathbb{CP}}^r$, its integral cup length is $r$.

For a cohomology class $a$, its rth cup power is its r-fold product in the cohomology ring. Naturality gives $f^{\ast }(a^r)=(f^{\ast }a{)}^r$. A nilpotent class has some positive cup power zero; the nilpotence order is preserved by ring isomorphisms and can obstruct a homotopy equivalence or a nonzero mapping degree.

For global cohomology classes satisfying the local basis criterion for Leray-Hirsch classes, multiplication commutes with the Mayer–Vietoris sequence, including its connecting homomorphism: multiply cochains on the right by global cocycle representatives. Induct over a finite trivializing open cover. If $U$ is a union of $r-1$ patches and $V$ is the last patch, $U\cap V$ has a cover by at most $r-1$ trivialized intersections. The induction hypothesis on $U$, $V$ and $U\cap V$, followed by the Five lemma, proves the theorem on $U\cup V$. A compact base supplies the finite cover.

Interpretation of the removed locus. Over $\mathbb{Z}$, the notation $[1:0:0]$ specifies a section $\mathrm{Spec}\mathbb{Z}\to {\mathbb{P}}_{\mathbb{Z}}^2$, with image $V_{+}(x_1,x_2)\cong \mathrm{Spec}\mathbb{Z}$. We remove this entire closed subscheme. This is the interpretation forced by the asserted open cover $X=U_1\cup U_2$. Removing one closed point in a particular fiber would not give that open cover. The native PDF uses the same notation as the transcription.

Put $a=x_0/x_1$, $b=x_2/x_1$ on $U_1$. On $U_2$ put $c=x_0/x_2$, $d=x_1/x_2$. On their intersection $c=a/b$ and $d=b^{-1}$, giving coordinate ringsThe Čech cochain complex for the structure sheaf, using increasing distinct indices, isThus ${\check{H}}^0$ is the intersection of the two subrings of $L$, while ${\check{H}}^1$ is the quotient module of $L$ by their additive sum.

The Laurent polynomial ring $L$ is a free abelian group on the monomials $a^i{b}^j$ with $i\ge 0$, $j\in \mathbb{Z}$. The first subring has the basis $j\ge 0$. The second has the basisso its allowed exponents satisfy $j\le -i$. A monomial belongs to both precisely when $i=j=0$. The monomials belonging to neither satisfy $i\ge 2$ and $-i<j<0$. Since the subgroups are spanned by subsets of the same basis, their sum introduces no other relations. ThereforeHere the square brackets denote cohomology classes, and the direct sums are as abelian groups. In particular, $[a^2{b}^{-1}]=[x_0^2/(x_1{x}_2)]$ is a nonzero cohomology class. This scheme is a separated scheme, being an open subscheme of projective space over $\mathbb{Z}$, and is a quasi-compact scheme by (d). Both charts and their intersection are affine schemes, so this Čech cohomology equals sheaf cohomology. The nonzero first sheaf cohomology of the quasi-coherent structure sheaf rules out an affine scheme:

For integral singular cochains $a\in C^p(X)$ and $b\in C^q(X)$, define the cup product on an ordered simplex byThe same construction works over any commutative coefficient ring. Expanding the coboundary as the alternating sum of faces gives the Leibniz identityIndeed faces deleting vertices before the split occur in the first sum, faces after the split occur in the second, and the two terms at the common vertex cancel with opposite signs. Thus the product of cocycles is a cocycle. Replacing $a$ by $a+\delta c$ changes the product with a cocycle $b$ by $\delta (c⌣b)$; replacing $b$ by $b+\delta d$ changes the product with a cocycle $a$ by $(-1{)}^p\delta (a⌣d)$. Hence the product is independent of representatives and defines a well-defined product on cohomology classes.

With ${\mathbb{F}}_2={\mathbb{Z}}_2$ coefficients, the real-projective-space cohomology ring isFor $n>m\ge 1$, the pullback of the degree-one generator is either zero or $t$. The second choice would send the zero class $t^{m+1}$ to the nonzero class $t^{m+1}\in H^{m+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)$, contradicting the naturality of the cup product. Thus the generator pulls back to zero, and so do all its positive powers. For $m=0$ the target's reduced cohomology is already zero. Therefore every such map induces zero on reduced mod-two cohomology.

