# Magnifying Elements in Semigroups of Fixed Point Set Transformations Restricted by an Equivalence Relation

Authors: Thananya Kaewnoi; Ronnason Chinram; Montakarn Petapirak
Journal: Journal of Mathematics
 (2021)

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## Abstract

LetX be a nonempty set and ρ be an equivalence relation on X. For a nonempty subset S of X, we denote the semigroup of transformations restricted by an equivalence relation ρ fixing S pointwise by EFSX,ρ. In this paper, magnifying elements in EFSX,ρ will be investigated. Moreover, we will give the necessary and sufficient conditions for elements in EFSX,ρ to be right or left magnifying elements.


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## Body

## 1. Introduction

Magnifying elements of a semigroup were first mentioned in 1963 by Ljapin in [1]. An element a of a semigroup S is called a right (left) magnifying element if there exists a proper subset M of S such that Ma=S (aM=S). Many studies on magnifying elements in semigroups were conducted by many authors in various aspects between 1971 and 2003. In 1971, Migliorini [2] introduced the new notion of a minimal subset M relative to a magnifying element a of S. By means of this, he constructed an infinite chain of minimal subsets properly contained in the preceding ones, which gives rise to an infinite number of magnifying elements of the form an, where a∈S is a magnifying element and n is any positive integer. Migliorini also investigated the structure of a semigroup S with minimal subsets in [3]. According to Tolo [4], if the proper subset M of S relative to a magnifying element a is a subsemigroup of S, then a is called a strong magnifying element. It was shown that if a semigroup S contains a strong magnifying element, then S is factorizable, i.e., S=AB for some proper subsemigroups A,B of S. In other words, the existence of strong magnifying elements plays an important role in factorizing a semigroup. In 1992, Catino and Migliorini [5] provided the example of semigroups with nonstrong magnifying elements which are factorizable and determined the existence of magnifying elements in simple, bisimple, and regular semigroups by improving Tolo’s results. Moreover, they proved that the magnifying elements of the natural partial-order semigroups are maximal. Two years later, Magill [6] provided the necessary and sufficient conditions for elements in a semigroup with identity to be left or right magnifying elements and applied the result to the semigroup of linear transformations of a vector space and the semigroup of all continuous self-maps of a topological space. In 1996, Gutan [7] constructed the semigroup containing both strong and nonstrong magnifying elements, which turned out to be a positive answer to the question, posed in [5, 6], of whether or not there is a semigroup containing both strong and nonstrong magnifying elements. A year later, he showed in [8] the necessary and sufficient conditions for a semigroup containing a magnifying element to be factorizable. The results point out that every semigroup containing magnifying elements, which is not necessarily strong, is factorizable; and this improved the results by Tolo, and Catino and Migliorini. In 1999, Gutan characterized semigroups containing left strong magnifying elements with a minimal subsemigroup and proposed the method for obtaining such a semigroup in [9]. In 2000, Gutan [10] introduced the definition of very good magnifying elements in a semigroup. If such a set M relative to a magnifying element a is a ring ideal, then a is a very good magnifying element. Furthermore, he established a characterization of semigroups in which all the left magnifying elements are very good. In 2003, Gutan and Kisielewicz [11] presented the notion of primitive semigroups and further constructed semigroups having both good and bad magnifying elements. In addition, some general properties of semigroups containing magnifying elements with its minimal subsemigroup were established. At that time, many researchers focused on the minimal subsets relative to magnifying elements. Recently, some researchers have paid attention to magnifying elements in various transformation semigroups. For instance, Luangchaisri et al. [12] generalized Magill’s results in partial transformation semigroups, and Prakitsri [13] investigated magnifying elements in linear transformation semigroups with infinite nullity and in those with infinite co-rank. He showed that linear transformation semigroups with infinite nullity have no right magnifying elements. However, all left magnifying elements in these semigroups are strong magnifying elements. Contrarily, linear transformation semigroups with infinite co-rank have no left magnifying elements but all right magnifying elements are strong. In [14], the conditions for elements in the semigroup of transformations with a fixed point set to be magnifying elements have been established by Petapirak, Kaewnoi, and Chinram. In this paper, efforts have been made to extend the results obtained in [14] by showing the necessary and sufficient conditions for elements in the semigroup of transformations restricted by an equivalence relation with a fixed point set to be right or left magnifying elements.


