Past exam of the topology course of the University of Oxford / 2023 / 2 / a

The quotient set $X/\sim$ consists of the equivalence classes $[x]=\{y\in X:y\sim x\}$. Its collapsing map is the canonical projection of a quotient setEquip this set with the quotient topology:This is a topology because preimages preserve arbitrary unions, finite intersections, the empty set, and the whole set. The resulting topological space is the quotient space. The displayed rule immediately implies that the preimage of every open set is open, so $p$ is continuous; it is also surjective.

Use the equivalence relation induced by a function:Equality in $Y$ supplies reflexivity, symmetry, and transitivity, so this is an equivalence relation. Define $\bar{f}:X/\sim \to Y$ by $\bar{f}([x])=f(x)$. It is well-defined; it is injective by the definition of $\sim$ and surjective because $f$ is surjective.

For an open set $W\subseteq Y$,is open by continuity of $f$. The quotient topology therefore makes ${\bar{f}}^{-1}(W)$ open, proving that $\bar{f}$ is continuous.

The continuous image of a compact space theorem applied to $p$ makes $X/\sim$ compact. If $C$ is closed in $X/\sim$, it is compact, and $\bar{f}(C)$ is compact in the Hausdorff space $Y$, hence closed. Thus $\bar{f}$ is a closed map. Its inverse function is continuous, since its preimage of $C$ is precisely $\bar{f}(C)$. We have proved that $\bar{f}$ is a homeomorphism, including the compact-to-Hausdorff continuous bijection theorem used in the argument.

Here $D^2$ must denote the closed disc $\{(x,y):x^2+y^2\le 1\}$, as its boundary points participate in the identification. For $r=\sqrt{x^2+y^2}$, defineThe Pythagorean trigonometric identity gives $\Vert F(x,y){\Vert }^2=1$, so the image lies in the sphere $S^2$. Away from the origin the formula is continuous. At the origin, the first two coordinates have absolute value at most $|\sin (\pi r)|\to 0$ and the third tends to $1$, proving continuity there as well.

Every boundary point maps to $(0,0,-1)$, while the origin maps to $(0,0,1)$. For a point $(u,v,w)\in S^2$ with $-1<w<1$, there is a unique $r\in (0,1)$ with $\cos (\pi r)=w$, and its unique preimage isThis point has radius $r$, because $u^2+v^2=1-w^2={\sin }^2(\pi r)$. The two poles have the preimages already described. Thus $F$ is surjective and its fibers are exactly the prescribed equivalence classes.

The closed disc is compact by the Heine-Borel theorem, and $S^2$ is Hausdorff with the subspace topology from Euclidean space. The preceding result gives the homeomorphism

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is finite and cannot be all of $\mathbb{R}$.

Suppose the quotient space had distinct points with disjoint open neighbourhoods $U,V$. Since the quotient map $p$ is surjective, both preimages $p^{-1}(U),p^{-1}(V)$ would be nonempty. Since $p$ is continuous, these would be disjoint open subsets of $\mathbb{R}$ with its cofinite topology, contradicting the observation above. Thus any such quotient space with at least two points is not Hausdorff.