In the article "Categories and Modalities" from the book Categories for the Working Philosopher Kohei Kishida starts by describing the following three adjoint functors, each of the top functors giving rise to a free structure.
The simplest one is $\mathrm{Sn_τ}$ which takes any set to the free algebra of that set in the category of τ algebras. We can follow the usual mapping of an algebraic signature τ containing operators of various arities, to an endofunctor with the formula:
$P_\tau: \mathbf{Sets} \to \mathbf{Sets}:: S \mapsto \sum_{*\in \tau}S^{ar(*)} $
where $ar(*)$ returns the arity of each connective * in $\tau$.
We have as objects of $\mathbf{\tau Alg}$ the algebra given by the functions $f: P_\tau(X) \to X$ and as morphism between these objects the homomorphisms between these algebras. I.e. given two algebras $f: F(X) \to X$ and $g: F(Y) \to Y$ we have a function $m: X \to Y$ that makes the diagram commute, i.e. $f;m = F(m);g$.
For a Boolean algebra with operators {⋀,⋁,¬,⊤,⊥} the functor would be
${P_\tau}: Sets \to Sets:: X \mapsto X \times X + X \times X + X + 1 + 1$
for the binary operators $\land, \lor$ the unary operator $\lnot$ and the constants $\top$ and $\bot$. The Free functor $Sn_\tau$ then just takes a set to all the sentences that can be constructed by combining those operators on the set.
The question I have is with the next functor. It takes these algebras to another category $\mathbf{BpA}$ of algebras. The way to do that is to take the natural entailment relation $\vdash$ satisfying the laws of ...
Ok, but then what type of structure is that? How is the ordering encoded in the types here? Does each X have to be some pre-ordered thing?
I suppose I don't really understand what the Functor that determines the category of algebras BpA is.
Initially I was not quite sure if the turnstile ⊢ was meant to be part of the BpA Algebra.
Following conversation with Alex Kruckman below, it is clear that BpA adds the turnstile ⊢. That makes it easier to phrase the question.
An algebraic category has to be built on an endofunctor, or else there could not be a function $f: P(X) \to X$
It does not seem right to just adjust the previous algebra Functor by adding space for a turnstile at the end, as that would allow turnstiles to be placed anywhere between two normal operators in the free algebra it seems.
${P_{BpA}}: Sets \to Sets:: X \mapsto X \times X + X \times X + X + 1 + 1 + X \times X$
Perhaps it could be and endofunctor on $\mathrm{Sn_\tau}$?
${P_{BpA}}: \mathrm{Sn_τ} \to \mathrm{Sn_τ}:: X \times X$?
so that it just takes the sentences from $\mathrm{Sn_τ}$ and adds the turnstile? Though in that case it may need to also have the crossed out turnstile ⊬ to tell when something does not follow? But I am not sure if that would work recursively, as one could then end up with sentences with turnstiles on both sides of a central turnstile...
I found an answer to this in Institution Independent Modal Theory in the section "PreOrder Algebra (PAO)", which agrees with the answer by Alex Kruckman in the discussion to this question. This answer is more generalas it deals with multisorted algebras, where the question asked here was about single sorted pre-algebras.
Still it looks like this would work here as a functor for Boolean Pre Algebras, with the following operations and their arities
${ \lor ↦ 2, \land ↦ 2, \lnot ↦ 1, \top ↦ 0, \bot ↦ 0}$
and so the following functor for each of those operations
${P_{BpA}}: \Bbb{P}re \to \Bbb{P}re:: X \mapsto X \times X + X \times X + X + 1 + 1$
So that a Boolean Pre Algebra category (BpA) is made up of functions of the type $f: {P_{BpA}}(X) \to X$, for each X in $\Bbb{P}re$, the initial such algebra being the free one, that orders sentences according to logical implication in the PreOrder.
The PreOrder rules are those of boolean logic but without equality, that is we have the order as $$ a \land b \vdash b \land a \\ a \lor (b \lor c) \vdash (a \lor b) \lor c \\ a \land (b \land c) \vdash (a \land b) \land c \\ a \land a \vdash a, a \lor a \vdash a \\ a \lor (a \land c) \vdash a, a \land (a \lor c) \vdash a \\ x \land \bot \vdash \bot \\ x \lor \top \vdash x \\ x \land \lnot x \vdash \bot, x \lor \lnot x \vdash \top \\ $$
And the same with the turnstiles turned around.