Zero-section pullback is the Euler class

Both [F1] and [F2] use the same disk/sphere pair and the same fiber generators, so Thom uniqueness identifies their classes: the interface class of [F1] is the normalized class fixed by [F2]. Applying the relative-to-absolute map $j^{\ast }$ and then the zero-section map $s^{\ast }$ gives exactly $e(E)$, rather than an ill-typed direct pullback of a relative class.

For rank zero with standard unit orientation $u_E=1$ and $j^{\ast },s^{\ast }$ are identities, so $e(0_B)=1$; a different supplied orientation $o$ gives $e(0_B,o)=o$, the componentwise unit. For a positive-rank trivial bundle, choose a continuous unit-length section $s_1$ of the disk bundle (normalize a nowhere-zero section given by the supplied trivialization). The sections $s_t=t{s}_1$, $0\le t\le 1$, are homotopic in the disk bundle, so they pull $j^{\ast }{u}_E$ back to the same class; at $t=1$ the section factors through $S(E)$, where $i^{\ast }{j}^{\ast }{u}_E=0$ by exactness of the pair sequence. Hence the Euler class of a positive-rank trivial bundle is zero. Mod two the same definition requires no chosen orientation, and AC is inherited from the general AT suppliers.