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When an oriented rank-$d$ vector bundle $E$ has a closed oriented $d$-dimensional base $B$, its Euler number is the integer obtained by evaluating the Euler class of a vector bundle on the fundamental class of $B$. A nonzero Euler number obstructs a nowhere-zero section and therefore obstructs a framing.

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

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.

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

For an oriented real rank-$r$ vector bundle $\pi :E\to X$, let $u_E\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. If $s:X\to D(E)$ is the zero section, the Euler class of a vector bundle is

Let $u$ generate $H^{2k}(S^{2k};\mathbb{Z})$. The Poincaré-Hopf theorem givesFor every integer $m$, choose a map $f_m:S^{2k}\to S^{2k}$ of degree $m$. Naturality of the Euler class givesThus every even class belongs to ${\mathcal{E}}_{2k}(S^{2k})$.

More generally, every class $a\in H^{2k}(M;\mathbb{Z})$ on a closed $2k$-manifold is $f^{\ast }u$ for some map $f:M\to S^{2k}$, by the realization of top-dimensional cohomology by a sphere map. Pulling back $T{S}^{2k}$ givesso $2H^{2k}(M;\mathbb{Z})\subseteq {\mathcal{E}}_{2k}(M)$. The inclusion can be strict: every complex line bundle on $S^2$ has an underlying oriented real two-plane bundle, and these bundles realize every integral Euler class. Hence

The odd-dimensional analogue fails. The Euler class of an oriented odd-rank vector bundle is two-torsion, whereas $H^{2k+1}(S^{2k+1};\mathbb{Z})\cong \mathbb{Z}$ is torsion-free. Thus every oriented rank-$(2k+1)$ bundle on $S^{2k+1}$ has zero Euler class, and the nonzero subgroup $2\mathbb{Z}$ cannot lie in ${\mathcal{E}}_{2k+1}(S^{2k+1})$.

Finally, if $E$ and $F$ are oriented bundles of ranks $k$ and $l$, orient $E\oplus F$ by the ordered sum. The Whitney product formula for Euler classes statesTherefore cup product restricts to the asserted map ${\mathcal{E}}_k(X)\otimes {\mathcal{E}}_l(X)\to {\mathcal{E}}_{k+l}(X)$.

It need not be injective. For $X=S^2$ and $k=l=2$, one has ${\mathcal{E}}_2(S^2)=\mathbb{Z}$, so the source contains $\mathbb{Z}\otimes \mathbb{Z}\cong \mathbb{Z}$, but $H^4(S^2)=0$ and the map is zero. It need not be surjective either. For $X=S^2$ and $k=l=1$, every oriented real line bundle is trivial, so both degree-one Euler-class sets vanish, while ${\mathcal{E}}_2(S^2)=\mathbb{Z}$.

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 $M$.

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 $X=S(L\otimes \lambda )$.