## 2. Preliminaries

In this section, we first provide the reader with some basic but essential definitions.Asemigroup is a system S,⋅ consisting of a nonempty set S together with the binary associative operation ⋅, i.e., a⋅b belongs to S and a⋅b⋅c=a⋅b⋅c for all elements a,b,c∈S. For convenience, we write S instead of S,⋅ and let ab stand for a⋅b for any a,b∈S. A subset T of a semigroup S is called a subsemigroup of S if T is a semigroup under the operation of S. A nonempty set T of a semigroup S is a subsemigroup of S if ab∈T for all a,b∈T. The intersection of any set of subsemigroups of S is either an empty set or a subsemigroup of S.The next finding is an initial factor in this study.Theorem 1 (see [1], pp. 118–119).
The following statements are true:(1)
No element of a semigroup is simultaneously a left and a right magnifying element.(2)
Finite semigroups have no magnifying elements.(3)
Commutative semigroups have no magnifying elements.(4)
Semigroups with two-sided cancellation have no magnifying elements, and hence groups do not contain magnifying elements.Therefore, we will focus our attention on infinite noncommutative semigroups without two-sided cancellation.LetTX be the set of all functions from a nonempty set X into itself. As is well known, the composition of functions TX is closed, and the associative law holds. Therefore, TX is a semigroup under the composition of functions. We then call TX the full transformation semigroup. Throughout this paper, the identity function on X is denoted by idX. The range of a function α is denoted by ranα for all elements α∈TX. Moreover, we write functions from the right, xα rather than αx, and compose from the left to the right, xαβ rather than β∘αx for all elements α,β∈TX.LetΔX be an identity relation on X, and let ρ be an equivalence relation on X. For each x∈X, we denote the equivalence class of ρ containing x by xρ=y∈X∣x,y∈ρ and X/ρ=xρ∣x∈X. For any ∅≠S⊆X, as in [14], we set TFSX=α∈TX∣xα=xforallx∈S. The conditions for elements in TFSX to be magnifying elements are demonstrated as follows.Theorem 2 (see [14], Theorem 2.5).
A functionα∈TFSX is a left magnifying element if and only if α is one-to-one but not onto.Theorem 3 (see [14], Theorem 2.14).
A functionα∈TFSX is a right magnifying element if and only if α is onto but not one-to-one.Denote the transformation semigroup restricted by an equivalence relationρ by EX,ρ=α∈TX∣∀x,y∈X;x,y∈ρimpliesxα=yα. It is widely known that EX,ρ is a subsemigroup of TX. Let EFSX,ρ be the intersection of EX,ρ and TFSX. Evidently, if EFSX,ρ is a nonempty set, then EFSX,ρ is a subsemigroup of EX,ρ and TFSX. We then call EFSX,ρ the semigroup of transformations restricted by an equivalence relation ρ fixing S pointwise.


