why $g(x, y) = [t(x, y), \sqrt{x^2 + y^2}]$ ? why not $g(x, y) = [t(x, y), 1-\sqrt{x^2 + y^2}]$?

Question

I have some confusion on 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?

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 $g(x, y) = [t(x, y), \sqrt{x^2 + y^2}]$ ? why not $g(x, y) = [t(x, y), 1-\sqrt{x^2 + y^2}]$ ?

My thinking : Here $\left \{ S^1 \times [0,1] \right \}/\left \{ S^1 \times \left \{ 0 \right \} \right \} $ denote the cone $CS^1 $ of a topological space $S^1$

where $C= \{(x,y,z) \in \mathbb{R}^3 |z= 1-\sqrt{x^2+y^2} ., 0 \le z \le 1\}$

That means $ g(x,y)=[(t(x,y) , 1-\sqrt{x^2+y^2}]$

Answer 1

If you already know that $f$ is a contiuous bijection, there is no need to write down the inverse. The domain of $f$ is compact and the range is Hausdorff, thus $f$ is a homeomorphism.

Here is remark concerning the definition of $f$:

Its seems that you take $S^1 = [0,2\pi]/\{0,2\pi\}$, otherwise the expression $\rho(\cos \theta, \sin \theta)$ does not make sense. But you certainly know that we also may take $S^1 = \{(x,y) \in \mathbb R^2 \mid x^2+y^2 = 1\}$. Then simply define $f((x,y),\rho) = \rho \cdot (x,y)$.