Is $f:(-1,1)\to \{y\in\mathbb{C}\mid |y| =1\}\setminus\{-1\}:x\mapsto e^{i\pi x} $ a homeomorphism?

Question

Consider $f:(-1,1)\to \{y\in\mathbb{C}\mid |y| =1\}\setminus\{-1\}:x\mapsto e^{i\pi x} $. I think $f$ is a homeomorphism but I still need to show that the inverse $f^{-1}$ is continuous. For the inverse I calculated if $0\leq\theta<\pi$ that $f^{-1}(e^{i\theta})= \frac{\theta}{\pi} $ and if $\pi<\theta<2\pi$ then $\frac{\theta-2\pi}{\pi}$. How do I go on for the continuity?

Answer 1

I guess that to do it the way you wanted we could use the identification

$$ \{ y \in \mathbb{C} : \lvert y \rvert = 1 \} \setminus \{ -1 \} = \{ e^{i \theta} \in \mathbb{C} : \theta \in (- \pi , \pi) \}. $$

This way we get simply $f^{-1} (e^{i \theta}) = \frac{\theta}{\pi}$. To prove continuity we can take an open set of the form

$$ U = \{ e^{i \theta} \in \mathbb{C} : \theta \in (a,b) \subset (-\pi,\pi) \}$$

and we get

$$f^{-1}(U) = \left(\frac{a}{\pi},\frac{b}{\pi}\right). $$

Answer 2

Let $e^{i\theta_n} \to e^{i\theta}$. To show that $\theta_n \to \theta$ it is enough to show that $\theta$ is the only limit point of $(\theta_n)$. [This is because $(\theta_n)$ is a bounded sequence of real numbers]. If $\theta'$ is a limit point of this sequence then $\pi \leq \theta' \leq 2\pi$ and $e^{i\theta'}=e^{i\theta}$. Can you finish?