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The Euler class of $T{S}^{2n}$ evaluates to $2$. Its Gysin sequence of a sphere bundle givesAll products of positive-degree classes vanish.

The cell structure gives one integral cohomology generator in each degree divisible by four through $4n$. The Gysin sequence of a sphere bundle for the quaternionic tautological line bundle makes multiplication by its Euler class an isomorphism between successive such degrees. Choose $u$ to be the negative of that Euler class; its powers generate the displayed cohomology ring.

For an oriented proper smooth map $f:M^m\to N^n$ of oriented manifolds, the Gysin homomorphism is a degree-shifting pushforward $f_{!}:H^k(M)\to H^{k+n-m}(N)$, defined using Poincare duality and the appropriate pushforward on homology. For an oriented sphere bundle it is integration along the fiber and participates in the Gysin sequence of a sphere bundle. Oriented embeddings give the corresponding Thom-class construction.

A Gysin sequence of a sphere bundle is the long exact sequence induced by the Gysin homomorphism and the Euler class of a vector bundle.

For the unit sphere bundle $S(E)\to B$ of an oriented rank-$d$ vector bundle, the Gysin sequence contains

Let $R=M{U}^{\ast }(\mathrm{pt})$ be the coefficient ring of complex cobordism. We use its standard coefficient theorem: $R\cong \mathbb{Z}[b_1,b_2,\dots ]$ with $|b_j|=-2j$. In particular $R$ is concentrated in nonpositive even degrees. Let $H$ be the quaternionic tautological line bundle on ${\mathbb{HP}}^2$, regarded as a complex rank-two bundle by right multiplication by a fixed imaginary quaternion. SetWe will proveThe argument must establish the multiplication as well as the additive groups, since collapse of a spectral sequence alone does not settle ring extensions.

The quaternionic projective space has one cell in each dimension $0,4,8$, giving integral cohomology $\mathbb{Z}$ in exactly these three degrees. The sphere bundle of $H$ is the quaternionic Hopf fibrationWrite $u=e(H_{\mathbb{R}})=c_2(H)$ for its ordinary degree-four class, fixing its sign by this equality. The Gysin sequence of a sphere bundle gives surjections, in fact isomorphisms,Indeed the corresponding neighboring cohomology groups of $S^{11}$ vanish. Both source and target groups are $\mathbb{Z}$, so these maps make $u$ and $u^2$ generators. There are no higher cohomology groups. This proves the cohomology ring of quaternionic projective space in the form $H^{\ast }({\mathbb{HP}}^2;\mathbb{Z})=\mathbb{Z}[u]/(u^3)$.

Apply the multiplicative Atiyah-Hirzebruch spectral sequence for complex cobordism:The skeletal construction obtains this sequence from the long exact sequences of successive skeleta; the quotients are wedges of spheres, and the first differential is the ordinary cellular differential with coefficient groups $R^t$. Its multiplication on the associated graded group is the ordinary cup product with these coefficients. The finite eight-dimensional skeleton makes the filtration finite, so there is no convergence ambiguity.

Only columns $s=0,4,8$ and even rows $t$ are nonzero. If $r$ is odd, the target column $s+r$ is odd and zero. If $r$ is even, the target coefficient degree $t-r+1$ is odd and zero. The same observations exclude incoming differentials. Consequently $E_{\infty }=E_2$, with associated graded $R$-module freely generated by $1,u,u^2$ in filtrations $0,4,8$.

The canonical map from complex cobordism to integral cohomology preserves the chosen complex orientation and hence sends $q$ to $u$. The class $q$ vanishes on the zero-dimensional skeleton since $R^4=0$, and its leading skeletal term is $u$ in filtration four. Multiplicativity makes the leading term of $q^2$ the generator $u^2$ in filtration eight. Thus $1,q,q^2$ lift the associated graded basis.

This resolves all module extensions explicitly. For a class of total degree $d$, subtract its coefficient $a_0\in R^d$ in filtration zero. From the remainder subtract $a_{1}q$, where $a_1\in R^{d-4}$ is its leading coefficient in filtration four, and then subtract $a_2{q}^2$ with $a_2\in R^{d-8}$. The remaining filtration is beyond dimension eight and is zero. Conversely a relation $a_0+a_{1}q+a_2{q}^2=0$ has zero leading coefficient in each of the three successive filtrations, forcing all three $a_j$ to vanish. ThereforeFinally $q^3$ has degree twelve. Every associated graded contribution in that total degree would have coefficient degree $12-s>0$ for $s=0,4,8$, and all such coefficients vanish. Thus $M{U}^{12}({\mathbb{HP}}^2)=0$ and $q^3=0$. The polynomial map $R[t]/(t^3)\to M{U}^{\ast }({\mathbb{HP}}^2)$ taking $t$ to $q$ is surjective by the basis argument and injective by its uniqueness. This proves the complex cobordism ring of the quaternionic projective plane as a graded ring over the stated coefficient ring, including the absence of any additional coefficient-dependent relation.

