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 $I=[0,1]$.
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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: 
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 $[0, 1]$.
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](https://math.stackexchange.com/questions/1493993/universal-property-in-quotient-topology)

$$ f : \mathbb S^1/\sim \to [0,1]$$
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 $y= \pm \sqrt{1-x^2}$.