Abstract
Two important algebraic structures are -acts and Boolean algebras. Combining these two structures, one gets -Boolean algebras, equipped with a compatible right action of a monoid which is a special case of Boolean algebras with operators. In this article, we considered some category-theoretic properties of the category Boo-S of all -Boolean algebras with action-preserving maps between them which also preserve Boolean operations. The purpose of the present article is to study certain categorical and algebraical concepts of the category Boo-S, such as congruences, indecomposable objects, coproducts, pushouts, and free objects.
1. Introduction and Preliminaries
Two important algebraic structures in many fields such as in computer science are -acts and Boolean algebras and then a combination of these two structures called -Boolean algebras, introduced in [1], are essential. In this article, some category-theoretic properties of the category Boo-S of all -Boolean algebras, with action and operations preserving maps between them are considered. Let us explain our motivation and eagerness to do this article.
The study of categories in category which is different from the category set of sets are always of interest, for example, topological groups, topological rings, topological semigroups or the category of general (universal) algebras in an arbitrary category (see [2]). In special, where is the category Act-S of sets with an action of a monoid on them has been considered in [1, 3–6]. Ebrahimi and Mahmoudi, in [1, 4, 5], have investigated the category Boo of Boolean algebras in the category Act-S. They have studied some of their properties such as internal injectivity and completion in Boo-S. Jónsson and Tarski in [7, 8] introduced the concept of Boolean algebras with operators. All -Boolean algebras are special instances of Boolean algebras with operators that the set of these operators forms a monoid . So, we are persuaded to get some categorical and algebraic structures and concepts such as congruences, coproducts, and free objects, which were not obtained in [1, 4, 5, 7, 8].
On the other hand, acts over a semigroup or monoid , namely, -acts, and also Boolean algebras are extended in many applications such as in theoretical computer science, algebraic automata theory, combinatorial problems, theory of machines, and graph theory. A comprehensive survey of -acts was published by Kilp et al. in [9] and of Boolean Algebras by Koppelberg in [10] and Givant and Halmos in [11]. Recently, the study of the connection between actions and algebraic structures (for example, vector spaces, modules, -acts, -posets, and M-algebras) has been of interest for some authors (see for example [2, 3, 12, 13]). In [2], Ebrahimi introduced the concept of -algebras which is the action of a monoid on (universal) algebras. After that, the concepts of -acts, -poset, and soft -act were introduced, respectively, in [3, 12, 13]. Inspired by these studies, in this article, we identified congruences and some limits and colimits such as products, coproducts, equalizers, pullbacks, and pushouts in the category Boo-S. Also the existence of free objects on a set is shown and some adjoint situations are obtained. We should mention that the constructions in the category Boo-S are not mostly relevant to their counterparts in Act-S and Set.
Section 2 is devoted to definition and some elementary properties of -Boolean algebras. In Section 3, we widely investigated congruences and simple, subdirectly irreducible, and indecomposable algebras in Boo-S. Coproducts and pushouts in Section 4 and free -Boolean algebras in Section 5 are studied.
Throughout the article, will denote a given monoid. A (right) -act is a set on which acts unitarily from the right with the usual properties, that is, if there is an -action , denoting , such that and , where 1 denotes the identity of . In fact, an -act is a universal algebra where each is a unary operation on such that for each and . Thus, all objects of the category Act-S form an equational class. On the other hand, considering as a one-object category whose morphisms are the elements of , the functor category is isomorphic to Act-S. Hence, since any functor category , for a small category , is a topos, the category Act-S is a topos (see [12]). An element of an -act is called a fixed element if for all . The set of all fixed elements of is denoted by . Note that one can always adjoin a fixed element to and get an act with a fixed element 0. For two -acts and , a map is called -homomorphism (or -map), if for each . An equivalence relation on an -act is called an -act congruence on if implies for all and . If is a congruence on , then the factor set , with the action given by , for and , is clearly an -act, called the factor act of by . Considering to be a one element monoid, the category Act-S is equivalent to the category Set and so the categorical constructions are obtained similarly for sets, for more information and the notions not mentioned here such as monomorphism, epimorphism, isomorphism, product, coproduct, equalizer, coequalizer, pullback, pushout, and free objects, about the category Act-S, see [9, 12]. Also, the interested reader is referred to [10, 11] for some required definitions and basic categorical ingredients of Boolean algebras needed in the sequel.
