I haven’t received any response from the user So I'm posting this question
I have some confusion in this answer
Question : Using the polar coordinates we can establish a map $f: \left \{ S^1 \times [0,1] \right \}/\left \{ S^1 \times \left \{ 0 \right \} \right \} \rightarrow D^2$ defined as $f(\theta, \rho)= \rho(\cos \theta, \sin \theta)$.
This map is continuos, injective and surjective, but
How to prove that is a homeomorphism?
I have some confusion in this answer
Here is the outline of the answer
To show $f$ is a homeomorphism without using compactness, you can explicitly construct the inverse of $f$, namely $$ g(x, y) = [t(x, y), \sqrt{x^2 + y^2}] $$ where brackets denote equivalence class, and $$ t(x, y) = \begin{cases} 0 & (x, y) = (0,0) \\ \frac{\pi}{2} & x = 0, y > 0\\ \frac{3\pi}{2} & x = 0, y < 0\\ \arctan(y/x) & x > 0 \\ \pi + \arctan(y/x) & x < 0 \end{cases} $$ where each of these values is to be considered a number (mod $2\pi$), hence an element of $S^1$.
The continuity of $g$ is pretty clear everywhere except along the $y$-axis. For those, you'll need to show that $g^{-1}(U)$ is open whenever $U$ is open in the domain, which will mean writing down a way to describe open sets in the quotient, which is a pain in the neck, but you're welcome to do it.
My confusion : Im not getting why $g(x, y) = [t(x, y), \sqrt{x^2 + y^2}]$ ? I mean, How $y=[\sqrt{x^2 + y^2}]?$
My thinking : Here $f(\theta, \rho)= \rho(\cos \theta, \sin \theta)= te^{i \theta} $ where $t \in [0,1]$
$(\theta, \rho) = f^{-1}(te^{i \theta})$
According to the solution,$f^{-1}(te^{i \theta})=f^{-1}(t\cos\theta , t \sin \theta )=g(x,y)$ ,
$\implies x= t \cos \theta ,y=\ t\sin \theta $
$\theta =[t(x,y)]$ and $\rho=[\sqrt{x^2 + y^2}]$