Extending maps to profinite completions in finitely generated quasivarieties

Abstract

We consider the problem of extending maps from algebras to their profinite completions in finitely generated quasivarieties. Our developments are based on the construction of the profinite completion of an algebra as its natural extension. We provide an extension which is a multi-map and we study its continuity properties, and the conditions under which it is a map.

Appendix A: Profinite completions of Boolean products

The generalization of Theorem 3.7 to Boolean products depends on the possibility to express emptiness in the dual space in terms of formulas in the algebra, as seen in the next result. Recall the following notation: if $a\in {\mathbf{A}}$ and $m\in {\mathbf{M}}$ we denote by [a : m] the set $\{\psi \in \mathcal {A}({\mathbf{A}}, {\mathbf{M}})\mid \psi (a)=m\}$. The family $\{[a:m]\mid a\in {\mathbf{A}}, m\in {\mathbf{M}}\}$ is a basis of clopen subsets of ${\mathbf{A}}^*$.

The following theorem generalizes the developments in Hansoul and Vrancken-Mawet (1984) about Boolean products of bounded distributive lattices.

Theorem A.1

Assume that $\mathop {M}\limits _{\sim }$ yields a logarithmic duality for $\mathcal {A}$ and that ${\mathbf{M}}$ is of finite type. Let ${\mathbf{A}}$ be a Boolean product of the family $({\mathbf{A}}_i)_{i\in I}$ of algebras of $\mathcal {A}$. If for every $n\in {\mathbb {N}}$ and every $m_1, \ldots , m_n\in M$ there is an open formula $\phi (x_1, \ldots , x_n)$ in the language of ${\mathbf{M}}$ such that for every $i \in I$ and every $a_1, \ldots , a_n \in {\mathbf{A}}_i$, it holds

$$\begin{aligned} {\mathbf{A}}_i^*=\bigcup _{\lambda =1}^{n}[a_\lambda : m_\lambda ]\quad \iff \quad {\mathbf{A}}_i\models \phi (a_1, \ldots , a_n), \end{aligned}$$

then ${\mathbf{A}}^\delta$ is $\mathcal {A}_\iota$-isomorphic to $\prod _{i\in I}{\mathbf{A}}_i^\delta$.

Proof

Let $f: {\mathbf{A}}\hookrightarrow \prod _{i\in I} {\mathbf{A}}_i$ be a Boolean representation of the family $({\mathbf{A}}_i)_{i\in I}$ of algebras of $\mathcal {A}$. For every $i\in I$ we denote by $\rho _i$ the embedding $(\pi _i)^*:A_i^* \hookrightarrow \amalg \{{\mathbf{A}}_i^* \mid i \in I\}$ where $\pi _i$ denotes the projection map from $\prod _{i\in I}A_i$ onto its i-th factor ${\mathbf{A}}_i$, i.e., $\rho _i$ is the map defined by $\rho _i(\psi )=\psi \circ \pi _i$. Let X be the set $\bigcup \{\rho _i({\mathbf{A}}_i^*)\mid i \in I\}$. Since $\mathop {M}\limits _{\sim }$ yields a logarithmic duality for $\mathcal {A}$, it is not difficult to see that $\bigcup \{\rho _i({\mathbf{A}}_i^*) \mid i \in J\}$ is isomorphic to $\amalg \{{\mathbf{A}}_i^* \mid i \in J\}$ for every finite subset J of I. It follows that X is a (not necessarily closed) substructure of $\amalg \{A_i^*\mid i \in I\}$ (such a verification involves only finitely many terms $\rho _i({\mathbf{A}}_i^*)$). In particular, X can be seen as

$$\begin{aligned} X=\amalg \{(A_i^*)^\flat \mid i \in I\}. \end{aligned}$$

(14)

We are going to prove that X can be equipped with a Boolean topology $\tau$ to obtain a topological structure that is isomorphic to $A^*$ and that is embeddable into $\amalg \{{\mathbf{A}}_i^*\mid i \in I\}$.

We define the topology $\tau$ on X as the topology generated by the sets

$$\begin{aligned}{}[a:m]=\bigcup \left\{ [\pi _i(f(a)):m]\mid i \in I\right\} , \quad a \in {\mathbf{A}}, m \in {\mathbf{M}}. \end{aligned}$$

The topology $\tau$ is clearly finer than the topology induced on X by $\amalg \{{\mathbf{A}}_i^* \mid i \in I\}$. Let us show that $\langle X,\tau \rangle$ is Boolean. It suffices to prove that it is compact. Assume that $X=\bigcup \{[a_\lambda :m_\lambda ]\mid \lambda \in L\}$ for some $a_\lambda \in {\mathbf{A}}$ and $m_\lambda \in {\mathbf{M}}$. For every $i \in I$, the family $\{[\pi _i(f(a_\lambda )): m_\lambda ]\mid \lambda \in L\}$ is an open covering of $\rho _i({\mathbf{A}}_i^*)$ and there is a finite subset $L_i$ of L such that

$$\begin{aligned} \rho _i(A_i^*)=\bigcup \left\{ [\pi _i(f(a_\lambda )): m_\lambda ]\mid \lambda \in L_i\right\} . \end{aligned}$$

(15)

By hypothesis, for every $i\in I$ there is an open formula formula $\phi _{in_i}$ with $n_i$ variables (where $n_i$ denotes $\vert L_i \vert$) such that identity (15) is equivalent to

$$\begin{aligned} {\mathbf{A}}_i \models \phi _{in_i}\left( (\pi _i(a_{\lambda }))_{\lambda \in L_i}\right) . \end{aligned}$$

(16)

Now, for every $i\in I$ let $\Omega _i$ be defined by

$$\begin{aligned} \Omega _i=\left\{ j \in I \mid {\mathbf{A}}_j \models \phi _{in}((\pi _j(f(a_{\lambda })))_{\lambda \in L_i})\right\} . \end{aligned}$$

The family $\{\Omega _i\mid i \in I\}$ is an open covering of I. By compactness, there is a finite subset J of I such that

$$\begin{aligned} I=\bigcup \{\Omega _j \mid j \in J\}. \end{aligned}$$

(17)

By combining (16) and (17), we obtain,

$$\begin{aligned} X=\bigcup \left\{ \bigcup \Big \{[a_\lambda :m_\lambda ]\mid \lambda \in L_j\Big \}\mid j \in J\right\} , \end{aligned}$$

which is a finite open covering of X extracted from $\{[a_\lambda : m_\lambda ]\mid \lambda \in L\}$.