2. Boolean Algebras with Semigroup Operations
In this section, we give a brief account of some basic definitions and elementary properties about -Boolean algebras needed in the sequel. As we say in Section 1, recalling the general notion of an algebra in a category, we discussed Boolean algebras in the category Act-S. More generally (see, [2]), an -algebra is an (ordinary) algebra of type which is also an -act such that each operation is an -map, or equivalently, each -action defined by is an algebra homomorphism. Thus, the category Boo-S of right -Boolean algebras is an example of -algebras. In other words, we have the following definition:
Definition 1 (see [4]). Let be a monoid. A right -Boolean algebra is a (possibly empty) Boolean algebra which is also an -act whose Boolean algebra operations are equivariant, that is, , , , and for each and .
Let be a monoid. Then, every Boolean algebra can be considered as a right -Boolean algebra, by trivial action define as for all and all . Thus, the category Boo is a full subcategory of Boo-S.
Lemma 1. Let a Boolean algebra be an -act. The following are equivalent:(i) is an -Boolean algebra.(ii)For every and (iii)For every and
Proof. Clear. Let and . Then, The proof of this part is similar to .
From now on we use Lemma 1, part or for the definition of -Boolean algebra. Also, one can define an order on an -Boolean algebra as follow: if and only if . It is easy to check that if , then .
By an -Boolean algebra map (or homomorphism), we mean a map between right -Boolean algebras which preserves binary operation (or ), unary operation and the action.
Remark 1. For a homomorphism and ,(i) and (ii)if , then Let be a family of right -Boolean algebras. The product of is their Cartesian product, with component wise action and operations. Clearly, the category Boo- is product complete (i.e., for every family of S-boolean algebras, the product exists in Boo-) and determined up to isomorphism. In particular, the terminal S-boolean algebra (the product of the empty family of S-Boolean algebras) is a one element object. Now, we are ready to explain a class of right -Boolean algebras.
Example 1. Consider the Boolean algebra and the group .(1)Let . It is not difficult to check that by the action satisfying and is an -Boolean algebra.(2)Let be the Boolean algebra with cardinality . Then, which is an -act by the action given by , for each and . Using part (1), and . Thus, is a right -Boolean algebra.(3)Let be a Boolean algebra with elements. Then, using part (1) and (2) by induction we can easily show that is an -Boolean algebra.
Example 2. For a -Boolean algebra , consider to be the set of all Boolean endomorphisms on . By the binary operation given by, for each , , clearly is a monoid. Define an action on as follow, for each and , . Now, it is not difficult to check that is an -Boolean algebra.
By definition, a sub-Boolean algebra of a Boolean algebra is a subset of which is closed under and . Also, a subset of an -act is called an -subact, if for each and , . So, a subset of an -Boolean algebra is said to be a sub -Boolean algebra of , if it is a sub Boolean algebra of as well as an -subact. Note that in which case binary and nullary operations are compatible with the -action. For a subset , a sub -Boolean algebra of generated by , is a smallest sub -Boolean algebra of containing . For an -Boolean algebra and , denote . An element is called an elementary product on X, if in which and . By the properties of right -Boolean algebras we can easily show the following theorem:
Theorem 1. For a right -Boolean algebra and a subset , the subsetis the least sub -Boolean algebra of , called the sub -Boolean algebra of generated by .
We consider the set of all covariant functors from the one-object category to the category Boo and natural transformations between them, as a category which called functor category and denoted by .
Theorem 2. The category S-Boo of left -Boolean algebras is isomorphic to the category .
Proof. Consider the functor S-Boo as follows:For each , with an action and for any natural transformation in S-Boo, we define to be the only component of . Also, we consider the functor : S-Boo , defined by and , for each S-Boo and . Note that each -Boolean homomorphism , is the natural transformation whose only component is . Now, it is not difficult to show that and .
