= Solution
Collapse the equator $E=\{(u,v,t)\in S^2:t=0\}$ to one point and identify no other points:
$$
\boxed{p\mathcal R q\iff p=q\ \text{or}\ (p\in E\text{ and }q\in E).}
$$
Each closed hemisphere is a disk; collapsing its boundary gives a sphere. The two boundaries are the same equator, so the resulting spheres meet at the single collapsed point. This is the <sphere with its equator collapsed>.
An explicit verification avoids relying only on the picture. Away from the poles, use $p=(\sqrt{1-t^2}\cos\theta,\sqrt{1-t^2}\sin\theta,t)$ and define
$$
f(p)=\begin{cases}(1-\cos(\pi t),\ \sin(\pi t)\cos\theta,\ \sin(\pi t)\sin\theta),&t\geq0,\\
(-1+\cos(\pi|t|),\ \sin(\pi|t|)\cos\theta,\ \sin(\pi|t|)\sin\theta),&t\leq0.
\end{cases}
$$
At the north and south poles use $(2,0,0)$ and $(-2,0,0)$, respectively. The angular terms tend uniformly to zero at both poles, making these extensions continuous. Both hemisphere formulas agree at the equator, with value $(0,0,0)$. Their images lie on the two specified unit spheres and cover them; away from the common origin the height parameter and angle are recovered uniquely. Thus the fibres are exactly the stated equivalence classes. The preceding compact-to-Hausdorff argument gives $\boxed{S^2/\mathcal R\cong Y}$.
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