Past exam of the options II course of the University of Oxford / 2004 / E (Topology)

A space is Hausdorff if distinct points have disjoint open neighbourhoods. In symbols, for $x\ne y$ there are open $U\ni x$ and $V\ni y$ with $U\cap V=\varnothing$.

A space is compact if every open cover has a finite subcover. The finite-intersection formulation says that every family of closed sets with the finite intersection property has nonempty total intersection.

The nested closed sets have the finite intersection property: the intersection of finitely many is the one with the largest index and is nonempty. Compactness gives $\boxed{Y_{\infty }\ne \varnothing }$.

For an open neighbourhood $W$ of the intersection, suppose no $Y_n$ were contained in $W$. Then $Z_n=Y_n\cap (X\setminus W)$ would be nested, nonempty and closed. The same argument gives a point in all $Z_n$, contradicting $Y_{\infty }\subseteq W$. Hence

For connectedness, suppose $Y_{\infty }=A\cup B$ were a separation into two nonempty relatively closed sets. Since $Y_{\infty }$ is compact, $A,B$ are disjoint compact subsets of $X$. The supplied separation fact gives disjoint open neighbourhoods $U_A,U_B$. Their union contains $Y_{\infty }$, so some $Y_n$ lies in that union. But $Y_n$ meets both neighbourhoods, since it contains $A$ and $B$. Its intersections with the two neighbourhoods are then a separation, contradicting its connectedness. Thus the nested intersection is connected.

For the quotient map $\pi$, a subset $U$ of the quotient is open exactly when ${\pi }^{-1}(U)$ is open in $X$. This definition immediately makes $\pi$ continuous. If $g$ is continuous then $g\circ \pi$ is continuous. Conversely, continuity of the composition gives, for every open $V\subseteq Z$, that ${\pi }^{-1}(g^{-1}(V))$ is open. By the defining criterion $g^{-1}(V)$ is open, so $g$ is continuous.

A map $h$ constant on each equivalence class defines a unique map $\bar{h}([x])=h(x)$. It is well-defined by the assumed class constancy, and its composition with $\pi$ is $h$. The preceding criterion proves that $\bar{h}$ is continuous.

For the integer lattice quotient, defineThis continuous map is onto the product of the two unit circles, and its fibres are exactly the lattice equivalence classes. It therefore induces a continuous bijection $\bar{F}:{\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{T}}^2$. Every class has a representative in $[0,1{]}^2$, so the quotient is the continuous image of that compact square and is compact. The target is Hausdorff as a subspace of ${\mathbb{C}}^2$. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, provingFinally, if $U\subseteq {\mathbb{R}}^2$ is open, then ${\pi }^{-1}(\pi (U))={\bigcup }_{m\in {\mathbb{Z}}^2}(U+m)$ is open, being a union of translates. The quotient criterion makes $\pi (U)$ open. Thus the quotient map is an open map.