For a rank-$r$ real vector bundle with a fiber metric, the Thom isomorphism theorem says that if the bundle is oriented over the coefficient ring $R$, there is a Thom class $u\in H^r(D(E),S(E);R)$ restricting to the chosen generator on every fiber pair, andis an isomorphism. For $R={\mathbb{F}}_2$, every real vector bundle has the required orientation. The disk bundle retracts to $B$, so apply the pair long exact sequence and the Thom isomorphism to get the Gysin sequence of a sphere bundleThe first map is multiplication by the Euler class of a vector bundle, the pullback to the zero section of the absolute image of $u$. The connecting map followed by inverse Thom isomorphism gives ${\pi }_{!}$.

Apply this over ${\mathbb{F}}_2$ to the tautological real line bundle $\gamma$ on ${\mathbb{RP}}^n$. Its unit sphere bundle is $S^n$, with projection sending a unit vector to its line. Put $a=e_2(\gamma )=w_1(\gamma )\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$, the mod-two Euler class of a real line bundle. For $n\ge 1$ the degree-zero pullback $H^0({\mathbb{RP}}^n)\to H^0(S^n)$ is an isomorphism. Exactness therefore makes the subsequent transfer zero, and multiplication by $a$ from $H^0$ into $H^1$ injective. In intermediate degrees the sphere cohomology vanishes, making each successive multiplication by $a$ an isomorphism. At degree $n$, the map from $H^{n-1}$ into $H^n$ is injective; the transfer $H^n(S^n)\to H^n({\mathbb{RP}}^n)$ is surjective because $H^{n+1}({\mathbb{RP}}^n)=0$. Since $H^n(S^n)={\mathbb{F}}_2$, that top group is one-dimensional too, and $a^n$ is its nonzero generator. The same endpoint reasoning covers $n=1$. This gives the Gysin proof of mod-two projective-space cohomology and the mod-two cohomology ring of real projective space:For $n=0$ this says simply $H^{\ast }({\mathbb{RP}}^0)={\mathbb{F}}_2$.

For a self-map of ${\mathbb{RP}}^4$, the induced map has $f^{\ast }a=\lambda a$ with $\lambda \in {\mathbb{F}}_2$, so $f^{\ast }{a}^j={\lambda }^j{a}^j$. The mod-two alternating trace isfor either value of $\lambda$. The alternating trace on homology equals that on the finite cellular chain complex; consequently this is the reduction modulo two of the integral Lefschetz number. That integer is nonzero, and the Lefschetz fixed-point theorem forces a fixed point. Thus every self-map of the real projective four-space has a fixed point, a case of the Lefschetz fixed-point property of even-dimensional real projective space.

For the odd-dimensional example, let $J(x_1,x_2,x_3,x_4)=(-x_2,x_1,-x_4,x_3)$, so $J^2=-I$. Its projectivization isIt is well-defined and continuous on ${\mathbb{RP}}^3$. A fixed projective point would have $Jv=tv$ for a nonzero real vector and real $t$, forcing $-v=J^{2}v=t^{2}v$ and therefore $t^2=-1$, impossible. This is a fixed-point-free complex-structure map on odd-dimensional real projective space.

Work with integral cohomology. For a frame $v=(v_1,\dots ,v_k)$ in the complex Stiefel manifold, defineThese fibres form the complement bundle on a complex Stiefel manifold, a complex vector subbundle of the trivial rank-$n$ bundle. Indeedis a smooth orthogonal projector of constant complex rank $n-k$; locally projecting suitable fixed basis vectors and applying Gram-Schmidt process provides bundle frames. The underlying real vector bundle has rank $r=2(n-k)$ and is oriented by its complex structure: a complex change of frame has real determinant equal to the positive number $|{\det }_{\mathbb{C}}A{|}^2$.

A point of its sphere bundle is $(v_1,\dots ,v_k,w)$ with $\Vert w\Vert =1$ and $w$ orthogonal to the frame. Appending $w$ gives a diffeomorphismwhose inverse forgets the last vector while retaining it as the unit vector in the complement. Thus the asserted homotopy equivalence is actually an identification of sphere bundles, for $0\le k<n$.

