I'm having trouble finishing this homework assignment. I did the first part by showing that the orbits are invariant: every element from the same $(S^1(z_1, z_2) \in S^3/S^1)$ is mapped to the same point in $\mathbb{R}$.
For the second part I found the following equations. For $(z_1, z_2), (w_1, w_2) \in S^3$:
$z_1 z_2 = w_1 w_2$
$\bar{z_1} \bar{z_2} = \bar{w_1} \bar{w_2}$
$z_1 \bar{z_1} = w_1 \bar{w_1} = r$ (for some $r$)
$z_2 \bar{z_2} = w_2 \bar{w_2} = 1 - r$
But I don't think that's enough to prove $(z_1, z_2) = (w_1, w_2)$, is there any more information I left out?
For the third part I showed that $(f, g, h)$ maps to $S^2$ since $||(f,g,h)(z_1, z_2) = 1||^2$ for $(z_1, z_2) \in S^3$. I think that because it is well defined and point-separating we know certainly know that it's image of $X$ is the whole of $S^2$ but I can't grasp why. Don't I still have left to prove that every element of $S^2$ has an inverse in $S^3/S^2$?
To do part $(2),$ note that if you know the values of the first two functions, you know $\Re z_1 z_2$ and $\Im z_1 z_2,$ so you know $z_1 z_2 = C.$ So, you know that $z_2 = C/z_1.$ you can then solve $|z_1|^2 + |z_2|^2 = 1; |z_1|^2 - |z_2|^2 = D$ (where $D$ is what ever $\tilde{h}$ tells you) for $|z_1|$ so, you know $z_1$ up to multiplication by $\exp(i \theta)$ for some theta. Since you know $z_1 z_2,$ to get the same $C,$ you have to multiply $z_2$ by $\exp(-i \theta) = \overline{\exp(i \theta).}$ so that tells you that the the three functions separate points.
For the third part: Since $X$ is compact, the image of $X$ is a compact subset of $S^2.$ The quotient $X$ is a manifold (you should check that), and the map is a smooth non-singular map (check that too). Therefore, by invariance of domain, the image is open in $S^2.$ The only open and closed subsets of $S^2$ are itself and the empty set, since $S^2$ is connected.