Suppose a continuous $f:S^n\to {\mathbb{R}}^n$ had $f(x)\ne f(-x)$ everywhere. Thenis a continuous odd map to $S^{n-1}$ and descends to $\bar{h}:{\mathbb{RP}}^n\to {\mathbb{RP}}^{n-1}$. For $n\ge 2$, a path from $x$ to $-x$ projects to a loop detected by the degree-one generator of the double cover. Its image lifts to a path from $h(x)$ to $-h(x)$, so is detected by the target's double cover as well. Thus ${\bar{h}}^{\ast }$ is nonzero in degree one, contradicting the preceding calculation. For $n=1$, an odd continuous map from the connected circle to $S^0$ is already impossible. The case $n=0$ is immediate. HenceThis proves the Borsuk-Ulam theorem rather than assuming it.

For a closed cover $A_1,\dots ,A_{n+1}$, assume no member contains an antipodal pair. Apply the just-proved theorem to $F(x)=(d(x,A_1),\dots ,d(x,A_n))$, using any metric on the sphere. Empty sets can be assigned a constant positive coordinate. It gives equal distances at $x,-x$. If either point belongs to an $A_i$ with $i\le n$, zero distance and closedness place both points in it, a contradiction. Hence neither lies in any of the first $n$ sets, and both must lie in $A_{n+1}$, another contradiction. This proves the Lusternik-Schnirelmann-Borsuk theorem for closed sets.

Four closed sets on $S^2$ need not contain an antipodal pair. Take the four unit vertices $v_i$ of a regular tetrahedron centered at zero and define $C_i=\{x:x\cdot v_i\ge x\cdot v_j\text{ for every }j\}$. These closed sets cover the sphere by maximizing the four scalar products. If $x$ and $-x$ belonged to $C_i$, both inequalities would force all four scalar products to be equal. Since $\sum v_i=0$, they would all be zero, and because the vertices span ${\mathbb{R}}^3$, $x$ would be zero. This is impossible on the sphere. Thus these four sets give the antipodal-free closed cover from simplex Voronoi cells counterexample.

Work in the homotopy category of simply connected CW complexes. This is one of the permitted assumptions on the fundamental group; it makes every coefficient system below untwisted. A rational space is a simply connected space whose homotopy groups are vector spaces over $\mathbb{Q}$. A rationalization of a topological space is a map $r:X\to X_{\mathbb{Q}}$ to such a space which inducesIts purpose includes a universal mapping property, which will be described after the construction.

An Eilenberg–MacLane space $K(A,n)$ has exactly one nonzero homotopy group, ${\pi }_n=A$, where $A$ is abelian when $n\ge 2$. The representability theorem says $[B,K(A,n)]\cong H^n(B;A)$, naturally in $B$ and $A$. More generally, a principal fibration is a homotopy pullback of a path-space fibration $PC\to C$ along a map from its base to $C$; the fibre $\Omega C$ acts on the fibre paths by concatenation. For the Postnikov tower construction take $C=K(A,n+1)$. A principal Eilenberg–MacLane fibration with fibre $K(A,n)$ is the homotopy pullback of the path fibration of $K(A,n+1)$. Equivalently, its total space is the homotopy fibre of a map $k:B\to K(A,n+1)$. It is classified by the cohomology class $[k]\in H^{n+1}(B;A)$; there is no nontrivial action of the base fundamental group on the fibre coefficients in this principal description.

The Postnikov tower theorem supplies spaces $P_{n}X$ and compatible mapswith ${\pi }_i(P_{n}X)={\pi }_i(X)$ for $i\le n$ and zero for $i>n$. For a simply connected $X$, the fibre of $P_{n}X\to P_{n-1}X$ is $K({\pi }_{n}X,n)$, and the fibration is principal. Its Postnikov invariant isrepresented by $P_{n-1}X\to K({\pi }_{n}X,n+1)$, whose homotopy fibre is $P_{n}X$. Thus the homotopy groups alone are not the tower data: the classes $k_n$ specify how each new group is attached. For instance a zero class gives a product with $K({\pi }_{n}X,n)$.

