Proof. In fact ${\operatorname {\mathrm {\mathbf {Rel}}}^{\mathbf {G}}}$ can be presented as the category of separated presheaves on a site $(\mathbf {C}, J)$ as follows (that separated presheaves are a quasi-topos is proved in [Reference Johnstone27, Section A2.6]). There are two kinds of objects in $\mathbf {C}$ : for each vertex $\alpha \in \mathbf {G}$ , there is an object $[\alpha ]$ in $\mathbf {C}$ ; for each edge in $\mathbf {G}$ , there is an object $[k]$ in $\mathbf {C}$ . For each edge in $\mathbf {G}$ , there are two arrows $[\alpha ]\longrightarrow [k]$ and $[\beta ]\longrightarrow [k]$ in $\mathbf {C}$ . These are the only non-identity arrows. The topology J consists of all the identity coverings, plus the coverings of the form $[\alpha ]\longrightarrow [k]$ and $[\beta ]\longrightarrow [k]$ for each edge in $\mathbf {G}$ . These coverings satisfy the axioms for a Grothendieck topology (see [Reference Johnstone27, Definition 2.1.9]).

A presheaf $X : \mathbf {C}^{\mathrm {op}} \longrightarrow \mathbf {Set}$ is separated iff $X([k])\subseteq X([\alpha ])\times X([\beta ])$ for each edge in $\mathbf {G}$ . This means that X is given by a family of sets $X([\alpha ])$ , plus a relation $X([k])$ for every edge k in $\mathbf {G}$ .

Proof. Trivially, limits and factorizations are all preserved because they are computed pointwise in powers of $\mathbf {Set}$ . $ \mathcal {M}$ -conservativity (but not conservativity!) is ensured by Proposition 4.2: if $S\buildrel {s}\over \hookrightarrow X$ is in $ \mathcal {M}$ but is not an iso, then $S\neq \top $ in $\operatorname {\mathrm {Sub}}_{ \mathcal {M} }(X)$ , hence there is an X-type P such that $S\not \in P$ by Lemma 4.1. If P is realized in M by some $a\in M(X)$ , we have that $a\not \in \lbrack {\kern-1pt}\lbrack S\rbrack {\kern-1.3pt}\rbrack _M$ , which means that the image of $M(s)$ is not $M(X)$ . Thus $ev_{({\mathbf {E}}, \mathcal {M})}(s)$ is not iso because its M-component is not such.

Proof. The proof is a simple check, using the stability of $ \mathcal {M}$ under pullbacks and the fact arrows in $ \mathcal {M}$ compose.

Proof. This is due to the continuity and subspace conditions of Definition 3.1. Suppose in fact that we have $a {\tilde {R}}b$ so that there exists the data mentioned in the statement of Lemma 4.7. Suppose that $b\in \lbrack {\kern-1pt}\lbrack U\rbrack {\kern-1.3pt}\rbrack _N$ for some $U\buildrel {u}\over {\hookrightarrow } Y$ . From $f(b')=b$ we obtain

$$ \begin{align*}b' \in f_N^*\lbrack {\kern-1.3pt}\lbrack U\rbrack {\kern-1.3pt}\rbrack_N = \lbrack {\kern-1.3pt}\lbrack f^* U\rbrack {\kern-1.3pt}\rbrack_{N}; \end{align*} $$

thus

$$ \begin{align*}s(b') \in \exists_{s_N} \lbrack {\kern-1.3pt}\lbrack f^* U\rbrack {\kern-1.3pt}\rbrack_{N}= \lbrack {\kern-1.3pt}\lbrack \exists_{s} f^* U\rbrack {\kern-1.3pt}\rbrack_{N}. \end{align*} $$

Since R satisfies (16) and $s(a')~R~s(b')$ , we obtain

$$ \begin{align*}s(a') \in \lbrack {\kern-1.3pt}\lbrack \Diamond\exists_{s} f^* U\rbrack {\kern-1.3pt}\rbrack_{M} \end{align*} $$

and

$$ \begin{align*}a' \in {s}_M^* \lbrack {\kern-1.3pt}\lbrack \Diamond\exists_{s} f^* U \rbrack {\kern-1.3pt}\rbrack_{M}= \lbrack {\kern-1.3pt}\lbrack s^*\Diamond\exists_{s} f^* U \rbrack {\kern-1.3pt}\rbrack_{M}. \end{align*} $$

By subspace and continuity conditions from Definition 3.1, we have

$$ \begin{align*}a' \in \lbrack {\kern-1.3pt}\lbrack s^* \Diamond\exists_{s} f^* U\rbrack {\kern-1.3pt}\rbrack_{M}= \lbrack {\kern-1.3pt}\lbrack \Diamond f^* U\rbrack {\kern-1.3pt}\rbrack_{M} \subseteq \lbrack {\kern-1.3pt}\lbrack f^*\Diamond U\rbrack {\kern-1.3pt}\rbrack_{M}= f_M^* \lbrack {\kern-1.3pt}\lbrack \Diamond U\rbrack {\kern-1.3pt}\rbrack_{M}. \end{align*} $$

Now, $f(a')=a$ yields $a\in \lbrack {\kern-1pt}\lbrack \Diamond U\rbrack {\kern-1.3pt}\rbrack _{M}$ , as desired.

