On unit-regular elements in various monoids of transformations
aa r X i v : . [ m a t h . G R ] F e b UNIT-REGULAR ELEMENTS IN THE FULLTRANSFORMATION MONOID AND MONOIDS OFTRANSFORMATIONS THAT PRESERVE A PARTITION
MOSAROF SARKAR AND SHUBH N. SINGH
Abstract.
Let T ( X ) be the full transformation monoid on a nonempty set X . In this paper, we prove that an element of T ( X ) is unit-regular if and onlyif it is semi-balenced. For a partition P of X , we characterize the unit-regularelements in monoid T ( X, P ) = { f ∈ T ( X ) | ( ∀ X i ∈ P )( ∃ X j ∈ P ) X i f ⊆ X j } under composition and its submonoids Σ( X, P ) = { f ∈ T ( X, P ) | ( ∀ X i ∈P ) Xf ∩ X i = ∅} , Γ( X, P ) = { f ∈ T ( X, P ) | ( ∀ X i ∈ P )( ∃ X j ∈ P ) X i f = X j } , Ω( X, P ) = { f ∈ T ( X, P ) | ( ∀ x, y ∈ X, x = y )( x, y ) ∈ E = ⇒ xf = yf } ,and T E ∗ ( X ) = { f ∈ T ( X ) | ( x, y ) ∈ E ⇐⇒ ( xf, yf ) ∈ E } , where E is theequivalence relation induced by P . We also characterize the regular elementsin Σ( X, P ) and Ω( X, P ). Introduction
We assume that the reader is familiar with the basic terminology of algebraicsemigroup theory. An element x of a monoid M is said to be unit-regular in M ifthere exists a unit u ∈ M such that xux = x ; and monoids consisting entirely ofunit-regular elements are called unit-regular . The concept of unit-regularity, whichis a stronger form of the regularity, was first appeared in the context of rings [12].There are several natural and motivating examples of such monoids. The unit-regularity in a monoid has received wide attention by a number of authors (see,e.g., [4, 5, 6, 7, 15, 16, 19, 20, 32]).Throughout this paper, let X be a nonempty set, P = { X i | i ∈ I } be a partitionof X , E be the equivalence relation induced by P , and T ( X ) be the full trans-formation monoid on X . It is well-known that T ( X ) is regular (cf. [18, p. 63,Exercise 2.6.15]). It is proven in [7, Proposition 5] that T ( X ) is unit-regular if andonly if X is finite. Pei [24, Corollary 2.3] characterized the regular elements in thesubmonoid T ( X, P ) = { f ∈ T ( X ) | ( ∀ X i ∈ P )( ∃ X j ∈ P ) X i f ⊆ X j } = { f ∈ T ( X ) | ( x, y ) ∈ E = ⇒ ( xf, yf ) ∈ E } of T ( X ), and concluded in [24, Proposition 2.4] that T ( X, P ) is regular if and only if P is trivial. The authors [29, Theorem 3.4] characterized the unit-regular elementsin T ( X, P ) for finite X . Several interesting results of T ( X, P ) and its submonoidshave also been investigated (see. e.g., [1, 2, 3, 8, 9, 10, 11, 13, 14, 21, 22, 23, 24,25, 26, 28, 30, 31]). Mathematics Subject Classification.
Key words and phrases.
Transformation semigroups, Set Partitions, Regular elements, Unit-regular elements.
