Pullback of the Thom class along a transverse section computes the Euler class

Pair comparison and bounding. Choose a smooth metric by Every smooth vector bundle admits a smooth bundle metric. The radial retraction of the punctured disk bundle to the sphere bundle, together with Long exact sequence of a pair in singular cohomology, makes the restriction $H^r(D(E),D(E)\setminus s_0(M);R)\to H^r(D(E),S(E);R)$ an isomorphism. Let ${\tilde{u}}_E$ be the inverse image of $u_E$; in rank zero both removed subspaces are empty. Replace $s$ by $\rho \circ s$ with $\rho (v)=v\mathrm{/}\sqrt{1+\mathrm{\mid }v{\mathrm{\mid }}^2}$; this is smooth and lands in the open unit disc bundle, has the same zero locus $Z$ because ${\rho }^{-1}(0)=\{0\}$, and $d{\rho }_0=\mathrm{i}\mathrm{d}$, so transversality of $s$ to the zero section at every point of $Z$ is unchanged. Thus $s$ is a map of pairs from $(M,M\setminus Z)$ to $(D(E),D(E)\setminus s_0(M))$, and below $s^{\ast }(u_E)$ abbreviates the precisely typed $s^{\ast }({\tilde{u}}_E)$. Take a tubular chart $\mathrm{\Phi }:\mathrm{\Omega }\subseteq {\nu }_Z\to U$ from [F1]. Its vertical differential followed by the quotient $q_z:T_{z}M\to {\nu }_{Z,z}$ gives $B_z=q_z\circ d{\mathrm{\Phi }}_{0_z}{\mathrm{\mid }}_{{\nu }_{Z,z}}$. Since $\mathrm{\Phi }$ fixes $Z$, its tangent differential is the identity on $T_{z}Z$; the invertibility of $d{\mathrm{\Phi }}_{0_z}$ therefore makes $B_z$ invertible. In local bundle coordinates its matrix consists of smooth first derivatives, and the inverse matrix is smooth by the cofactor formula. Thus $B$ is a smooth bundle automorphism. Replace $\mathrm{\Phi }$ by $\mathrm{\Psi }=\mathrm{\Phi }\circ B^{-1}$ on $B(\mathrm{\Omega })$; this is a tubular chart and $q_z\circ d{\mathrm{\Psi }}_{0_z}{\mathrm{\mid }}_{{\nu }_{Z,z}}=\mathrm{i}\mathrm{d}$.

Local normalization. On a trivializing chart $U$ write $s(x)=(x,f(x))$ with $f:U\to {\mathbb{R}}^r$, $f(0)=0$ and $d{f}_0$ surjective; $Z\cap U=f^{-1}(0)$ and $T_{0}Z=\ker d{f}_0$. The normalized Thom class restricts on $(D^r,S^{r-1})$ to the orientation generator [F2], so the local fibre component represents the fibre orientation class in the punctured-fibre pair; the invertible normal derivative permits computing its pullback on a small normal disk by a local diffeomorphism. Under the normal identification ${\nu }_Z\cong E{\mathrm{\mid }}_Z$ of [F6], the composite ${\nu }_{Z,0}\stackrel{d{f}_0}{\to }{\mathbb{R}}^r$ is an orientation-preserving isomorphism, the vertical part of $d{f}_0$ being exactly the comparison defining the induced orientation; so the local generator is the normal generator. This uses a tubular chart with normal differential the identity; an arbitrary orientation-reversing fibre reparametrization would reverse the integral generator.

Global identification. The tube of [F1] identifies a neighbourhood of $Z$ with a neighbourhood of the zero section of ${\nu }_Z$. On each sufficiently small normal disk at $z\in Z$, the fibre component of $s$ has its only zero at $z$ and has invertible derivative there by [F6]. Choice-free smooth inverse function theorem in Euclidean space gives a local diffeomorphism. Its differential preserves the supplied $R$-orientation by [F6] and the normalization in step 1.1, so its pullback carries the fibre orientation generator to the prescribed normal generator; over ${\mathbb{F}}_2$ the local degree is $1$ regardless of sign. Thus step 1.2 says precisely that the relative pullback restricts to the prescribed generator on every normal fibre pair. Thom uniqueness [F3] identifies it on the tube with $u_{{\nu }_Z}$. Excision [F4] identifies $H^r(M,M\setminus Z;R)$ with the relative cohomology of that tube and its complement of $Z$; choose a small metric disk subbundle inside the chart domain (possible uniformly because $Z$ is compact), and use excision to restrict to its interior. The punctured-disk comparison from step 1.1 identifies the resulting group with $H^r(D({\nu }_Z),S({\nu }_Z);R)$ after rescaling the metric. Hence $s^{\ast }(u_E)$ is the image of $u_{{\nu }_Z}$ under the tubular identification.

Euler class. Let $s_0:M\to D(E)$ be the zero section and $u_E^{\mathrm{a}\mathrm{b}\mathrm{s}}=j^{\ast }{u}_E\in H^r(D(E);R)$ its absolute restriction. The forget-support map $H^r(M,M\setminus Z;R)\to H^r(M;R)$ factors the pullback, so the image of $s^{\ast }(u_E)$ is $s^{\ast }(u_E^{\mathrm{a}\mathrm{b}\mathrm{s}})$. The affine homotopy $(x,t)\mapsto (1-t)s(x)$ stays inside the disc bundle and is a homotopy from $s$ to $s_0$ through sections, so Homotopic maps induce equal maps in singular cohomology gives $s^{\ast }(u_E^{\mathrm{a}\mathrm{b}\mathrm{s}})=s_0^{\ast }(u_E^{\mathrm{a}\mathrm{b}\mathrm{s}})$. By [F5] this is $e(E)$. Empty $Z$ and rank zero follow from the same formulas; AC is inherited from the Thom and tubular suppliers.