Start the induction at $V_0({\mathbb{C}}^n)=\{\mathrm{point}\}$, with cohomology the exterior algebra on no generators. SupposeThis is a free abelian group in every degree, and for $k\ge 1$ its smallest positive degree is $2n-2k+1=r+1$. Hence $H^r(V_k;\mathbb{Z})=0$. This vanishing holds at $k=0$ as well, since the base is a point and $r=2n>0$. The Euler class $e(E)$ belongs to this zero group and is therefore zero.

For this oriented rank-$r$ vector bundle, the Gysin sequence of a sphere bundle has segmentBecause $e(E)=0$, this becomesIn particular $H^{\ast }(V_{k+1};\mathbb{Z})$ is torsion-free. At $j=r-1$, choose a class $a\in H^{r-1}(V_{k+1};\mathbb{Z})$ with ${\pi }_{!}a=1$. Its restriction to the sphere fibre is an integral generator. The projection formula for sphere bundle integration, with the conventional fibre-integration sign, isIt can be obtained from the relative cohomology of the disk/sphere pair and the Thom isomorphism theorem. Consequently every cohomology class of $V_{k+1}$ is uniquely expressible as ${\pi }^{\ast }c+{\pi }^{\ast }b⌣a$: subtract the second term using its fibre integral, then use exactness to identify the remaining term; injectivity of ${\pi }^{\ast }$ and the projection formula prove uniqueness. This establishes an additive splitting by the old monomials and the old monomials multiplied by $a$.

To establish the ring, $|a|=r-1$ is odd, so graded commutativity of the cup product gives $a^2=-a^2$, and hence $2a^2=0$. The torsion-freeness just proved now gives $a^2=0$. Graded commutativity also gives the required anticommutation with each old odd generator, and the old relations persist under the injective ring map ${\pi }^{\ast }$. The additive splitting proves that there are no additional relations. This is exactly Gysin ring splitting for an odd-dimensional sphere bundle, and supplies a new exterior generator of degreeInduction therefore proves the full ring calculationOver $\mathbb{Z}$, graded commutativity by itself only kills twice an odd square; torsion-freeness is the essential extra step in this integral cohomology of a complex Stiefel manifold proof.

Finally an ordered orthonormal $n$-frame is a unitary matrix, so $V_n({\mathbb{C}}^n)=U(n)$. Each exterior generator may be included once or omitted from a basis monomial. The rank of degree $j$ is therefore the number of subsets of the generator degrees $1,3,\dots ,2n-1$ summing to $j$. Multiplication of their two-term generating factors givesThe highest exponent is $1+3+\cdots +(2n-1)=n^2$, agreeing with the real dimension of the unitary group, and the total rank is $2^n$.

The complex tautological line bundle over Complex projective space isOn the standard chart $U_i=\{[z_0:\cdots :z_n]:z_i\ne 0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The mapis a local trivialization; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial complex line bundle.

Its unit sphere bundle is $S^{2n+1}$: a unit vector $v\in {\mathbb{C}}^{n+1}$ corresponds to $([v],v)$. The projection is the Hopf fibration. Regard $\lambda$ as an oriented real rank-two vector bundle using its complex orientation, and put $t=e(\lambda )\in H^2({\mathbb{CP}}^n;\mathbb{Z})$. The Gysin sequence of a sphere bundle gives, for $2\le q\le 2n$,as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1({\mathbb{CP}}^n)=0$. The CW complex structure has one cell in dimensions $0,2,\dots ,2n$ and none above $2n$, so these groups and products giveThe Euler class of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb{Z}$.

Now let $\alpha$ generate $H^2(S^2;\mathbb{Z})$ and let $u_i$ be its pullback from the $i$th factor of $(S^2{)}^n$. The Künneth theorem gives $H^2((S^2{)}^n;\mathbb{Z})={⨁}_i\mathbb{Z}{u}_i$. If a map $f$ were invariant under all factor permutations, write $f^{\ast }\alpha ={\sum }_i{a}_i{u}_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta :S^2\to (S^2{)}^n$ be the diagonal. ThenThe second condition is $f\circ \Delta =\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. For $n>1$ this is impossible in $\mathbb{Z}$, and $\boxed{\text{no such continuous map exists}.}$ This is the diagonal-degree obstruction to a symmetric sphere retraction.