Also, clearly the category Boo-S is isomorphic to the category -Boo ( is the dual monoid of ). In particular, if is a commutative monoid, then the categories Boo-S and S-Boo are isomorphic, and hence the category Boo-S is isomorphic to the category and this means that, as we said before, the category Boo-S is a topos.
3. Congruences on -Boolean Algebras
The action of a semigroup on lattices, so called -lattices, was defined by Luo, [14]. The author specially studied the -lattice congruences of -lattices. In this section, we introduced congruences of right -Boolean algebras. Also, some characterizations of congruences generated by a subset are investigated. Two kinds of congruence characterizations (Proposition 1 and Theorem 3) are given here. Then, relations between congruences and ideals are investigated. Finally, some results about simple, subdirectly irreducible, and indecomposable right -Boolean algebras based on congruences are obtained.
Definition 2. An equivalence relation on an -Boolean algebra is said to be a congruence relation on , if is a sub -Boolean algebra of , or equivalently, for each and , implies , , and .
Note that, in Definition 2, implies . Now, Demorgan’s low deduces that . The set of all congruence relations on is denoted by . Also, each homomorphism induces the congruence relation, , on , defined by if and only if . As usual for a congruence and , the congruence class of is denoted by , or , and . By the action given by and operations , , and , is an right -Boolean algebras. It is easy to check that the canonical map , defined by , is a homomorphism. Also, implies . Note that for denotes the congruence generated by (i.e., the smallest congruence on containing ). We denote . For every semigroup without an identity one can adjoin an identity 1 by setting for all and get an -act denoted by . By placing , one can consider as a monoid.
Now, we are ready to explain the first congruence characterization.
Proposition 1. Let and . Then, for , one has if and only if either or for there exist and , such thatwhere , , and .
Proof. It is not difficult to check that is an equivalence relation on . Let and such that . The expression is obtained by adding to the right side of all terms of the chain. Using De Morgan’s laws, we get and also since the action of on the Boolean algebra operations of is equivariant, we get . Now, we prove that is an smallest congruence containing . Let be a congruence containig and . So or the chain holds. For each , since and is a congruence, and also . By continuing this argument . Thus, which implies .
In what follows (the second congruence characterization), we shall often use a more explicit version of the previous proposition.
Theorem 3. Let be a right -Boolean algebra and H . Consider the binary relation on by for all , if and only if or there exist and such thatwhere . Then, .
Proof. It is easy to check that is an equivalence relation. Suppose that , , and . If , then , and . Otherwise, there exist and such that holds. Clearly, and .
It suffices to show that . By the assumption, one gets the following equation:For each , deduces that and similarly . So, we have the following equation:Thus, which deduces that is a congruence relation an A. Suppose that , consider , , , , , , , and . So, , and hence . Now, let be a congruence relation containing and . Thus, or the holds. For each , . Thus, , and hence . Therefore, and then is the smallest congruence containing .
Corollary 1. For a right S-Boolean algebra and , if , then there exist , , and such that(i)(ii)
Corollary 2. Let and , the union of domain and image of . If , then there exist such that and
Definition 3. An -ideal of a right -Boolean algebra is a Boolean ideal of which is an -sub act as well. The set of all -ideals of an -Boolean algebra is denoted by .
Proposition 2. Let be an S-ideal in an S-Boolean algebra and . Then, the following are equivalent:(i)(ii) for some .(iii) (iv), where
Proof. (i) (ii) Let . Using Corollary 1 in which .
(ii) (iii) .
(iii) (ii) .
(ii) (i) Consider , , , and . So, .
(iv) (ii) It is easy to check that .
(iii) (iv) . Thus, , and therefore .
Corollary 3. Let be an -ideal in an -Boolean algebra and . Then,(i)(ii) is the greatest congruence on having as a whole class.
Proof. (i)It is clear that . For the converse, let . Then, there exists such that . Thus, , and hence .(ii)Let , , and . Then, which implies . So, , and hence . Since , by using Proposition 2 (ii), . Thus, .Using Corollary 3, one has . So, we have the following theorem.