In recent years, the submonoidsΣ( X, P ) = { f ∈ T ( X, P ) | ( ∀ X i ∈ P ) Xf ∩ X i = ∅} , Γ( X, P ) = { f ∈ T ( X, P ) | ( ∀ X i ∈ P )( ∃ X j ∈ P ) X i f = X j } ,T E ∗ ( X ) = { f ∈ T ( X ) | ( ∀ x, y ∈ X )( x, y ) ∈ E ⇐⇒ ( xf, yf ) ∈ E } of T ( X, P ) have also been considered. Moreover, the regular elements in Γ( X, P )and T E ∗ ( X ) have been characterized (cf. [29, Theorem 5.5] and [8, Theorem 3.1],respectively). For finite X , the authors characterized in [29, Proposition 3.5] theunit-regular elements in Σ( X, P ) and proved in [29, Proposition 5.8] that everyregular element in Γ( X, P ) is unit-regular.Let us defineΩ( X, P ) = { f ∈ T ( X, P ) | ( ∀ x, y ∈ X, x = y )( x, y ) ∈ E = ⇒ xf = yf } . It is routine to verify that Ω( X, P ) is a submonoid of T ( X, P ). Clearly, if E = { ( x, x ) | x ∈ X } , then Ω( X, P ) = T ( X ); and if E = X × X , then Ω( X, P ) = { f ∈ T ( X ) | f is injective } . The principle object of this paper is to characterize theunit-regular elements in T ( X ), T ( X, P ), Σ( X, P ), Γ( X, P ), Ω( X, P ), and T E ∗ ( X ).The rest of this paper is structured as follows. In the next section, we pro-vide the basic definitions, results, and notation used throughout the paper. InSection 3, we prove that an element of T ( X ) is unit-regular if and only if it issemi-balenced. We also characterize the regularity of the submonoids Ω( X ) = { f ∈ T ( X ) | f is injective } and Γ( X ) = { f ∈ T ( X ) | f is surjective } of T ( X ). In Sec-tion 4, we characterize the unit-regular elements in T ( X, P ). In Section 5, we firstcharacterize the regular elements in Σ( X, P ) and then characterize the unit-regularelements in Σ( X, P ), Γ( X, P ), and T E ∗ ( X ). In Section 6, we finally characterizethe regular and unit-regular elements in Ω( X, P ).2. Preliminaries and Notation
Let X be a nonempty set. The cardinality of X is denoted by | X | . For A, B ⊆ X ,we write A \ B to denote the set of elements of A that are not in B . A partition of X is a collection of nonempty disjoint subsets, called blocks , whose union is X .A partition is called trivial if it has only singleton blocks or a single block. Wedenote by ∆( X ) the equivalence relation { ( x, x ) | x ∈ X } on X . A cross-section ofan equivalence relation E on X is a subset of X which intersects every equivalenceclass of E in exactly one point. The letter I will be reserved for an indexing set.We write maps to the right of their arguments, and compose them accordingly.The composition of maps will be denoted simply by juxtaposition. Let f : X → Y bea map. We denote by Af the image of a subset A ⊆ X under f . The preimage of asubset B ⊆ Y under f is denoted by Bf − . We write dom( f ), codom( f ), and im( f )for the domain, codomain, and image of f , respectively. The defect of f , denotedby d( f ), is defined by d( f ) = | Y \ im( f ) | . The kernel of f , denoted by ker( f ), isan equivalence relation on X defined by ker( f ) = { ( x, y ) ∈ X × X | xf = yf } . Wedenote by π ( f ) the partition of X induced by ker( f ), namely π ( f ) = { yf − | y ∈ Xf } . Note that a map f is injective if and only if ker( f ) = ∆( X ). We write T f to denote any cross-section of ker( f ). Note that | T f | = | im( f ) | . The collapse of f ,denoted by c( f ), is defined by c( f ) = | X \ T f | (cf. [17, p. 1356]). If A ⊆ X and NIT-REGULAR ELEMENTS IN TRANSFORMATION MONOIDS 3 B ⊆ Y such that Af ⊆ B , then there is a map g : A → B defined by xg = xf forall x ∈ A and, in this case, we say that the map g is induced by f .Let M be a monoid with identity e . An element x ∈ M is called unit if thereexists y ∈ M such that xy = yx = e . The set U ( M ) of all units in M formsa subgroup of M , and is called the group of units of M . An element x ∈ M is said to be regular in M if there exists y ∈ M such that xyx = x . Such anelement y is called an inner inverse for x , and it need not be unique. Denote U ( x ) = { y ∈ U ( M ) | x = xyx } . If in addition, the inner inverse y is a unit, then x is said to be unit-regular in M . By reg( M ) and ureg( M ), we mean the sets ofall regular and unit-regular elements in M , respectively. If reg( M ) = M (resp.ureg( M ) = M ), we say that the monoid M is regular (resp. unit-regular ). Notethat U ( M ) ⊆ ureg( M ) ⊆ reg( M ), and if M is unit-regular then any submonoid of M containing U ( M ) is also unit-regular.Let X be a nonempty set. A selfmap on X is a map from X into X . A permutation of X is a bijective selfmap on X . We say that a selfmap f on X preserves a partition P of X if for every X i ∈ P , there exists X j ∈ P such that X i f ⊆ X j . We denote by S ( X ) the symmetric group on X . We write Ω( X ) (resp.Γ( X )) to denote the submonoids of T ( X ) consisting of all injective (resp. surjective)selfmaps on X . Note that the monoids T ( X ), Ω( X ), and Γ( X ) have the symmetricgroup S ( X ) as their group of units; and Ω( X ) = Γ( X ) = S ( X ) if and only if X isfinite.We refer the reader to the standard book [18] for additional information fromalgebraic semigroup theory.3. Unit-regular elements in T ( X )In this section, we characterize the unit-regular elements in T ( X ). We thenprovide an alternative proof of Proposition 5 in [7] which characterizes the unit-regularity of T ( X ). We also characterize the regularity of Ω( X ) and Γ( X ). Webegin by proving the following lemma. Lemma 3.1.
