Show the map $g:\mathbb{S}^1\to\mathbb{S}^1$ defined by $g(\cos(\theta), \sin(\theta)) = (\cos(a\theta), \sin(a\theta))$ is open where $a\in\Bbb N$.

Question

For a function $g: \mathbb{S}^1 \to \mathbb{S}^1$ defined by $g(\cos(\theta), \sin(\theta)) = (\cos(a\theta), \sin(a\theta))$ where $a$ is an integer, how do I show $g$ is an open map? I know that the idea is to show that for all open sets $U$ of $\mathbb{S}^1, g(U)$ is open. Any help would be appreciated!

Answer 1

HINT: For $\theta_0<\theta_1$ let $I(\theta_0,\theta_1)=\{\langle\cos\theta,\sin\theta\rangle:\theta_0<\theta<\theta_1\}$; the sets $I(\theta_0,\theta_1)$ are a base for the topology of $\Bbb S^1$, and it suffices to show that each $g[I(\theta_0,\theta_1)]$ is open in $\Bbb S^1$. What is $g[I(\theta_0,\theta_1)]$?

Answer 2

The map $g$ is the restriction to $S^1$ of the map$$\begin{array}{rccc}G\colon&\Bbb C&\longrightarrow&\Bbb C\\&z&\mapsto&z^a,\end{array}$$which is an open map, by the open mapping theorem. Therefore, $g$ is open too. You can also say that the restriction of $G$ to $\Bbb C\setminus\{0\}$ is open since $G'$ is never $0$.

Answer 3

The map $f(z)=z^a$ defined on $\mathbb{C}$ is open for non zero integers $a$ by open mapping theorem.Restrict it to $S^1$ now and use the fact that the standard topology on $S^1$ is the subspace topology endowed by the usual Euclidean topology on $\mathbb{C}$.