Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 10G (Analysis and Topology)

A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

Let $X$ be compact and $q:X\to X/R$ the quotient map. If $(U_i{)}_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i){)}_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.

The Hausdorff property need not survive. On the real line, defineThe quotient $\mathbb{R}/\mathbb{Q}$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb{R}$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the non-Hausdorff quotient of the real line by rational translation.

Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a closed map. For every closed $C\subseteq X$,is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

DefineThis is continuous and takes values on the unit sphere. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the universal property of the quotient topology gives a continuous map

For $0<y<1$, the third coordinate $\cos (\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline{f}$ is injective. The spherical-coordinate formula also shows that it is surjective.

The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now makes $\overline{f}$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the suspension of a topological space $S^1$, which is $S^2$. This proves the square quotient model of the two-sphere.