Let us denote by g the restriction of $f^*$ to X. Hence, for any $\rho _i(\psi )\in \rho _i({\mathbf{A}}_i)$, we have $g(\rho _i(\psi ))=\psi \circ \pi _i\circ f$. We aim to prove that g is an $\mathcal {X}$-isomorphism between $\langle X,\tau \rangle$ and ${\mathbf{A}}^*$.

First we prove that g is a $\mathcal {X}^\flat$-embedding. We have to prove that if r represents an n-ary relation or the graph of a (partial) operation in the language of $\mathop {M}\limits _{\sim }$ and if $\psi _1, \ldots , \psi _n \in X$, we have the following equivalence

$$\begin{aligned} (\psi _1, \ldots , \psi _n) \in r^{X} \Leftrightarrow (g(\psi _1), \ldots , g(\psi _n))\in r^{{\mathbf{A}}^*}. \end{aligned}$$

(18)

Let J be a finite subset of I such that $\{\psi _1, \ldots , \psi _n\}\subseteq \bigcup \{\rho _j({\mathbf{A}}_j^*)\mid j \in J\}$. Let us denote by Y the latter set. We have already noted that Y, considered as a substructure of $\amalg \{A_i^*\mid i \in I\}$ is isomorphic to $\amalg \{A_j^*\mid j \in J\}$. Since $f:A\hookrightarrow \prod _{i \in I}A_i$ is a Boolean representation of A, the map $f_J:A\rightarrow \prod _{j\in J}A_j:a \mapsto (\pi _j(a))_{j \in J}$ is onto. Hence, the dual map $f_J^*:Y \rightarrow A^*$ is an embedding and is clearly equal to the restriction of g to Y. Then, it follows successively

$$\begin{aligned} (\psi _1, \ldots , \psi _n)\in r^X\Leftrightarrow & {} (\psi _1, \ldots , \psi _n)\in r^Y\\\Leftrightarrow & {} (f_J^*(\psi _1), \ldots , f_J^*(\psi _n))\in r^{{\mathbf{A}}^*}\\\Leftrightarrow & {} (g(\psi _1), \ldots , g(\psi _n))\in r^{{\mathbf{A}}^*}, \end{aligned}$$

which establishes equivalence (18), as required.

Finally, since g is the restriction on X of a continuous map, it is a continuous map for the induced topology on X. From the fact that $\tau$ is finer than the induced topology we eventually conclude that $g:\langle X,\tau \rangle \rightarrow A^*$ is an $\mathcal {X}$-embedding. We deduce that $\langle X,\tau \rangle \in \mathcal {X}$.

For the last part of the proof, we show that the evaluation map

$$\begin{aligned} h:{\mathbf{A}}\rightarrow \mathcal {X}(X, \mathop {M}\limits _{\sim }):a \mapsto h(a): \rho _i(\psi )\mapsto \psi (\pi _i(f(a))) \end{aligned}$$

is an isomorphism. It is clearly a homomorphism. Moreover, if $a, b \in {\mathbf{A}}$ and $a\ne b$ then there is an $i \in I$ such that $\pi _i(f(a))\ne \pi _i(f(b))$, i.e., such that $e_{{\mathbf{A}}_i}(\pi _i(f(a)))\ne e_{{\mathbf{A}}_i}(\pi _i(f(b)))$. Let $\psi \in {\mathbf{A}}_i^*$ with $e_{{\mathbf{A}}_i}(\pi _i(f(a)))(\psi )\ne e_{{\mathbf{A}}_i}(\pi _i(f(b)))(\psi )$. It means that $\psi (\pi _i(f(a)))\ne \psi (\pi _i(f(b)))$ which proves that h is one-to-one. Moreover, since $h^*=g$ and since g is an embedding, we deduce that h is onto and so, is an isomorphism.

Hence, it follows successively that

$$\begin{aligned} {\mathbf{A}}^\delta \simeq \mathcal {X}^\flat ({\mathbf{A}}^*, \mathop {M}\limits _{\sim })\simeq \mathcal {X}^\flat (X, \mathop {M}\limits _{\sim })\simeq \mathcal {X}^\flat (\amalg _{i\in I}({\mathbf{A}}_i^*)^\flat , \mathop {M}\limits _{\sim }), \end{aligned}$$

where we have used (14) to obtain the latter isomorphism. Then, we obtain

$$\begin{aligned} \mathcal {X}^\flat \Bigg (\amalg _{i\in I}({\mathbf{A}}_i^*)^\flat , \mathop {M}\limits _{\sim }\Bigg ) \simeq \prod _{i\in I}\mathcal {X}^\flat \Big ({\mathbf{A}}_i^*, \mathop {M}\limits _{\sim }\Big )\simeq \prod _{i\in I}{\mathbf{A}}_i^\delta \end{aligned}$$

where the first isomorphism is obtained by partnership duality (Davey et al. 2012, Theorem 2.4) and is also an $\mathcal {A}_\iota$-isomorphism. $\square$