Suppose \((X,\tau)\) is a topological space and let \((X,\tau')\) be the same set with the finite complement topology. Then \(X\) is \(T_1\) if and only if \(\operatorname{id}:(X,\tau)\to(X,\tau')\) is continuous.
Solution
Let \(\tau_f\) denote the finite-complement topology on \(X\).
Suppose first that \((X,\tau)\) is \(T_1\). Then every finite subset of \(X\) is closed, being a finite union of closed singletons. Hence every cofinite subset is \(\tau\)-open. Therefore \[ \tau_f\subseteq\tau, \] which is exactly the condition that \[ \operatorname{id}:(X,\tau)\to(X,\tau_f) \] be continuous.
Conversely, suppose this identity map is continuous. For every \(x\in X\), the set \(X\setminus\{x\}\) is open in the finite-complement topology, so its inverse image under the identity—namely itself—is \(\tau\)-open. Thus \(\{x\}\) is \(\tau\)-closed. Hence \((X,\tau)\) is \(T_1\).