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I want to prove that $f: \mathbb{C}^{*} \to \mathbb{C}^{*} $ given by $f(z)=z^n$ is a covering map.

My attempt:

If $z_0 \in \mathbb{C}^{*}$ then for some $z\in \mathbb{C}^{*}$ such that $f(z)=z_0$. Implies that $z^n=z_0$. Now by changing this equation to polar form we get, $r^ne^{in\theta}=r_0e^{i\theta_1}$. Implies $r=(r_0)^{1/n}$ and $n\theta=\theta_1 +2k\pi$. Hence $n$ solution of equation $z^n=z_0$ are $z_k=r_0^{1/n}(\cos(\theta_1/n+2k\pi/n)+\sin(\theta_1/n+2k\pi/n))$ where $k=0,1,\cdot\cdot\cdot,n-1$.

I am stuck now on how to construct an open set $U$ around the point $z_0$ and how to take its inverse image and write it as a disjoint union of open sets $\{v_\alpha \}$ such that each one is mapped homeomorphic to $U$.

Thanks, for any help!!

Lalit Tolani
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Manoj
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1 Answers1

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Thinking graphically: the map $z \mapsto z^n$ "wraps" the punctured plane around itself. Circles of radius $r$ are taken to circles of radius $r^n$ and lines of angle $\theta$ are taken to lines of angle $n\theta$.

So let's try to use polar rectangles to do this, ie, sets of the form $$R(r_1,r_2,\theta_1,\theta_2) = \{ re^{i\theta}\ |\ r \in (r_1,r_2), \theta\in(\theta_1,\theta_2)\}.$$

Exercise: Recalling that the preimages of a point $re^{i\theta}$ are of the form $r^{\frac{1}{n}}e^{i\frac{\theta}{n} + \frac{2\pi}{n}}$, characterize the preimage of $R(r_1,r_2,\theta_1,\theta_2)$

Fix $z=0 = r_0e^{i\theta_0}$. Consider the small open polar rectangle $R(r_0-\epsilon, r_0 + \epsilon, \theta_0 - \eta, \theta_0 + \eta)$ for some small numbers $\epsilon, \eta$.

Exercise: Show that $\epsilon$ and $\eta$ can be chosen small enough in terms of $r_0$ and $\theta_0$ and $n$ to solve the problem.

Extra credit: Show that $\epsilon$ and $\eta$ cannot be chosen independently of $z_0$

Neal
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