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A complex rank-$r$ vector bundle has a canonical real orientation, and its Euler class equals its top Chern class. For a complex line this is $e(L_{\mathbb{R}})=c_1(L)$. The splitting principle for complex vector bundles gives an injective cohomology pullback on which the bundle splits into lines. The Whitney product formula for Euler classes and Whitney sum formula for Chern classes then identify both sides with the product of those first Chern classes, proving the general equality.

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