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

Fix $x_0,y_0\in Z$. For a continuous function $f:Z\times Z\to \mathbb{R}$, every coordinate slice is continuous, because the maps $x\mapsto (x,y)$ and $y\mapsto (x_0,y)$ into the product topology are continuous. Their target $\mathbb{R}$ is Hausdorff. Part (a)(i) givesTherefore every such $f$ is a constant function. This is the two-factor case of continuous maps from products of cofinite spaces to Hausdorff spaces are constant.

The second coordinate projection ${\pi }_2:Z\times \mathbb{R}\to \mathbb{R}$ is continuous and nonconstant. If a homeomorphism $H:Z\times Z\to Z\times \mathbb{R}$ existed, ${\pi }_2\circ H$ would be a nonconstant continuous function, contradicting the preceding conclusion. Hence

For the half-plane, put $H_{+}=[0,\infty )$ as a set, so $K=\mathbb{R}\times H_{+}$. Its two ambient topologies give different answers:In $Z\times Z$, if $K$ were closed, its inverse image under the continuous map $y\mapsto (x_0,y)$ would be closed in $Z$. That inverse image is $H_{+}$, which is neither a finite set nor all of $Z$, so it is not closed in the cofinite topology. Thus $K$ is not closed.

To establish compactness, note that the subspace topology on any subset $A$ of a topological space with the cofinite topology is again cofinite. An open cover of a nonempty such $A$ has a nonempty member omitting at most finitely many points; finitely many more members cover those points. Hence every subset of a cofinite space is compact. In particular both $Z$ and $H_{+}$ with its cofinite subspace topology are compact. The subspace topology on $K\subseteq Z\times Z$ is their product topology, as one checks by tracing basis rectangles. The product of two compact spaces is compact, so $K$ is compact here.

For clarity, the finite-product compactness result needs no Hausdorff condition. Given an open cover of $A\times B$, cover each point by a basis rectangle lying in one covering member. For fixed $a\in A$, use compactness of $B$ to select finitely many rectangles whose second factors cover $B$, and intersect their first factors to get an open neighbourhood $U_a$ of $a$. Those finitely many original members cover $U_a\times B$. Finally finitely many $U_a$ cover $A$, producing a finite subcover of $A\times B$.

In $Z\times \mathbb{R}$, the second coordinate projection hasSince the ray is closed in the usual real line, $K$ is closed. It is not compact: its image under a continuous map under ${\pi }_2$ is $[0,\infty )$, which is not compact. For an explicit obstruction, the open intervals $(-1,n)$, $n\ge 1$, cover this ray but have no finite subcover.

Finally consider the diagonal $W$ in $Z\times \mathbb{R}$. DefineThis is a bijection. The identity function from the usual real line to $Z$ is continuous: each nonempty cofinite open set is the complement of a finite set, and finite sets are closed in the usual real line. Both coordinates of $\phi$ are therefore continuous, so $\phi$ is continuous for the product topology and the subspace topology of $W$. Its inverse is the restricted second coordinate projection, which is continuous. ThereforeFor the second conclusion, $Z$ is compact whereas $W\cong \mathbb{R}$ is not compact; homeomorphisms preserve compactness. The diagonal in a product with cofinite and Euclidean topologies thus inherits the usual topology, despite one cofinite factor.

Write $Q=(Z\times I)/\mathcal{R}$. Since $Q$ is homeomorphic to the circle, it is Hausdorff. For fixed $t\in I$, the functionis continuous, being a coordinate slice of the quotient map. Part (a)(i) therefore makes it constant. Choose $x_0\in Z$ and setThen $g$ is continuous andSince the quotient map $p$ is a surjection, so is $g$. This is Hausdorff factorization through a cofinite product factor: a continuous map into a Hausdorff space cannot distinguish the points of the coordinate with the cofinite topology.

If $g$ were injective, it would be a continuous bijection from the compact closed interval $I$ to the Hausdorff space $Q$. Part (a)(ii) would make it a homeomorphism, contradicting the fact that the circle and interval are not homeomorphic.

To recall that obstruction explicitly, remove the interior point $1/2$ from $I$. The resulting topological subspace is separated by the nonempty relatively open sets $[0,1/2)$ and $(1/2,1]$, so it is not connected. The circle with any one point removed is connected: after a rotation, it is the image under a continuous map of the connected open interval $(0,2\pi )$ under $t\mapsto (\cos t,\sin t)$. A homeomorphism and its restriction preserve connectedness, giving a contradiction.

Thus $g$ is not injective. Choose distinct $a,b\in I$ with $g(a)=g(b)$. ThenIn particular the requested $x,a,b$ exist. The argument proves the stronger conclusion that the same two interval coordinates work for every $x$.