Let f : X → Y and g : Y → X be maps. If f gf = f , then im ( f g ) isa cross-section of the equivalence relation ker( f ) .Proof. Let yf − ∈ π ( f ). It suffices to show that | ( Xf ) g ∩ yf − | = 1. Since y ∈ Xf ,there exists x ∈ X such that xf = y . Write yg = z . Clearly z = yg ∈ ( Xf ) g . Since f gf = f , we obtain zf = ( yg ) f = ( xf ) gf = xf = y. It follows that z ∈ yf − , so | ( Xf ) g ∩ yf − | ≥ z , z ∈ ( Xf ) g ∩ yf − . Then z f = y = z f . As z , z ∈ ( Xf ) g , there exist y , y ∈ Xf such that y g = z and y g = z . Since y , y ∈ Xf , there exist x , x ∈ X such that x f = y and x f = y . Since f gf = f , we obtain y = x f = ( x f ) gf = ( y g ) f = z f = y. Similarly, we obtain y = y . It follows that y = y , so z = z . Hence | ( Xf ) g ∩ yf − | = 1. This completes the proof. (cid:3) The following corollary is a direct consequence of Lemma 3.1 by replacing Y with X . MOSAROF SARKAR AND SHUBH N. SINGH
Corollary 3.2.
Let f, g ∈ T ( X ) . If f gf = f , then im ( f g ) is a cross-section of theequivalence relation ker( f ) . Before we begin to prove one of the main theorems, we recall an importantdefinition from [17].
Definition 3.3. [17, p. 1356] An element f of T ( X ) is said to be semi-balenced ifc( f ) = d( f ) .The following theorem gives a characterization of the unit-regular elements in T ( X ). Theorem 3.4.
Let f ∈ T ( X ) . Then f ∈ ureg ( T ( X )) if and only if f is semi-balenced.Proof. Suppose first that f ∈ ureg( T ( X )). Then there exists g ∈ S ( X ) such that f gf = f . Therefore, by Corollary 3.2, the set im( f g ) is a cross-section of theequivalence relation ker( f ). Write im( f g ) = T f . Since g is bijective, we obtain( X \ Xf ) g = Xg \ ( Xf ) g = X \ T f . It follows that c( f ) = | X \ T f | = | X \ Xf | = d( f ), so f is semi-balenced.Conversely, suppose that f is semi-balenced. Note that | T f | = | Xf | and | T f ∩ yf − | = 1 for all yf − ∈ π ( f ). Therefore, we choose a bijection g : Xf → T f such that yg = x whenever xf = y and x ∈ T f . Since | X \ T f | = | X \ Xf | , wearbitrarily choose any bijection g : X \ Xf → X \ T f . Using these two bijections,we now define a map g : X → X by yg = ( yg if y ∈ Xf,yg if y ∈ X \ Xf .
Clearly g ∈ S ( X ). One can also verify in a routine manner that f gf = f . Hence f ∈ ureg( T ( X )). (cid:3) The following simple lemma will be useful in the sequel.
Lemma 3.5.
Let f : X → Y be a map. Then (i) f is injective if and only if c ( f ) = 0 . (ii) f is surjective if and only if d ( f ) = 0 .Proof. (i) The statement follows directly from the fact that the map f is injective ifand only if ker( f ) = ∆( X ).(ii) The statement follows directly from the definition of a surjective map. (cid:3) If X is finite, then every element of T ( X ) is semi-balenced (cf. [17, p. 1356]).However, we have the following lemma. Lemma 3.6.
Every element of T ( X ) is semi-balenced if and only if X is finite.Proof. Suppose that every element of T ( X ) is semi-balenced. Assume, to the con-trary, that X is an infinite set. Choose an arbitrary element x of X . Note that thesets X and X \ { x } have the same cardinality. Therefore there exists is a bijectivemap g : X → X \ { x } , so the map g : X → X is injective but not surjective. Then NIT-REGULAR ELEMENTS IN TRANSFORMATION MONOIDS 5 we get c( g ) = 0 and d( g ) ≥ X is finite.Conversely, suppose that the set X is finite and let f ∈ T ( X ). Note that T f , Xf ⊆ X and | T f | = | Xf | . Since X is finite, we obtainc( f ) = | X \ T f | = | X | − | T f | = | X | − | Xf | = | X \ Xf | = d( f ) . Hence f is semi-balenced. Since f is an arbitrary element, the proof is complete. (cid:3) From Lemma 3.6 and Theorem 3.4, we have the corollary below that characterizesthe unit-regularity of T ( X ). Corollary 3.7.
