Inverting the Hopf map from $\mathbb{CP}^1$ to $\mathbb{S}^2$ Let
$$
\alpha=e^{i\varepsilon}\cos\left(\frac{\theta}{2}\right)\qquad\beta=e^{i\kappa}\sin\left(\frac{\theta}{2}\right)
$$
be 2 generic complex numbers, we could find a Hopf map from $\mathbb{CP}^1$ to $\mathbb{S}^2$:
$$
\mathcal{H}\begin{bmatrix}\alpha\\\beta\end{bmatrix}
=
\begin{bmatrix}2\text{Re}(\bar\alpha\beta) \\ 2\text{Im}(\bar\alpha\beta)\\|\alpha|^2-|\beta|^2\end{bmatrix}
$$
I wonder is this a bijection? It looks like given a vector in $\mathbb{S}^2\subset\mathbb{R}^3$, we could solve for $\alpha$ and $\beta$. However, I'm not pretty sure how we could express this inverse map that goes from $\mathbb{S}^2$ to $\mathbb{CP}^1$. Thanks!
 A: The Hopf map is first defined as a map
$$ \tilde H : \mathbb S^3 \to \mathbb S^2$$
and it descends to a map on quotient
$$ \mathcal H : \mathbb S^3/ \mathbb S^1 \to \mathbb S^2$$
and $\mathbb{CP}^1$ is identified as $\mathbb S^3/ \mathbb S^1$.
To find the inverse, write $\gamma=\beta/\alpha$. Then we have $\bar\alpha \beta = |\alpha|^2 \gamma$. Using also that $|\alpha|^2 +|\beta|^2=1$, one has $|\alpha|^2 (1+|\gamma|^2)=1$. Thus we have
$$\mathcal H \begin{pmatrix} \alpha \\ \beta\end{pmatrix} = \begin{pmatrix}  \frac{ 2\operatorname{Re} \gamma}{1+|\gamma|^2} \\ \frac{ 2\operatorname{Im} \gamma}{1+|\gamma|^2} \\ \frac{1-|\gamma|^2}{1+|\gamma|^2} \end{pmatrix}$$
and the RHS is the stereogrpahic projection
$S:\mathbb C \to \mathbb S^2$, the inverse has a nice formula
$$S^{-1} \begin{pmatrix} x\\ y\\z \end{pmatrix} =\frac x {1+z} + i \frac y {1+z}$$
So the inverse can be written as
$$\mathcal H ^{-1} \begin{pmatrix} x\\ y\\ z\end{pmatrix}= \begin{bmatrix}1+z \\ x+iy\end{bmatrix} $$
or
$$(2(1+z))^{-1/2} \begin{pmatrix}1+z \\ x+iy\end{pmatrix} $$
if you want the image to be in $\mathbb S^3$ (we used $x^2 + y^2+z^2 = 1$ here).
Note that the above map is defined only when $z\neq -1$: to find $\alpha, \beta $ explicitly as a function of $x, y ,z$ is the same as finding a section of the fibration $\mathbb S^3\to \mathbb S^2$. But there is not a continuous section in this case.
