For a talk about Euler and Chern classes I need the Thom isomorphism with $\mathbb{Z}$ coefficients at some point (when proving the Gysin exact sequence). Recall that for an oriented real $n$-plane bundle $\xi$ with projection $\pi: E \to B$ the Thom isomorphism is given by:
$$\Phi: H^k\left(B\right)\to H^{n+k}\left(E,E_0\right), \; x \mapsto\pi^{*}\left(x\right) \cup u$$
where $u$ denotes the fundamental class of $H^n \left(E,E_0\right)$.
I don't have enough time to prove or even sketch the proof of this fact but I want to explain why this makes sense in 2 or 3 sentences. I'll prove the fact that $\pi$ induces an isomorphism in cohomology anyway, so I'd like to understand the cup product part of this isomorphism. So far I consulted Milnor's Characteristic classes but without success. I'd be glad if someone could give me a reference or explain this to me.
Best Answer
$u$ is the Thom class, it is Kronecker dual to those classes "orthogonal" to $E_0$. That is, think of those simplices $\nu$ which intersect $E_0$ in one point, then $u(\nu)=1$. Now $E_0$ is homeomorphic to $B$, so if we have a simplex $\tau$ in $B$ we can think of it is being in $E_0$. Now take $\tau\times \nu$, then the image of the $f$ where $f$ is dual to $\tau$ (that is $f(\tau)=1$) under the Thom isomorphism will be the function dual to $\tau\times \nu$, so $\Phi(f)( \tau\times \nu)=1$. Cup product comes from the external product operation $\times$, thus that is how it enters into the formula.
In Milnor, he proves Thom, via a lemma which is basically "Thom for $\mathbb{R}^n$", this is the local picture for the Thom map so I would go back and study that lemma. One can also try to analyse the theorem in terms of Poincare duality, in which case you are taking an intersection with $E_0$ and then dualising.