 Solution

Collapse the equator $E=\{(u,v,t)\in S^2:t=0\}$ to one point and identify no other points: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 defineAt 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 $\boxed{S^2/\mathcal{R}\cong Y}$.