State also the standard rational comparison results used in constructing the tower: $K(A,n)\to K(A\otimes \mathbb{Q},n)$ is a rational homology equivalence; a map between simply connected spaces is a rational homology equivalence if and only if it induces isomorphisms on all rational homotopy groups; and a rational homology equivalence induces isomorphisms in cohomology with any constant rational vector-space coefficients. The last assertion uses the universal coefficient theorem over the field $\mathbb{Q}$. These results are used here without proof, as allowed for this essay.

Construct rational stages inductively, with $P_1^{\mathbb{Q}}=\ast$. Suppose $r_{n-1}:P_{n-1}X\to P_{n-1}^{\mathbb{Q}}$ has already been constructed and is a rational homology equivalence. Put $V_n={\pi }_{n}X\otimes \mathbb{Q}$ and change coefficients in $k_n$ to obtain ${\overline{k}}_n\in H^{n+1}(P_{n-1}X;V_n)$. The comparison result gives an isomorphismLet $k_n^{\mathbb{Q}}$ be its unique inverse image and defineEquality of the pulled-back classes gives a homotopy-commuting square of classifying maps, and hence a map $r_n:P_{n}X\to P_n^{\mathbb{Q}}$ over $r_{n-1}$. On the fibre it is the map induced by ${\pi }_{n}X\to V_n$. The homotopy exact sequences therefore show that $r_n$ tensors all homotopy groups through degree $n$ with $\mathbb{Q}$, while the higher groups of both stages vanish. The rational comparison theorem makes $r_n$ a rational homology equivalence, allowing the induction to continue.

Take the homotopy inverse limit of the rational tower:The convergence theorem for a Postnikov tower gives $X\simeq {\mathrm{holim}}_n{P}_{n}X$ and the compatible stage maps induce $r:X\to X_{\mathbb{Q}}$. To see what convergence says here, fix a homotopy degree $i$. From stage $n\ge i$ onwards the group and its transition maps are constant; the inverse-limit group is ${\pi }_{i}X\otimes \mathbb{Q}$ and the associated derived-limit obstruction is zero. Thus this construction has exactly the claimed rational homotopy groups. Taking a homotopy limit, rather than an arbitrary set-theoretic inverse limit, is part of the construction.

The rational localization theorem says that this map has the universal propertyIn other words every map from $X$ to a rational space factors through $r$, and the factorization is unique up to homotopy. This concerns homotopy classes, not strict uniqueness of point-set maps. The theorem can be formulated as localization of the simply connected homotopy category at rational homotopy equivalences; rational spaces are its local objects. It also implies uniqueness of rationalization of a topological space: apply the property to two candidate rational spaces to obtain inverse equivalences under $X$. In particular the construction forgets torsion in homotopy while retaining precisely the information needed for maps to rational targets.

The same theory extends to nilpotent spaces, where the fundamental group is nilpotent and its action on higher homotopy groups is nilpotent. That version requires localization of the fundamental group and compatible treatment of its coefficient action. The simply connected assumption chosen here avoids those additional issues and supplies the full construction and mapping property requested.

It is enough to assume that $h_k:{\pi }_k(X)\otimes \mathbb{Q}\to H_k(X;\mathbb{Q})$ is surjective for every $k\ge 2$; an integrally surjective Hurewicz homomorphism certainly has this property. First give the minimal-model argument suggested by the hint. Let $A=(\Lambda V,d)$ be a simply connected Sullivan minimal model, with $V$ in degrees at least two. Put $A^{+}={⨁}_{k>0}{A}^k$. Its indecomposable quotient of an augmented algebra isMinimality makes the induced differential on $V$ zero, so projection is a chain map $q:A^{+}\to (V,0)$. The Sullivan model identification says that the induced mapis dual to rational Hurewicz homomorphism. Under the usual finite-type duality convention, the assumed surjectivity makes $q_{\ast }$ injective.