Proof. Suppose that R satisfies (13). The only element of $\mathbf {1}_M \times \mathbf {1}_N$ (here $\mathbf {1}$ is the terminal object of ${\mathbf {E}}$ ) is in $\tilde {R}$ because it is in R. The fact that $\tilde {R}$ is closed under binary products is seen by observing that the product of two diagrams of the shape presented in Lemma 4.7 is again of the same type, because $s_1 \times s_2 \in {\mathcal {M}}$ whenever $s_1, s_2 \in {\mathcal {M}}$ .

Suppose that R is a singleton and contains only $(a,b) \in X_M \times X_N$ . Let $R_p$ be the closure of R under (13). We first show that $R_p \subseteq \tilde {R}$ . Indeed, for any $n\geq 0$ , the image of $(a,b)$ by the diagonal map $X_M \times X_N \longrightarrow X_M^n \times X_N^n$ is in $\tilde {R}$ , and $R_p$ consists precisely of all these elements. Since $R_p \subseteq \tilde {R}$ , we obtain that $\tilde {R_p} = \tilde {R}$ because $\tilde {R_p} \subseteq \tilde {\tilde {R}} = \tilde {R}$ . Hence $\tilde {R_p}$ satisfies (13) by the first part of this lemma.

Proof. We must prove that the evaluation functor is $ \mathcal {M}$ -conservative and preserves all the operations. $ \mathcal {M}$ -conservativity is proved in the same way as in Theorem 4.3.

$\boldsymbol {ev}_{({{\mathbf{E}}}, \mathcal {M})}$ preserves finite limits. For any coherent model M, the functor ${X \mapsto M(X)}$ preserves finite limits. For any modal transformation $R \subseteq M\times N$ , the functor ${X \mapsto R(X)}$ preserves finite limits. Since limits in $\mathbf {Rel^{G({\mathbf {E}}, \mathcal {M})}}$ are computed pointwise, this shows that $ev_{({\mathbf {E}}, \mathcal {M})}$ preserves finite limits.

$\boldsymbol {ev}_{({{\mathbf{E}}}, \mathcal {M})}$ preserves the left class of the factorization system. The left class of the factorization system of $\mathbf {Rel^{G({\mathbf {E}}, \mathcal {M})}}$ consists of all the morphisms whose M-component is surjective for each coherent model M. If $f : X\longrightarrow Y$ is a morphism in the left class of ${\mathbf {E}}$ , then $M(f) : M(X) \longrightarrow M(Y)$ is surjective for each coherent model M. We get that $ev_{({\mathbf {E}}, \mathcal {M})}(f)$ is in the left class of $\mathbf {Rel^{G({\mathbf {E}}, \mathcal {M})}}$ .

$\boldsymbol {ev}_{({{\mathbf{E}}}, \mathcal {M})}$ preserves the right class of the factorization system. The right class of the factorization system of $\mathbf {Rel^{G({\mathbf {E}}, \mathcal {M})}}$ consists of all the morphisms $g : V\longrightarrow W$ such that $g_M : V_M \longrightarrow W_M$ is injective for each coherent model M, and moreover such that for each modal transformation $R \subseteq M\times N$ between coherent models, ${V_R \subseteq V_M \times V_N}$ is the restriction of $W_R \subseteq W_M \times W_N$ via $g_M \times g_N$ . If $f:X\longrightarrow Y$ is a morphism in the right class $ \mathcal {M}$ of ${\mathbf {E}}$ , then $f_M:X_M \longrightarrow Y_M$ is injective for each coherent model M. Moreover, the reflection condition (15) for R imposes that $X_R \subseteq X_M\times X_N$ is the restriction of $Y_R \subseteq Y_M\times Y_N$ via $f_M\times f_N$ . We obtain that $ev_{({\mathbf {E}}, \mathcal {M})}(f)$ is in the right class of $\mathbf { Rel^{G({\mathbf {E}}, \mathcal {M})}}$ .

$\boldsymbol {ev}_{({{\mathbf{E}}}, \mathcal {M})}$ preserves the modality. One side of the preservation condition is (16), the other side is a converse of (16), namely

(17) $$ \begin{align} a\in \lbrack {\kern-1.3pt}\lbrack \Diamond S\rbrack {\kern-1.3pt}\rbrack_M ~~\Rightarrow~~{\text{there are}}~N, b, R~\mathrm{s.t.}~ aRb ~\mathrm{and}~b\in \lbrack {\kern-1.3pt}\lbrack S\rbrack {\kern-1.3pt}\rbrack_N \end{align} $$

for every M, X, $S\in \operatorname {\mathrm {Sub}}_{ \mathcal {M}}(X)$ and $a \in X_M$ . To prove it, pick such $a\in \lbrack {\kern-1pt}\lbrack \Diamond S\rbrack {\kern-1.3pt}\rbrack _M$ and consider the following subsets of $\operatorname {\mathrm {Sub}}_{ \mathcal {M}}(X)$ :

(18) $$ \begin{align} \Gamma~=~\{ S\}, \qquad \Delta~=~\{ T\mid a\not\in \lbrack {\kern-1.3pt}\lbrack \Diamond T\rbrack {\kern-1.3pt}\rbrack_M\}. \end{align} $$

Clearly $\Gamma $ is closed under finite meets and $\Delta $ is closed under finite joins (because $\Diamond $ commutes with joins) and downward-closed (because $\Diamond $ preserves the lattice ordering); in addition $\Gamma \cap \Delta =\emptyset $ since $a\in \lbrack {\kern-1pt}\lbrack \Diamond S\rbrack {\kern-1.3pt}\rbrack _M$ . By Proposition (4.2), there are a model N and $b\in X_N$ realizing a prime filter extending $\Gamma $ and disjoint from $\Delta $ . This means in particular that $b\in \lbrack {\kern-1pt}\lbrack S\rbrack {\kern-1.3pt}\rbrack _N$ and that the family of relations ${R=\{ R(Y)\subseteq Y_M\times Y_N\}_Y}$ containing just the singleton pair $(a, b) \in R(X)$ satisfy condition (16). Now it is sufficient to apply Lemmas 4.8 and 4.9.

