Today, I am going to consider a topology problem: If \(X\) is a compact normal topological space, and if \(\sim\) is an equivalence relation on \(X\), under what conditions is the quotient space \(X/\sim\) of equivalence classes also compact normal? Here is one possible solution.
Assume \((X,\tau)\) is a compact normal \((T_4+T_1)\) topological space, \((Y,\kappa)\) be a \(T_1\) topological space, and let \(\pi\colon X\to Y\) be a surjective continuous map. Then \(Y\) is also compact normal.
For example this works when each equivalence class is closed.
Assume \(K\subseteq X\) is compact, and consider \(\pi( K )\subseteq Y\). Take an open cover \(\mathcal{U}\) of \(\pi( K )\). Then \[ \pi^{-1}\mathcal{U} = \left\{\pi^{-1}(U)\mid U\in\mathcal{U} \right\} \] is an open cover of \(\pi^{-1}\pi( K )\). However, since \(K\subseteq \pi^{-1}\pi( K )\), we also have that \(\pi^{-1}\mathcal{U}\) is an open cover of \(K\). But \(K\) is compact. So, there is a finite subcover \(\pi^{-1}\mathcal{U}'\). Since \(\pi\pi^{-1}(U)\subseteq U\) for every \(U\), we get that \(\mathcal{U}'\) is also a finite subcover of \(\pi( K )\).
Easily follows from the fact that \(X\) is compact and \(\pi(X)=Y\).
Assume \(\mathcal{U}\) is an open cover of \(K\). Then \[\mathcal{U}' = \mathcal{U}\cup\left\{K^c\right\}\] is an open cover of \(X\), but \(X\) is compact. So, \(\mathcal{U}'\) has a finite subcover \(\mathcal{U}"\). Then \(\mathcal{U}"\setminus\left\{K^c\right\}\) is a finite subcover of \(K\).
Fix \(y\in K^c\). Since \(X\) is \(T_2\) any point \(a\in K\) and \(y\in K^c\) can be separated by open neighborhoods which means for all \(a\in K\), there are open subset \(U_a,V_a\in \tau\) where \(a\in U_a\) and \(y\in V_a\) and \(U_a\cap V_a\) is empty. Then \(\mathcal{U} = \left\{U_a \right \}_{a\in K}\) is an open cover of \(K\). Since \(K\) is compact, \(\mathcal{U}\) has a finite subcover. Then there are finitely many points \(a_1,\ldots,a_n\in K\) such that \[ K\subseteq \bigcup_{i=1}^n U_{a_i}\quad \text{ and } \quad y\in V = \bigcap_{i=1}^n V_{a_i} \] and \(V\cap K\) is empty. Thus \(K^c\) is open, or equivalently, \(K\) is closed.
In other words, for every closed subset \(K\subseteq X\), the image \(\pi( K )\subseteq Y\) is closed. This easily follows from the fact that \(K\subseteq X\) is closed if and only if \(K\) is compact, and \(\pi\) sends compact sets to compact sets.
Assume \(A_1\) and \(A_2\) are two disjoint closed subsets of \(Y\). Then \(\pi^{-1}(A_i)\) for \(i=1,2\) are two disjoint closed subsets of \(X\). Since \(X\) is normal, there are two disjoint open subsets \(U_i\) for \(i=1,2\) such that \(\pi^{-1}(A_i)\subseteq U_i\). Then since \(\pi\) is surjective we get \(A_i^c = \pi\pi^{-1}(A_i^c) \supseteq \pi(U_i^c)\). Since \(\pi\) is a closed map, \(\pi(U_i^c)\) is closed, and therefore, \(\pi(U_i^c)^c\) is open with the property that \(A_i\subseteq \pi(U_i^c)^c\). All that remains is to check the disjointness of \(\pi(U_1^c)^c\) and \(\pi(U_2^c)^c\). We get \[ \pi(U_1^c)^c \cap \pi(U_2^c)^c = \left(\pi(U_1^c)\cup \pi(U_2^c)\right)^c = \left(\pi(U_1^c\cup U_2^c)\right)^c = \pi(X)^c = \emptyset \] since \(U_1\cap U_2\) is the empty set.
We can do with less. In fact, the whole proof relies on the fact that \(\pi\) is a closed surjection. So, here is one possible extension: Assume \((X,\tau)\) is \(T_4+T_1\) and \((Y,\kappa)\) is \(T_1\). Assume also that \(\pi\colon X\to Y\) is a closed continuous surjection. Then \(Y\) is also a \(T_4+T_1\) space.