 Solution

The set $X/\mathcal{R}$ consists of the equivalence classes $[x]$ for $\mathcal{R}$. The canonical surjection $q:X\to X/\mathcal{R}$, $q(x)=[x]$, defines its quotient topology:The quotient space includes this topology, not just the collection of equivalence classes.

Regard the circle as $S^1=\{z\in \mathbb{C}:|z|=1\}$ and use the continuous surjectionDefine $z\mathcal{R}w$ precisely when $h(z)=h(w)$. This is an equivalence relation, being equality of function values. Equivalently $w=z$ or $w=\overline{z}$, so the construction is the circle modulo reflection.

Since $h$ is constant on each equivalence class, it induces a bijectionIt is continuous by the universal property of the quotient topology, because $\overline{h}\circ q=h$ is continuous. The circle is compact by the Heine-Borel theorem, and its quotient space is a continuous image of a compact space, hence compact. The closed interval is Hausdorff. Part (a)(ii) therefore provesGeometrically, the equivalence relation folds the upper and lower semicircles together, retaining the two real endpoints separately.