If \(E\) is the product bundle \(B \times D^n\) with \(B\) a CW complex, show that the Thom space \(T(E)\) is the \(n\)-fold reduced suspension \(\Sigma^n(B_+)\), where \(B_+\) is the union of \(B\) with a disjoint basepoint, and that the Thom isomorphism specializes to the suspension isomorphism \(\tilde{H}^i(B; R) \approx \tilde{H}^{n+i}(\Sigma^n B; R)\) given by the reduced cross product in §3.2.
Solution
For the product disk bundle \[ E=B\times D^n, \qquad E'=B\times S^{n-1}, \] the Thom space is \[ T(E)=\frac{B\times D^n}{B\times S^{n-1}}. \] Adjoining a disjoint basepoint to \(B\), this quotient is naturally \[ B_+\wedge(D^n/S^{n-1}) \cong B_+\wedge S^n =\Sigma^n(B_+). \] Hence \[ \boxed{T(E)\cong\Sigma^n(B_+).} \]
Let \(u\in H^n(D^n,S^{n-1};R)\) be the fiber orientation class. The Thom class of the product bundle is \[ 1\times u\in H^n(B\times D^n,B\times S^{n-1};R). \] The Thom isomorphism sends \[ a\in H^i(B;R) \] to the relative cross product \[ a\times u. \] Under the quotient identification above this is exactly the reduced suspension isomorphism \[ \widetilde H^i(B_+;R) \xrightarrow{\cong} \widetilde H^{i+n}(\Sigma^nB_+;R). \] Restricting to reduced cohomology of \(B\) gives the usual \[ \boxed{\widetilde H^i(B;R)\cong\widetilde H^{i+n}(\Sigma^nB;R)} \] induced by the reduced cross product.