This is the proof to the proposition 6.1 in John Lee's ITM.
However, for the function given in the proof, I cannot show how it is surjective and makes the same identifications as the closed disk modulo the equivalence relation generated by $(x,y) \sim (-x,y)$ for $(x,y) \in \partial \bar{B}^2$. I would greatly appreciate any help with this part.
We have $\pi(0,\pm1) = (0,0,\pm1)$. Thus it remains to check that each $(a,b,c) \in S^2$ with $-1 < c < 1$ is in the image of $\pi$. We must find $x$ such $(x,c) \in \overline{B}^2$ and $\pi(x,c) = (a,b,c)$. With abbreviations $r = \sqrt{1-c^2}$ and $\xi = \dfrac{\pi x}{r}$ this means $a = -r\cos \xi, b = -r\sin \xi$ which is equivalent to $-\dfrac{a}{r} = \cos \xi, -\dfrac{b}{r} = \sin \xi$. But $a^2 + b^2 = r^2$, thus $(-\dfrac{a}{r}, -\dfrac{b}{r})$ lies on the unit circle and there exists $\xi \in [-\pi,\pi]$ such that $(-\dfrac{a}{r}, -\dfrac{b}{r}) = (\cos \xi,\sin \xi)$. But now we have $x = \dfrac{r\xi}{\pi} \in [-r,r]$, thus $x^2 + c^2 \le r^2 + (1 - r^2) = 1$, i.e. $(x,c) \in \overline{B}^2$.
Let $\pi(x_1,y_1) = \pi(x_2,y_2)$. Then clearly $y_1= y_2 =y$. If $y = \pm 1$, then $x_1 = x_2 = 0$ and hence $x_1 = -x_2$. Now consider the case $-1 < y < 1$. W.l.o.g. we amy assume $x_1 \le x_2$. With $r = \sqrt{1-y^2}$ we get $x_i^2 \le 1 - y^2 = r^2$, i.e. $x_i \in [-r,r]$. With $\xi_i = \dfrac{\pi x_i}{r}$ we see that $\xi_i \in [-\pi,\pi]$ and $(\cos \xi_1,\sin \xi_1) =(\cos \xi_2,\sin \xi_2)$. This implies that $\xi_1 = \xi_2$ or $\xi_1 = -\pi, \xi_2 = \pi$. The latter shows that $x_1 = - x_2$.
Conversely, consider $(x,y)$ in the boundary. This means $x = \pm \sqrt{1-y^2}$, thus $\dfrac{\pi x}{\sqrt{1-y^2}} = \pm \pi$ and we see that $\pi(x,y) = \pi(-x,y)$.
Remark: The minus-signs in the first two coordinates of $\pi(x,y)$ are unnecessary.