A product of two positive cohomology classes has a decomposable cocycle representative, so its image under $q_{\ast }$ is zero. Injectivity therefore givesThis deduction uses the hypothesis; zero cup products by themselves would not yet prove formality of a topological space.

Choose a homogeneous cohomology basis $h_{\lambda }$ and cocycle representatives $z_{\lambda }\in A$. Their indecomposable parts $q(z_{\lambda })$ are linearly independent. Extend them to a graded basis of $V$ and replace the corresponding generators by the actual cocycles $z_{\lambda }$. This is a valid change of free algebra generators: each $z_{\lambda }$ differs from its indecomposable part by products of generators of strictly smaller degree, so degree induction constructs the inverse substitution. Call these closed generators $C$, and the remaining generators $N$.

Define a map of graded algebrasIts target has zero differential. Every differential in $A$ is decomposable, and all products of positive target classes vanish, so $\varphi$ is a chain map. If $z$ is any positive cocycle, its indecomposable part is in the span of the $q(z_{\lambda })$, because these form the image of $q_{\ast }$. Write $q(z)={\sum }_{\lambda }{c}_{\lambda }q(z_{\lambda })$. Then $\varphi (z)={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$, while injectivity of $q_{\ast }$ gives $[z]={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$. Thus $\varphi$ induces the identity on cohomology and is a quasi-isomorphism. This proves that $X$ is a formal space.

The rational wedge sum assertion has a direct topological proof, which also avoids finite-dimensional duality restrictions in the preceding model argument. Choose a homogeneous basis $b_{\lambda }$ of ${\tilde{H}}_{\ast }(X;\mathbb{Q})$. By surjectivity, each basis class is the image under the Hurewicz homomorphism of an element of ${\pi }_{n_{\lambda }}(X)\otimes \mathbb{Q}$. In $X_{\mathbb{Q}}$ this element is represented by an actual sphere map $S^{n_{\lambda }}\to X_{\mathbb{Q}}$. Simple connectivity and $H_1=0$ ensure $n_{\lambda }\ge 2$. The wedge sum of these maps givesand induces an isomorphism of rational homology by the choice of basis. The rational Whitehead theorem now makes it a rational homotopy equivalence, givingA wedge sum of spheres is a formal space. For each individual sphere, its augmented minimal model maps by a quasi-isomorphism to its sphere cochain algebra, and also to its cohomology algebra, by killing the extra odd generator in the even-sphere case. Take based simplicial models of the spheres. Their wedge sum is a simplicial pushout over the common basepoint, so the algebra of rational polynomial differential forms is the corresponding fibre product of augmented sphere algebras. Taking fibre products of these model maps gives a quasi-isomorphism zigzag to the square-zero cohomology algebra. This also works for infinitely many sphere summands: the positive parts are products of the individual complexes, and products of vector-space exact sequences over a field are exact. In the finite case its minimal resolution is the minimal model of a wedge of simply connected spheres. Thus the topological proof establishes both conclusions without imposing finite rank on the sphere basis. For the zero reduced-homology case the wedge sum is a point. Both the minimal-model construction and the explicit sphere map explain why the surjectivity hypothesis is decisive.

Geometric quantization starts from a classical symplectic manifold $(P,\omega )$ and seeks a Hilbert space and quantum observables that reflect its geometry. It has two distinct stages: prequantization, which represents the full Poisson algebra on sections of a line bundle, and a choice of polarization in geometric quantization, which reduces the excessive phase-space dependence of those sections.

Use the conventions ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. A prequantum line bundle is a Hermitian line bundle $L\to P$ with unitary connection of curvatureExistence requiresfor every closed integral two-cycle $\Sigma$, equivalently integrality of the symplectic cohomology class. Necessity follows from the integral curvature periods of a unitary line bundle; conversely an integral class defines such a bundle, and one adjusts its connection to give the specified curvature. This is a restriction on the classical system, not a choice of local coordinates. For an exact cotangent symplectic form $\omega =-d\theta$, with $\theta =p_{j}d{q}_j$, the bundle can be trivial and a connection is $\nabla =d-i\theta /\hslash$. On a sphere with symplectic area $4\pi s$, the condition becomes $2s/\hslash \in \mathbb{Z}$. The sign of the curvature convention changes if one changes the Hamiltonian-vector-field convention or uses the dual bundle; the integrality condition is unaffected.