Theorem 4. For an -Boolean algebra , there is a one to one correspondence between the sets and .
For an -Boolean algebra , the congruences and are called diagonal congruence and universal congruence, respectively. A congruence is called trivial, if it is universal or diagonal. An -Boolean algebra is called simple if . As a consequence of Theorem 4, we have is simple if and only if .
In the following lemma, -ideals generated by subsets of an -Boolean algebra are constructed.
Lemma 2. Let be an S-Boolean algebra and . The set is the smallest S-ideal containing .
Proof. Clearly, and implies that . Also, is closed under and which is an -sub act of , which is a subset of any S-ideal of containing , so it is the smallest S-ideal containing .
Using Lemma 2, we have the following proposition:
Proposition 3. An -Boolean algebra is simple, if and only if for each there exist such that .
In view of Proposition 3, we get the following example. It shows that for every Boolean algebra one can consider a monoid and an action of on , such that the induced -Boolean algebra is simple.
Example 3. Let be a Boolean algebra. The set of all Boolean endomorphisms on endowed with composition of morphisms, denoted by , is a monoid. It is not difficult to check that by the left -action, for and , is a left -Boolean algebra. Using [11], Th. 20.12, for each , there exists a proper maximal ideal of containing . The map is defined by , if and if is a Boolean endomorphism. So, for each , there exists an endomorphism such that . Now, Proposition 3 shows that is a simple left -Boolean algebra.
If and are congruences on an -Boolean algebra , then the relational product is the binary relation on defined by , if and only if there is such that and . Also, the pair is called a pair of factor congruences on , if , , and .
Definition 4. (i)An -Boolean algebra is said to be indecomposable if is not isomorphic to a direct product of two nontrivial right -Boolean algebras(ii)An -Boolean algebra is called subdirectly irreducible, if is trivial or there is a minimum congruence in (i.e., the intersection of all nontrivial congruences is nontrivial)By [15], Corollary II.7.7, an -Boolean algebra is indecomposable if and only if the only factor congruences on are and . Unlike the case of Boolean algebras, in which every simple, subdirectly irreducible, and indecomposable algebras are equal to , in the category Boo-S, these mentioned concepts are not necessary to be equal (example 2 and 3). Also, example 1 shows that, we can construct a simple -Boolean algebra of each Boolean algebra.
Theorem 5. An -Boolean algebra is indecomposable if and only if .
Proof. Let be an indecomposable -Boolean algebra and a fixed element such that . We define the binary relation and as follows:Then, . We only show the transitive property for . Let and . So, , and similarly . Thus, .
It is clear that and . Thus, . If , then and hence . Thus, . Now, we consider ; the congruence generated by . For each we have the following equation:Thus, , by transitivity. So, . Next, we showed that . Let . Then, there exists such that , and hence and . Consider . So, and , which implies and similarly, . Hence, and . In a similar way, we concluded . Therefore, is a pair of factor congruences on , which is a contradiction. Thus, .
For the converse, let be a decomposable -Boolean algebra. So, there exists a pair of nontrivial factor congruences on . By [15], Th. II.5.9, Corollary II.7.7, . Thus, there exists nontrivial element with . Hence, for each , which deduces . Thus, , a contradiction.
As we mentioned before, each indecomposable Boolean algebra is isomorphic to 2 (see [15]). Theorem 5 is a generalization for that results. Indeed, if we consider the Boolean algebra , then by the note just after Definition 1, is a right -Boolean algebra with trivial action. Clearly, in this -Boolean algebra each element is a fix element. So, Theorem 5 implies that is an indecomposable -Boolean algebra iff .
Using [15], Theorem. II.8.5, each simple algebra is subdirectly irreducible and each subdirectly irreducible algebra is indecomposable. In what follows, we introduce two examples to show the converse is not generally true.
Let be a subset of a poset . Then, and . Moreover, if , then we use instead of , for simplicity.
