A necessary condition for S4-completeness?

Question

It is well-known that the modal logic S4 is complete with respect to the class of all finite quasi-trees (where we interpret the $\Box$ modality as topological interior, and topologize a quasi-tree with the up-set topology). It is also well-known that p-morphisms (open, continuous surjections) preserve modal validity. Thus, for any space $X$, the existence of p-morphisms from $X$ onto every finite quasi-tree is a sufficient condition for $X$ to be S4-complete. This technique can be used to establish, for example, McKinsey and Tarski's famous result that S4 is the logic of any dense-in-itself, metrizable space.

My question is:

Is this condition also necessary? Said differently: is there a space $X$ and a finite quasi-tree $Q$ such that $X$ is S4-complete but there exists no p-morphism $\rho: X \to Q$?

This seems like a natural question to ask, but I haven't had much luck in finding any discussion about it. Even just a pointer to the right body of literature would be very much appreciated.

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Addendum

Here I'll define my terms a little more carefully, and spell out the translation of my question in terms of the more standard Kripke semantics.

Recall that quasi-orders are sets equipped with reflexive, transitive binary relations, which is precisely the class of Kripke frames corresponding to S4. A quasi-order $Q = (Q,\leq)$ is called a quasi-tree if $Q/\sim$ is a tree, where $\sim$ is the equivalence relation on $Q$ defined by

$$x \sim y \iff x \leq y \textrm{ and } y \leq x.$$

As mentioned in the comments, there is a correspondence between quasi-orders and Alexandrov spaces, one direction of which is given by topologizing quasi-orders with the up-set topology. There is also a notion of a p-morphism between quasi-orders, nicely outlined by Wikipedia. A p-morphism between quasi-order corresponds to an open, continuous map between the corresponding Alexandrov spaces.

I use the phrase "$X$ is S4-complete" (perhaps somewhat idiosyncratically?) to mean that every formula validated by $X$ is provable in S4; equivalently, $X$ refutes all non-theorems of S4. It is known that if $Q$ is any quasi-order and for each finite quasi-tree $Q_{t}$ there exists a surjective p-morphism $\rho_{t}: Q \to Q_{t}$, then $Q$ is S4-complete. One can then ask:

Is the converse true? Does every S4-complete quasi-order Q admit maps $\rho_{t}$ as above?

If not, then a counter-example can be "lifted" into the topological setting, thus answering my original question. However, a positive answer to this question does not immediately resolve the topological version since the quantification in the topological version is over all spaces, rather than just the Alexandrov spaces. Nonetheless, I would be interested in an answer (or even a hint at an answer) to either question.

Answer 1

S4 is complete with respect to a Kripke frame or general frame or topological frame $F$ if and only if $F$ is an S4-frame and for every finite rooted S4-frame $G$, there exists a p-morphism of a generated subframe of $F$ onto $G$.

The left-to-right implication follows from the existence of Fine’s frame formulas: there is a formula $\alpha_G$ such that for any K4-frame $H$, $\alpha_G$ is refutable in $H$ if and only if there exists a p-morphism of a generated subframe of $H$ onto $G$. One way of constructing $\alpha_G$ is as follows. Assume that $r$ is a root of $G$, and let $R$ be the accessibility relation of $G$. We put

$$\alpha_G=\Box^+\biggl(\bigwedge_{\substack{i,j\in G\\\\i\ne j}}(p_i\to\neg p_j)\land\bigwedge_{\substack{i,j\in G\\\\i\mathrel Rj}}(p_i\to\Diamond p_j)\land\bigwedge_{i\in G}\Bigl(p_i\to\Box\bigvee_{i\mathrel Rj}p_j\Bigr)\biggr)\to\neg p_r,$$

where $\Box^+\phi=\phi\land\Box\phi$ (which is equivalent to $\Box\phi$ in S4; the formula above works for K4 as well). Let $\models$ be a valuation in $H$ such that $x\not\models\alpha_G$ for some $x\in H$. Let $H_x$ be the generated subframe of $H$ rooted at $x$. For every $y\in H_x$, there exists a unique $i\in G$ such that $y\models p_i$; put $f(y)=i$. Then $f\colon H_x\to G$ is a p-morphism such that $f(x)=r$. Conversely, given such a p-morphism, one can construct a valuation refuting $\alpha$ by reversing the process.

Now, if $G$ is an S4-frame, it is a p-morphic image of itself, hence $\alpha_G$ is not valid in $G$, and a fortiori it is not an S4-tautology. Thus, any $H$ wrt which S4 is complete must also refute $\alpha_G$, hence there exists a p-morphism from a generated subframe of $H$ onto $G$.

Since every finite rooted S4-model is a p-morphic image of a finite quasi-tree, one can restrict attention to such $G$’s.

I’m not quite familiar with the topological semantics of S4, but I suppose the criterion translates to something to the effect of: S4 is complete wrt a space $X$ iff for every finite quasi-tree $G$, there exists an open subset $U\subseteq X$ and a p-morphism (whatever that means when the space is not ordered) of $U$ onto $G$.

Answer 2

The converse does not hold in general. Thanks to Nick and Guram Bezhanishvili and David Gabelaia for coming up with a counterexample.

Let $X$ denote the topological sum of $\mathbb{R}$ and a single point space, $\{x\}$. Since $\mathbb{R}$ sits as an open subspace of $X$ and $\mathbb{R}$ is S4-complete, $X$ is also S4-complete. On the other hand, since $x$ is isolated in X, there is no hope of finding an open map which sends $x$ to a 2-element cluster (since neither point in such a cluster is itself isolated).

In terms of Kripke frames (i.e. quasi-orders), an analogous counter-example can be constructed by starting with some S4-complete quasi order $Q$ (for example, the infinite binary tree), and adjoining to it a new element $x$ which is incomparable to any of the original elements of $Q$.

There is a sense in which these counter-examples are unsatisfying, because they work essentially by exhibiting objects which have two entirely separate "pieces": one piece to satisfy S4-completeness, and the other to block the existence of certain open, continuous maps. A refinement of the question may therefore be in order, though such a refinement is still at the "pencil and paper" stage, and not quite the "post on MO" stage.