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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.

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.

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.

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.

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.

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.

Every second-countable $T_0$ space is the 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.

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 it 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 $f(n,x) = e(n,g(x))$ 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 the 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.

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.

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.