Example 4. There is a four elements subdirectly irreducible right -Boolean algebras, which is not simple. Consider an arbitrary semigroup and the Boolean algebra by the action and . It is not difficult to show that has only one nontrivial -ideal . So, by Theorem 4, has only one nontrivial congruence. Thus, is subdirectly irreducible which is not simple.
Example 5. If a Boolean algebra has at least two atoms, then there is an action on such that is an indecomposable and not subdirectly irreducible -Boolean algebra. Let be an atom. The subsets and are a partition of . Consider two elements left zero semigroup with 1 an identity and define an action . By this action is a right -Boolean algebra. Now, is indecomposable by using Theorem 5. For each atom , the subset is a right ideal of . Let be the set of all atoms of . Since , is not subdirectly irreducible.
4. Coproduct
In [16], R. Lagrange studied the Amalgamation and epimorphisms in the category of -complete Boolean algebras. After that, Banaschewski [17] introduced the structure of strong amalgamations of Boolean algebras. In this section, we showed, based on [17], the coproduct (free product) of a family of right -Boolean algebras exists and consequently, the pushout of two -Boolean homomorphisms with same domain, exists as a qoutient of their coproduct.
Using [17], for Boolean algebras and , we consider the lattice by the following requirements:for all finite and . Also, the mapsare bounded lattice embeddings so that and are sublattices of consisting of complemented elements.
Coproduct of Boolean algebras and is a sublattice of a lattice generated by in which both and are embedded.
Furthermore, for each and , . So and . Also, if or , thenand if and are not bottom and , then and are not bottom as well as and .
Now, if and are right -Boolean algebras, for each , which are not bottoms and , define and if or , then . Also, for each in which , we defined . Using these actions, in the following, we deduced that is an -Boolean algebra which is a coproduct of and , and both and are -Boolean monomorphisms.
The abovementioned actions are well defined. Assume that , , , and . Let and are not bottom. Then, and are not bottom and and . Thus, and , which implies that . If or are bottom, then or are bottom too and hence . Thus, the action is well defined.
We will show the action is well defined. Let and in which . Thus, such that and . Using the fact that, , we have the following equation:and in a similar way . If one of the elements , , , and is bottom, then . Now, let the mentioned elements are not bottoms. Then, and and by using the previous paragraph and the fact that and are right -Boolean algebras, we have the following equation:
Similar calculations show that and . Therefore, is an -Boolean algebra.
Also is an -Boolean coproduct of and . For, let and be two -Boolean maps. Since is a coproduct in the category of Boolean algebras, there exists Boolean homomorphism such that and . Consider and in which . If , then and similarly for the case . So . Thus, is an -Boolean homomorphism. Also, is unique.
As a corollary of the existence of coproducts, it is natural to consider the pushouts in the category of right -Boolean algebras.
Lemma 3. Pushout of the following diagram exists.
Proof. Let be the coproduct of the pair and be a right ideal of generated by the set . We consider , in which is a congruence generated by and , where is a canonical map. Now, it is not difficult to show that is a pushout of the diagram. For , . Thus, . To prove the universal property of pushouts, let be homomorphisms such that . So, there is a homomorphism satisfying . On the other hand, . Thus, , and hence there is a homomorphism such that . It follows that . Moreover, is uniquely determined, which implies that is a pushout of the diagram in Figure 1.
In [17], Banacshewski has shown that, in the category of Boolean algebras, the pushout of the previous diagram is a qoutient of the coproduct as in which is an ideal generated by the set . It is not difficult to show that , and hence . Thus, is also a coproduct of the diagram in Figure 1.
Finally, we showed that pushouts preserve monomorphisms, that is, in the pushout diagram of Lemma 3, is a monomorphism whenever is a monomorphism. Let and . Then, and by Corollary 3.7, there exists such that which implies . Hence, , and by [17], there exists such that . So, by [17], comparison principle, and which deduces . Thus, and since is a monomorphism, . Hence, , a contradiction, showing that is a monomorphism.
