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Put $F={\mathbb{F}}_2$. Under the usual bundle convention of a paracompact base, choose a fiber metric and let $D(E),S(E)$ be the disk bundle and sphere bundle of the rank-$n$ real vector bundle. A mod-two Thom class is a relative cohomology classwhose restriction to every fiber pair $(D^n,S^{n-1})$ is the nonzero generator. Equivalently it is a class in $H^n(E,E\setminus B;F)$, where $B$ is identified with the zero section. No orientation choices are needed over $F$.

The Thom isomorphism theorem asserts that this class exists and that, for every $r$, the mapis an isomorphism. The disk bundle retracts to the zero section $s:B\to D(E)$, so $H^r(D(E);F)\cong H^r(B;F)$. Define the mod-two Euler class $e_2(E)=s^{\ast }u\in H^n(B;F)$, where the pullback includes the passage from relative to absolute cohomology. Under the Thom identification, the map from relative to absolute cohomology sends $a$ to $a⌣e_2(E)$. Substituting into the pair's long exact sequence in cohomology gives the Gysin sequence of a sphere bundle:Here ${\pi }_{!}$ is the connecting homomorphism followed by ${\Phi }^{-1}$. Thus the sequence and the cup-product map have been derived from the Thom theorem, not just stated. The Thom/Gysin constructions are also discussed in Hatcher's Vector Bundles and K-Theory, §3.2.

The original PDF specifies the tautological bundle on ${\mathbb{RP}}^n$; the occurrences of ${\mathbb{R}}^n$ in this part of the TeX transcription have lost the projective-space symbol. Let $L\to {\mathbb{RP}}^n$ be this real line bundle. Its sphere bundle consists of pairs $(\ell ,v)$ with $v$ a unit vector in the line $\ell$, so $S(L)\cong S^n$, with projection the antipodal double covering. Write $t=e_2(L)$.

The additive mod-two groups, obtainable from the one-cell-in-each-dimension cellular chain complex, areFor $n\ge 1$, both the base and $S^n$ are connected, so ${\pi }^{\ast }:H^0({\mathbb{RP}}^n;F)\to H^0(S^n;F)$ is an isomorphism. Exactness makes the following ${\pi }_{!}$ zero and multiplication by $t$ injective from $H^0$ to $H^1$. For $1\le r<n$, the term $H^r(S^n;F)$ vanishes, so the same Gysin sequence makesinjective. Since these groups are one-dimensional, the maps are isomorphisms. Hence $1,t,\dots ,t^n$ are the respective nonzero generators, while $t^{n+1}=0$ by dimension. The mod-two cohomology ring of real projective space is thereforeFor $n=0$ this is simply $F$, with $t=0$.

For integral coefficients, the additive groups areIndeed, the cellular homology of real projective space has boundary multiplication by $2$ in even positive chain degrees and zero in odd degrees, and the integral cochain differential is its transpose. It remains to determine the products; the additive groups alone do not do so.

For $n\ge 2$, the coefficient sequence $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to F\to 0$ has a Bockstein homomorphism $\beta$. Because $H^1({\mathbb{RP}}^n;\mathbb{Z})=0$ and $H^2({\mathbb{RP}}^n;\mathbb{Z})=\mathbb{Z}/2$, exactness makes $a=\beta (t)$ the nonzero integral degree-two class. Reduction $\rho :H^2({\mathbb{RP}}^n;\mathbb{Z})\to H^2({\mathbb{RP}}^n;F)$ is injective: its kernel is the image of multiplication by $2$, which is zero. ThusMore generally, reduction is injective in each positive even degree, and its compatibility with the cup product gives $\rho (a^j)=t^{2j}$. Consequently $a^j$ is the nonzero generator whenever $2j\le n$. Also $2a=0$, and powers beyond the dimension vanish.

If $n=2m$ is even, these powers and the unit account for all groups, so the integral cohomology ring of real projective space isIf $n=2m+1$ is odd, add an integral top-dimensional orientation class $b$ of degree $2m+1$. Its reduction is the nonzero class $t^{2m+1}$. Every product $ab$ and $b^2$ is zero by dimension, and $b$ has infinite additive order. HenceThese formulas include $m=0$: the even space is a point with ring $\mathbb{Z}$, and the odd space is a circle with ring $\mathbb{Z}[b]/(b^2)$, $|b|=1$. The polynomial presentations are interpreted with the displayed grading and products; no extra positive-degree products are left unspecified.

