Thom isomorphism

Remarks

The relative group $H^{k+n}(E, E_0)$ is the cohomology analogue of a relative homology group, and $\smile$ is the relative cup product. The proof treats a trivial bundle $B \times \R^n$ first, where the result is the Künneth formula with $H^\bullet(\R^n, \R^n \setminus \set{0})$, which is $\Z$ in degree $n$ only. It then glues local pieces by Mayer–Vietoris arguments and passes to general $B$ by a limit argument. Without an orientation, the same statement holds with $\Z/2$ coefficients. For paracompact $B$ one may choose a metric, and then $H^\bullet(E, E_0)$ is the reduced cohomology of the Thom space of the bundle.

Example: for the trivial bundle $B \times \R^n$, $u$ is the pullback of the generator of $H^n(\R^n, \R^n \setminus \set{0})$, and the isomorphism is $H^k(B) \cong H^{k+n}(B \times (\R^n, \R^n \setminus \set{0}))$.

Going deeper

For a student

The Thom isomorphism is the source of the Euler class: pull the Thom class back along the zero section to get $e \in H^n(B)$. It leads to the Gysin sequence of a sphere bundle and to Poincaré duality type results for normal bundles. Thom used it to compute the cohomology of Thom spaces and from that the cobordism ring. Versions exist for any multiplicative cohomology theory, given a suitable orientation of the bundle.