Construct a homeomorphism between $S^1/\rho$ and $S^1$ (the unit circle)
where $S^1=\{(x,y)\in \mathbb{R}^2|x^2+y^2=1\}$ and the equivalence relation is $$(x',y')\rho(x'',y'') \iff y''\leq 0 \text{ and } y'\leq 0.$$
I get it intuitively, I know this equivalence relation is identifying the part of the circle below and on the $x$-axis as a single class in the quotient, so is like shrinking them to single point, so that the upper part of the circle closes in the quotient, and therefore it should be homeomorphic to a circle $S^1$.
Now I am having trouble finding the expression for the homeomorphism
I think I could do some like $f(t)=(cos(2t), sin(2t)))$, with $t\in [0,\pi]$, to parametrize $S^1$ where $t$ is the arc-length of the upper half of $S^1$ , but I need to relate it to the quotiented space and write everything in cartesian coordinates.
Can someone shed some light?
The expression for f should be such that
$f(x',y')=f(x'',y'') \iff (x',y')\rho (x'',y'')$.
Best Answer
Instead of using polar coordinates and trying double the angle for positive $y$-coordinates, you might consider the following projection:
Inside the unit circle $C_1$ place another circle $C_2$ with center $(0,\tfrac 1 2)$ and radius $\tfrac 1 2$. Now given any $(x,y)\in C_1$ the line segment between $(x,y)$ and the origin intersects $C_2$ in a unique point. See this illustration using GeoGebra.
The projected point on $C_2$ stays stationary when $y\le 0$, which is exactly what you need to define your homeomorphism.
The unit circle $C_1$ is given by $x^2+y^2=1$, the second circle $C_2$ is given by $x^2+(y-\frac 1 2)^2 = \frac 1 4$. The line segment between $(x,y)$ and the origin consists of the points $(tx,ty)$ for $t\in[0,1]$.
Hence, the intersection point satisfies $$ (xt)^2+\left(yt-\frac 1 2\right)^2 = \frac 1 4, $$ which is equivalent to $$ \underbrace{(x^2 + y^2)}_{=1} t^2 - yt = 0. $$ When $y\ge 0$, solving for $t$ yields $t=0$ (the origin) and $t=y$, corresponding to the point $(xy,y^2)$.
Now shifting and scaling yields a homeomorphism $C^2\to S^1$ given by $(x,y)\mapsto (2x,2y-1)$ that sends our intersection point on $C_2$ to $$ \left( 2xy, 2y^2 - 1 \right). $$
Hence, the map \begin{align*} S^1 &\longrightarrow S^1 \\ (x,y) &\longmapsto \begin{cases} (2xy, 2y^2 - 1) & y > 0, \\ (0,0) & y \le 0 \end{cases} \end{align*} induces the desired homeomorphism $S^1/\rho \to S^1$.