The Kostant-Souriau prequantum operator for a real observable isIts derivative part describes the classical Hamiltonian flow and the multiplication term corrects the bracket. Indeed,Since $X_f(g)=-\{f,g\}$, expanding the operator commutator givesWith the Hermitian bundle metric and Liouville measure, these operators are formally symmetric for real $f$ because Hamiltonian flows preserve symplectic volume. On noncompact spaces, domains and self-adjointness still require attention; the algebraic identity initially holds on an appropriate common smooth domain.

Prequantization alone is too large. Sections depend on $2n$ phase-space variables, and the canonical commutation relations are represented reducibly. A polarization is an involutive maximal isotropic distribution $\mathcal{F}\subset T_{\mathbb{C}}P$ of complex rank $n$. Polarized sections satisfyThe curvature vanishes on pairs of polarization vectors because $\omega {|}_{\mathcal{F}}=0$, and involutivity makes the equations locally compatible. A real polarization is tangent to a Lagrangian foliation. A complex polarization in a Kähler manifold chooses one complex tangent type. With the curvature sign used here, choose the holomorphic tangent distribution, giving antiholomorphic wavefunctions. The conjugate line-bundle and sign convention gives the more usual holomorphic description.

For $T^{\ast }Q$, the vertical real polarization is spanned locally by ${\partial }_{p_j}$. Since the canonical one-form has no vertical component, its polarized sections are functions of $q$ alone. In the above trivialization,One does not simply integrate these states over every momentum to retain the prequantum norm: they are constant along noncompact polarization leaves and that integral would diverge. The polarized inner product instead uses the configuration-space measure or an intrinsic half-density description. This recovers the Schrödinger realization.

Only observables whose Hamiltonian flows preserve $\mathcal{F}$ act directly on the polarized space. For example, the free-particle flow generated by $p^2/2$ sends vertical tangent vectors into directions with a configuration-space component, so its raw prequantum operator does not preserve the vertical polarization. To obtain the usual kinetic operator one needs an additional procedure, such as transporting the polarization and pairing back, rather than pretending every classical observable restricts automatically. The Blattner-Kostant-Sternberg pairing and related projection methods address this problem. The impossibility of quantizing all classical observables with every desired algebraic property is another reason to specify the admissible observable algebra.

A metaplectic correction tensors $L$ with a square root of the polarization's canonical determinant bundle when such a square root exists. These half-forms make the pairing more intrinsic and supply corrections to operator transport. They account for familiar Maslov and zero-point shifts; their existence is an extra topological condition. For a flat phase plane, take $z=q+ip$ and replace $\theta =pdq$ by the gauge-equivalent ${\theta }_s=(pdq-qdp)/2$. ThenThe antiholomorphic function therefore has Gaussian norm, the complex-conjugate Bargmann-Fock space. This explicit calculation fixes which polarization matches our curvature sign. Choosing the opposite tangent type with the same sign would give growing Gaussian sections, not the desired Hilbert space. On compact positive Kähler examples, the corresponding holomorphic description (or its conjugate in these conventions) gives finite-dimensional state spaces.

Finding a suitable polarization is therefore a substantial part of the construction. A globally smooth real Lagrangian foliation may fail to exist; for example a regular real polarization on $S^2$ would require a line field that its tangent topology forbids. Singular foliations and noncompact leaves complicate both states and inner products. Although the restricted prequantum connection is locally flat on a Lagrangian leaf, a global nonzero parallel section requires trivial holonomy. In an integrable system this selects the Bohr-Sommerfeld quantization leaves; including the half-form correction gives the familiar action conditionin the usual Maslov-index convention. Complex polarizations require compatible global complex geometry and positivity. Different choices can lead to quite different realizations, and a natural unitary comparison is not automatic. Thus integral curvature provides prequantization; polarization and its global compatibility determine the actual quantum state space.

