A nowhere-zero section forces the Euler data to vanish

Class level. The bundle and the nowhere-zero section meet precisely the positive-rank Thom hypotheses of [F1], so $e(E)=0$ in $H^r(M;R)$.

Numerical consequences. A section disjoint from the zero section is vacuously transverse to it with empty zero locus, so step 1.1 and [F2] give $e(E)\cap [M]=0$, and when $r=n$ the Kronecker evaluation of the zero class is $0$; part (i) follows. For part (ii), scale the nowhere-zero smooth normal field by a positive constant into the tube (possible by compactness of $A$). Its section $s$ has $Z(s)=\mathrm{\varnothing }$, and the push-off $A_s$ is disjoint from $A$; By The self-intersection number of a complementary-dimensional oriented submanifold the disjoint transverse count is zero modulo two, so $A{\cdot }_{2}A=0$, in agreement with [F4]. Under the additional integral orientations of $M$ and $A$, the induced normal orientation meets [F3], and the same empty signed count gives $A\cdot A=⟨e({\nu }_A),[A]⟩=0$. The class-level assertion of step 1.1 uses the AT Euler construction [F5]. No converse is asserted: vanishing of the Euler data does not in general produce a nowhere-zero section, as the clutching witness below shows.

For the failure of the converse, take the oriented rank-three bundle $E_{\rho }\to S^4$ clutched by quaternion conjugation $\rho :S^3\to SO(3)$; The quaternion double cover generates the third homotopy group of SO(3) proves it is nontrivial. Its Euler class is zero because Homology of spheres and Topological universal coefficient short exact sequence for cohomology give $H^3(S^4;\mathbb{Z})=0$ (both the Hom and Ext inputs are zero). A nowhere-zero section would span a trivial line; a bundle metric and Short exact sequences of numerable vector bundles split would give $E_{\rho }\cong {\epsilon }^1\oplus F$ with $F$ oriented of rank two. By Oriented clutching classifies oriented bundles over spheres, $F$ is clutched by a map $S^3\to SO(2)$. Sending a rotation matrix to its first column identifies $SO(2)$ with the circle. Since $S^3$ is simply connected by $S^n$ is simply connected for every $n\ge 2$, $\mathbb{R}\to \mathbb{R}\mathrm{/}\mathbb{Z}$ is a universal covering and Lifting criterion for maps from path-connected locally path-connected spaces lift that map to $\mathbb{R}$, where straight-line contraction makes it nullhomotopic. Thus $F$ and then $E_{\rho }$ would be trivial, a contradiction. This retains the general failure of the converse without making an A-page theorem depend on a B-page example.