The monoid T ( X ) is unit-regular if and only if X is finite.Proof. Suppose that T ( X ) is unit-regular. Then, by Theorem 3.4, every elementof T ( X ) is semi-balenced. Hence the set X is finite by Lemma 3.6.Conversely, suppose that X is finite and let f ∈ T ( X ). Then f is semi-balencedby Lemma 3.6. It follows from Theorem 3.4 that f ∈ ureg( T ( X )). Since f is anarbitrary element, the monoid T ( X ) is unit-regular. (cid:3) If X is a finite set and f ∈ T ( X ), the next proposition gives the cardinality ofthe set U ( f ). Proposition 3.8.
Let X be a finite set and f ∈ T ( X ) . Then | U ( f ) | = d ( f )! Y x ∈ Xf | xf − | . Proof.
Let f ∈ T ( X ). Since X is finite, there exists g ∈ S ( X ) such that f gf = f by Corollary 3.7. Then im( f g ) is a cross-section of ker( f ) by Corollary 3.2. Write T f = im( f g ). Note that | T f | = | Xf | and T f depends on g . Observe that any suchbijection g ∈ S ( X ) can be only of the following form: yg = ( yg if y ∈ Xf,yg if y ∈ X \ Xf , where g : Xf → T f is a bijection defined by yg = x whenever xf = y and x ∈ T f ;and g : X \ Xf → X \ T f is any bijection. Note that for every cross-section T f ofker( f ) there is unique such bijection g and d( f )! number of such bijections g .By the multiplication principle, the total number of possible cross-sections T f ofker( f ) is Q y ∈ Xf | yf − | . Hence | U ( f ) | = d( f )! Y y ∈ Xf | yf − | by the multiplication principle. (cid:3) We now characterize the regularity of Ω( X ) in the following proposition. Proposition 3.9.
The monoid Ω( X ) is regular if and only if Ω( X ) = S ( X ) .Proof. Suppose that Ω( X ) is regular. Since S ( X ) is the group of units of Ω( X ),we have S ( X ) ⊆ Ω( X ). For the reverse inclusion, let f ∈ Ω( X ). Then c( f ) = 0 byLemma 3.5(i). Since f ∈ reg(Ω( X )), there exists g ∈ Ω( X ) such that f gf = f . Itfollows from Corollary 3.2 that im( f g ) is a cross-section of ker( f ). Since c( f ) = 0 MOSAROF SARKAR AND SHUBH N. SINGH and im( f g ) is a cross-section of ker( f ), it follows that ( Xf ) g = X and therefored( f g ) = 0. Since g is injective, we subsequently have d( f ) = 0 by [17, Theorem2.2(iv)]. Hence f is surjective by Lemma 3.5(ii) and consequently f ∈ S ( X ).The converse is immediate. (cid:3) The next proposition provides a characterization of the regularity of Γ( X ). Proposition 3.10.
The monoid Γ( X ) is regular if and only if Γ( X ) = S ( X ) .Proof. Suppose that Γ( X ) is regular. Since S ( X ) is the group of units of Γ( X ), wehave S ( X ) ⊆ Γ( X ). For the reverse inclusion, let f ∈ Γ( X ). Since f ∈ reg(Γ( X )),there exists g ∈ Γ( X ) such that f gf = f . It follows from Corollary 3.2 that im( f g )is a cross-section of ker( f ). Since f, g ∈ Γ( X ), we obtain im( f g ) = ( Xf ) g = Xg = X . But im( f g ) is a cross-section of ker( f ), we therefore have c( f ) = 0. Hence f isinjective by Lemma 3.5(i) and consequently f ∈ S ( X ).The converse is immediate. (cid:3) The following corollary is obvious from Proposition 3.9 and 3.10 .
Corollary 3.11.
Let M = Ω( X ) , Γ( X ) . If M is regular, then M is unit-regular. Unit-regular elements in T ( X, P )In this section, we characterize the unit-regular elements in T ( X, P ). Recall thatthe authors [28, Theorem 3.4] characterized the unit-regular elements in T ( X, P )for finite X . We begin by recalling an important definition of a map correspondingto a map in T ( X, P ) from [27]. Definition 4.1. ([27, p. 220]) Let P = { X i | i ∈ I } be a partition of X and let f ∈ T ( X, P ). The character of f , denoted by χ ( f ) , is a map χ ( f ) : I → I definedby iχ ( f ) = j whenever X i f ⊆ X j . In case of finite X , the map χ ( f ) has also been studied with the notation f (see,e.g., [1, 10, 11]). The authors gave a characterization of the elements of Σ( X, P ) and T E ∗ ( X ) in terms of χ ( f ) in [28, Theorem 3.2] and [28, Theorem 3.4], respectively.Note that the monoid T ( I ) is regular (cf. [18, p. 63, Exercise 2.6.15]), but neednot be unit-regular (cf. [7, Proposition 5]). We now prove the following lemma. Lemma 4.2.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ T ( X, P ) . If f ∈ ureg ( T ( X, P )) , then χ ( f ) ∈ ureg ( T ( I )) .Proof. If f ∈ ureg( T ( X, P )), then there exists g ∈ S ( X, P ) such that f gf = f .Note that χ ( g ) ∈ S ( I ) by [28, Theorem 5.8]. We also have from [27, Lemma 2.3]that χ ( f ) χ ( g ) χ ( f ) = χ ( fgf ) = χ ( f ) . Hence χ ( f ) ∈ ureg( T ( I )). (cid:3) Before proving Theorem 4.4, we recall a crucial lemma from [28].
