Exercise 27, § 8

Let $X$ be a topological space. Show that there exists a Kolmogoroff (resp. accessible, Hausdorff, regular) space $Y$ and a continuous mapping $\varphi : X \to Y$ which has the following universal property: for every continuous mapping $f : X \to Z$ of $X$ into a Kolmogoroff (resp. accessible, Hausdorff, regular) space $Z$, there is a unique continuous mapping $g : Y \to Z$ such that $f = g \circ \varphi$ (Set Theory, Chapter IV, § 3, no. 1). Prove that:

a) For Kolmogoroff spaces, we may take $Y$ to be the quotient space of $X$ by the equivalence relation $\{ \bar{x} \} = \{ \bar{y} \}$.

b) For accessible spaces, we may take $Y$ to be the quotient space of $X$ by the following equivalence relation $R$: the graph of $R$ is the intersection of all graphs $G$ of equivalences on $X$ such that $G(x)$ is closed in $x$ for each $x \in X$; the graph of $R$ then contains the graph of the relation $\{ \bar{x} \} \cap \{ \bar{y} \} \neq \emptyset$ between $x$ and $y$.

c) For Hausdorff spaces and regular spaces, show that conditions (CU_I) to (CU_{III}) of Set Theory, Chapter IV, § 3, no. 2, are satisfied, by using in particular Exercise 6 a). Give examples in which $X$ is not regular and the mapping $\varphi : X \to Y$ of $X$ into the corresponding universal regular space is bijective [cf. Exercise 20 b)].

d) If $X$ is a topological space which satisfies axiom $(O_{III})$, show that the universal Kolmogoroff space corresponding to $X$ is canonically homeomorphic to the universal regular space corresponding to $X$.