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

A topological space is Hausdorff when any two distinct points have disjoint open neighbourhoods.

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is a finite union of finite sets and cannot be all of $Z$. Thus $Z$ is a hyperconnected space.

If the continuous map $f$ took distinct values at $s,t\in Z$, choose disjoint open neighbourhoods $G,H$ of $f(s),f(t)$ in the Hausdorff space $Y$. By continuity, $f^{-1}(G)$ and $f^{-1}(H)$ would be nonempty disjoint open sets of $Z$, a contradiction. ThereforeThe proof establishes the general fact that continuous maps from an infinite cofinite space to a Hausdorff space are constant.

A homeomorphism is a bijection for which both the function and its inverse are continuous maps.

Let $f:X\to Y$ satisfy the given assumptions. For a closed subset $F$ of $X$, the closed subset of a compact space theorem makes $F$ compact. The continuous image of a compact space theorem makes $f(F)$ compact. Because $Y$ is Hausdorff, its compact subset $f(F)$ is closed. Thus $f$ is a closed map.

For $g=f^{-1}:Y\to X$, the inverse image of a closed set $F\subseteq X$ iswhich is closed in $Y$. The closed-set criterion for continuity therefore makes $g$ continuous. Hence $f$ is a homeomorphism, proving the compact-to-Hausdorff continuous bijection theorem.