Lemma 4.3. [28, Lemma 5.2]
Let P = { X i | i ∈ I } be a partition of X and let f ∈ T ( X ) . Then f ∈ T ( X, P ) if and only if there exists a unique family B ( f, I ) = { f i | f i is induced by f and dom ( f i ) = X i ∀ i ∈ I } . The following theorem characterizes the unit-regular elements in T ( X, P ). NIT-REGULAR ELEMENTS IN TRANSFORMATION MONOIDS 7
Theorem 4.4.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ T ( X, P ) . Then f ∈ ureg ( T ( X, P )) if and only if (1) χ ( f ) ∈ ureg ( T ( I )) ; (2) for each j ∈ Iχ ( f ) there exists i ∈ I such that (i) | X i | = | X j | , (ii) X i f = X j ∩ Xf , and (iii) c ( f i ) = d ( f i ) where f i ∈ B ( f, I ) .Proof. Suppose first that f ∈ ureg( T ( X, P )). Then there exists g ∈ S ( X, P ) suchthat f gf = f .(1) Lemma 4.2 takes care of this part.(2) Let j ∈ Iχ ( f ) and let jχ ( g ) = i .(i) Since g ∈ S ( X, P ) and jχ ( g ) = i , we get | X i | = | X j | by [28, Lemma3.6(ii)].(ii) We first show that X i f ⊆ X j ∩ Xf . Since X i f ⊆ Xf , it suffices toprove that X i f ⊆ X j . Since j ∈ Iχ ( f ) , there exists k ∈ I such that kχ ( f ) = j and so X k f ⊆ X j by definition of χ ( f ) . Since jχ ( g ) = i , wealso have X j g = X i by definition of χ ( g ) and [28, Lemma 3.6]. Then X k f = ( X k f ) gf ⊆ ( X j g ) f = X i f . This implies that X k f ⊆ X j ∩ X i f .Since f preserves P , we thus get X i f ⊆ X j .For the reverse inclusion, let y ∈ X j ∩ Xf . Since y ∈ Xf , thereexists x ∈ X such that xf = y . Also y ∈ X j and jχ ( g ) = i , wesee that yg ∈ X i . Then y = xf = ( xf ) gf = ( yg ) f ∈ X i f . Hence X j ∩ Xf ⊆ X i f as required.(iii) Since jχ ( g ) = i , we have from [28, Lemma 3.6] that X j g = X i andso X j g j = X i where g j ∈ B ( g, I ). Also, by (ii), we have X i f = X j ∩ Xf and so X i f i ⊆ X j where f i ∈ B ( f, I ). Since f gf = f ,we see that f i g j f i = f i . Then ( X i f i ) g j ⊆ X i is a cross-section ofker( f i ) by Lemma 3.1. Write ( X i f i ) g j = T f i . Since g ∈ S ( X, P ),the map g j ∈ B ( g, I ) is bijective by [28, Theorem 5.8]. Therefore( X j \ X i f i ) g j = X j g j \ ( X i f i ) g j = X i \ T f i and hence c( f i ) = d( f i ).Conversely, suppose that the given conditions hold. By (2), let T χ ( f ) ⊆ I be across-section of ker( χ ( f ) ) such that the preimage j ( χ ( f ) ) − ∩ T χ ( f ) of every element j ∈ Iχ ( f ) satisfies all the three conditions of (2). By (1), let χ ( g ) ∈ S ( I ) be thefollowing bijection that satisfies χ ( f ) χ ( g ) χ ( f ) = χ ( f ) . jχ ( g ) = ( i ∈ T χ ( f ) if j ∈ Iχ ( f ) and iχ ( f ) = j,i ∈ I \ T χ ( f ) if j ∈ I \ Iχ ( f ) and | X i | = | X j | . Using the above bijection χ ( g ) , we now construct a bijection of S ( X, P ) whosecharacter is equal to χ ( g ) .Let j ∈ Iχ ( f ) . Then there exists i ∈ T χ ( f ) such that iχ ( f ) = j . By (2), we thenhave | X i | = | X j | , X j ∩ Xf = X i f , and c( f i ) = d( f i ) where f i ∈ B ( f, I ). Therefore,we can choose a bijection g j : X j → X i such that yg j = x whenever x ∈ T f i and xf i = y . MOSAROF SARKAR AND SHUBH N. SINGH