The quaternionic projective space ${\mathbb{HP}}^n$ is the space of one-dimensional right quaternion subspaces of ${\mathbb{H}}^{n+1}$. Equivalently it is the quotient of the unit sphere $S^{4n+3}$ by simultaneous right multiplication by unit quaternions. Its coordinate filtration has one open cell ${\mathbb{H}}^k\cong {\mathbb{R}}^{4k}$ in each dimension $4k$, for $0\le k\le n$. Hence its cellular cohomology is $\mathbb{Z}$ in those dimensions and zero otherwise.

Let $\mathcal{H}$ be the quaternionic tautological line bundle. Its unit sphere bundle is $S^{4n+3}\to {\mathbb{HP}}^n$, with fibre $S^3$. The Gysin sequence of a sphere bundle shows that multiplication by its Euler class $e\in H^4$ is an isomorphism from $H^{4k}$ to $H^{4k+4}$ for $k<n$. Choose the generator $u=-e$. Its powers generate every nonzero positive degree, giving the cohomology ring of quaternionic projective spaceThis also accounts for $n=0$.

First take $n=2$. Under the coordinate inclusion $i:{\mathbb{CP}}^2\to {\mathbb{HP}}^2$, the pulled-back quaternionic line is the quaternionic extension of the complex tautological line $L$. As a complex rank-two bundle it is $L\oplus \overline{L}$: a transition scalar $c\in \mathbb{C}$ acts on the two complex coordinates of a quaternion by $c$ and $\overline{c}$. Put $x=c_1(L^{\ast })$, the degree-two generator of the cohomology ring of complex projective space. The Whitney sum formula for Chern classes givesHere the Euler class of a complex vector bundle is its top Chern class, using the complex orientation. In particular this degree-four pullback has coefficient one; it is not a multiple of larger absolute value. The compatible tautological bundles on the projective filtrations give the same equality for every $n$, and multiplicativity then determines the whole ring map:It is zero whenever $2k>n$. These facts are the complex inclusion into quaternionic projective space.

For an odd prime $p$, the Steenrod reduced powers are natural stable cohomology operationsThey satisfy $P^0=1$, the Cartan formula, $P^{i}a=0$ when $2i>|a|$, and $P^{i}a=a^p$ when $|a|=2i$. In particular, on ${\mathbb{CP}}^{\infty }$ one has $P^{1}x=x^p$ and $P^{i}x=0$ for $i>1$. The Cartan formula and the binomial theorem givePass to the infinite projective spaces, where $i^{\ast }:H^{\ast }({\mathbb{HP}}^{\infty };{\mathbb{F}}_p)\to H^{\ast }({\mathbb{CP}}^{\infty };{\mathbb{F}}_p)$, $u\mapsto x^2$, is injective. The equality just obtained determines the Steenrod powers on quaternionic projective space; restricting to the finite spaces giveswith coefficients modulo $p$ and powers above $n$ set to zero. In particular $P^{1}u=2u^{(p+1)/2}$, $P^{2}u=u^p$, and $P^{i}u=0$ for $i>2$. The infinite-space argument matters: the finite inclusion cannot detect those degrees for which $x^{2k}=0$.

Finally put $Q={\mathbb{HP}}^{15}/{\mathbb{HP}}^{12}$ and $S={\Sigma }^{48}{\mathbb{HP}}^3$. Both have reduced cohomology $\mathbb{Z}$ in degrees $52,56,60$ and zero otherwise. Choose integral generators $a_{13},a_{14},a_{15}$ for $Q$ whose pullbacks under the quotient map are $u^{13},u^{14},u^{15}$, and suspended integral generators $b_1,b_2,b_3$ for $S$ from $u,u^2,u^3$.

Use $p=5$. Naturality for the quotient and the formula above giveStability under the suspension isomorphism instead givesAny homotopy equivalence $h:Q\to S$ would induce isomorphisms on the rank-one integral groups, so $h^{\ast }{b}_1=\epsilon a_{13}$ and $h^{\ast }{b}_3=\delta a_{15}$ with $\epsilon ,\delta \in \{1,-1\}$. Reducing modulo five and commuting $h^{\ast }$ with $P^1$ would require $\epsilon =2\delta$ in ${\mathbb{F}}_5$. Neither $1$ nor $-1$ equals $2$ or $-2$ modulo five. ThereforeThe essential point is that integral generator signs constrain Steenrod comparisons. Arbitrary changes of basis over ${\mathbb{F}}_5$ could rescale these two nonzero coefficients into agreement; a genuine equivalence must also preserve the integral lattices, where only the two signs are available.