Proof. (SS $\Rightarrow $ IS): Suppose that $h:X\longrightarrow Y$ is in ${\mathcal {M}}^\bot $ , is mono and stably open. Take Z and $S\in \operatorname {\mathrm {Sub}}_ {\mathcal {M}} (X\times Z)$ ; we have

$$ \begin{align*}\Diamond S ~=~\Diamond (h\times 1_Z)^* \exists_{h\times 1_Z} S~=~(h\times 1_Z)^*\Diamond \exists_{h\times 1_Z} S \end{align*} $$

by first-order reasoning and stable openness. Thus $h\in {\mathcal {M}}$ by (SS) and consequently h is iso.

(IS $\Rightarrow $ SS): Let $m:A\longrightarrow X$ be a brittle subspace; let us factor it as $A \buildrel q\over \longrightarrow Q \buildrel i \over \longrightarrow X$ with $i\in \mathcal {M}$ and $q\in {\mathcal {M}}^\bot $ . We show that q is stably open (so that (IS) applies, q is iso and then $m\in \mathcal {M}$ ). Take Z and $S\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Q\times Z)$ ; we have by (6)

(first-order reasoning) $$ \begin{align} (q\times 1_Z)^*\Diamond S &= (q\times 1_Z)^*(i\times 1_Z)^*\Diamond \exists_{i\times 1_Z} S \nonumber\\&= (m\times 1_Z)^*\Diamond \exists_{i\times 1_Z} S \nonumber\\&= (m\times 1_Z)^*\Diamond \exists_{i\times 1_Z} \exists_{q\times 1_Z} (q\times 1_Z)^* S \end{align} $$

(by the hypothesis on m) $$\begin{align} &= (m\times 1_Z)^*\Diamond \exists_{m\times 1_Z} (q\times 1_Z)^* S \ \ \ \qquad\qquad\quad \nonumber\\&= \Diamond (q\times 1_Z)^* S. \end{align} $$

This shows that $q\times 1_Z$ is open.

(FS $\Rightarrow $ IS): Let $h:X\longrightarrow Y$ be a brittle isomorphism; consider R defined as

$$ \begin{align*}R:= X\buildrel{(h, 1_X)}\over\longrightarrow Y\times X. \end{align*} $$

Notice that $\pi _Y\circ R=h$ is mono and belongs to ${\mathcal {M}}^\bot $ by our hypotheses on h, so R is a functional relation; if we manage to prove that R is also continuous, then by functional saturation there is $f:Y\longrightarrow X$ such that $R=(h,1_X) \simeq (1_Y,f)$ , which means that h is iso with inverse f.

Now notice that, by first-order reasoning, for every Z and $S\in \operatorname {\mathrm {Sub}}_{ \mathcal {M}}(X\times Z)$ we have that $R\circ S$ is equal to $\exists _{h\times 1_Z}(S)$ . As a consequence, the continuity of our R can be expressed as the “Barcan formula” for $h\times 1_Z$

$$ \begin{align*}\Diamond \exists_{h\times 1_Z}S \leq \exists_{h\times 1_Z} \Diamond S, \end{align*} $$

which is an immediate consequence of the stable openness of h and the fact that h (and so also $h\times 1_Z$ ) satisfies the hypotheses of Lemma 5.2 below.

(IS $\Rightarrow $ FS): Let $R\buildrel {r}\over \hookrightarrow X\times Y$ be functional and continuous. We show that $R\buildrel {r}\over \hookrightarrow X\times Y \buildrel {\pi _X}\over \longrightarrow X$ is stably open: then (IS) applies and $\pi _X\circ r$ has an inverse $h:X\longrightarrow R$ witnessing the isomorphism of R with the graph of $\pi _Y\circ r\circ h$ . In order to show that $\pi _X\circ r$ is stably open, we use Lemma 5.2 and prove the “Barcan formula” for $(\pi _Y\circ r) \times 1_Z$ for every Z. To this aim, let $S\in \operatorname {\mathrm {Sub}}_{ \mathcal {M}}(R\times Z)$ ; we have

$$ \begin{align*}\Diamond \exists_{(\pi_X\circ r)\times 1_Z} S~=~ \Diamond \exists_{\pi_{X\times Z}}\exists_{r\times 1_Z} S~ =~ \Diamond \exists_{\pi_{X\times Z}} (\pi_{X\times Y}^*R\wedge \exists_{r\times 1_Z} S) \end{align*} $$

by first-order reasoning and

$$ \begin{align*}\Diamond \exists_{\pi_{X\times Z}} (\pi_{X\times Y}^*R\wedge \exists_{r\times 1_Z} S) ~\leq~ \exists_{\pi_{X\times Z}} (\pi_{X\times Y}^*R\wedge \Diamond \exists_{r\times 1_Z} S) \end{align*} $$

by Lemma 5.3 below. We now continue by first-order reasoningFootnote ⁴ and obtain

$$ \begin{align*}\exists_{\pi_{X\times Z}} (\pi_{X\times Y}^*R\wedge \Diamond \exists_{r\times 1_Z} S) = \exists_{\pi_{X\times Z}} \exists_{r\times 1_Z}({r\times 1_Z})^*\Diamond \exists_{r\times 1_Z} S, \end{align*} $$

which is equal to $\exists _{(\pi _X\circ r)\times 1_Z}\Diamond S$ , as required.