Let j ∈ I \ Iχ ( f ) . Then we arbitrarily choose a bijection h j : X j → X jχ ( g ) . Usingthese bijections, we define a map g : X → X by xg = ( xg j if x ∈ X j where j ∈ Iχ ( f ) ,xh j if x ∈ X j where j ∈ I \ Iχ ( f ) . Since χ ( g ) ∈ S ( I ) and each map of B ( g, I ) is bijective, it follows from [28,Theorem 5.8] that g ∈ S ( X, P ). We can also verify in a routine manner that f gf = f . Hence f ∈ ureg( T ( X, P )). (cid:3) Unit-regular elements in Σ( X, P ) , Γ( X, P ) , and T E ∗ ( X )In this section, we first give a characterization of a regular element f in Σ( X, P )in term of its character χ ( f ) . We then characterize the unit-regular elements inΣ( X, P ), Γ( X, P ), and T E ∗ ( X ), respectively. Proposition 5.1.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ Σ( X, P ) .Then f ∈ reg (Σ( X, P )) if and only if χ ( f ) ∈ reg (Γ( I )) .Proof. Suppose first that f ∈ reg(Σ( X, P )). Then there exists g ∈ Σ( X, P ) suchthat f gf = f . Therefore χ ( f ) = χ ( fgf ) = χ ( f ) χ ( g ) χ ( f ) by [27, Lemma 2.3]. Since f, g ∈ Σ( X, P ), we have χ ( f ) , χ ( g ) ∈ Γ( I ) by Theorem[28, Theorem 3.4] and so χ ( f ) ∈ reg(Γ( I )).Conversely, suppose that χ ( f ) ∈ reg(Γ( I )). Then there exists h ∈ Γ( I ) such that χ ( f ) hχ ( f ) = χ ( f ) . Since χ ( f ) ∈ reg(Γ( I )), it follows from Proposition 3.10 that χ ( f ) ∈ S ( I ) and so h = ( χ ( f ) ) − .For each i ∈ I , let x i be a fixed element of X i . Moreover, for each x ∈ Xf , let x ′ be a fixed element of xf − . Define a map g : X → X by xg = ( x ′ if x ∈ X i ∩ Xf ,x i ( χ ( f ) ) − if x ∈ X i \ Xf .
Observe that g ∈ T ( X, P ) and χ ( g ) = h ∈ Γ( I ). Therefore g ∈ Σ( X, P ) by [28,Theorem 3.2]. Since g maps each element x of Xf to a fixed element x ′ of xf − ,we can verify in a routine manner that f gf = f . Hence f ∈ reg(Σ( X, P )). (cid:3) We now have the following corollary from Proposition 3.10 and Proposition 5.1.
Corollary 5.2.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ Σ( X, P ) . If f ∈ ureg (Σ( X, P )) , then χ ( f ) ∈ S ( I ) .Proof. If f ∈ ureg(Σ( X, P )), then f ∈ reg(Σ( X, P )). Therefore χ ( f ) ∈ reg(Γ( I ))by Proposition 5.1 and so χ ( f ) ∈ S ( I ) by Proposition 3.10. (cid:3) Let us now make the following trivial remark which will be used in the sequel.
Remark . Let M be a monoid and let N be a submonoid of M . Then(i) ureg( N ) ⊆ ureg( M ).(ii) if U ( M ) ⊆ N , then N ∩ ureg( M ) ⊆ ureg( N ).Recall that, for finite X , the authors [29, Proposition 3.5] gave a characterizationof the unit-regular elements in Σ( X, P ). The following proposition characterizes theunit-regular elements in Σ( X, P ) for arbitrary X . NIT-REGULAR ELEMENTS IN TRANSFORMATION MONOIDS 9
Proposition 5.4.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ Σ( X, P ) .Then f ∈ ureg (Σ( X, P )) if and only if (i) χ ( f ) ∈ Ω( I ) , (ii) | X i | = | X j | whenever iχ ( f ) = j , and c ( f i ) = d ( f i ) where f i ∈ B ( f, I ) .Proof. Suppose first that f ∈ ureg(Σ( X, P )).(i) Then χ ( f ) ∈ S ( I ) by Corollary 5.2. Since S ( I ) ⊆ Ω( I ), it follows that χ ( f ) ∈ Ω( I ).