Thom isomorphism theorem

Let $\xi \to X$ be a $d$-dimensional vector bundle over a topological space $X$, and let $h^{*}$ be a multiplicative generalized cohomology theory, such as ordinary cohomology. Let $\tau \in h^d(D(\xi ),S(\xi ))$ be a Thom class for $\xi$, where $D(\xi )$ and $S(\xi )$ are the associated disk and sphere bundles of $\xi$.

Since $h^{*}$ is a multiplicative theory, there is a generalized cup product map

$$h^{*}(D(\xi )){\otimes }_{h^{*}}{h}^{*}(D(\xi ),S(\xi ))\to h^{*}(D(\xi ),S(\xi )),$$

where the tensor product is over the coefficient ring $h^{*}(\mathrm{pt})$ of the theory. Using the isomorphism $p^{*}:h^{*}(X)\cong h^{*}(D(\xi ))$ induced by the homotopy equivalence $p:D(\xi )\to X$, we obtain a homomorphism

$$T:h^n(X)\to h^{n+d}(D(\xi ),S(\xi ))\cong {\tilde{h}}^{n+d}(X^{\xi })$$

taking $\alpha$ to $p^{*}(\alpha )\cdot \tau$. Here $X^{\xi }$ stands for the Thom space $D(\xi )/S(\xi )$ of $\xi$.

Thom isomorphism theorem $T$ is an isomorphism $h^{*}(X)\cong {\tilde{h}}^{*+d}(X^{\xi })$ of graded modules over $h^{*}(\mathrm{pt})$.

Remark 1

When $\xi$ is a trivial bundle of dimension $1$, this generalizes the suspension isomorphism. In fact, a typical proof of this theorem for compact $X$ proceeds by induction over the number of open sets in a trivialization of $\xi$, using the suspension isomorphism as the base case and the Mayer-Vietoris sequence to carry out the inductive step.

Remark 2

There is also a homology Thom isomorphism ${\tilde{h}}_{*+d}(X^{\xi })\cong h_{*}(X)$, in which the map is given by cap product with the Thom class rather than cup product.