Quotients in categories of metric spaces

Question

There are several categories whose objects are metric (or pseudo-metric) spaces. Natural choices of morphisms are continuous, uniformly continuous, Lipschitz or short (= non-expansive or contractive) maps.

For an equivalence relation $E\subseteq X\times X$ on the underlying set $X$, a quotient in one of the categories is a coequalizer of the restrictions $p_1,p_2:E\to X$ of the projections $\pi_1,\pi_2: X\times X\to X$ to $E$ (where, in each of the categories above, $X\times X$ gets the maximum (pseudo-) metric $((x,y),(a,b))\mapsto\max\{d(x,a),d(y,b)\}$ which makes the cartesian product a categorical product).

Explicitly, this means that a morphism $q:X\to Q$ with $q\circ p_1=q\circ p_2$ (i.e., $q(x)=q(y)$ whenever $x,y$ are equivalent) is a quotient if, for all morphisms $f:X\to Y$ with $f\circ p_1=f\circ p_2$, there is a unique morphism $\tilde f:Q\to Y$ with $f=\tilde f\circ q$.

A nice example is the category $\Psi$MetLip of pseudo-metric spaces and Lipschitz maps where quotients always exist, namely the pseudo-metric on the set-theoretic quotient described in Eric Wofsey's answer to Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't?. The answer of Włodzimierz Holsztyński to Quotient of metric spaces shows that for the equivalence relation $E=\{(x,y)\in [0,1]^2:f(x)=f(y)\}$ for the Cantor staircase function $f$, the quotient in MetLip of $[0,1]$ is trivial (a singleton, because every Lipschitz function which is constant on the equivalence classes is constant) whereas the quotient in MetCon is the interval $[0,1]$.

Here, I am interested in the categories MetCon and $\Psi$MetCon with continuous functions as morphisms. These are full subcategories of HTop and Top of (Hausdorff) topological spaces with continuous functions.

A simple example of an equivalence relation on $\mathbb R$ where quotients in Top and $\Psi$MetCon both exist but are different is given by $x\sim y$ is $x=y=0$ or $x,y\neq 0$ which has only two equivalence classes. The quotient in Top is $\{0,1\}$ with the Sierpinski topology $\{\emptyset,\{0,1\},\{1\}\}$ whereas the quotient in $\Psi$MetCon is $\{0,1\}$ with the pseudo-metric $d=0$ (that the quotients are different should mean that the latter induces a different topology).

For the equivalence relation on $\mathbb R$ collapsing $\mathbb Z$ to a point, the quotient in HTop exists (it is not first countable) but no quotient exists in MetCon.

Is there an example of an equivalence relation in a metric space having a quotient in MetCon but whose topology differs from the one of the quotient in HTop?

If you prefer theorem proving instead of constructing examples you should try to prove:

An equivalence relation on a metric space has a quotient in MetCon if and only if the quotient in HTop is metrizable.

Answer 1

$\newcommand{cr}{\operatorname{cr}}$ Start by letting $X_T$ be any second-countable, functionally Hausdorff space which is not regular. Edit: We'll need to add an additional assumption here, the necessity of which I'd like to thank Jochen Wengenroth for pointing out. There are many possible formulations, and there is certainly something easier than the following. Recall that a subset of $X_T$ is z-compact if any cozero covering of it has a finite subcover.

Assumption $(\ast)$ Each point of $X_T$ has a cozero neighbourhood $V$ which is contained in in a z-compact subset.

This extra assumption won't play a role until further on, so we start with the following observation. Since $X_T$ is first-countable, it is the image of a metrisable space $X$ by a continuous open mapping $p:X\rightarrow X_T$. This is a classical characterisation of first-countability due to Ponomarev.

Since every open mapping is quotient, $X_T$ is the quotient of $X$ by the relation $x\sim y$ if and only if $p(x)=p(y)$. The universal property enjoyed by $p$ states that for any space $Y$ and any map $f:X\rightarrow Y$ respecting the quotient relation $\sim$, there is a unique continuous function $f_T:X_T\rightarrow Y$ such that $f_T\circ p=f$.

If we assume that $Y$ is metrisable, then there is a further factorisation of $f$. To discuss this we recall that the full subcategory of completely regular spaces is bireflective in $\mathrm{Top}$ (I do not assume that completely regular implies Hausdorff). The reflector, often called the completely regular modification, sends a space $Z$ to the space $\cr(Z)$ which has the same underlying set, and which has the topology generated the family of all cozero sets in $Z$. This topology is smaller than the original topology, and the identity function induces a continuous natural bijection $r_Z:Z\rightarrow \cr(Z)$. The space $\cr(Z)$ is always completely regular, and for any completely regular space $Y$, any map $f:Z\rightarrow Y$ factors uniquely through $r_Z$.

