Past exam of the options I course of the University of Oxford / 2010 / E / E2 / b

For each continuous map to a Hausdorff target, equality of its values is reflexive, symmetric and transitive. The relation $R$ requires all these equalities simultaneously. Explicitly, $f(x)=f(x)$ gives reflexivity; $f(x)=f(y)$ implies $f(y)=f(x)$; and $f(x)=f(y)=f(z)$ gives $f(x)=f(z)$. These arguments apply to every map in the defining family, proving $R$ is an equivalence relation.

If $x\in A=g^{-1}(B)$ and $yRx$, then $g(y)=g(x)\in B$, so $y\in A$. Thus $A$ contains every equivalence class it meets: it is a saturated subset. Always $A\subseteq p^{-1}(p(A))$; conversely, a point of the latter set is equivalent to some $x\in A$ and hence belongs to $A$. Therefore

Take distinct classes $p(x)\ne p(y)$. The points are not related by $R$, so by the defining negation there is a Hausdorff space $Z$ and a continuous map $f:X\to Z$ with $f(x)\ne f(y)$. Choose disjoint open neighborhoods $U$ and $V$ of those two values. Their inverse images $A=f^{-1}(U)$ and $C=f^{-1}(V)$ are disjoint open subsets of $X$ and are saturated by part (ii).

Hence $p^{-1}(p(A))=A$ and $p^{-1}(p(C))=C$, making $p(A)$ and $p(C)$ open in the quotient topology. They contain the chosen classes and are disjoint: an equivalence class meeting both would contradict saturation and $A\cap C=\varnothing$. Thus $X/R$ is Hausdorff. This quotient is the Hausdorff reflection by continuous maps; every continuous map from $X$ to a Hausdorff space factors uniquely through it.