I'm searching an homeomorphism between $S^2$ and the space $A=\dfrac{S^2/\{\pm 1\}}{S^1}$, where the numerator is the quotient space given by the group action by multiplication of $\{\pm 1\}$ on $S^2$, and then we collapse $S^1$ to a point.
Intuitively, I understand what's happening: $S^2/\{\pm 1\}$ is a "semisphere", and if we collapse its base (that is homeomorphic to $S^1$) to a point, we obtain $S^2$ again. The problem is that I don't know how to formalize it: I tried $$S^2 \xrightarrow{\pi}S^2/\{\pm1\} \xrightarrow{\pi}A$$ but it doesn't work because is not injective. Then I tried to find an identification between $S^2/\{\pm 1\} \rightarrow A$ that respect the equivalence relation given by the collapsing of $S^1$, but it didn't go anywhere.
Using spherical coordinates $\theta, \phi$, where $\theta$ is the polar angle:
$$ f(\theta, \phi) = \left\{ \begin{array}{rcr} (2\theta, \phi) & \;\;\mbox{when}\;\; & 0\leq \theta \leq \frac{\pi}{2} \\ (2\pi - 2\theta, \pi+\phi) & \;\;\mbox{when}\;\; &\frac{\pi}{2} \leq \theta \leq \pi \end{array} \right\} $$
The equator $\theta=\frac{\pi}{2}$, a.k.a. $S_1$, gets mapped to the South pole $(\pi,0)$, and the opposite points $(\theta,\phi)$ and $(\pi-\theta, \pi+\phi)$ are mapped to the same point $(2\theta, \phi) = (2\pi+2\theta, 2\pi+\phi)$ thus ensuring bijectivity between $S^2 / \{\pm 1\} / S^1$ and $S^2$.