Is every T0 2nd countable space the quotient of a separable metric space?

Question

Suppose the space $X$ has a countable basis and $X$ is $T_{0}$. Must there exist a separable metrizable space $Y$ and a quotient map q:$Y \rightarrow X$?

(Some surrounding facts:

Every metrizable space is 2nd countable iff it's separable.

Every 2nd countable space is 1st countable and hence Frechet and hence sequential and hence the quotient of a locally compact metrizable space. ( However in the canonical proof, $Y$ is the disjoint union of a typically very large collection of convergent sequences [Franklin] and usually not separable, even if $X$ itself is a separable metric space).

If $X$ is $T_{0}$ and regular and 2nd countable then $X$ is metrizable (Urysohn metrization)).

For a non $T_{0}$ counterexample let $X$ have cardinality larger than the real numbers and employ the indiscrete topology.)

If the answer is `no' can a counterexample $X$ be $T_1$ or even $T_2$?

(Edit: the answer is `yes' and Francois Dorais and Andrej Bauer provide two explicit solutions below and also point out relevant references.

The $T_{1}$ case was settled by Paul Strong. As shown below similar tactics settle the $T_{0}$ case. The question is relevant to topological domain theory. For example

``The similarity between our definitions and results and those of Schroder was first observed by Andrej Bauer, who proved that the sequential spaces with admissible representations are exactly the T0 (quotients of countably based) spaces...''.

From the paper `Topological and Limit-space Subcategories of Countably Based Equilogical Spaces.' by Menni and Simpson.

homepages.inf.ed.ac.uk/als/Research/Sources/subcats.pdf

Answer 1

Every second-countable $T_0$ space is a quotient of a separable metric space and this essentially follows from the proof of the $T_1$ case given by Paul Strong in Quotient and pseudo-open images of separable metric spaces [Proc. Amer. Math. Soc. 33 (1972), 582-586].

A continuous map $f:X \to Y$ is sequence covering if for every convergent sequence $y_n \to y$ in $Y$ there is a convergent sequence $x_n \to x$ in $X$ such that $y_n = f(x_n)$ for all $n$ and $y = f(x)$. (Since the spaces under consideration are not necessarily Hausdorff, when I say "convergent sequence" I always mean a convergent sequence together with a choice of limit.)

Fact 1. If $f:X \to Y$ is a sequence covering continuous map and $Y$ is sequential then $f:X \to Y$ is a quotient map.

To prove this, we need to show that a map $g:Y \to Z$ is continuous if (and only if) $g \circ f:X \to Z$ is continuous. Since $Y$ is sequential, it is enough to show that if $y_n \to y$ is a convergent sequence in $Y$, then $g(y_n) \to g(y)$ is a convergent sequence in $Z$. Since $f:X \to Y$ is sequence covering, we can find a convergent sequence $x_n \to x$ in $X$ that maps to $y_n \to y$. Assuming only that $g \circ f:X \to Z$ is continuous, it follows that $g(f(x_n)) = g(y_n) \to g(f(x)) = g(y)$ is indeed a convergent sequence in $Z$.

Let $\omega+1 = \{0,1,\dots,\omega\}$ be the one-point compactification of $\omega$. The space $Y^{\omega+1}$ with the compact-open topology consists of all convergent sequences in $Y$ (along with a choice of limit). Since $\omega+1$ is compact Hausdorff, the evaluation map $e:(\omega+1)\times Y^{\omega+1}\to Y$ is continuous. Moreover, this map is obviously sequence covering. Even better:

Fact 2. If $g:X \to Y^{\omega+1}$ is a continuous surjection then the continuous map $f:(\omega+1)\times X \to Y$ defined by $f(n,x) = e(n,g(x)) = g(x)(n)$ is sequence covering.

Combining these facts, we see that if $Y^{\omega+1}$ is sequential and the continuous image of a separable metric space $X$ then $Y$ is a quotient of the separable metric space $(\omega+1)\times X$.

Now if $Y$ is $T_0$ and second-countable then so is the space $Y^{\omega+1}$. Therefore, it suffices to prove:

Fact 3. Every second-countable $T_0$ space $Y$ is the continuous image of a separable metric space.