Proof. Trivial, by first-order reasoning, taking in mind that $\exists _h$ is the inverse correspondence of $h^*$ .

Proof. Observe that, by first-order reasoning or directly by applying the Beck–Chevalley condition to a suitable pullback square, we have

(24) $$ \begin{align} \exists_{\pi_{X\times \tilde Z}}((\pi_{X\times Y}^*R)\wedge U) = (\Delta_X \times 1_{\tilde Z})^*(R\circ U) \text{.} \end{align} $$

Thus we get

(by continuity of R) $$\begin{align} \Diamond \exists_{\pi_{X\times \tilde Z}} (\pi_{X\times Y}^*R\wedge U) &= \Diamond (\Delta_X \times 1_{\tilde Z})^*(R\circ U) \nonumber\\ &\leq (\Delta_X \times 1_{\tilde Z})^* \Diamond (R\circ U) \nonumber\\ &\leq (\Delta_X \times 1_{\tilde Z})^* (R \circ \Diamond U) \\ &= \exists_{\pi_{X\times \tilde Z}} (\pi_{X\times Y}^*R\wedge \Diamond U).\nonumber\end{align} $$

Proof. The right to left direction is trivial. For the converse, let $\Phi $ be $ \mathcal {M}$ -conservative. Notice that, for $S,T\in \operatorname {\mathrm {Sub}}_{ \mathcal {M}_1}(X)$ , we have that “ $S\leq T$ and $\Phi (S)\simeq \Phi (T)$ imply $S\simeq T$ ”: this is because of Lemma 2.4 and $ \mathcal {M}$ -conservativity. Suppose now that m is mono and that $\Phi (m)$ is iso. Then $m\in {\mathcal {M}}^\bot $ because the inequality $\exists _m (\top ) \leq \top $ becomes an equality in ${\mathbf {E}}_2$ , so that $\exists _m (\top ) \simeq \top $ . For the same reason, $m\times 1_Z$ is open for all Z because $\Phi (m)\times 1_{\Phi (Z)}$ is open being an iso. Thus m is a brittle isomorphism and so it is iso since $({\mathbf {E}}_1, \mathcal {M}_1)$ is saturated.

Proof. We find first $a'$ , $b'$ and finally $R'$ .

Construction of $\mathbf {a'}$ . Let $\mathcal S$ be a lex subfunctor of M such that $\sum _{Z\in {\mathbf {E}}} \mathcal {S}(Z)$ has cardinality at most $\mathrm {card}{\,{\mathbf {E}}}$ and comprises all c such that there is $c'$ with $cR c'$ . Given $f:Z\longrightarrow V$ , $A\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times Z)$ and $B \in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times V)$ we write

$$\begin{align*}F(f,A,B) = (q \times 1_Z)^*(A) \land (1_X\times f)^*\Diamond(q\times 1_V)^*(B) \text{.} \end{align*}$$

For $c\in \mathcal {S}(Z)$ , let $\Theta _c$ be the set of the subobjects $S\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(X\times Z)$ such that there exist $f:Z\longrightarrow V$ , $d\in V_N$ , $A\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times Z)$ , $B\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times V)$ such that $f(c)Rd$ , $S=F(f,A,B)$ , $(a,c)\in \lbrack {\kern-1pt}\lbrack A\rbrack {\kern-1.3pt}\rbrack _M$ and $(b,d)\in \lbrack {\kern-1pt}\lbrack B\rbrack {\kern-1.3pt}\rbrack _N$ . This defines an X-pretype $\Theta $ over $\mathcal {S}$ because:

1. 1. $\Theta _c$ is a prefilter on $\operatorname {\mathrm {Sub}}_{\mathcal {M}}(X\times Z)$ and (PT0) is satisfied: in fact, if we have $f_i:Z\longrightarrow V_i$ , $d_i\in N_{V_i}$ , $A_i\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times Z)$ , $B_i\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times V_i)$ such that $f_i(c)Rd_i$ , $S_i= F(f_i, A_i,B_i)$ , $(a,c)\in \lbrack {\kern-1pt}\lbrack A_i\rbrack {\kern-1.3pt}\rbrack _M$ and $(b,d_i)\in \lbrack {\kern-1pt}\lbrack B_i\rbrack {\kern-1.3pt}\rbrack _N$ (for $i=1,2)$ , then taking $V:=V_1\times V_2$ , $f:=(f_1,f_2)$ ,

   $$ \begin{align*}~A:=A_1\wedge A_2,~ B:=(\pi_{Y\times V_1})^*B_1\wedge (\pi_{Y\times V_2})^*B_2,\end{align*} $$
   we have by continuity of pullbacks along projections
   $$ \begin{align*}F(f,A,B) \leq F(f_1,A_1,B_1) \wedge F(f_2,A_2,B_2) \end{align*} $$
   and moreover $f(c)=(f_1(c),f_2(c))R(d_1,d_2)$ by (13), $(a,c)\in \lbrack {\kern-1pt}\lbrack A\rbrack {\kern-1.3pt}\rbrack _M$ and $(b,d_1,d_2)\in \lbrack {\kern-1pt}\lbrack B\rbrack {\kern-1.3pt}\rbrack _N$ .

