Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / E / E4 / b

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.

If $X$ is connected, its continuous surjective image under $q$ is connected by (a)(ii). Conversely suppose $X/R$ is connected and $A\subseteq X$ is clopen. By (i) both $A$ and its complement are saturated. The quotient definition then makes $q(A)$ open, and its complement $q(X\setminus A)$ open too. Thus $q(A)$ is clopen in the connected quotient, so is empty or the whole quotient. Pulling back gives $A=\varnothing$ or $X$. By the clopen criterion $X$ is connected. Hence $\boxed{X\text{ connected}⟺X/R\text{ connected}}$.