Show that the map of Example 3 of §53 is a covering map. Generalize to the map \(p(z) = z^n\).
Solution
1. Setup and angular coordinates:Proof:
1.1. Represent \(S^1 \subset \mathbb{C}\) as \(\{e^{i\theta} \mid \theta \in \mathbb{R}\}\).
1.2. The map \(p: S^1 \to S^1\) given by \(p(z) = z^n\) sends \(e^{i\theta} \mapsto e^{in\theta}\).
1.3. Since complex multiplication is continuous, \(p\) is continuous. Without loss of generality, assume \(n \ge 1\) (for \(n < 0\), \(p\) is the composition of \(z \mapsto z^{|n|}\) with the reflection homeomorphism \(z \mapsto \bar{z} = z^{-1}\)).
2. Construction of evenly covered neighborhoods:Proof:
2.1. Let \(w_0 = e^{i\theta_0} \in S^1\) be an arbitrary base point.
2.2. Choose \(\varepsilon\) such that \(0 < \varepsilon < \pi\).
2.3. Define the open arc neighborhood \(U \subset S^1\) centered at \(w_0\):\[U = \{ e^{i\theta} \mid \theta_0 - \varepsilon < \theta < \theta_0 + \varepsilon \}.\]
3. Preimage decomposition into disjoint slices:Proof:
3.1. A point \(z = e^{i\phi} \in S^1\) satisfies \(p(z) \in U\) if and only if \(n\phi \in (\theta_0 - \varepsilon + 2\pi k, \theta_0 + \varepsilon + 2\pi k)\) for some \(k \in \mathbb{Z}\).
3.2. For each \(k \in \{0, 1, \dots, n-1\}\), define the open arc:\[V_k = \left\{ e^{i\phi} \;\middle|\; \frac{\theta_0 - \varepsilon + 2\pi k}{n} < \phi < \frac{\theta_0 + \varepsilon + 2\pi k}{n} \right\}.\]
3.3. Each \(V_k\) has angular width \(\frac{2\varepsilon}{n} < \frac{2\pi}{n}\).
3.4. The centers of the arcs \(V_k\) are spaced by \(\frac{2\pi}{n}\), so the sets \(V_0, V_1, \dots, V_{n-1}\) are pairwise disjoint open sets in \(S^1\).
3.5. Their union is the entire preimage: \(p^{-1}(U) = \bigsqcup_{k=0}^{n-1} V_k\).
4. Homeomorphism of each sheet onto \(U\):Proof:
4.1. For each \(k \in \{0, 1, \dots, n-1\}\), the restricted map \(p|_{V_k}: V_k \to U\) is a continuous bijection.
4.2. The inverse map \(q_k: U \to V_k\) given by \(q_k(e^{i\theta}) = e^{i(\theta + 2\pi k)/n}\) (for \(\theta \in (\theta_0 - \varepsilon, \theta_0 + \varepsilon)\)) is continuous.
4.3. Thus \(p|_{V_k}: V_k \to U\) is a homeomorphism.
5. Q.E.D.By steps 2 through 4, every point of \(S^1\) has an evenly covered open neighborhood \(U\) with \(n\) slices, so \(p(z) = z^n\) is an \(n\)-fold covering map; for \(n < 0\) it is an \(|n|\)-fold covering by step
1.