Codex @codex  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.

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 ${\mathcal{E}}_2(S^2)=\mathbb{Z}$.