5. Free -Boolean Algebras and Adjoint Situations
By an right -Boolean algebra on a right -act (Boolean algebra) we mean an -Boolean algebra with an -map (Boolean homomorphism) which has the following universal property. For each -Boolean algebra and an -map (Boolean homomorphism) there exists a unique right -Boolean homomorphism such that . In particular, a right -Boolean algebra on a set is an act-free by taking the set to be an -act with trivial action (each element is fixed).
Lemma 4. If is a right -act, the power set Boolean algebra with the following action is a left -Boolean algebra. For each and , .
Note that, by a similar argument, Lemma 4 is also true for a left -act.
Lemma 5. Let be an infinite right -act. Then, there exists an act-free -Boolean algebra on .
Proof. Consider a map defined by, . By twice using Lemma 4, is a right -act. At first, we showed that is an -act monomorphism. It is clear that is well defined. Note that, for each and ,If , then and hence which implies . Thus, is a right -act monomorphism. Also, , sinceWe showed that , the sub--Boolean algebra of generated by , is a free -Boolean algebra generated by . By Theorem 1, each is of the form in which each such that . For an -Boolean algebra and an -map , we considered the map , defined by . To show is well defined, we proved the following steps:
Step 1. For every different elements , we have . Otherwise, consider . If , then and hence , which is a contradiction. So, we can partite the set in to two subsets and such that is a nonempty set and . Thus,Obviously , which is a contradiction and hence . We concluded that if , then there exists such that and .
Step 2. Let . Then, and hence and . Applying Theorem 1, consider in which each and in which each andSince , we have for each . Thus,and hence for each , which , and , we obtained the following equation:By Step 1, there exist and such that and . Thus, and hence . By distributive low, . So, and in a similar way which implies . Therefore, is well defined.
Routine calculations show that is an -Boolean homomorphism and unique under the universal property of free objects.
Proposition 4. If is a right -act, then there exists an act-free -Boolean algebra on .
Proof. If is an infinite -act, then we are done by using Lemma 5. Let be a finite -act. Consider , the free word semigroup over . The semigroup by the action, for each and , , is an -act. Let be an -Boolean algebra and an -map. The map defined by is an -act homomorphism. So there exists a unique -Boolean algebra homomorphism such that . Consider , the sub -Boolean algebra of generated by and defined by . It is clear that . Let be an -Boolean homomorphism such that . Using the -Boolean homomorphism defined in the proof of Lemma 5,and on the other hand,which means . Therefore, is a free -Boolean algebra on the -act .
Using the fact that, for each nonempty set the -act , with the action , is a set-free -act on , we have the following theorem:
Theorem 6. Let be a set and an act-free -Boolean algebra on the -act . Then, is a set-free -Boolean algebra on .
In what follows, we give adjoined pairs between the category Boo-S and categories Act-S, Boo, and Set. Applying Proposition 4 and Theorem 6, we have Theorem 7.
Theorem 7. (i)The act-free functor F: Act-S (ii)The set-free functor F : Set Boo-S is a left adjoined to the forgetful functor U : Boo-S SetLet be a set and a set-free Boolean algebra on , which exists by [11]. Also consider to be a set-free -Boolean algebra on . The following theorem shows that we can consider as a Boolean-free -Boolean algebra on .
Theorem 8. For a set , is a Boolean-free -Boolean algebra on .
Proof. Suppose that and are set-free extensions. Since is a Boolean algebra and is set-free Boolean algebra, there is a Boolean homomorphism such that . Now, let be an -Boolean algebra and a Boolean homomorphism. Since is set-free on , there exists such that and hence . Obviously, is unique. Thus, is a Boolean-free -Boolean algebra on .
Theorem 9. For each Boo and Boo-S, there is a one to one correspondence between the sets of -Boolean algebra homomorphisms and of Boolean homomorphisms.
Proof. Suppose that and are set free and Boolean-free extensions, respectively. The set map , given by , is an inverse of the set map , given by , defined in Theorem 8.
Data Availability
No data were used to support this study.
Disclosure
This study was a part of the Tafresh University.
Conflicts of Interest
The author declares no conflicts of interest.