Past exam of the options I course of the University of Oxford / 2013 / E (Topology)

A basis of a topology $\mathcal{T}$ is a collection $\mathcal{B}\subseteq \mathcal{T}$ such that every open set is a union of members of $\mathcal{B}$. Equivalently, for each open $U$ and every $x\in U$, there exists $B\in \mathcal{B}$ with $x\in B\subseteq U$. The empty union represents the empty set. In particular, the basis covers $X$, and any intersection of two basis elements is itself a union of basis elements.

The following open rectangles form a basis of a topology for the product topology:These open rectangles in a finite product cover $X\times Y$, and intersections remain rectangles since $(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)$. One may equivalently take products of elements from a chosen basis of a topology in each factor.

Choose $v_0\in V$, using its nonemptiness. For each $u\in U$, the openness of $U\times V$ in the product topology supplies open $O_u\subseteq X$, $W_u\subseteq Y$ such thatSince $v_0\in W_u$, every point of $O_u$ belongs to $U$. Thus $U={\bigcup }_{u\in U}{O}_u$ is open. Interchanging the factors, using a fixed $u_0\in U$, proves $V$ open. Therefore $\boxed{U\in {\mathcal{T}}_X,V\in {\mathcal{T}}_Y}$. Nonemptiness matters: an empty product is open whatever the other factor is.

Yes: at least one factor topology must be indiscrete. Suppose instead that both have a nonempty proper open set, say $\varnothing \ne U⊊X$ and $\varnothing \ne V⊊Y$. Thenis open in the product topology. Both coordinate projections of $W$ are the full factors, because $U,V$ are nonempty. If $W$ were a rectangle $A\times B$, this would force $A=X$, $B=Y$, hence $W=X\times Y$. But choosing $x\notin U$ and $y\notin V$ gives $(x,y)\notin W$, a contradiction. Thus both factors cannot have nontrivial open sets. This is rectangularity of every product-open set.

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