Collapse pulls the Thom class back to the Poincaré dual

For $r>0$ and nonempty $S$, choose $0<a<b<\rho$ inside the supplied tube and put $K=\mathrm{\Phi }(D_a(\nu ))\subset U$. This is compact. Excision identifies $H^r(U,U\setminus K;R)$ with $H^r(D_b(\nu ),D_b(\nu )\setminus D_a(\nu );R)$. The outer annulus retracts onto $S_b(\nu )$, so the natural pair sequence identifies this group with $H^r(D_b(\nu ),S_b(\nu );R)$. Scaling gives the normalized Thom class, hence a supported class $u\in H_c^r(U;R)$. In $Y=\mathrm{Th}(\nu )$, the complement of the image of $D_{a\mathrm{/}\rho }(\nu )$ contracts radially to the basepoint, fixing that point; thus the same pair-sequence argument lifts $u_{\nu }$ uniquely to that support pair. The collapse pulls this lift back to $(M,M\setminus K)$, and its restriction to $U$ is the class just constructed. Excision for the open inclusion $U\subset M$, with compact support $K\subset U$, therefore gives $c^{\ast }{u}_{\nu }=e(u)$ after forgetting the disjoint basepoint. The rank-zero collapse instead extends the Thom multiplier from the clopen tube $U=S$, giving the same equality directly.

The zero-section inclusion $z:S\to U$ is a homotopy equivalence, with inverse the bundle projection $p$ and homotopy $(x,v)\mapsto (x,tv)$. Put $b=p_{\ast }{D}_U(u)\in H_s(S;R)$. To find its image in $H_s(S,S\setminus \{x\};R)$, localize the relative cap calculation [F3] over an oriented trivializing ball about $x$. Write its coordinates in normal-first order $N\times T$. Thom uniqueness identifies the local class with the pullback of a normal cocycle $\eta$, normalized to evaluate to $1$ on the oriented normal relative cycle $a$; let $d$ be the tangent relative orientation cycle. The product orientation is represented by the signed shuffle $\mathrm{sh}(a\otimes d)$. For any product chain $z_0$, the front-evaluation formula gives ${\mathrm{pr}}_{T\mathrm{\#}}({\mathrm{pr}}_N^{\ast }\eta \cap z_0)=(\eta \otimes \mathrm{i}\mathrm{d}){\mathrm{AW}}_{r,s}(z_0),$ where contraction is zero on tensor summands of normal degree other than $r$. On the excisive disk-product triad, relative naturality in [F6] gives $\mathrm{AW}\mathrm{sh}\simeq \mathrm{i}\mathrm{d}$. Contracting that homotopy by the cocycle $\eta$ leaves equal relative homology classes, so the displayed cap sends the product orientation to $\eta (a)d=d$. This computes the image of $b$ as the chosen local orientation generator of $S$; no restriction of ordinary homology to an open set is used. The normal-first order accounts for the positive sign.

By [F4] a compact oriented manifold's fundamental class is the unique class with these local restrictions, so $b=[S]$ componentwise, also when $S$ is disconnected. Since $z_{\ast }{p}_{\ast }$ is the identity on $H_s(U;R)$, step 2.1 gives $D_U(u)=z_{\ast }[S]$. Open-extension naturality [F2] now gives $c^{\ast }{u}_{\nu }\cap [M]=D_{M}e(u)=j_{\ast }{D}_U(u)=j_{\ast }{z}_{\ast }[S]=i_{\ast }[S]$. The isomorphism $D_M$ makes its cohomological reformulation unique.

Over ${\mathbb{F}}_2$ every fiber and tangent orientation has its canonical generator, and the same local computation proves the formula without orientability. For $r=0$, $S$ is a union of components and the collapse extends the componentwise orientation multiplier; cap sends it to the specified $[S]$. Empty $S$ gives the zero class and empty manifolds give zero groups. AC is used only through the general Thom and duality suppliers. This proves the collapse application, retaining AT ownership of those suppliers.