Show that $\Bbb S^1 / {\sim}$ is homeomorphic to $I=[0,1]$.

Question

Let $\Bbb S^1=\{(x,y) \mid x^2 +y^2 =1 \} \subset \Bbb R$. Define an equivalence relation $\sim$ such that $(x,y) \sim (a,b) \iff |x|=|a|, |y|=|b|$. Show that $\Bbb S^1 / {\sim}$ is homeomorphic to $I=[0,1]$.

What is the approach to take here? Can I define some continuous $f : \Bbb S^1 \to I$ and use $\pi :\Bbb S^1 \to \Bbb S^1 /{\sim}$ somehow to get a homeomoprhism $f^* : \Bbb S^1 / {\sim} \to I$?

I think there was a definition that this $f^*$ is a homeomorphism if and only if $f$ is a quotient map?

Answer 1

Using your observation, define the mapping

$$ F : \mathbb S^1 \to [0,1], F(x, y) = |x|. $$

Since $F(\pm x, \pm y) = |y| = F(x, y)$, which means that if $(x, y) \sim (a, b)$ then $F(x, y) = F(a,b)$. In particular, $F$ descend to a continuous mapping

$$ f : \mathbb S^1/\sim \to [0,1]$$ with $f\circ \pi = F$.

To show that $f$ is a homeomorphism, we find explicitly the inverse: let $G : [0,1] \to \mathbb S^1$ be given by $G(t) = (t,\sqrt{1-t^2})$ and let $g = \pi \circ G$. Then $g$ is continuous $ f(g(t)) = t$ and

$$ g f([x, y]) = g(|x|) = [|x|, \sqrt{1-x^2}] = [x,y]$$

since $y= \pm \sqrt{1-x^2}$.

Answer 2

Consider the absolute value of the first projection $|\pi_1| : \Bbb S^1 \to [0, 1]$ that maps $(s, t) \mapsto |s|$.

Notice that $\Bbb S^1 / \sim_{|x| = |a|, |y| = |b|}$ is precisely the set of all fibers above $|\pi_1|$. You can check this yourself and it is tedious but not too hard. You can use the following fact to get started: if $(x, y), (a, b) \in \Bbb S^1$ such that $|x| = |a|$, then $$|y| = \sqrt{1 - |x|^2} = \sqrt{1 - |a|^2} = |b|$$

Also, notice that $|\pi_1|$ is an open surjective continuous map. This is because the projection $\pi_1 : \Bbb S^1 \to [-1, 1]$ is an open surjective continuous map and the restricted absolute value map $| \cdot | : [-1, 1] \to [0, 1]$ is also an open surjective continuous map (the absolute value map is not open in general but in this restricted case it is).

Hence, $|\pi_1|$ is a quotient map since open surjective continuous maps are quotient maps.

Thus, by a standard result about quotient maps (the "definition" you talk about in your last paragraph, cf. Corollary $22.3$ in Munkres), $|\pi_1|$ induces a homeomorphism $|\pi_1|^\ast : \Bbb S^1 / \sim_{|x| = |a|, |y| = |b|} \to [0, 1]$ via $\pi : \Bbb S^1 \to \Bbb S^1 / \sim_{|x| = |a|, |y| = |b|}$ such that $|\pi_1| = |\pi_1|^\ast \circ \pi$.

Note: By symmetry, you could have equally used $|\pi_2| : \Bbb S^1 \to [0, 1]$ instead of $|\pi_1| : \Bbb S^1 \to [0, 1]$ and gotten a different homeomorphism from $\Bbb S^1 / \sim_{|x| = |a|, |y| = |b|}$ to $[0, 1]$.