Let \(M\) be a connected topological/smooth manifold. Prove that if the fundamental group \(\pi_1(M)\) has no subgroup of index 2, then \(M\) is orientable.
Solution
1. A connected manifold \(M\) has an orientation character\[ w:\pi_1(M,x_0)\longrightarrow\{\pm1\}\cong\mathbb Z/2, \]which records whether transport of a local orientation around a loop preserves or reverses it.
2. The manifold is orientable if and only if \(w\) is trivial.
2.1. If \(M\) is orientable, a global orientation is preserved along every loop, so \(w=1\).
2.2. Conversely, if \(w\) is trivial, a choice of local orientation at \(x_0\) transports independently of path to every point, producing a global orientation.
3. Suppose \(M\) were nonorientable. Then \(w\) would be nontrivial. Since its codomain has order \(2\), nontriviality makes \(w\) surjective, and therefore\[ \ker w\le\pi_1(M) \]has index \(2\).
4. This contradicts the hypothesis that \(\pi_1(M)\) has no subgroup of index \(2\). Hence \(M\) is orientable.
5. Equivalently, \(\ker w\) is the subgroup corresponding to the connected orientation double cover of a nonorientable manifold.