In my notes:
It is claimed that $f^{-1}_{+}\left ( x \right )=\left ( x,\sqrt{1-x^{2}} \right )$. I am unable to understand how this is derived. Equally so, I am unable to understand why the range of the map is restricted to the open interval of 0 and 1.
Any explanation is appreciated.
Thanks in advance.
\begin{equation} f^{-1}_{+}\left ( x \right )=\left ( x,\sqrt{1-x^{2}} \right ) \end{equation}
because $f_{+}$ is a projection, it takes points of $S^1$ (actually points in the upper half circumference) and it projects them on the x-axis between $(-1,1)$; $f^{-1}_{+}$ takes $x\in(-1,1)$ and it brings them in the upper half circumference, so $f^{-1}_{+}$ has the upper half circumference as range and every point of this range has $y=\sqrt{1-x^2}\in(0,1]$. You can obtain $f^{-1}_{+}$ from the equation of the circumference $x^2+y^2=1$:
\begin{equation} y^2=1-x^2\\ y=\pm\sqrt{1-x^2} \end{equation}
You choose $+$ because $f^{-1}_{+}$ has the upper half circumference as range.