Euler Class construction of Chern Class

Question

So I'm reading up on how to construct a Chern class inductively via the Euler class and Wikipedia (and the text I'm reading) makes the claim that the Gysin sequence gives that $p_0^*$ is an isomorphism. But I honestly am not seeing how it could be possible that the Gysin sequence is giving this as an isomorphism (although by the way it's written, I feel like it's obvious). My first thoughts are that the cohomology groups to the "sides" of our two groups are zero, but that seems like it would imply that the ith cohomology is zero for all spaces, which seems absurd.

Answer 1

Let's introduce some notation. Let $p: E \to B$ be a complex vector bundle of rank $n$, and let $E_0 = E \setminus B$ denote the complement of the zero section in $E$. The space $E_0$ is essentially a $S^{2n-1}$-bundle over $B$, so we have the Gysin sequence $$\cdots \to H^{i-2n}(B) \xrightarrow{\wedge e} H^i(B) \xrightarrow{p_0^*} H^i(E_0) \to H^{i-2n+1}(B) \to \cdots.$$

For $i < 2n - 1$, it is clear that the two groups on either side is zero, so $p_0^*$ is an isomorphism in this degree. (Note: I am not making any assertions when $i \geq 2n$.) But this is more than what we need (i.e., $i < n$) for the inductive definition of Chern classes.

One intuitive heuristic to see why $p_0^*$ is an isomorphism in this range is to assume that $E_0$ is a trivial sphere bundle, i.e, $E_0 \cong B \times S^{2n-1}$. Then by the Kunneth formula one sees that the cohomologies of $B$ and $E_0$ agree up to degree $< 2n-1$, since $\tilde{H}^i(S^{2n-1}) = 0$ for $i < 2n-1$.