Show that if a disk bundle \(D^n \to E \to B\) has a Thom class with \(\mathbb{Z}\) coefficients, then it is orientable.
Solution
Let \[ (D^n,S^{n-1})\longrightarrow(E,E')\longrightarrow B \] be the disk-bundle pair and let \[ c\in H^n(E,E';\mathbb Z) \] be a Thom class. For each \(b\in B\), its restriction \[ c_b\in H^n(D_b^n,S_b^{n-1};\mathbb Z)\cong\mathbb Z \] is a generator, hence determines an orientation of the fiber.
Let \(\gamma\) be a path in \(B\). Fiber transport along \(\gamma\) gives a homotopy equivalence of pairs \[ L_\gamma:(D^n_{\gamma(0)},S^{n-1}_{\gamma(0)}) \longrightarrow(D^n_{\gamma(1)},S^{n-1}_{\gamma(1)}). \] Since \(c\) is one global relative cohomology class, naturality of restriction gives \[ L_\gamma^*(c_{\gamma(1)})=c_{\gamma(0)}. \] Thus every transport map preserves the chosen generator of the top relative cohomology, so it has degree \(+1\). In particular transport around every loop preserves orientation. Therefore the bundle is orientable: \[ \boxed{\text{an integral Thom class forces orientability}.} \]