Past exam of the options I for mathematics and philosophy course of the University of Oxford / 2013 / E / 2

Let $q:X\to X/\mathcal{R}$ send each point to its equivalence class. The quotient topology isIt is the finest topology on the quotient set making the canonical surjection $q$ continuous. The preimage test, rather than a choice of representatives, defines its open sets.

Use the equivalence relation of belonging to the same fibre:The map $\bar{f}:X/\mathcal{R}\to Y$, $[x]\mapsto f(x)$, is well defined, injective by the definition of the relation, and surjective because $f$ is surjective. It is continuous by the quotient topology: for open $V\subseteq Y$, $q^{-1}({\bar{f}}^{-1}(V))=f^{-1}(V)$ is open.

The quotient space is compact as the continuous image of compact $X$, and $Y$ is Hausdorff. The compact-to-Hausdorff continuous bijection theorem makes $\bar{f}$ a homeomorphism. Indeed, images of closed subsets of a compact space are compact and hence closed in a Hausdorff space, so the inverse is continuous.

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 $\boxed{S^2/\mathcal{R}\cong Y}$.