Show that deleting a point from a manifold of dimension greater than 1 does not affect orientability of the manifold.
Solution
1. Characterization of orientability via the orientation double cover:
1.1. An \(n\)-manifold \(M\) is orientable if and only if its 2-sheeted orientation covering space \(p: \widetilde{M} \to M\) is trivial (i.e. \(\widetilde{M}\) is disconnected into two homeomorphic copies of \(M\)).
Hatcher Section 3.3 (The Orientation Double Cover).
1.2. For any point \(x_0 \in M\), let \(M' = M \setminus \{x_0\}\). The orientation double cover of \(M'\) is the restriction:\[ \widetilde{M'} = p^{-1}(M') = \widetilde{M} \setminus p^{-1}(x_0). \]
naturality of covering space constructions under restriction to open subsets.
1.3. Since \(p: \widetilde{M} \to M\) is a 2-to-1 covering, \(p^{-1}(x_0)\) consists of exactly two points \(\{y_1, y_2\} \subset \widetilde{M}\).
covering degree 2.
2. Connectedness and point removal in dimension \(n \ge 2\):
2.1. Removing a finite set of points from a connected \(n\)-dimensional manifold with \(n \ge 2\) preserves connectedness:Let \(N\) be any connected \(n\)-manifold with \(n \ge 2\). For any two points \(y_1, y_2 \in N\), \(N \setminus \{y_1, y_2\}\) is connected.
in a Euclidean chart \(\mathbb{R}^n\) (\(n \ge 2\)), \(\mathbb{R}^n \setminus \{0\}\) is connected, and any path between points in \(N \setminus \{y_1, y_2\}\) can be perturbed around isolated points.
2.2. If \(M\) is orientable:\(\widetilde{M} = M_1 \amalg M_2\) is a disjoint union of two connected components homeomorphic to \(M\).Since \(n \ge 2\), removing \(y_1\) from \(M_1\) and \(y_2\) from \(M_2\) leaves each component connected:\[ \widetilde{M'} = (M_1 \setminus \{y_1\}) \amalg (M_2 \setminus \{y_2\}). \]Thus \(\widetilde{M'}\) is disconnected, so \(M'\) is orientable.
1 applied to each component.
2.3. If \(M\) is non-orientable:\(\widetilde{M}\) is connected.Since \(n \ge 2\), by 1 removing the two points \(\{y_1, y_2\}\) leaves \(\widetilde{M'} = \widetilde{M} \setminus \{y_1, y_2\}\) connected.Since its orientation cover is connected, \(M'\) is non-orientable.
1 applied to the connected cover \(\widetilde{M}\).
3. Conclusion:\(M \setminus \{x_0\}\) is orientable if and only if \(M\) is orientable for all \(n \ge 2\). Q.E.D.
1 and 2.