(ii) Let iχ ( f ) = j . Since f ∈ ureg(Σ( X, P )) and Σ( X, P ) ⊆ T ( X, P ), we have f ∈ ureg( T ( X, P )) and χ ( f ) ∈ S ( I ) by Remark 5.3(i) and by Corollary 5.2,respectively. Then, by Theorem 4.4, we get | X i | = | X j | and c( f i ) = d( f i )where f i ∈ B ( f, I ).Conversely, suppose that the given conditions hold. Since S ( X, P ) ⊆ Σ( X, P ) ⊆ T ( X, P ), by Remark 5.3(ii), it suffices to show that f ∈ ureg( T ( X, P )).Since f ∈ Σ( X, P ), we have χ ( f ) ∈ Γ( I ) by [28, Theorem 3.2]. By (i), weknow that χ ( f ) ∈ Ω( I ) and so χ ( f ) ∈ Γ( I ) ∩ Ω( I ) = S ( I ). This implies that χ ( f ) ∈ ureg( T ( I )).Let iχ ( f ) = j . Then X i f ⊆ X j by definition of χ ( f ) . Recall that χ ( f ) ∈ S ( I ).Therefore X i f = X j ∩ Xf . By (ii), we also have | X i | = | X j | and c( f i ) = d( f i )where f i ∈ B ( f, I ). Hence f ∈ ureg( T ( X, P )) by Theorem 4.4 as required. (cid:3) For finite X , it is proven in [29, Proposition 5.8] that reg(Γ( X, P )) = ureg(Γ( X, P )).The following proposition provides a characterization of the unit-regular elementsin Γ( X, P ) for arbitrary X . Proposition 5.5.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ Γ( X, P ) .Then f ∈ ureg (Γ( X, P )) if and only if (i) χ ( f ) ∈ ureg ( T ( I )) , (ii) for each j ∈ Iχ ( f ) there exists i ∈ I such that the map f i ∈ B ( f, I ) from X i to X j is injective.Proof. Suppose first that f ∈ ureg(Γ( X, P )). Then f ∈ ureg( T ( X, P )) by Remark5.3(i).(i) Then Lemma 4.2 takes care of this part.(ii) Clearly f ∈ reg(Γ( X, P )). Hence, by Theorem [29, Theorem 5.5], for each j ∈ Iχ ( f ) there exists i ∈ I such that the map f i ∈ B ( f, I ) from X i to X j is injective.Conversely, suppose that the given conditions hold. Since f ∈ Γ( X, P ), we knowfrom [29, Remark 5.3] that every map of B ( f, I ) is surjective. It follows from (ii)that for each j ∈ Iχ ( f ) there exists i ∈ I such that the map f i ∈ B ( f, I ) from X i to X j is bijective. Therefore, for the preimage i ∈ j ( χ ( f ) ) − , we get | X i | = | X j | , X i f = X j ∩ Xf , and c( f i ) = d( f i ) where f i ∈ B ( f, I ). Thus f ∈ ureg( T ( X, P ))by Theorem 4.4. Since S ( X, P ) ⊆ Γ( X, P ), it follows from Remark 5.3(ii) that f ∈ ureg(Γ( X, P )). (cid:3) The following proposition characterizes the unit-regular elements in T E ∗ ( X ). Proposition 5.6.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ T E ∗ ( X ) .Then f ∈ ureg ( T E ∗ ( X )) if and only if (i) χ ( f ) ∈ Γ( I ) , (ii) | X i | = | X j | whenever iχ ( f ) = j , and c ( f i ) = d ( f i ) where f i ∈ B ( f, I ) .Proof. Suppose first that f ∈ ureg( T E ∗ ( X )). Then f ∈ ureg( T ( X, P )) by Remark5.3(i).(i) It follows that χ ( f ) ∈ ureg( T ( I )) by Lemma 4.2. Since f ∈ T E ∗ ( X ), wehave χ ( f ) ∈ Ω( I ) by [28, Theorem 3.4]. Since S ( I ) ⊆ Ω( I ), it follows fromRemark 5.3(ii) that χ ( f ) ∈ ureg(Ω( I )) and so χ ( f ) ∈ S ( I ) by Proposition3.9. Hence χ ( f ) ∈ Γ( I ).(ii) Let iχ ( f ) = j . From (i) , we see that χ ( f ) ∈ S ( I ). Recall that f ∈ ureg( T ( X, P )). Therefore, by Theorem 4.4, we get | X i | = | X j | and c( f i ) =d( f i ) where f i ∈ B ( f, I ).Conversely, suppose that the given conditions hold. Since S ( X, P ) ⊆ T E ∗ ( X ) ⊆ T ( X, P ), it suffices to show that f ∈ ureg( T ( X, P )).Since f ∈ T E ∗ ( X ), we have χ ( f ) ∈ Ω( I ) by Theorem [28, Theorem 3.4]. It followsfrom (i) that χ ( f ) ∈ Γ( I ) ∩ Ω( I ) = S ( I ) and so χ ( f ) ∈ ureg( T ( I )).Again, since χ ( f ) ∈ S ( I ), we see that X j ∩ Xf = X i f whenever iχ ( f ) = j .Combining these with condition (ii), Theorem 4.4 concludes that f ∈ ureg( T ( X, P ))as required. (cid:3) The Submonoid Ω( X, P )In this section, we characterize the regular and unit-regular elements in Ω( X, P ),respectively. Let us first make the following useful remark.
