Inverse of function of projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$? Is it possible to explicitly define the inverse function of this function of the projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$?
$$f([x,y]) = \left( \dfrac{2xy}{x^2+y^2}, \dfrac{x^2-y^2}{x^2+y^2}\right).$$
 A: I guess you define $\mathbb RP^1 = (\mathbb C \setminus \{0\})/ \sim$, where $z \sim z'$ iff $z' = tz$ for some $t \in \mathbb R \setminus \{0\}$. Note that I identified $\mathbb R^2$ with $\mathbb C$. Then your map $f$ is essentially given by $f([z]) = \dfrac{z^2}{\lvert z \rvert^2}$ (note that I exchanged the first and second coordinate of your original $f([z])$ which does not play a role conceptually because the "flip-map" $\tau(x+iy) = i \overline z = y + ix$ is a homeomorphism on $S^1$).
This map is a well-defined continuous bijection, hence a homeomorphism. You ask for its inverse $f^{-1} : S^1 \to \mathbb RP^1$. It is well-known that for each complex number $w \ne 0$ we can choose a square root $\sqrt w$ (there are two possible choices which differ by a factor $-1$, but we shall not be specific how to choose $\sqrt w$). Note that $\sqrt w \in S^1$ for $w \in S^1$. We claim that
$$f^{-1}(w) = [\sqrt w] .$$
Note that the RHS does not depend on the choice of $\sqrt w$ because $\sqrt w \sim -\sqrt w$. But now we have $f([\sqrt w]) = \dfrac{\sqrt w ^2}{\lvert \sqrt w \rvert^2} = w$ which proves our claim.
Concerning square roots it may be interesting to have a look at my answer to Analytic Functions - Entire Function .