2. 2. If $(c',c)\in {\mathcal S}(Z'\times Z)$ and $F(f,A,B)\in \Theta _c$ with $f:Z\longrightarrow V$ and the required data, we have that $(1_X \times \pi _Z)^* F(f,A,B) = F(\pi _Z f , \pi _{Y\times Z}^* A,B)\in \Theta _{c,c'}$ with the obvious data, so that (PT1) is satisfied.

3. 3. If $C \in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Z)$ and $c\in \lbrack {\kern-1pt}\lbrack C\rbrack {\kern-1.3pt}\rbrack _M$ , then $\pi _Z^*C \in {\uparrow } \Theta _c$ : to see this, take ${f: Z\longrightarrow \mathbf {1}}$ , $A:=\pi _Z^*C,$ and $B=\top $ . Thus (PT2) is satisfied.

The consistency of $\Theta $ is established as follows: if we have (under the above mentioned data)

$$ \begin{align*}F(f,A,B) = (q \times 1_Z)^*(A) \land (1_X\times f)^*\Diamond(q\times 1_V)^*(B) \leq \bot \end{align*} $$

then applying $\exists _{q\times 1}$ to this inequality, using the Frobenius condition, the Beck–Chevalley condition and finally Definition 6.1, we get

$$ \begin{align*}A \land (1_Y\times f)^*\Diamond B \leq \bot, \end{align*} $$

which is impossible since $(a,c)\in \lbrack {\kern-1pt}\lbrack A\rbrack {\kern-1.3pt}\rbrack _M$ , $(b,d)\in \lbrack {\kern-1pt}\lbrack B\rbrack {\kern-1.3pt}\rbrack _N$ , $aRb$ , $f(c) R d,$ and so $(a,f(c))\in \lbrack {\kern-1pt}\lbrack \Diamond B\rbrack {\kern-1.3pt}\rbrack _M$ , because R satisfies (13) and (16).

Thus $\Theta $ is realized in M by some $a'\in M(X)$ . Taking in particular f to be the identity arrow and A to be the equality over Y in the above data, we see that $ q(a')=a$ and that whenever we have $c R d$ and $(b,d)\in \lbrack {\kern-1pt}\lbrack B\rbrack {\kern-1.3pt}\rbrack _N$ we have also $(a',c)\in \lbrack {\kern-1pt}\lbrack \Diamond (q\times 1)^* B\rbrack {\kern-1.3pt}\rbrack _M$ .

Construction of $\mathbf {b'}$ . Let now $\mathcal S'$ be a lex subfunctor of N such that $\sum _{Z\in {\mathbf {E}}} \mathcal {S}'(Z)$ has cardinality at most $\mathrm {card}{\,{\mathbf {E}}}$ and comprises all d such that there is c with $cR d$ . Given $f:Z\longrightarrow V$ and $A\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(X\times V)$ , $B\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times Z)$ we write

$$\begin{align*}G(f,A,B) = (1_X \times f)^*(A) \land (q\times 1_Z)^*(B) \text{.} \end{align*}$$

For $d\in \mathcal {S}'(Z)$ , let $\Psi _d$ be the set of subobjects $S\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(X\times Z)$ such that there exist $f:Z\longrightarrow V$ , $c\in V_M$ , $A\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(X\times V)$ , $B\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times Z)$ such that $cRf(d)$ , $S=G(f,A,B)$ , $(a',c)\in \lbrack {\kern-1pt}\lbrack \Box A\rbrack {\kern-1.3pt}\rbrack _M$ and $(b,d)\in \lbrack {\kern-1pt}\lbrack B\rbrack {\kern-1.3pt}\rbrack _N$ . This defines an X-pretype $\Psi $ over $\mathcal {S}'$ because:

1. 1. $\Psi _d$ is a prefilter on $\operatorname {\mathrm {Sub}}_{\mathcal {M}}(X\times Z)$ and (PT0) is satisfied: in fact, if we have $f_i:Z\longrightarrow V_i$ , $c_i\in M_{V_i}$ , $A_i\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(X\times V_i)$ , $B_i\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Y\times Z)$ such that $c_iRf_i(d)$ , $S_i=G(f_i,A_i,B_i)$ , $(a',c_i)\in \lbrack {\kern-1pt}\lbrack \Box A_i\rbrack {\kern-1.3pt}\rbrack _M$ and $(b,d)\in \lbrack {\kern-1pt}\lbrack B_i\rbrack {\kern-1.3pt}\rbrack _N$ (for $i=1,2$ ), then taking $f:=(f_1,f_2)$ ,

   $$ \begin{align*}~A:=(\pi_{X\times V_1})^*A_1\wedge (\pi_{X\times V_2})^*A_2,~ B:=B_1\wedge B_2,\end{align*} $$
   we have $G(f,A,B)\leq G(f_1,A_1,B_1) \wedge G(f_2,A_2,B_2)$ and moreover $(c_1,c_2)Rf(d)$ , $(b,d)\in \lbrack {\kern-1pt}\lbrack B\rbrack {\kern-1.3pt}\rbrack _N$ and finally $(a',c_1,c_2)\in \lbrack {\kern-1pt}\lbrack \Box A\rbrack {\kern-1.3pt}\rbrack _M$ because
   $$ \begin{align*}(\pi_{X\times V_1})^* \Box A_1\wedge (\pi_{X\times V_2})^*\Box A_2 \leq \Box ((\pi_{X\times V_1})^*A_1\wedge (\pi_{X\times V_2})^*A_2) \end{align*} $$
   by continuity.

