Prove that the map which winds the plane three times on itself given, in terms of complex numbers, by $ z\mapsto z^3 $ is a closed map.
I was trying to solve problem 20 in chapter 2 from Basic Topology by Armstrong.
I have already proved that the map $ z\mapsto z^3 $ is an open map, how am I supposed to show that it is also a closed map?
Can I directly show that $ \mathbb{C}\to\mathbb{C}: z\mapsto z^3 $ is a homeomorphism thus it is also a closed map?
Edit: A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.
$z \to z^{3}$ is not one-to-one so it is not a homeomorphism. Let $C$ be a closed set and consider the image $\{z^{3}:z \in C\}$. Suppose $z_n^{3} \to \zeta$. We have to show that $\zeta$ can be written as $z^{3}$ for some $z$. Now $\{z_n^{3}\}$ is a bounded sequence which implies that $\{z_n\}$ is also bounded. There is a subsequence of this sequence converging to some $z$ and continuity of the map $z \to z^{3}$ tells you that $\zeta =z^{3}$.
As requested by the OP I am providing a second proof that does not use compactness. There an open sector containing $\zeta$ in which $z \to z^{3}$ is a homeomorphism. [In any sector of the type $|\arg(\theta) - \arg(\theta_0)| <\pi /3$ the given map is a homeomorphism]. We can apply the inverse map $ z \to z^{1/3} $ to conclude that $z_n \to \zeta ^{1/3}$ so $\zeta ^{1/3} \in C$.