Lemma Let $Z$ be a space. The following statements hold.

1. $\cr(Z)$ is Hausdorff if and only if $Z$ is functionally Hausdorff.
2. If $Z$ is functionally Hausdorff and second-countable, then $\cr(Z)$ has a $G_\delta$-diagonal.
3. If $Z$ is functionally Hausdorff, then $\cr(Z)$ is locally compact if and only if $Z$ satisfies $(\ast)$.

Proof $(1)$ If $\cr(Z)$ is Hausdorff, then it is Tychonoff. The Tychonoff property is not preserved by topological refinement, but the functional Hausdorff property is.

$(2)$ This is a consequence of a more general result proved by Taras Banakh here.

$(3)$ Since it is Tychonoff, $\cr(Z)$ is locally compact if and only if each point has a cozero neighbourhood contained inside a compact set. By the Alexander Subbase Lemma, compactness in $\cr(Z)$ is equivalent to each cozero covering having a finite subcovering. Since the cozero sets in $Z$ and $\cr(Z)$ are the same, the previous condition on the points of $\cr(Z)$ spells out as property $(\ast)$ holding in $Z$. $\quad\blacksquare$

Returning now to the opening discussion, we have a metric space $X$ and its functionally Hausdorff, non-metrisable, quotient $X_T$. Let $X_M=\cr(X_T)$ be the completely regular modification of $X_T$ and $r:X_T\rightarrow X_M$ the universal bijection.

Then for any metric space $Y$ and any continuous map $f:X\rightarrow Y$ respecting the quotient relation, there is a unique map $f_T:X_T\rightarrow Y$ with $f_T\circ p=f$. Since $Y$ is completely regular, there is a unique map $f_M:X_M\rightarrow Y$ satisfying $f_M\circ r = f_T$.

Proposition $X_M$ is metrisable.

Proof Since $X_M$ is the one-to-one image of the second-countable space $X_T$, it is Lindelöf. As it is Tychonoff, it is paracompact. Due to the Lemma, $X_M$ is locally compact and has a $G_\delta$-diagonal. Since any compact Hausdorff space with a $G_\delta$-diagonal is metrisable, $X_M$ is locally metrisable. Since it is paracompact, it is metrisable. $\quad\blacksquare$

Now, writing $q=r\circ p$, for $x,y\in X$ we have $x\sim y$ if and only if $p(x)=p(y)$ if and only if $q(x)=q(y)$, since $r$ is bijective. We have shown that for any metrisable space $Y$ and any continuous mapping $f:X\rightarrow Y$ respecting the quotient relation $\sim$, there is a unique continuous mapping $f_M:X_M\rightarrow Y$ such that $f_M\circ q=f$. It follows that $X_M$ is a quotient of $X$ in $\mathrm{MetCon}$.

It remains to show that the class of spaces satisfying all the assumptions is nonempty. A list of second-countable, functionally-Hausdorff, non-regular spaces is found here. Following Jochen Wengenroth's suggestion, we'll single out Smirnov's deleted sequence topology on the real line. This is a copy of the real line given the topology in which a point $x\neq 0$ has a neighbourhood base consisting of the sets $$(x-1/n,x+1/n)\qquad n\in\mathbb{N},$$ and $0$ has a neighbourhood base consisting of the sets $$(-1/n,1/n)\setminus A,\qquad n\in\mathbb{N},\;A\subseteq\{1/n\mid n\in\mathbb{N}\}.$$ The topology is finer than the Euclidean topology, so is functionally Hausdorff. Since $\{0\}$ cannot be separated from the closed set $\{1/n\mid n\in\mathbb{N}\}$, the topology is not regular. It's clear that the topology is locally compact at all points $x\neq 0$. On the other hand, property $(\ast)$ holds at $0$, since $(-1/n,1/n)$ is a local base for the cozero sets here.

Thus if $X_T$ is Smirnov's deleted line, then $X_M=\cr(X_T)$ is a locally compact metric space bijecting onto the real line. It's easy to see that this implies that $X_M$ is in fact homeomorphic to the real line.

There is a metric space $X$ and an equivalence relation $\sim$ on $X$ whose Hausdorff quotient $X_T$ in $Top_2$ is Smirnov's deleted sequence topology, and whose metrisable quotient $X_M$ in $\mathrm{MetCon}$ is the real line.