2. 2. If $(d',d)\in {\mathcal S}(Z'\times Z)$ and $G(f,A,B)\in \Psi _d$ with $f:Z\longrightarrow V$ and the required data, then $(1_X\times \pi _Z)^*G(f,A,B) = G(\pi _Z f,A,\pi _{Y\times Z}^* B)\in \Psi _{d,d'}$ with the obvious data, so that (PT1) is satisfied.

3. 3. If $D {\kern-1pt}\in{\kern-1pt} \operatorname {\mathrm {Sub}}_{\mathcal {M}}(Z)$ and $d{\kern-1pt}\in{\kern-1pt} \lbrack {\kern-1pt}\lbrack D\rbrack {\kern-1.3pt}\rbrack _M$ , then $\pi _Z^*D \in {\uparrow } \Psi _d$ : to see this, take ${f {\kern-1pt}:{\kern-1pt} Z \longrightarrow \mathbf {1}}$ , c the unique element of $M_{\mathbf {1}}$ , $A=\top $ and $B=\pi _Z^*D$ . Thus (PT2) is satisfied.

The consistency of $\Psi $ is established as follows: if we have (under the above mentioned data)

$$\begin{align*}(1_X \times f)^*(A) \land (q\times 1_Z)^*(B) \leq \bot \end{align*}$$

then applying $\exists _{1_X\times f}$ and using the Frobenius and the Beck–Chevalley conditions, we get

$$ \begin{align*}A\wedge (q\times 1_V)^*\exists_{1_Y\times f} B\leq \bot \end{align*} $$

and hence

$$ \begin{align*}\Box A\wedge \Diamond (q\times 1_V)^*\exists_{1_Y\times f} B\leq \bot.\end{align*} $$

From $(b,d)\in \lbrack {\kern-1pt}\lbrack B\rbrack {\kern-1.3pt}\rbrack _N$ we get $(b,f(d))\in \lbrack {\kern-1pt}\lbrack \exists _{1_Y\times f} B\rbrack {\kern-1.3pt}\rbrack _N$ and from $cRf(d)$ and the construction of $a'$ , we obtain $(a',c)\in \lbrack {\kern-1pt}\lbrack \Diamond (q\times 1_V)^*\exists _{1_Y\times f} B\rbrack {\kern-1.3pt}\rbrack _M$ . This, together with $(a',c)\in \lbrack {\kern-1pt}\lbrack \Box A\rbrack {\kern-1.3pt}\rbrack _M$ yields a contradiction.

Thus $\Psi $ is consistent and realized in N by some $b'$ . Taking in particular B to be the equality over Y and f to be the identity in the above data, we see that $ q(b')=b$ and $(b',d) \in \lbrack {\kern-1pt}\lbrack A\rbrack {\kern-1.3pt}\rbrack _N$ implies $(a',c)\in \lbrack {\kern-1pt}\lbrack \Diamond A\rbrack {\kern-1.3pt}\rbrack _M$ whenever $cRd$ (remember that $\Diamond =\neg \Box \neg $ ).

Construction of $\mathbf {R'}$ . Let $R'$ be the family of relations defined as follows, for every Z and $c\in Z_M$ , $d\in Z_N$ :

$c R' d$ iff there exist $g:X\times Z'\longrightarrow Z, c'\in Z^{\prime }_M, d'\in Z^{\prime }_N$ such that $c' R d'$ and $c=g(a',c'), d=g(b',d')$ .

It is immediate to see that $R'$ satisfies conditions (13) and (16), so it is the desired $R'$ .

Proof. Pick a well ordering of the triples $(a,b,q)$ (where $q:X\longrightarrow Y$ is in $\mathcal {Q}$ , $a\in Y_M$ , $b\in Y_N$ and $aRb$ ) by some ordinal $\kappa $ (notice that the cardinality of $\kappa $ does not exceed $\mathrm {card}{\,{\mathbf {E}}}$ ). We define a sequence $(R_\xi \subseteq M\times N)_{\xi \leq \kappa }$ by:

• • $R_0$ is given;

• • $R_{\xi +1}$ is obtained by applying Lemma 6.5 to $R_\xi $ and the $\xi $ -th element of our well ordering;

• • $R_\lambda = \bigcup _{\xi <\lambda } R_\xi $ when $\lambda $ is a limit ordinal.

By induction, we have $\mathrm {card}{\,R_\xi } \leq \mathrm {card}{\,{\mathbf {E}}}$ for all $\xi $ . We then define $R^{\prime }_1$ as $\tilde {R_\kappa }$ , i.e., we apply the construction of Lemma 4.7 to ${R_\kappa }$ . By referring to the construction presented in Lemma 4.7, we see that $\mathrm {card}{\,R^{\prime }_1} \leq \mathrm {card}{\,{\mathbf {E}}}$ . Since $R_\kappa $ satisfies (13) and (16), we obtain that $R^{\prime }_1$ is a modal transformation satisfying all conditions (13)–(16) and also (26), but the latter only when $(a,b)$ is taken in $R_0 \subseteq R^{\prime }_1$ . To get the desired $R'$ it is now sufficient to repeat $\omega $ -times the above construction leading from $R_0$ to $R^{\prime }_1$ , and take the union.

