Exercise 27, § 8

Let X be a topological space. Now prove that there exists a Kolmogoroff (resp. accessible, separated, regular) space Y and a continuous mapping $\varphi : X \to Y$ satisfying the following universal property: for every continuous mapping $f : X \to Z$ from X into a Kolmogoroff (resp. accessible, separated, regular) space Z, there exists one and only one continuous mapping $g : Y \to Z$ such that $f = g \circ \varphi$ (E, IV, p. 22). We shall prove successively that: a) For Kolmogoroff spaces, one can take for Y the quotient space of X by the equivalence relation $\{ \overline{x} \} = \{ \overline{y} \}$. b) For accessible spaces, one can take for Y the quotient space of X by the following equivalence relation R: the graph of R is the intersection of the graphs G of equivalences in X such that G(x) is closed in X for every $x \in X$; G then contains the graph of the relation $\{ \overline{x} \} \cap \{ \overline{y} \} \neq \varnothing$ between x and y. c) For separated spaces and regular spaces, now prove that the conditions (CU₁) to (CU₃) of E, IV, p. 23 are verified, using in particular Exer. 6 a) of I, p. 101. Give examples where X is not regular and where the mapping $\varphi : X \to Y$ in the corresponding universal regular space is bijective (cf. I, p. 103, Exer. 20 b)). d) If X is a space satisfying the axiom (O₃), now prove that the corresponding universal Kolmogoroff space of X is canonically homeomorphic to the corresponding universal regular space of X.

e) If one takes for X the accessible quasi-compact space of Example 2 of I, p. 60, now prove that the corresponding universal separated space is reduced to a single point.