Thom isomorphism for oriented vector bundles

Choose the metric by [F1]. With the orientation supplied, [F2] gives a normalized class $u_{\xi }$ and identifies the only relative Serre row with $H^k(B;R)$. Its edge is the displayed cup-product map and is an isomorphism in every degree.

For the unoriented calculation, use the $E_2$ formula and finite convergence stated in [F2]. By [F3], every row is zero except $q=n$, and that row is exactly ${\mathcal{O}}_R(\xi )$ before any orientation is selected. Thus the sequence collapses with one filtration quotient and gives the canonical isomorphism $H^k(B;{\mathcal{O}}_R(\xi ))\cong H^{k+n}(D(\xi ),S(\xi );R)$. A global untwisted Thom class is neither chosen nor asserted in this clause.

If $u$ and $u^{\mathrm{\prime }}$ are normalized, step 1.1 writes $u-u^{\mathrm{\prime }}={\pi }^{\ast }a⌣u_{\xi }$ for a unique $a\in H^0(B;R)$. Fiber restriction gives $a(b)o_b=0$ for every $b$; since $o_b$ is a free rank-one generator, $a(b)=0$. Thus $a=0$ componentwise and $u=u^{\mathrm{\prime }}$.

Under a supplied finite trivializing cover, [F4] constructs the same unique normalized class and cup isomorphism by finite Mayer–Vietoris. Uniqueness from step 2.1 identifies it with the Serre class, so this is genuinely a special case and not an additional hypothesis on the general theorem.

For $n=0$, $u=1$ and both maps are identities; on an empty base they are the unique maps of zero groups. Point and disconnected bases, the zero ring, degree-zero/negative input, the only Serre row and both filtration endpoints are covered by [F2]. AC is used exactly through [F1], [F2], and the local-coefficient cohomology interface [F3]; the finite-cover proof [F4] uses none.