An orientation of a vector bundle of real rank $d$ over $R$ is a coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. For a complex bundle of rank $m$, a complex basis gives the real basis $(v_1,i{v}_1,\dots ,v_m,i{v}_m)$. A complex change of basis has positive real determinant $|{\det }_{\mathbb{C}}A{|}^2$, so these local choices agree and give the canonical orientation of a complex vector bundle over $\mathbb{Z}$, hence over every commutative ring $R$.

Let $D(E)$ and $S(E)$ be the disk and sphere bundles of an $R$-oriented rank-$d$ bundle over compact $B$. A Thom class $u_E\in H^d(D(E),S(E);R)$ restricts to the chosen generator on every fibre. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$ for the zero section $s$. Substituting the Thom isomorphism into the long exact sequence of the pair $(D(E),S(E))$ gives the Gysin sequence of a sphere bundle

Apply this to the Hopf fibration and induct on $n$. If $x=c_1(\mathcal{O}(1))$ has degree two, the result is the cohomology ring of complex projective space

The Künneth theorem over a principal ideal domain gives a natural short exact sequence with tensor-product and Tor terms; the sequence splits, though not naturally. Here all groups are free, so the Tor term vanishes andIf a graded-ring automorphism sends $x$ to $ax+by$, then $(ax+by{)}^{n+1}=0$. The coefficients of the nonzero mixed monomials force $ab=0$; the same applies to the image of $y$. Invertibility then forces a signed permutation matrix on the basis $x,y$. Conversely, complex conjugation on either factor changes the sign of its degree-two generator, and swapping the factors exchanges $x$ and $y$. Thus the realizable group isthe group of all signed $2$ by $2$ permutation matrices, as described by the cohomology automorphisms of a product of two complex projective spaces.

An $R$-orientation of a rank-$r$ vector bundle $E\to B$ is a coherent choice of generator infor every fiber. Equivalently, it is a Thom class $u_E\in H^r(D(E),S(E);R)$ restricting to those generators. The Euler class of a vector bundle iswhere $j:(D(E),\varnothing )\to (D(E),S(E))$ forgets the subspace and $s$ is the zero section.

The Thom isomorphism theorem isInsert these isomorphisms into the long exact cohomology sequence of $(D(E),S(E))$ and use the deformation retraction $D(E)\simeq B$. This gives the Gysin sequence of a sphere bundleTo verify the labelled maps, represent $a$ under the Thom isomorphism by ${\pi }^{\ast }a⌣u_E$. Its image in $H^{\ast }(D(E))$ pulls back along the zero section toThus multiplication by the Euler class is exactly the map from relative to absolute cohomology in the pair sequence.

Now let $M^m\subset S^n$ and let $V$ be its tubular neighborhood. The normal bundle has rank $r=n-m$. With ${\mathbb{F}}_2$ coefficients it is automatically oriented. Since $n>2m+1$, one has $r>m+1$, so its Euler class lies above the dimension of $M$ and vanishes. The Gysin sequence consequently splits into short exact sequences of vector spaces and givesOn the other hand, Alexander duality givesOver the field ${\mathbb{F}}_2$, the latter is naturally dual to ${\tilde{H}}^{n-q-1}(M;{\mathbb{F}}_2)$, which expresses the complement cohomology entirely in terms of that of $M$.

For unit vectors, the defining inequality for $Y$ is equivalent to $v\cdot w\ge 0$. WriteThen $t=(1-\Vert u{\Vert }^2{)}^{1/2}$, sois a homeomorphism $D(T{S}^{2n})\to Y$. Replacing $v$ by $-v$ gives the same description of $Y^{\prime }$. Their intersection isthe unit tangent bundle, equivalently a Stiefel manifold. Since $e(T{S}^{2n})$ evaluates to $\chi (S^{2n})=2$, its Gysin sequence gives the cohomology of the unit tangent bundle of an even-dimensional sphere:Every product of positive-degree classes vanishes.

Let $G$ consist of the eight signed permutations of the two factors:It is the group of signed permutation matrices in dimension two, hence isomorphic to the dihedral group $D_8$, and every element preserves the equation $v\cdot w=0$. If $a,b$ generate $H^{2n}(X;\mathbb{Z})$ from the two sphere factors, these maps act by the corresponding signed permutation matrix, because the antipodal map of the even-dimensional sphere has degree $-1$. The eight actions are distinct, so only the identity can be homotopic to $1_X$.

