Let \(M\) be a connected topological manifold. (1) Prove that if \(\widetilde{M}\) is an orientable connected manifold and \(p: \widetilde{M} \to M\) is a \(k\)-fold covering map onto a non-orientable manifold \(M\), then the covering degree \(k\) must be even (or infinite). (2) Prove that if \(\pi_1(M)\) has no subgroup of index \(2\), then \(M\) must be orientable.
Solution
Let \[ w:\pi_1(M)\longrightarrow\{\pm1\}\cong\mathbb Z/2 \] be the orientation character. Since \(M\) is connected, it is orientable exactly when \(w\) is trivial.
1. Suppose \(M\) is non-orientable and \(p:\widetilde M\to M\) is a connected covering with \(\widetilde M\) orientable. Put\[ H=p_*\pi_1(\widetilde M)\le\pi_1(M). \]Then\[ H\subseteq\ker w. \]
A loop in \(\widetilde M\) preserves the orientation of \(\widetilde M\). Its projection therefore has trivial orientation monodromy in \(M\), so its class lies in \(\ker w\).
2. Since \(M\) is non-orientable, \(w\) is surjective, so\[ [\pi_1(M):\ker w]=2. \]Thus, when the covering has finite degree,\[ k=[\pi_1(M):H] =[\pi_1(M):\ker w]\,[\ker w:H] =2[\ker w:H], \]so \(k\) is even. If \([\ker w:H]\) is infinite, then the covering has infinitely many sheets.
3. If \(\pi_1(M)\) has no subgroup of index \(2\), then \(M\) is orientable.
If \(M\) were non-orientable, the surjective orientation character would have kernel of index \(2\), contradicting the hypothesis.