I'll offer a necessary and sufficient condition for the (complete) metrisability of a Suslin space. There are many approaches to this problem, and I'll limit myself here to discussing a single condition which has a similar flavour to the definition of Suslin spaces. I'll assume familiarity with Engelking's book [Eng].
Recall that a surjective mapping $f:X\rightarrow Y$ is perfect if it is closed and has compact fibres.
Theorem A Suslin space is (completely) metrisable if and only if it admits a perfect mapping onto a (complete) metric space.
To prove this we'll need to introduce some auxiliary notions.
A network for a space $X$ is a family of its subsets $\mathcal{N}$ such that every open subset of $X$ can be represented as a union of sets in $\mathcal{N}$. A base is exactly a network consisting of open sets. Thus every base is a network, but networks are more general. For instance, the family of singletons is a network for any space.
If $\mathcal{N}$ is a network on $X$ and $f:X\rightarrow Y$ is surjective, then $f(\mathcal{N})=\{f(N)\mid N\in\mathcal{N}\}$ is a network on $Y$. Since every base is a network and every separable metric space has a countable base, it follows that every Suslin space has a countable network. In fact, there is a converse to this.
Proposition A $T_0$ space $X$ has a countable network if and only if it is the continuous image of a separable metric space
Proof Sufficiency has been dealt with above so suppose that $\mathcal{N}$ is a countable network on $X$. Then the family $\mathcal{B}$ of all finite intersections of members of $\mathcal{N}$ is a base for a new, larger topology on $X$. Let $X'$ be the underlying set of $X$ equipped with this new topology. Since $\mathcal{B}$ is countable, $X'$ is second-countable. Since $X$ is $T_0$ and $\mathcal{N}$ is a network, $X'$ is $T_0$. Thus by a result of Ponomarev [Eng, Ex.4.2.D], $X'$ is the image of a separable metric space. Since $X'$ maps bijectively onto $X$, $X$ is the image of a separable metric space. $\quad\blacksquare$
The metrisability theorem I've stated above will actually be valid for any Hausdorff space with a countable network.
Recall that the weight $w(Y)$ of a space $Y$ is the least cardinality of a base for $Y$.
Proposition Let $X$ be a functionally Hausdorff space and $\mathcal{N}$ a network for $X$. Then there is a Tychonoff space $Y$ of weight $w(Y)\leq|\mathcal{N}|$ and a continuous bijection $f:X\rightarrow Y$.
Proof Let
$$\mathcal{P}=\{(M,N)\in\mathcal{N}\times\mathcal{N}\mid M,N\;\text{are completely separated}\}.$$
Since $X$ is functionally Hausdorff, distinct points have disjoint cozero neighbourhoods, so $\mathcal{P}$ is nonempty. For each $(M,N)\in\mathcal{P}$ let $\varphi_{(M,N)}:X\rightarrow I$ be a continuous function such that $M\subseteq\varphi_{(M,N)}^{-1}(0)$ and $N\subseteq\varphi_{(M,N)}^{-1}(1)$. Then the family $\mathcal{F}=\{\varphi_{(M,N)}\mid (M,N)\in\mathcal{P}\}$ separates the points of $X$, so the canonical mapping $X\rightarrow\prod_{\varphi\in\mathcal{F}}I$ is a continuous injection. Letting $Y$ be its image grants $w(Y)\leq w(\prod_{\varphi\in\mathcal{F}}I)=|\mathcal{F}|=|\mathcal{N}|$. $\qquad\blacksquare$
The Urhysohn Metrisation Theorem states that a regular Hausdorff space with a countable base is metrisable. Thus we have the following result.
Corollary A functionally Hausdorff space with a countable network condenses onto a separable metric space. $\quad\blacksquare$
Turning to other aspects of the problem, we'll use the following.
Proposition If $Y$ is a paracompact space and $f:X\rightarrow Y$ a perfect mapping, then $X$ is paracompact.
Proof The proof is brief and given in [Eng, Th.5.1.35]. $\quad\blacksquare$
As a consequence, any Hausdorff space mapping onto a paracompact space via a perfect map will be paracompact, hence normal and functionally Hausdorff.
Finally, we need to mention the property of Čech-completeness. A Tychonoff space is Čech-complete if it is a $G_\delta$-subset of its Stone-Čech compactification. Other equivalent characterisations are given in [Eng, \S3.9], where the property is studied extensively. We need only two results from this section.
Theorem ([Eng, Th.4.3.26]) A metrisable space is completely metrisable if and only if it is Čech-complete. $\quad\blacksquare$
Theorem ([Eng, Th.3.9.10]) If $f:X\rightarrow Y$ is a perfect mapping between Tychonoff spaces, then $X$ is Čech-complete if and only if $Y$ is. $\quad\blacksquare$
Proof of the theorem We need only discuss sufficiency. Thus let $X$ be a Suslin space and suppose that $f:X\rightarrow M$ is a perfect mapping onto a metric space. By the corollary above, there is a metric space $N$ and a continuous injection $g:X\rightarrow N$. Let $\phi:X\rightarrow M\times N$ be the diagonal mapping $\phi(x)=(f(x),g(x))$. Then $\phi$ is a perfect mapping [Eng, Th.3.7.10], and hence a closed embedding. Since $M,N$ are metrisable, it follows that $X$ is metrisable.
In case $M$ is completely metrisable, then it is Čech-complete, and hence so is $X$ by the Theorem given above. But then $X$ is Čech-complete and metrisable, hence completely metrisable. $\quad\blacksquare$
Corollary A Suslin space $X$ is (completely) metrisable if and only if it embeds as a closed subspace in a product $K\times M$, where $K$ is compact Hausdorff and $M$ is separable and (completely) metrisable.
Proof If $X$ embeds in $K\times M$ as a closed subspace, then the composite $X\subseteq K\times M\xrightarrow{proj}M$ is perfect. $\quad\blacksquare$
I'll end with a few remarks. Morita's M-spaces are characterised by sequences of open covers. Another class of relevant spaces are Arkhangel'skii's p-spaces, which can be characterised either by sequences of open covers or by how they are situated in their Stone-Čech compactifications.
The classes of M-spaces and p-spaces are distinct, but it's known that a Hausdorff space is a paracompact M-space if and only if it is a paracompact p-space if and only if it is the perfect preimage of a metrisable space. Similarly, a Tychonoff space is Čech-complete and a paracompact if and only if it is the perfect preimage of a completely metrisable space [Eng, Ex.5.5.9]. Thus we have the following.
Corollary The following statements hold.
- A regular Suslin space is metrisable if and only if it is an M-space if and only if it is a p-space.
- A regular Suslin space is completely metrisable if and only if it is Čech-complete.
Proof The only thing we need to do is point that every Suslin space is Lindelöf, and that a regular Lindelöf space is paracompact [Eng, Th.5.1.2]. $\quad\blacksquare$
The following statements are more easily proved directly using the fact that a countably compact paracompactum is compact, and that a locally metrisable paracompactum is metrisable. I'll include them here as corollaries to indicate how the statements above are natural generalisations of more direct metrisation results.
Corollary Let $X$ be a Suslin space.
- Any regular, countably compact subspace of $X$ is compact and metrisable.
- Any locally compact subspace of $X$ is metrisable.
Proof $1.$ A countably compact space is an M-space. $2.$ A locally compact space Hausdorrff space is Čech-complete. $\quad\blacksquare$
References
[Eng] R. Engelking, General Topology, Revised edition, Sigma series in pure mathematics 6, Heldermann (1989).
