A simple homeomorphism between unit circle and quotiented unit circle 
I want to show $S^1/\mathscr{R}$ and $S^1$ are homeomorphic
where $S^1$ is the unit circle and the equivalence relation is $$(x',y')\mathscr{R}(x'',y'') \iff y''\leq 0 \text{ and } y'\leq 0.$$

Now I am having trouble finding the expression for the homeomorphism
I found the same question here
Construct a homeomorphism between $S^1/\rho$ and $S^1$
but the solution is too long and complicated, this problem was part of a short exam and the solution is not expected to be that long and complicated
A partner suggested
$f(x,y)=(\cos(2\arccos(x)), \sin(2\arccos(y)))$ for $y>0$ and
f(x,y)=(1,0) for $ y\leq 0$
The professor didn't say it was wrong nor did he say it was right.

*

*Is this suggestion correct? or is there a simpler one?


*We need to map the half the unit circle to a whole unit circle and the other half to a single point and then prove surjectivity, injectivity of the funtion for $y>0$ so that the whole function passes to the quotient as injectivite,  and continuity .The homeomorphism will follow from the fact that once we know it is bijective and continuous the function will go from a compact space to a T2 space
I guess continuity ad surjectivity follows just by inspection,How will I go about proving injectivity for $y>0$?
 A: It is much easier to think about this in polar coordinates since a point on the unit circle can be uniquely identified by an angle $\theta\in (0,2\pi]$. So we identify $S^1$ with the space $(0,2\pi]$ under the topology given by "circular intervals" that possibly wrap around $2\pi$. (There is an explicit homeomorphism $\theta\mapsto (\cos \theta,\sin\theta)$ that you can use to translate back and forth.) It's more convenient to use $2\pi$ for "angle $0$" in this case.
Then you can view $S^1/\mathscr{R}$ as $(0,\pi)\cup\{\ast\}$ where $\ast$ is the single point corresponding to the class of angles in $[\pi,2\pi]$.
The map your partner suggests (modulo an issue discussed below) is $f:S^1/\mathscr{R}\to S^1$ where $f(t)=2t$ for $0<t<\pi$ and $f(\ast)=2\pi$. Described out loud, this map doubles all angles in the upper half circle $(0,\pi)$ so that they cover everything in $(0,2\pi)$, and then sends $\ast$ to $2\pi$ to fill the last point.
Written this way, it is quite clear that the map is a bijection. Continuity is not hard to prove either. A basic open set is a set of the form $(r,s)$  or $(0,r)\cup (s,2\pi]$ for $0<r<s\leq 2\pi$. In the first case the preimage is $(r/2,s/2)$. In the second case the preimage is $(0,r/2)\cup (s/2,\pi)\cup\{\ast\}$ (which is open in the quotient topology since its preimage under the quotient map back to $S^1$ is $(0,r/2)\cup (s/2,2\pi]$.)

I will add some further details: I have identified $(0,2\pi]$ and $S^1$ to make the problem easier. Formally, the underlying map in this identification is $\theta\mapsto (\cos\theta,\sin\theta)$. I am not saying that this is the homeomorphism you ask for in the main question. The homeomorphism is the map $f$ which is the same as your partners (modulo issue) up to this identification between $S^1$ and $(0,2\pi]$.
If you want to unwrap the identification to get a function in Cartesian coordinates then you just have to go through the trig. Start with $(x,y)$ in the upper half circle. Go polar by taking $\theta=\arccos(x)$ (which is in $(0,\pi)$). Apply $f$ to get $2\arccos(x)$. Now go back to Cartesian coordinates to get $(\cos(2\arccos(x)),\sin(2\arccos(x))$. We have our function.
Actually, going through this I can see either there is a typo or your partner didn't quite get it right: $\arccos(y)$ should be $\arccos(x)$. If it looks strange to have the map depend only on $x$ then recall that a point in the upper half of the unit circle is completely determined by its $x$ coordinate. So really this is a "one parameter" situation, which is all the more reason to go polar.
