Show that every covering space of an orientable manifold is an orientable manifold.
Solution
Let \(p:\widetilde M\to M\) be a covering map and suppose \(M\) is orientable. For each \(x\in M\), choose an orientation class \[ \mu_x\in H_n(M,M-\{x\};\mathbb Z) \] that varies locally constantly with \(x\).
If \(\widetilde x\in\widetilde M\) and \(x=p(\widetilde x)\), choose an evenly covered neighborhood \(U\) of \(x\) and let \(\widetilde U\) be the sheet containing \(\widetilde x\). The restriction \[ p:\widetilde U\longrightarrow U \] is a homeomorphism, hence induces an isomorphism on local homology. Define the local orientation at \(\widetilde x\) to be the unique class \[ \widetilde\mu_{\widetilde x}\in H_n(\widetilde M,\widetilde M-\{\widetilde x\};\mathbb Z) \] that maps to \(\mu_x\) under this local identification.
On a fixed sheet over an evenly covered neighborhood, these classes vary locally constantly because the classes \(\mu_x\) do. Thus the collection \(\widetilde\mu_{\widetilde x}\) defines an orientation of \(\widetilde M\). Therefore every covering space of an orientable manifold is orientable.