Every element of $G$ acts trivially on $H^{\ast }(Z;\mathbb{Z})$. The degree-$2n$ group is $\mathbb{Z}/2$, so its automorphism is forced to be the identity. On the top class, regard $Z$ as the Stiefel manifold of two-frames. The signed permutations are the right action of $O(2)$. The determinant-one component is connected, while a determinant-minus-one element is homotopic within that component to $(v,w)\mapsto (v,-w)$, the antipodal map on the $S^{2n-1}$ fibers. That map has degree $+1$, so the top class is also fixed.

The subgroup preserving $Y$ iswhere $A(v,w)=(-v,w)$, $B(v,w)=(v,-w)$, and $S(v,w)=(w,v)$. It consists exactly of the elements with ${\epsilon }_1{\epsilon }_2=1$. Under the deformation retraction $Y\simeq S^{2n}$ onto the diagonal, $1$ and $S$ act as the identity, while $AB$ and $ABS$ act as the antipodal map. They therefore act on $H^{2n}(Y;\mathbb{Z})$ by $+1,+1,-1,-1$, respectively.

The Thom isomorphism theorem identifies $H^{\ast }(Y,Z)$ with a degree-$2n$ shift of $H^{\ast }(S^{2n})$. On the Thom class in degree $2n$, the same four elements act by $+1,+1,-1,-1$: the factor swap reverses each normal vector, which preserves the orientation because the normal rank $2n$ is even, while the simultaneous antipodal map reverses the oriented tangent fiber. On the degree-$4n$ relative class every element acts by $+1$, since the base and fiber signs for the last two elements cancel.

When $n=1$, $Z=V_2({\mathbb{R}}^3)\cong SO(3)$. Every signed permutation of $(v,w)$ extends, after sending the third frame vector $v\times w$ to the required sign, to right multiplication by an element of $SO(3)$. Since $SO(3)$ is path-connected, every right translation is homotopic to the identity. Therefore

A rank-$d$ vector bundle $\pi :E\to B$ is $R$-oriented when its fiber groups $H^d(E_b,E_b\setminus \{0\};R)$ admit a locally coherent choice of generator. Equivalently, it has a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to that generator on every fiber. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$, where $s$ is the zero section. The Gysin sequence of a sphere bundle is

Over ${\mathbb{F}}_2=\mathbb{Z}/2$, the required rings areas in the cohomology ring of complex projective space and the mod-two cohomology ring of real projective space. By the Künneth theorem, the base has ring ${\mathbb{F}}_2[x,y]/(x^{n+1},y^{m+1})$.

Let $L$ be the underlying real plane bundle of the complex tautological bundle on ${\mathbb{CP}}^n$, and let $\lambda$ be the real tautological line bundle on ${\mathbb{RP}}^m$. Under the splitting principle, write the formal Stiefel-Whitney roots of $L$ as $a,b$, so $a+b=w_1(L)=0$ and $ab=w_2(L)=x$. Tensoring with $\lambda$ adds $y=w_1(\lambda )$ to each root. Thus the mod-two Euler class of $L{\otimes }_{\mathbb{R}}\lambda$ is

For $n=m=2$, put $B={\mathbb{CP}}^2\times {\mathbb{RP}}^2$. Multiplication by $e=x+y^2$ onhas ranks $1,1,2,1,1,0,0$ from degrees $0$ through $6$. More explicitly, it is injective through degree three; in degree four its kernel is generated by $x^2+x{y}^2$, and all of degrees five and six lie in its kernel. The Gysin sequence therefore yieldsfor the unit sphere bundle $X=S(L\otimes \lambda )$.

For the sphere bundle of an oriented rank-$r$ real vector bundle, the integration map in the Gysin sequence lowers degree by $r-1$. Its compatibility with the cup product is the displayed projection formula. The construction is the connecting map of the disk/sphere pair followed by the inverse Thom isomorphism theorem; the connecting map has its graded module sign, and the degree sign in the conventional fibre-integration map gives the displayed left-module formula.

For a real rank-$r$ vector bundle $E\to B$, the Thom isomorphism theorem with ${\mathbb{F}}_2$ coefficients identifies the relative cohomology sequence of its disk bundle and sphere bundle withThe multiplier is the top Stiefel–Whitney class, also called the mod-two Euler class. No orientation of $E$ is required. For the real tautological line bundle, the sphere bundle is the antipodal cover of Real projective space; this sequence proves that all powers of the degree-one generator through the dimension of the base are nonzero.