Homeomorphism between $\mathbb{S}^1/(x\sim -x)$ and $\mathbb{S}^1$

Question

I was reading this paper by Kim A. Frøyshov about the real projective plane and in part of the proof to show that it is homeomorphic to $\mathbb{S}^1$ they first showed it was homeomorphic to $\mathbb{S}^1/(x\sim-x)$. That part I understood but then they show that $\mathbb{S}^1/(x\sim-x)$ is homeomorphic to $\mathbb{S}^1$ by taking: $$\mathbb{S}^1=\{z\in\mathbb{C}:\|z\|=1\}$$

And then taking the map:

$$f:\mathbb{S}^1\rightarrow \mathbb{S}^1/(x\sim-x),\space f(z)=z^2$$

They then claim that since $\mathbb{S}^1/(x\sim-x)$ is compact, and that $\mathbb{S}^1$ is Hausdorf that $f$ is a homeomorphism. This part I have trouble with, how can $f$ be a homeomorphism if it is not injective? What am I missing? I see that it is continuous and surjective but where do they get injectivity from?

Answer 1

The map $f$ should be from $\Bbb S^1/(x \sim -x)$ to $\Bbb S^1$, given by $f([z]) = z^2$. Note that $[z] = \{z,-z\}$ for every $z\in \Bbb S^1$. $f$ is well-defined, for if $[z_1] = [z_2]$, then either $z_1 = z_2$ or $z_1 = -z_2$. In any case, $z_1^2 = z_2^2$, so $f([z_1]) = f([z_2])$.

We show that $f$ is a continuous bijection. Since $\Bbb S^1/(x\sim -x)$ is compact, and $\Bbb S^1$ is Hausdorff, $f$ is a homeomorphism.

1. $f$ is injective. Suppose $f([z_1]) = f([z_2])$. Then, $z_1^2 = z_2^2$, so $z_1 = z_2$ or $z_1 = -z_2$. By definition of $\sim$, $[z_1] = [z_2]$.
2. $f$ is surjective. For every $z\in \Bbb S^1$, there is an equivalence class in $\Bbb S^1/(x\sim -x)$, namely $[z^{\frac12}]$, such that $f([z^{\frac12}]) = z$.

I hope you can show the continuity of $f$ to conclude.

Answer 2

$\mathbb{RP}^1=\mathbb{R}\cup \{\infty \}$,so we can imagine $S^1$ is homeomorphic to $\mathbb{RP}^1$ by projection. I hope this will help you think about it.