Let $M$ be a compact Riemannian manifold without boundary. Let $\delta =d^{\ast }$ be the formal $L^2$ adjoint operator of the exterior derivative, and let ${\Delta }_H=d\delta +\delta d$ be the Hodge Laplacian. Write ${\mathcal{H}}^r(M)=\ker {\Delta }_H\cap {\Omega }^r(M)$ for the smooth harmonic differential forms. The Hodge decomposition theorem states that this space is finite-dimensional andan $L^2$-orthogonal direct sum. Thus every smooth $r$-form is uniquely the sum of a harmonic, an exact and a coexact form; its exact and coexact components are unique, though the primitives used to write those components need not be unique. AlsoThe isomorphism sends a harmonic form to its de Rham cohomology class. For example, a closed coexact component $\delta \gamma$ must vanish because $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$; an exact harmonic form also vanishes by orthogonality. These facts explain uniqueness of the harmonic representative of a cohomology class.

On an oriented $n$-manifold the codifferential is $\delta =(-1{)}^{n(r+1)+1}\ast d\ast$ on $r$-forms. The general formulation above also holds on nonorientable Riemannian manifolds: use the globally defined Riemannian volume density for the $L^2$ inner product and its formal adjoint, rather than require a global ordinary volume form. The closed-manifold hypothesis is essential unless suitable boundary conditions are added.

For an open subset $U\subseteq B$, let $E_U={\pi }^{-1}(U)$ and define the degree-preserving module mapUsing the prescribed fiber basis identifies its domain with the tensor product in the question. We prove that ${\Phi }_B$ is an isomorphism, giving the Leray-Hirsch theorem in this setting.

First suppose $U$ is a trivializing patch. The cohomological Künneth theorem over a field identifies $H^{\ast }(U\times F;\mathbb{Q})$ with the finite free graded $H^{\ast }(U;\mathbb{Q})$-module on a homogeneous basis of $H^{\ast }(F;\mathbb{Q})$. The required finiteness is supplied by the finite list $c_1,\dots ,c_v$ spanning the fiber's total cohomology. Express the restrictions of the $c_j$ in this product basis. A coefficient multiplying a fiber basis element of degree $k_i$ belongs to $H^{k_j-k_i}(U;\mathbb{Q})$; thus no coefficient can increase fiber degree. The degree-zero blocks are invertible on every path component, because restriction at every point gives a basis. Their inverses are again degree-zero cohomology classes.

After inverting those blocks, the remaining change-of-basis map is $I+N$, where $N$ strictly lowers fiber degree. There are only finitely many fiber degrees, so $N$ is nilpotent and $(I+N{)}^{-1}=I-N+N^2-\cdots$ is a finite sum. This proves the local basis criterion for Leray-Hirsch classes and therefore the isomorphism on every trivializing patch, without requiring the patch to be contractible.

Now use a finite trivializing open cover $U_1,\dots ,U_r$, supplied by compactness of $B$. Induct on $r$. Put $U=U_1\cup \cdots \cup U_{r-1}$ and $V=U_r$. The intersection $U\cap V$ is covered by the at most $r-1$ trivializing open sets $U_i\cap V$, so the induction hypothesis applies to $U$, $V$ and $U\cap V$.

Take the direct sum of shifted Mayer–Vietoris sequences for the base and compare it with the sequence for $E_U\cup E_V$. Restriction commutes with multiplication by each global cohomology class $c_j$. The connecting homomorphism also commutes: multiply the cochain representatives on the right by the restriction of a global cocycle representing $c_j$, and use $d(b⌣c_j)=db⌣c_j$. Consequently the maps $\Phi$ form a morphism of exact sequences. The Five lemma makes ${\Phi }_{U\cup V}$ an isomorphism, completing the finite-cover proof of the Leray-Hirsch theorem.

This is an additive graded module isomorphism. It does not assert that the fiber classes have the same multiplicative relations in the total space.

