Euler class by zero-section pullback of the Thom class

The definition is the published one. The composite $s^{\ast }{j}^{\ast }{u}_{\xi }$ has the same domain, the same maps and the same normalization data as the class of [F1]: the pair $(D(\xi ),S(\xi ))$, the relative-to-absolute map $j^{\ast }$, the zero section $s$, and the normalized Thom class $u_{\xi }$ of [F2]. Thus $e(\xi )$ is not a second Euler construction but the same class, and every property recorded in [F1] — naturality for orientation-preserving pullbacks, the sign under orientation reversal, and the orientation-dependent rank-zero value — applies to it verbatim. In particular the shorthand $s^{\ast }{u}_{\xi }$ in the statement means $s^{\ast }{j}^{\ast }(u_{\xi })$ and never the pullback of an absolute class along the zero section alone.

Coefficient and rank conventions. For $R={\mathbb{F}}_2$, [F3] supplies the canonical orientation, so $e_2(\xi )$ is defined for every real bundle in the Thom scope and, in characteristic two, reversing the orientation does not change the class. For $R=\mathbb{Z}$ the class depends on the supplied integral orientation and changes sign when that orientation is reversed. In rank zero, $D(\xi )=B$, $S(\xi )=\mathrm{\varnothing }$, and $j$ and $s$ are the identity maps. Fiberwise normalization [F2] says that $u_{\xi }$ restricts to the supplied orientation $o$ on every point, hence $u_{\xi }=o$ and the composite is $o\in H^0(B;R)$; it is $1$ only for the standard unit orientation. Over the empty base there is exactly one class, the zero class, and over the zero ring the unit and the zero class coincide. These conventions agree with the corresponding clauses of [F1].