Euler class of zero and trivial positive-rank bundles

The rank-zero value. Give $0_B$ the standard unit orientation. Its disk bundle is $B$, its sphere bundle is empty, and the zero section and relative-to-absolute map are identities; fiber normalization makes the Thom class the unit $1$ in degree zero. Hence $e(0_B,1)=s^{\ast }{j}^{\ast }(1)=1\in H^0(B;R)$ by [F1]. For another supplied rank-zero orientation $o$, the same calculation instead gives $e(0_B,o)=o$.

The trivial bundle of positive rank. For $n\ge 1$ the constant section $b\mapsto (b,e_1)$ of ${\epsilon }_B^n$ is nowhere zero and continuous. Its global product chart, together with the constant partition of unity $1$, makes the bundle numerable, while [F3] supplies its standard product $R$-orientation. The standing hypothesis places $B$ in the general Thom scope. Thus every hypothesis of [F2] holds, and $e({\epsilon }_B^n)=0\in H^n(B;R)$.

Boundary cases. For the empty base the unique cohomology class in every degree is zero, and the standard rank-zero convention reads $e(0,1)=1=0$ in the zero ring; the two displayed identities remain consistent. For a point base, ${\epsilon }^0$ with unit orientation contributes $e=1\in H^0(\ast ;R)=R$ and ${\epsilon }^n$ with $n\ge 1$ contributes $e=0\in H^n(\ast ;R)=0$. For the zero ring both classes coincide with the unique element, so the identities hold. The section of step 1.2 is a specified function and no choice is made in it; AC is used only through the Thom suppliers that give the Euler class its value.