The real tautological line bundle has total space $E=\{(\ell ,v):\ell \in {\mathbb{RP}}^n,v\in \ell \}$ and projection $(\ell ,v)\mapsto \ell$. On the open chart $U_i=\{[x]:x_i\ne 0\}$, the vector $s_i([x])=x/x_i$ is a continuous nowhere-zero section of a vector bundle, independent of the representative of $[x]$. The explicit local trivialization iswith inverse $([x],v)\mapsto ([x],v_i)$. Both maps are continuous and linear on each fiber. The charts cover the base, proving the existence of the required local trivializations.

Let $E\to B$ be a rank-$r$ vector bundle oriented over a commutative ring $R$, with $r\ge 1$. An $R$-orientation is a coherent choice of generator of each fiber's top relative cohomology, or equivalently a Thom class $u_E$. The Thom isomorphism theorem identifies $H^{q-r}(B;R)$ with $H^q(D(E),S(E);R)$ by $a\mapsto {\pi }^{\ast }a⌣u_E$. The disk bundle retracts to $B$, and the relative-to-absolute map becomes multiplication by the Euler class $e(E)$. The pair's long exact sequence therefore becomes the Gysin sequence of a sphere bundle:With real coefficients, the bundle must be oriented in the usual sense. For the computation here use $R={\mathbb{F}}_2$, over which every real vector bundle is oriented, including the nonorientable real tautological line bundle when $n\ge 1$.

The unit sphere bundle of ${\gamma }_n$ is $S^n$: a unit vector determines its line. Put $a=e_2({\gamma }_n)=w_1({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. For $n\ge 1$, the pullback $H^0({\mathbb{RP}}^n;{\mathbb{F}}_2)\to H^0(S^n;{\mathbb{F}}_2)$ is an isomorphism, so the next connecting homomorphism is zero and multiplication by $a$ is injective on $H^0$. Since the sphere's cohomology vanishes strictly between degrees zero and $n$, exactness makes multiplication by $a$ an isomorphism $H^{q-1}\to H^q$ for $1\le q<n$, and injective for $q=n$. For $q=1$ these assertions also use the just-noted vanishing of the connecting homomorphism from $H^0(S^n)$.

The standard $n$-dimensional CW complex structure of Real projective space gives $H^q=0$ for $q>n$. Consequently the segment $H^n(S^n;{\mathbb{F}}_2)\to H^n({\mathbb{RP}}^n;{\mathbb{F}}_2)\stackrel{⌣a}{\to }{H}^{n+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)=0$ makes $H^n$ at most one-dimensional. The preceding injectivity makes it exactly one-dimensional, generated by $a^n$. This includes $n=1$; $n=0$ is a point separately. We have derived the multiplication, not just the dimensions:

An odd map between spheres descends to $\bar{f}:{\mathbb{RP}}^n\to {\mathbb{RP}}^m$. The odd maps pull back the real tautological line bundle argument gives ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$: with $x$ a unit vector the fiber map is $tx\mapsto tf(x)$, unchanged when $(x,t)$ is replaced by $(-x,-t)$. Naturality of the first Stiefel–Whitney class gives ${\bar{f}}^{\ast }b=a$. If $n>m$, the relation $b^{m+1}=0$ would imply $a^{m+1}=0$ in a ring where this power is nonzero. Hence $\boxed{n\le m}$. The case $n=0$ is immediate, and the same argument rules out $m=0<n$.

For the map with separate oddness, pass to $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$. Let $a,b$ be the degree-one classes from its two domain factors and $c$ the target class. There is an isomorphismgiven on fibers by $(sx)\otimes (ty)\mapsto stg(x,y)$; flipping either unit representative leaves this map well defined. The First Stiefel–Whitney class of a tensor product of real line bundles gives ${\bar{g}}^{\ast }c=a+b$. By the Künneth theorem,Pulling back $c^{n+1}=0$ shows that $(a+b{)}^{n+1}=0$. The two end monomials already vanish, while every interior monomial $a^j{b}^{n+1-j}$, $1\le j\le n$, is a distinct nonzero basis element. Therefore all interior binomial coefficients in row $N=n+1$ are even. To identify such rows, write $N={\sum }_{i\in I}{2}^i$. In ${\mathbb{F}}_2[t]$ repeated squaring gives $(1+t{)}^N={\prod }_{i\in I}(1+t^{2^i})$. If $I$ has at least two elements, the coefficient of $t^{2^{\min I}}$ is one and its exponent is strictly between zero and $N$, a contradiction. Thus $N$ is a power of two, as in binomial coefficients with even interior terms, andThis is the necessary cohomological obstruction to separately odd sphere multiplication; it does not assert existence in every dimension of this form.

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 $M$.