Write $A=E\times \{p\}$, $B=\{p\}\times E$ and $D={\Delta }_E$. Their self-intersection numbers vanish: each factor curve has trivial normal bundle, and the self-intersection of the diagonal of a curve is $2-2g(E)=0$. Each pair meets transversely once. Their intersection pairing is thereforeIf a rational linear combination of their cycle classes vanished in $H^2(E\times E,\mathbb{Q})$, intersecting it with $A,B,D$ would multiply its coefficients by this invertible matrix and give zero. All coefficients must vanish, proving the three classes are linearly independent.

The cycle class map detects the Chow Künneth failure for an elliptic self-product. The codimension-one image of the external product of Chow classes consists of sums from ${\mathrm{CH}}^1(E)\otimes {\mathrm{CH}}^0(E)$ and ${\mathrm{CH}}^0(E)\otimes {\mathrm{CH}}^1(E)$. Here ${\mathrm{CH}}^0(E)=\mathbb{Z}[E]$, and every divisor's cohomology class is its degree of a divisor times the class of a point. Thus the image's cohomology is contained in the span of $[A],[B]$. The diagonal's independent class lies outside that span, so its Chow class lies outside the external-product image. Therefore the natural map of Chow groups is not surjective.

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.

For $n\ge 1$, let $\#$ denote concatenation in the first cube coordinate and let $\cdot$ denote pointwise multiplication of maps into the topological group. Both operations descend to the homotopy group, and the constant map is their common identity. They satisfy the interchange rulewhich follows by considering the two halves of the first coordinate. Using the identity class $0$ givesTherefore $[\alpha \beta ]=[\alpha ]+[\beta ]$. The equality is on based homotopy classes; the usual reparametrization homotopies justify the unit identities for concatenation. This is the Eckmann-Hilton argument, and proves homotopy-group addition in a topological group even for $n=1$.

For the unit quaternions, the explicit inverse isBoth compositions cancel in the displayed order, without commuting the quaternions. Multiplication, inversion and all integer powers are continuous, so this proves that ${\Phi }_{k,l}$ is a homeomorphism for all integers $k,l$.

Put $m=k+l$. Let $a,b$ be the standard generators of $H_3(S^3\times S^3;\mathbb{Z})$, represented by the first and second factors. The previous pointwise-multiplication result and the Hurewicz theorem imply that the mapping degree of $q\mapsto q^m$ on $S^3$ is $m$, including negative integers. Restricting $\Phi$ to the two factors therefore givesThe map on $H_0$ is the identity. For the top homology group, take the dual degree-three cohomology classes $x,y$. We have ${\Phi }^{\ast }x=mx+y$, ${\Phi }^{\ast }y=x$. Since these classes have odd degree, graded commutativity of the cup product givesThus ${\Phi }_{\ast }$ is multiplication by $-1$ on $H_6$, and all other homology groups of the product are zero.

For the gluing, regard $\Phi$ as the attaching identification from $\partial (D^4\times S^3)$ to $\partial (S^3\times D^4)$. In these coordinates its first input is the boundary coordinate of $D^4$, its second input the fibre coordinate; its second output is the boundary coordinate of the other $D^4$. This makes the two pieces the usual two trivializations of a three-sphere bundle over the four-sphere.

Let $A=S^3\times D^4$ and $B=D^4\times S^3$, and parametrize their common boundary $T=S^3\times S^3$ using the $B$ coordinates. Collar neighbourhoods give an open cover with the same homotopy types, so the Mayer–Vietoris sequence applies. Both pieces retract onto $S^3$. In degree three, the map into the homology of the pieces isIndeed, inclusion into $A$ retains the first output coordinate, while inclusion into $B$ retains the second input coordinate. Exactness now givesThe second relation eliminates the second generator of the cokernel, leaving a single generator with relation $m$ times that generator equal to zero. Thus $\mathrm{coker}L=\mathbb{Z}/m\mathbb{Z}$. The kernel is zero if $m\ne 0$ and is generated by $(1,0)$ if $m=0$.

The degree-six boundary class gives $H_7(X_{k,l})\cong H_6(T)=\mathbb{Z}$. All remaining positive-degree groups outside degrees $3,4,7$ vanish by the same Mayer–Vietoris sequence; connectedness gives $H_0=\mathbb{Z}$. Consequently the complete integral answer isHere $\mathbb{Z}/0\mathbb{Z}$ means $\mathbb{Z}$, and a negative $m$ gives the same cyclic group as $|m|$. In particular, $m=\pm 1$ gives the integral homology of a seven-sphere, while $m=0$ gives the integral homology of $S^3\times S^4$.

