Past exam of the topology course of the University of Oxford / 2017 / 2 / b

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.

The product topology on the Cartesian product $X\times Y$ has a basis of a topology consisting of all rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the smallest topology making the two coordinate projections continuous.

Take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, use the Hausdorff condition in $X$ to find disjoint open neighbourhoods $U_1,U_2$. Then $U_1\times Y$ and $U_2\times Y$ are disjoint open neighbourhoods of the two points. If $x_1=x_2$, then $y_1\ne y_2$; disjoint open neighbourhoods $V_1,V_2$ in $Y$ give $X\times V_1$ and $X\times V_2$. Thus $X\times Y$ is Hausdorff, as established by the forward implication of the Hausdorff property of a nonempty product. An empty Cartesian product is automatically Hausdorff.