- Show that if \(M^\text{orientable} \mapsvia{\pi_k} M^\text{non-orientable}\) is a \(k\dash\)fold cover, then \(k\) is even or \(\infty\).
Solution
1. Let \(p:\widetilde M\to M\) be a connected orientable covering of a connected nonorientable manifold, and let \(H=p_*(\pi_1\widetilde M)\le\pi_1(M)\).
The number of sheets is the index \([\pi_1(M):H]\).
2. The orientation character \(w_1:\pi_1(M)\to\mathbb Z/2\) is nontrivial and vanishes on \(H\).
Nonorientability makes \(w_1\) onto. A loop in \(\widetilde M\) preserves the chosen orientation there, so its projected loop has trivial orientation character.
3. Therefore \(H\subseteq\ker w_1\), and\[[\pi_1(M):H]=2[\ker w_1:H]\]whenever the latter index is finite.
The kernel of the surjective orientation character has index \(2\); use multiplicativity of subgroup indices.
4. Thus an orientable cover of a nonorientable manifold has even finite degree or infinite degree.
This follows directly from 3.