The exterior derivative is an $\mathbb{R}$-linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$, agrees with the differential $df$ of a smooth function, satisfies the graded Leibniz rule $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^p\alpha \wedge d\beta$ for $\alpha \in {\Omega }^p(M)$, and has $d^2=0$. These properties determine it locally and hence globally. To see locality directly, if a differential form $\alpha$ vanishes near $x$, choose a smooth bump function $\chi$ equal to one near $x$ and supported where $\alpha =0$. The identity $d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $(d\alpha {)}_x=0$. Thus global forms can be computed using local extensions in a manifold chart.

In coordinates write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forcesThis is the uniqueness of the exterior derivative from its axioms. The coordinate formula also establishes existence: it has the stated properties, and the chain rule shows that coordinate changes give the same operator.

The de Rham cohomology is the real quotient vector spaceThus a cohomology class records a closed differential form modulo an exact differential form. In degree zero there are no exact forms; closed functions are locally constant. The Poincaré lemma says that every closed form of positive degree on a star-shaped open subset of ${\mathbb{R}}^n$ is exact, and hence that positive-degree closed forms are locally exact on a smooth manifold.

For the first de Rham cohomology of the two-sphere, let $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with ${\mathbb{R}}^2$, so a closed one-form $\alpha$ has primitives $f_U,f_V$. On the connected overlap $U\cap V$, the derivative of $f_U-f_V$ is zero, so this difference is a constant. Subtracting that constant from $f_U$ makes the primitives agree. They glue to a global smooth primitive. Therefore $H_{\mathrm{dR}}^1(S^2)=0$.

Let $q:S^2\to {\mathbb{RP}}^2$ be the double covering map and $a$ the antipodal map. The pullback of a differential form identifies forms downstairs with $a$-invariant forms upstairs. Averaging $(\eta +a^{\ast }\eta )/2$ commutes with $d$. If an invariant form is exact upstairs, averaging its primitive proves it exact downstairs. Conversely an invariant cohomology class has an invariant representative by the same averaging. This proves the de Rham cohomology of a finite quotient identification $H_{\mathrm{dR}}^p({\mathbb{RP}}^2)\cong H_{\mathrm{dR}}^p(S^2{)}^{a^{\ast }}$. In degree zero the sphere is connected and $a^{\ast }$ fixes constants; degree one is zero; in degree two the granted action is multiplication by $-1$, whose invariant real subspace is zero. Forms of degree greater than two vanish. HenceThis is real de Rham cohomology; it does not detect the integral two-torsion of the real projective plane.

For a closed oriented smooth four-manifold, $b_2^{+}$ is the dimension of a maximal positive-definite subspace of its real intersection form. A symplectic form forces $b_2^{+}\ge 1$, because its cohomology class has positive square. Reversing orientation exchanges the positive and negative indices.

The $n$th Postnikov invariant is the cohomology class $k_n\in H^{n+1}(P_{n-1}X;{\pi }_{n}X)$ whose representing map has homotopy fibre $P_{n}X$. It specifies how the new homotopy group is attached to the previous stage. A zero invariant gives the product $P_{n}X\simeq P_{n-1}X\times K({\pi }_{n}X,n)$. Merely tensoring homotopy groups with $\mathbb{Q}$ does not describe rationalization of a topological space unless these classes are also transported.

The Whitehead square of a class $\alpha \in {\pi }_{m}Y$ is $[\alpha ,\alpha ]\in {\pi }_{2m-1}Y$. For the identity class of an even-dimensional sphere, the Hopf invariant is two with the standard product-cell orientation: folding the two sphere summands makes the middle cohomology class of the mapping cone pull back to $a+b$ on $S^m\times S^m$, and $(a+b{)}^2=2ab$. For odd $m$, graded symmetry makes the square two-torsion, hence rationally zero.