Kolmogorov quotient

Remarks

Equivalently, $x \equiv y$ if and only if $\overline{\set{x}} = \overline{\set{y}}$. The space $\mathrm{KQ}(X)$ is a T0 space. The quotient map $q \colon X \to \mathrm{KQ}(X)$ is open and closed, and $U \mapsto q(U)$ is a bijection from the open sets of $X$ to the open sets of $\mathrm{KQ}(X)$. If $X$ is already $T_0$, then $q$ is a homeomorphism.

Example: let $(X, d)$ be a pseudometric space with its topology. Here $x \equiv y$ if and only if $d(x, y) = 0$, because a point with $d(x, y) > 0$ has a ball around $x$ that misses $y$. So $\mathrm{KQ}(X)$ is the metric space obtained by identifying points at distance $0$.

Non-example: the identity map of a two-point set $\set{a, b}$ with the trivial topology is not a Kolmogorov quotient map. Its target is not $T_0$, since $a$ and $b$ have the same neighbourhoods; the Kolmogorov quotient of this space is a single point.