The Euler class of an oriented odd-rank bundle is two-torsion

The map $\phi =-{\mathrm{id}}_E:E\to E$, fiberwise multiplication by $-1$, is a bundle isomorphism over ${\mathrm{id}}_B$. In any oriented local frame its matrix is $-I_n$, whose determinant has sign $(-1{)}^n=-1$. By the orientation-cover action in [F1], $\phi$ therefore exchanges the two fiber orientations and carries the section $o$ to the opposite section $-o$. Thus it is an orientation-preserving bundle isomorphism from the oriented bundle $(E,o)$ to the differently oriented bundle $(E,-o)$ over the identity.

The orientation-sign law gives $e(E,-o)=-e(E,o)$, by the reversal clause of [F2] applied to the same underlying bundle with its two orientations.

Oriented naturality gives $e(E,o)=e(E,-o)$. Applying the naturality clause of [F2] to the isomorphism $\phi$ over the identity base map, the class of the source and the class of the target agree, which is the displayed identity and uses only that $\phi$ is orientation-preserving.

Combining gives $e(E,o)=e(E,-o)=-e(E,o)$, hence $2e(E,o)=0$ in the abelian group $H^n(B;\mathbb{Z})$. No assertion that $\phi$ is homotopic to the identity is made: the argument compares two orientations of one bundle through an orientation-preserving isomorphism, exactly as displayed.

Boundary cases. Rank one is included directly in steps 1.1--3.1, so no separate triviality or section claim is needed. Rank zero is excluded: $\det (-I_0)=1$, so the map does not carry an orientation to its negative. Even positive rank is excluded for the same determinant-sign reason: there $-\mathrm{id}$ is orientation-preserving on $(E,o)$ itself, so the argument gives no two-torsion conclusion. Over the empty base the group is zero and the identity is vacuous. The only choice principle used is the Thom-theoretic AC of [F2].