$f(z)=z^n$ with $f: \mathbb C^* \to \mathbb C^*$ is a covering map

Question

Consider the map $f:\mathbb{C} \setminus \{0\} \rightarrow \mathbb{C} \setminus \{0\}$ given by $$f(z)=z^n$$ for $n \in \mathbb{N}$. The book I am reading from claims that this is a covering map, but I would like to prove it and I am not sure how. By definition, I need an open cover $\{U_\alpha\}$ of $\mathbb{C} \setminus \{0\}$ and show that $f^{-1}(U_\alpha)$ is a disjoint union of open sets $V_{\alpha\beta}$ in $\mathbb{C} \setminus \{0\}$ and that the restriction of $f$ on each of these open sets is a homeomorphism $$f_{V_{\alpha\beta}}:V_{\alpha\beta} \rightarrow U_\alpha.$$ How can I show this rigurously? What open sets should I take for the cover?
Also, if I consider the same map defined on the whole complex plane $f:\mathbb{C} \rightarrow \mathbb{C}$, is it still a covering map?

Answer 1

As a preparation let us show that $F : S^1 \to S^1, F(z) = z^n$, is a covering map (with $n$ sheets). Note that $S^1 =\{ z \in \mathbb C \mid \lvert z \rvert = 1\} \subset \mathbb C^* = \mathbb C \setminus \{0\}$ is the standard unit circle.

It is well-known that the map $\phi : \mathbb R \to S^1, \phi(t) = e^{it}$, is a surjective open map. See for example my answers to Open sets on the unit circle $S^1$ . We have $$\phi(t) = \phi(s) \text{ if and only if } t - s = 2\pi k \text{ for some } k \in \mathbb Z . \tag{1} $$ As a consequence, for each open interval $(a,b)$ with $b - a \le 2\pi$ the restriction $\phi_{a,b} : (a,b) \to \phi((a,b))$ of $\phi$ is a homeomorphism onto the open subset $\phi((a,b))$ of $S^1$. To see this, note that $\phi_{a,b}$ is a continuous open surjection. It is moreover injective by $(1)$.

1. Let $z_0 \in S^1$. Then $U(z_0) = S^1 \setminus \{ -z_0 \}$ is an open neigborhood of $z_0$ in $S^1$. Choose $t_0 \in \mathbb R$ such that $\phi(t_0) = z_0$. Then $U(z_0) = \phi((t_0 - \pi, t_0 + \pi))$. To see this, first note that $\phi([t_0 - \pi, t_0 + \pi]) = S^1$ by (1): Given $z \in S^1$, pick $t \in \mathbb R$ such that $\phi(t) = z$. Then $t - t_0 \in [2 \pi k - \pi, 2 \pi k + \pi]$ for some $k \in \mathbb Z$ and thus $t \in [2 \pi k + t_0 - \pi, 2 \pi k + t_0 + \pi]$ which shows $z \in \phi([2 \pi k + t_0 - \pi, 2 \pi k + t_0 + \pi]) = \phi([t_0 - \pi, t_0 + \pi])$. But now $\phi(t_0 - \pi) = \phi(t_0+\pi) = -z_0$, and these are only two points of $[t_0 - \pi, t_0 + \pi]$ mapped to $-z_0$.

2. The open sets $U_j = \phi((\frac{t_0 + 2\pi j - 1}{n}, \frac{t_0 + 2\pi j + 1}{n})) \subset S^1$, $j = 0,\ldots, n-1$, are pairwise disjoint (which is obvious by $(1)$) and are mapped by $F$ homeomorphically onto $U(z_0)$. To see the latter, note that $s_j : (\frac{t_0 + 2\pi j - 1}{n}, \frac{t_0 + 2\pi j + 1}{n}) \to (t_0 - \pi, t_0 + \pi), s_j(t) = nt - 2 \pi j$, is a homeomorphism. Hence $$F_j = \phi_{t_0 - \pi, t_0 + \pi} \circ s_j \circ \phi_{\frac{t_0 + 2\pi j - 1}{n}, \frac{t_0 + 2\pi j + 1}{n}}^{-1} : U_j \to U(z_0)$$ is a homeomorphism. But for $t \in (\frac{t_0 + 2\pi j - 1}{n}, \frac{t_0 + 2\pi j + 1}{n})$ we have $$\phi_{t_0 - \pi, t_0 + \pi}(s_j(t)) = \phi(nt - 2 \pi j) = \phi(nt) = e^{int} = (e^{it})^n = F(\phi(t)) = F(\phi_{\frac{t_0 + 2\pi j - 1}{n}, \frac{t_0 + 2\pi j + 1}{n}}(t)) ,$$ i.e. $F_j(z) = F(z)$.

3. We have $F^{-1}(U(z_0)) = \bigcup_{j=0}^{n-1} U_j$ which completes the proof that $F$ is an $n$-sheeted covering map. The inclusion "$\supset$" is clear. So let $z \in F^{-1}(U(z_0))$, i.e. $F(z) \in U(z_0)$. Write $z = \phi(t)$. Then $F(z) = \phi(t)^n = \phi(tn) \in \phi((t_0 - \pi, t_0 + \pi))$. Hence $tn \in (t_0 - \pi + 2\pi k , t_0 + \pi +2\pi k)$ for some $k \in \mathbb Z$ and $t \in (\frac{t_0 + 2\pi k - 1}{n}, \frac{t_0 + 2\pi k + 1}{n})$. Write $k = rn + j$ with $r \in \mathbb Z$ and $j \in \{0,\ldots,n-1 \}$. Then $z = \phi(t) = \phi(t - 2\pi r)$, where $t -2 \pi r \in (\frac{t_0 + 2\pi j - 1}{n}, \frac{t_0 + 2\pi j + 1}{n})$. Thus $z \in U_j$.

Since $F$ is a covering map and $\mu : (0,\infty) \to (0,\infty), \mu(r) = r^n$, is a homeomorphism, also $F \times \mu : S^1 \times (0,\infty) \to S^1 \times (0,\infty)$ is a covering map.

The map $$h : S^1 \times (0,\infty) \to \mathbb C^*, h(z,r) = rz ,$$ is a homeomorphism (with inverse $h^{-1}(w) = (w/\lvert w \rvert, \lvert w \rvert)$). Therefore also $$h \circ (F \times \mu )\circ h^{-1} : \mathbb C^* \to \mathbb C^*$$ is a covering map. But we have $$(h \circ (F \times \mu)(z,r) = h(z^n,r^n) = r^nz^n = (rz)^n = f(rz) = f(h(z)) ,$$ hence $f = h \circ (F \times \mu )\circ h^{-1}$.

The map $\bar f : \mathbb C \to \mathbb C, \bar f(z) = z^n$, is a covering map if and only if $n = 1$. For $n = 1$ it is the identity map which is trivially a covering map. For $n > 1$ we have $\bar f^{-1}(0) = \{0\}$ whereas for $z \ne 0$ the set $\bar f^{-1}(z)$ has $n > 1$ elements.

For $n > 1$ it is a branched covering. See for example https://en.wikipedia.org/wiki/Branched_covering.