Solution
= Solution
Let $A$ be <clopen> and $C$ an <equivalence class>. The intersections $C\cap A$ and $C\setminus A$ are complementary relatively open subsets of $C$. <Connectedness> of $C$ means one must be empty. Thus every <equivalence class> is entirely in $A$ or entirely outside it, proving $\boxed{q^{-1}(q(A))=A}$. This is precisely saturation.