Proof. Let $\lambda =\mathrm {card}{\,{\mathbf {E}}}^+$ be the successor of the cardinal $\mathrm {card}{\,{\mathbf {E}}}$ . Let $\mathbf {G}_\lambda ({\mathbf {E}})$ be the graph of coherent $\lambda $ -saturated models of the f-coherent category $({\mathbf {E}}, {\mathcal {M}})$ and modal transformations which respect quotients. We will show that the canonical evaluation functor

$$ \begin{align*}ev_{({\mathbf{E}}, {\mathcal{M}})} : {\mathbf{E}} \longrightarrow \operatorname{\mathrm{\mathbf{Rel}}}^{\mathbf{G}_\lambda({\mathbf{E}})}\end{align*} $$

is a morphism of Boolean modal categories. We know that it is ${\mathcal {M}}$ -conservative, since there are enough $\lambda $ -saturated models (see the extension of Proposition 4.2 to $\lambda $ -saturated models).

The only non-immediate thing that we must check is that whenever $M \in \mathbf {G}_\lambda ({\mathbf {E}})$ , for each $S \subseteq X$ in ${\mathbf {E}}$ and each $a \in \lbrack {\kern-1pt}\lbrack \Diamond S\rbrack {\kern-1.3pt}\rbrack _M$ there is some $N \in \mathbf {G}_\lambda ({\mathbf {E}})$ , a modal transformation $R \subseteq M\times N$ which respects quotients and some $b \in \lbrack {\kern-1pt}\lbrack S\rbrack {\kern-1.3pt}\rbrack _N$ such that $(a,b) \in R$ . We start by building $N \in \mathbf {G}_\lambda ({\mathbf {E}})$ and $b \in X_N$ such that $b \in \lbrack {\kern-1pt}\lbrack S\rbrack {\kern-1.3pt}\rbrack _N$ and for any $S' \subseteq X$ such that $b \in \lbrack {\kern-1pt}\lbrack S'\rbrack {\kern-1.3pt}\rbrack _N$ we have $a \in \lbrack {\kern-1pt}\lbrack \Diamond S'\rbrack {\kern-1.3pt}\rbrack _M$ . The reasoning is exactly the same as in the case without quotients. Lemma 6.6 now shows that the pair $(a,b) \in X_M \times X_N$ is contained in a modal transformation $R \subseteq M\times N$ respecting quotients.

Proof. For $S\in \operatorname {\mathrm {Sub}}_{\mathcal {M}}(A+B)$ , we have $S=S_1\vee S_2$ , with $S_1\leq A$ and ${S_2\leq B}$ . Then $\iota _A^* \Diamond S=\iota _A^* \Diamond S_1 \vee \iota _A^* \Diamond S_2=\iota _A^* \Diamond S_1$ by (27). On the other hand $\Diamond \iota _A^* S= \iota _A^*\Diamond \exists _{\iota _A}\iota _A^* S=\iota _A^* \Diamond S_1$ , by (6). The argument showing openness of $\iota _B$ is the same.

Proof. Immediate by Theorem 6.7.

Proof. Let $S\buildrel {s}\over \hookrightarrow X$ be in ${\mathcal {M}}$ . If we let $\iota _1, \iota _2$ be the two injections $X\rightrightarrows X+X$ , then $(s\iota _1, s\iota _2):S \hookrightarrow (X+X)^2$ is in ${\mathcal {M}}$ by Lemma 2.2 and its reflexive closure, i.e., its join in $\operatorname {\mathrm {Sub}}_{\mathcal {M}}((X+X)^2)$ with the diagonal, is an equivalence relation. Take its quotient $q: X+X\twoheadrightarrow Q$ and then the equalizer $E \buildrel {e}\over \hookrightarrow X$ of $\iota _1q$ and $\iota _2q$ . By the universal property, there is k such that $ke=s$ . Now $k\in {\mathcal {M}}$ as the first component of an arrow in ${\mathcal {M}}$ (see Lemma 2.4). However, it is not difficult to see that in every ${\mathcal {M}}$ -conservative embedding $\eta $ of $({\mathbf {E}}, {\mathcal {M}})$ into a category of the kind $\operatorname {\mathrm {\mathbf {Rel}}}^{\mathbf {G}}$ , we have that $\eta (k)$ is iso (this is because coproducts and quotients of equivalence relations are computed set-wise in $\operatorname {\mathrm {\mathbf {Rel}}}^{\mathbf { G}}$ ). Thus k is iso by Theorem 6.11 and so s is an equalizer.

