Continuous bijection from $S^1$ to $\mathbb P^1(\Bbb R)$

Question

Consider the relation $\sim$, where $\sim$ on $\mathbb R^2 - \{0 \}$ defined so that $\vec x \sim \vec y$ when there is some non zero scalar so that $\vec x = \lambda \vec y$

Let $ P=\mathbb R^2 / \sim $, where $\sim$ is the equivalence relation and $\mathbb R^2$ has been given the standard topology. Find a bijective continuous function from $S'$ to $P$, where $S' = \{ (x,y) \in \mathbb R^2 : x^2+y^2=1 \}$.

My attempt: So $P=(\mathbb R^2 - \{0 \}/ \sim )$ with the quotient topology. Define $f: S' \to P$ as follows:

Let $(x,y) \in S'$ be represented as $(x,y) = (\cos \theta, \sin \theta), -\pi\le \theta \lt \pi $. For a point $(a,b) \in \mathbb R^2 - \{0 \}$, let $[(a,b)]$ be equivalence class in $P$ that contains $(a,b)$.

Let $f((x,y)) = f((\cos \theta, \sin \theta)) = [(\cos \frac{\theta}{2}, \sin \frac{\theta}{2} )]$. Since every element $[(a,b)]$ of $P$ can be represented by a unique angle $\alpha \in [\frac{\pi}{2}, \frac{\pi}{2}) $ as $[(a,b)] = [(\cos \alpha, \sin \alpha )] = [(\cos \frac{\theta}{2}, \sin \frac{\theta}{2} )], \theta \in [-\pi, \pi)$ $f$ is bijection. To show that $f$ is continuous: if $\theta_{0} \in (-\pi,\pi)$, $f$ is continuous at $\theta_{0}$. At $\theta_{0} = -\pi, f((-1,0))=[(0,-1)]$. A neighborhood of $[(0,-1)]$ contains an interval $\{ [(\cos \alpha, \sin \alpha )] : |\alpha - (- \frac{\pi}{2})| < \epsilon \}$ it is inverse image under $f$ is equal to $\{ [(\cos \theta, \sin \theta )] : |\theta - \pi| < \frac {\epsilon}{2} \}$

and is open in $S'$. Therefore, $f $ is a continuous bijection. Please advice.

Answer 1

Your proof has the right idea, but it has a gap.

Let $(x,y) \in S^1$ be represented as $(x,y) = (\cos \theta, \sin \theta), -\pi\le \theta \lt \pi $.

This is correct. You have a unique function $\phi : S^1 \to [-\pi,\pi)$ such that $(\cos(\phi(z)), \sin(\phi(z))= z$. Note I write $z = (x,y)$.

Then you define $f(z) = [(\cos(\phi(z)/2), \sin(\phi(z)/2)]$. Clearly this gives a bijection.

To show that $f$ is continuous you need to know that $\phi$ has the following two prooperties:

1. $\phi$ is continuous in all $z_0 \ne (-1,0)$.
2. If $(z_n)$ is a sequence in $S^1$ such that $z_n \to (-1,0)$, then $\lvert \phi(z_n) \rvert \to \pi$.

I think you should give an argument that this true (or at least a reference to some standard theorem). You can do it by methods of complex analysis (take the imaginary part of the complex logarithm on an adequate sliced plane) or by using the functions $\arcsin$ and $\arccos$ which are known to be continuous.