Thom-defined Euler class of an oriented vector bundle

Assume the general Thom theorem's AC hypothesis, and let $\xi \to B$ be an $R$-oriented rank-$n$ bundle with normalized Thom class $u_{\xi }$. Let $j^{\ast }:H^n(D(\xi ),S(\xi );R)⟶H^n(D(\xi );R)$ be the relative-to-absolute map in the pair sequence, and let $s:B\to D(\xi )$ be the zero section. The Thom-defined Euler class is $e_{\mathrm{T}\mathrm{h}}(\xi )=s^{\ast }{j}^{\ast }(u_{\xi })\in H^n(B;R)\mathrm{.}$

This definition uses no later characteristic-class page. Naturality of the pair sequence and of the Thom class gives $e_{\mathrm{T}\mathrm{h}}(f^{\ast }\xi )=f^{\ast }{e}_{\mathrm{T}\mathrm{h}}(\xi )$ for an orientation-preserving pullback. Reversing an integral orientation negates the class.

For rank zero, $j$ and $s$ are identities and $u=1$, so $e_{\mathrm{T}\mathrm{h}}(0_B)=1\in H^0(B;R)$. On an empty base the class is the unique zero class; over the zero ring it is zero (and also the unit). Point bases, identity pullbacks, the zero section and both maps of the pair all follow the displayed composite. The formula is choice-free once $u_{\xi }$ is supplied; AC is inherited only from general Thom existence and naturality.

The rank-zero assertion in the preceding paragraph is restricted to the standard unit orientation. For an arbitrary supplied rank-zero orientation $o\in H^0(B;R)$, fiber normalization gives $u=o$; since $j$ and $s$ are identities, $e_{\mathrm{T}\mathrm{h}}(0_B,o)=o$. Thus reversing the integral unit orientation gives $u=-1$ and $e_{\mathrm{T}\mathrm{h}}(0_B,-1)=-1$, consistently with the orientation-sign convention above.