Proof. (Ad (i)). Let

$$ \begin{align*}A\buildrel{\iota_A}\over \longrightarrow A+B \buildrel{\iota_B}\over\longleftarrow B \end{align*} $$

be a disjoint union and let

$$ \begin{align*}A\buildrel{f}\over \longrightarrow X \buildrel{g}\over\longleftarrow B \end{align*} $$

be further arrows as displayed. We must find a unique $h:A+B\longrightarrow X$ such that $h\circ \iota _A=f$ and $g\circ \iota _B=g$ . Uniqueness comes from the fact that the equalizer ${E\hookrightarrow A+B}$ of two such h’s must be equal to $\top $ , because it is easy to see, by first-order reasoning, that $E\wedge A=A$ and $E\wedge B=B$ . For existence, consider the relation $R \buildrel {r}\over {\hookrightarrow } (A+B)\times X$ given by

$$ \begin{align*}(\exists_{\iota_A\times 1_X}(f\times 1_X)^* =_X) \vee (\exists_{\iota_B\times 1_X}(g\times 1_X)^* =_X)~. \end{align*} $$

In the internal language of doctrines, this can be expressed by the formula

$$ \begin{align*}\exists a \,(\iota_A(a)=z \wedge f(a)=x) \vee \exists b \,(\iota_B(b)=z \wedge g(b)=x) \end{align*} $$

(with free variables z of type $A+B$ and x of type X). By first-order reasoning it is possible to see that the composite arrow

$$ \begin{align*}k:R\buildrel{r}\over{\hookrightarrow} (A+B)\times X \buildrel{\pi_{A+B}}\over\longrightarrow A+B \end{align*} $$

is both in ${\mathcal {M}}^\bot $ (i.e., that $\exists _k(\top )=\top $ ) and mono (i.e., that $k(z_1)=k(z_2)\longrightarrow z_1=z_2$ holds). Moreover, applying the characterization of disjoint unions in relational $\mathbf {G}$ -sets, in every conservative embedding $\eta $ of $({\mathbf {E}}, {\mathcal {M}})$ into a category of the kind $\operatorname {\mathrm {\mathbf {Rel}}}^{\mathbf {G}}$ , $\eta (k)$ turns out to be an isomorphism. By conservativity, k is an isomorphism and then

$$ \begin{align*}A+B\buildrel{k^{-1}}\over \longrightarrow R \buildrel{r}\over {\hookrightarrow} (A+B)\times X\buildrel{\pi_X}\over \longrightarrow X \end{align*} $$

gives the required h.

(Ad (ii)). Given a zero object O and any object X, we must find a unique ${h:O\longrightarrow X}$ . Uniqueness has been observed above. For existence, consider the projection $\pi _O:O\times X \longrightarrow O$ : this is mono and becomes iso in any representation into relational $\mathbf {G}$ -sets, so it is iso by conservativity. The inverse of $\pi _O$ composed with the projection into X gives the required h.

(Ad (iii)). Suppose we are given a quotient $q:X\twoheadrightarrow Y$ . We show that q is a regular epimorphism. Take $f:X\longrightarrow Z$ with $ker(q)\leq ker(f)$ . We need to find a unique $h:Y\longrightarrow Z$ such that $h\circ q=f$ . Uniqueness comes from the fact that q is epi, as any arrow in ${\mathcal {M}}^\bot $ (recall Proposition 2.3). Consider the relation $R\buildrel {r}\over {\hookrightarrow } Y\times Z$ given by

$$ \begin{align*}\exists_{\pi_{Y\times Z}}(\pi_{X\times Z}^* (q\times 1_Y)^* {=_Y} \land \pi_{X\times Z}^* (f\times 1_Z)^* {=_Z}). \end{align*} $$

In the internal language of doctrines, this can be expressed by the formula

$$ \begin{align*}\exists x\, (q(x)= y\wedge f(x)=z). \end{align*} $$

By first-order reasoning it is possible to see that the composite arrow

$$ \begin{align*}k:R\buildrel{r}\over{\hookrightarrow} Y\times Z \buildrel{\pi_Y}\over\longrightarrow Y \end{align*} $$

is both in ${\mathcal {M}}^\bot $ and mono. Moreover, applying the characterization of quotients in relational $\mathbf {G}$ -sets, in every conservative embedding $\eta $ of $({\mathbf {E}}, {\mathcal {M}})$ into a category of the kind $\operatorname {\mathrm {\mathbf {Rel}}}^{\mathbf {G}}$ , $\eta (k)$ turns out to be an isomorphism. By conservativity, k is an isomorphism and then

$$ \begin{align*}Y\buildrel{k^{-1}}\over \longrightarrow R \buildrel{r}\over {\hookrightarrow} Y\times Z\buildrel{\pi_X}\over \longrightarrow Z \end{align*} $$

gives the required h.

We have shown that any quotient q is the coequalizer of its kernel $ker(q)$ . Since any equivalence relation is of the form $ker(q)$ , it follows that, reciprocally, any regular epimorphism is a quotient.

(Ad (iv)). We show that the pullback of a regular epi $q: X\longrightarrow Q$ along any arrow is a regular epi. Let us factorize such a pullback $q'$ as

$$ \begin{align*}X'\buildrel{q"}\over \longrightarrow Q' \buildrel{m}\over \longrightarrow Q", \end{align*} $$

where $q"$ is a regular epi ( $q"$ is the quotient, aka the coequalizer, of the kernel of $q'$ ) and m is mono by the lemma below. In a conservative representation $\eta $ of $({\mathbf {E}}, \mathcal {M})$ into a relational $\mathbf {G}$ -sets category, it turns out $\eta (m)$ is an iso because relational $\mathbf {G}$ -sets are a regular category, hence m itself is iso by conservativity.

Proof. This can be established using first-order reasoning via the internal language. Alternatively, make a diagram chasing, using the fact that the class ${\mathcal {M}}^\bot $ is pullback-stable and is contained in the class of epis by Proposition 2.3.