The normal Thom class realizes the Poincare dual of a closed submanifold

Choose a metric by [F1]. In the derivative calculation and final quotient-coordinate transport of The tubular neighbourhood theorem in a smooth ambient manifold, the constructed derivative sends a tangent vector and a normal lift to their sum, hence induces the identity on the quotient at every zero vector. Compactness of $Z$ allows a uniform small metric disk inside its domain: cover $Z$ by finitely many smaller trivializing patches with compact closures, and take the minimum of their positive allowable radii. This also makes the closed disk compact, since on each such patch its fibre coordinates lie in a bounded closed ball. Rescale the metric so this disk is the unit disk. Radial retraction of $D(\nu )\setminus Z$ onto $S(\nu )$ and the pair sequence [F6] give an isomorphism $H^r(D(\nu ),D(\nu )\setminus Z;R)\to H^r(D(\nu ),S(\nu );R)$. Lift $u_{\nu }$ uniquely along it and use tubular excision to define $\alpha$. For $r=0$, the punctured bundle and sphere are both empty, so this comparison is the identity.

Let $U$ be the open tube and $K$ a smaller closed disk bundle inside it. The inclusion of the outer annulus $U\setminus K$ into the punctured tube is a fibrewise homotopy equivalence, by radial movement to a radius strictly between the inner and outer radii. Pair sequences therefore identify the Thom lift with a class $u_K\in H^r(U,U\setminus K;R)$. It defines $u_c\in H_c^r(U;R)$. Excision extends $u_K$ to $H^r(M,M\setminus K;R)$, and its absolute image is $\bar{\alpha }$. The support compatibility in [F3] gives $\bar{\alpha }\cap [M]=j_{\ast }{D}_U(u_c)$ for $j:U\hookrightarrow M$: represent $[M]$ and its restriction $[M{]}_K$ by the same chain, and cap with the cocycle vanishing outside $K$.

The projection $\pi :U\to Z$ and zero section $\zeta :Z\to U$ are homotopy inverses by fibrewise contraction. Put $b={\pi }_{\ast }{D}_U(u_c)\in H_z(Z;R)$. To compute its restriction at $x\in Z$, restrict to a trivializing product of a tangent ball $T$ and a normal disk $N$. This localization is legitimate at chain level: shrink a tangent ball about $x$, represent the Thom class with support inside a smaller normal disk, and subdivide the finitely many chains until small for the product neighbourhood and its complement. In the quotient modulo $Z\setminus \{x\}$, pieces projected outside the tangent ball vanish. The relative cap and small-chain comparison of [F6] therefore reduce the restriction of $b$ to the cap on this disk product.

Let $d$ and $a$ be the positive tangent and normal relative orientation cycles in this product, with degrees $z$ and $r$. Its ambient orientation cycle is the shuffle $S_{T,N}(d\otimes a)$: its restrictions have the prescribed tangent-first local orientation, so [F4] identifies it with that relative orientation class. The local Thom cocycle is pulled back from a normal cocycle $\eta$ with $\eta (a)=1$, by [F2]. For a product chain $c$, the cap definition gives ${\mathrm{pr}}_{T\mathrm{\#}}({\mathrm{pr}}_N^{\ast }\eta \cap c)=(\eta \otimes \mathrm{i}\mathrm{d}){\mathrm{AW}}_{N,T}({\tau }_{\mathrm{\#}}c),$ where $\tau :T\times N\to N\times T$ swaps the factors and only normal degree $r$ is contracted. In each path of [F5], exchanging the $z$ tangent and $r$ normal steps reverses the order of each of the $zr$ unlike pairs, so ${\tau }_{\mathrm{\#}}{S}_{T,N}(d\otimes a)=(-1{)}^{rz}{S}_{N,T}(a\otimes d)$; and ${\mathrm{AW}}_{N,T}{S}_{N,T}\simeq \mathrm{i}\mathrm{d}$. These identities remain valid on the product relative complexes: the model homotopies preserve the coordinate subspaces, and [F6] supplies the small-chain comparison for their union. Contracting a chain homotopy with the closed $\eta$ gives a boundary (with the cap boundary sign), so the resulting local homology class is $(-1{)}^{rz}\eta (a)d=(-1{)}^{rz}d$. Thus $b$ restricts at every $x$ to $(-1{)}^{rz}$ times the local orientation of $Z$.

By [F4], $b=(-1{)}^{rz}[Z]$, including disconnected $Z$. Fibrewise contraction and homotopy invariance give $D_U(u_c)={\zeta }_{\ast }b$, so step 2.1 yields $\bar{\alpha }\cap [M]=(-1{)}^{rz}{j}_{\ast }{\zeta }_{\ast }[Z]=(-1{)}^{rz}(i_Z{)}_{\ast }[Z]$. Empty $Z$ gives zero throughout. For rank zero the normal Thom class is the supplied orientation unit, and the same determinant comparison gives the oriented component classes; no assumption that that unit is always $+1$ is made. In characteristic two the sign is $1$. AC is used through [F1]–[F3], and no extra orientation selection is made.