Remark . Let f ∈ T ( X, P ). Then f ∈ Ω( X, P ) if and only if every map in B ( f, I ) is injective.The next theorem characterizes the regular elements in Ω( X, P ). Theorem 6.2.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ Ω( X, P ) . Then f ∈ reg (Ω( X, P )) if and only if for every j ∈ Iχ ( f ) there exists i ∈ I such that themap f i ∈ B ( f, I ) from X i to X j is surjective.Proof. Suppose first that f ∈ reg(Ω( X, P )), and let j ∈ Iχ ( f ) . Then there exists i ∈ I such that iχ ( f ) = j and so X i f ⊆ X j by definition of χ ( f ) . Then there is amap f i ∈ B ( f, I ) from X i to X j . If f i is surjective, then we are done. Otherwise,suppose that f i is not surjective.Recall that f ∈ reg(Ω( X, P )). There exists g ∈ Ω( X, P ) such that f gf = f .Since j ∈ Iχ ( f ) ⊆ I , there exists l ∈ Iχ ( g ) such that jχ ( g ) = l and so X j g ⊆ X l by definition of χ ( g ) . Then there is a map g j ∈ B ( g, I ) from X j to X l . Recall that X i f ⊆ X j , X j g ⊆ X l , and f gf = f . We obtain X i f = ( X i f ) gf ⊆ ( X j g ) f ⊆ X l f. Since f preserves P , we get X l f ⊆ X j . Therefore there is a map f l ∈ B ( f, I ) from X l to X j . We now claim that f l is surjective.Assume, to the contrary, that there exists y ∈ X j \ X l f l . Write yg j = z and zf l = w . Clearly z ∈ X l , w ∈ X j , and y = w . Since f, g ∈ Ω( X, P ), the maps g j NIT-REGULAR ELEMENTS IN TRANSFORMATION MONOIDS 11 and f l are injective by Remark 6.1 and so ( wg j ) f l = ( yg j ) f l . But ( yg j ) f l = zf l = w ,and ( wg j ) f l = ( wg j ) f = ( wg ) f = ( zf l ) gf = ( zf ) gf = zf = zf l = w which is a contradiction. Hence the map f l ∈ B ( f, I ) is surjective as required.Conversely, suppose that the given condition holds. By Remark 6.1, we knowthat every map in B ( f, I ) is injective. Then, by hypothesis, for every j ∈ Iχ ( f ) there exists i ∈ I such that the map f i ∈ B ( f, I ) from X i to X j is bijective. Denoteby h j the inverse map of f i : X i → X j . Note that the map h j : X j → X i is bijective.We now define a map g : X → X by xg = ( xh j if x ∈ X j where j ∈ Iχ ( f ) ,x otherwise . Clearly g ∈ Ω( X, P ). We can verify in a routine manner that f gf = f . Hence f ∈ reg(Ω( X, P )). (cid:3) The following proposition characterizes the unit-regular elements in Ω( X, P ). Proposition 6.3.
Let P = { X i | i ∈ I } be a partition of X and let f ∈ Ω( X, P ) .Then f ∈ ureg (Ω( X, P )) if and only if (i) χ ( f ) ∈ ureg ( T ( I )) , and (ii) for each j ∈ Iχ ( f ) there exists i ∈ I such that the map f i ∈ B ( f, I ) from X i to X j is surjective.Proof. Suppose first that f ∈ ureg(Ω( X, P )). Then f ∈ ureg( T ( X, P )) by Remark5.3(i).(i) Then Lemma 4.2 takes care of this part.(ii) Note that f ∈ reg(Ω( X, P )) and hence Theorem 6.2 takes care of this part.Conversely, suppose that the given conditions hold. Since f ∈ Ω( X, P ), everymap in B ( f, I ) is injective by Remark 6.1. Then, by hypothesis, for each j ∈ Iχ ( f ) there exists i ∈ I such that the map f i ∈ B ( f, I ) from X i to X j is bijective andso | X i | = | X j | , X i f = X j ∩ Xf , and c ( f i ) = d ( f i ). Hence f ∈ ureg( T ( X, P )) byTheorem 4.4. Since S ( X, P ) ⊆ Ω( X, P ) ⊆ T ( X, P ), we thus get f ∈ ureg(Ω( X, P ))by Remark 5.3(ii). (cid:3) References [1] J. Ara´ujo, W. Bentz, J. D. Mitchell, and C. Schneider. The rank of the semigroup of trans-formations stabilising a partition of a finite set.
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Department of Mathematics, Central University of South Bihar, Gaya, Bihar, India
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