Tho see this, fix a countable base $(U_n)_{n\lt\omega}$ for $Y$. Let $X \subseteq \omega^\omega$ consist of all $x:\omega\to\omega$ such that $(U_{x(n)})_{n \lt\omega}$ enumerates a neighborhood basis for some point of $Y$. Since $Y$ is $T_0$, a neighborhood basis for a point uniquely determines that point. Therefore, we obtain a natural surjection $f:X \to Y$, which is easily seen to be continuous.

Since a second-countable space is always sequential, we can combine the three facts above to conclude that every second-countable $T_0$ space is a quotient of a separable metric space.

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Actually, as pointed out in Andrej's answer the map in Fact 3 is already a quotient map (since it is an open mapping). Strong needed the more complicated construction since his assumptions are in terms of countable networks instead of countable bases.

Answer 2

Here is an elementary construction of such a quotient.

Let $X$ be a $T_0$-space and $B_0, B_1, B_2, \ldots$ a countable base for $X$. For any point $x \in X$ define $N(x) = \{i \in \mathbb{N} \mid x \in B_i\}$, the index set of basic neighborhoods of $x$. Given a sequence $\alpha : \mathbb{N} \to \mathbb{N}$ let $i(\alpha) = \{\alpha(k) \mid k \in \mathbb{N}\}$, the image of the sequence.

The Baire space $\mathbb{N}^\mathbb{N}$ is countably based and $0$-dimensional with the ultrametric $$d(\alpha, \beta) = 2^{-\min_k (\alpha_k \neq \beta_k)}.$$ Let $D$ be the subspace $$D = \{\alpha \in \mathbb{N}^\mathbb{N} \mid \exists x \in X . i(\alpha) = N(x)\},$$ which consists of those sequences that enumerate the index set of some point in $X$. Note that the point $x$ in the definition of $D$ is unique for a given $\alpha$, if it exists, because $X$ is $T_0$. Define the map $q : D \to X$ by $$q(\alpha) = \text{"the $x$ such that $i(\alpha) = N(x)$"}.$$ It is a basic exercise in topology to verify that $q$ is a quotient map. Thus, every countably based $T_0$-space is the quotient of a $0$-dimensional countably based ultrametric space.

Also note that the map $N : X \to \mathcal{P}(\mathbb{N})$ is an embedding when the codomain is equipped with the Scott topology.

Answer 3

The following canonical construction yields a `Yes' answer if $X$ is countable.

Each $x \in X$ is assigned a metric space (or spaces) as follows.

Let $U_{1},U_{2},...$ be a countable basis for $X$. For each $x \in X$ let $S_{x}$ denote the set of indices $n$ such that $x \in U_{n}$. Let $V(x,n)$ denote the intersection of the first $n$ basic open sets in the collection determined by $S_{x}$. Note $V(x,1)$,$V(x,2),...$ is a nested local basis at $x$ whose intersection is the closure of $x$.

Case 1. Suppose the sequence $V(x,1)$,$V(x,2),...$ is not eventually constant

Let $W(x,n)=V(x,n) \setminus V(x,n+1)$ with the discrete topology and distance $2^{-n}$ between distinct points. Notice we may extend the metric to $W_{x}$, (the union of the sets $W(x,n)$), so that all nonconstant convergent sequences converge to a single new point $x_{x}$ in the completion. Let $Y_{x}$ denote the completion of $W_{x}$ and observe there is a natural map $Y_{x} \rightarrow X$.

Case 2. Not case 1 means the closure of $x$ is isolated. For each $y$ such that $x \rightarrow y$ build a compact metric space $Y(x,y)$ consisting of a nontrivial convergent sequence. Note there is a natural surjective map $Y(x,y) \rightarrow \{x,y\}$. Perform this construction even if Case 1. applies.

Now let $Y$ be the disjoint union of all metric spaces constructed above. By construction we have natural surjective map $Y \rightarrow X$. To see this is a quotient, fix a nonclosed $A \subset X$. Since $X$ is a sequential space obtain a convergent sequence $a_{n} \in A$ with a limit $x \notin A$.

If for some fixed $n$ the constant sequence $a_{n} \rightarrow x$ then Case 2 shows the preimage of A is not closed. Otherwise Case 1 shows the premage of $A$ is not closed.

Thus $Y \rightarrow X$ is a quotient map. If $X$ is countable then by construction $Y$ is countable and hence separable.