## 3. Propositions forEFSX,ρ

In this section, we illustrate the following propositions which are characterizations ofEFSX,ρ.Proposition 1.
EX,ρ=TX,ρ if and only if ρ=ΔX.Proof.
Assume thatEX,ρ=TX,ρ. Let x,y∈X such that x,y∈ρ. Clearly, the identity function idX∈TX,ρ. By assumption, idX∈EX,ρ. Hence, x=xidX=yidX=y. This shows that ρ=ΔX. Conversely, assume that ρ=ΔX. Clearly, EX,ρ⊆TX,ρ. Let α∈TX,ρ. Then, for all x,y∈ρ, xα,yα∈ρ. So xα=yα since ρ=ΔX. This shows that α∈EX,ρ. Therefore, EX,ρ=TX,ρ.Note that ifρ=ΔX and α∈TX, then for all x,y∈ρ, xα,yα∈ρ. Hence, TX,ρ=TX if ρ=ΔX. By the proof of Proposition 1, EX,ρ=TX if and only if ρ=ΔX.Proposition 2.
EFSX,ρ=TFSX if and only if ρ=ΔX.Proof.
Assume thatEFSX,ρ=TFSX. Let x,y∈X be such that x,y∈ρ. Clearly, the identity function idX∈TFSX. By assumption, idX∈EFSX,ρ. Hence, x=xidX=yidX=y. This shows that ρ=ΔX. Conversely, assume that ρ=ΔX. It is clear that TX,ρ=TX. By Proposition 1, we have EX,ρ=TX. So we have EFSX,ρ=α∈EX,ρ∣xα=xforallx∈S=α∈TX∣xα=xforallx∈S=TFSX.Proposition 3.
The identity functionidX belongs to EFSX,ρ if and only if ρ=ΔX.Proof.
Assume that the identity functionidX belongs to EFSX,ρ. Let x,y∈X such that x,y∈ρ. Then, x=xidX=yidX=y. This shows that for all x,y∈X, if x,y∈ρ, then x=y, which implies that ρ is an identity relation on X. Conversely, assume that ρ=ΔX. By Proposition 2, EFSX,ρ=TFSX. It is clear that idX∈TFSX. So idX∈EFSX,ρ.Clearly,EFSX,ρ is a proper subset of EX,ρ. It is easy to verify that if ρ=X×X, then EX,ρ is the set of all constant functions, and if ρ=ΔX and S=X, then EFSX,ρ=idX.Proposition 4.
Ifρ=X×X and S>1, then EFSX,ρ is empty.Proof.
Assume thatρ=X×X and S>1. Then there are distinct elements x,y∈S such that x,y∈ρ. Suppose that there exists an element α∈EFSX,ρ. Then x=xα=yα=y, which is a contradiction. Therefore, EFSX,ρ is empty.Proposition 5.
Ifρ=X×X and S=1, then EFSX,ρ=1.Proof.
Assume thatρ=X×X and S=1. Then there is only one element s∈S. Hence, there is a function α∈EFSX,ρ defined by xα=s for all x∈X. Suppose that EFSX,ρ>1, and let β∈EFSX,ρ. By assumption, x,s∈ρ for all x∈X, and hence, xα=s=sβ=xβ for all x∈X. This shows that α=β. Therefore, EFSX,ρ=1.By Propositions4 and 5, left and right magnifying elements do not exist in EFSX,ρ if ρ=X×X.


## 4. Main Results

In this section, we will give the necessary and sufficient conditions for elements inEFSX,ρ to be right or left magnifying elements.


### 4.1. Right Magnifying Elements

By Proposition2, we obtain the next theorem.Theorem 4 (see [14]).
Suppose thatρ=ΔX and S≠X. A function α is right magnifying in EFSX,ρ if and only if α is onto but not one-to-one.Our next two results are related to the existence of magnifying elements inEFSX,ρ.Lemma 1.
IfS∩xρ>1 for some x∈X, then EFSX,ρ is empty.Proof.
Letx∈X. Assume that S∩xρ>1. Suppose that there is an element α belonging to EFSX,ρ. By assumption, there are two distinct elements s1,s2∈S∩xρ. Then s1,s2∈ρ. So s1=s1α=s2α=s2, which is a contradiction.Lemma 2.
IfS∩xρ=1 for all x∈X, then EFSX,ρ=1.Proof.
Ifρ=ΔX and S∩xρ=1 for all x∈X, then X=S, and hence, EFSX,ρ=idX. So EFSX,ρ=1. Next, assume that ρ≠ΔX and S∩xρ=1 for all x∈X. Then S≠X, and hence for each x∈X∖S, there is a unique element sx∈S such that x,sx∈ρ. Define a function α by(1)xα=x,ifx∈S,sx,ifx∈X∖S,for all x∈X. Clearly, α∈EFSX,ρ. Next, we will show that α is the only element in EFSX,ρ. Let β be a function in EFSX,ρ. For all s∈S, sα=s=sβ. By assumption, for each x∈X∖S, there is a unique sx∈S such that x,sx∈ρ, and hence, xβ=sxβ=sx=xα. Therefore, α=β and EFSX,ρ=1.By Lemmas1 and 2, if ρ≠ΔX and either S∩xρ=1 for all x∈X or S∩xρ>1 for some x∈X, then there exists no magnifying element in EFSX,ρ.Example 1.
ConsiderX=N, S=1,2, and x,y∈ρ if and only if x≡ymod2.
Clearly,ρ is an equivalence relation on X and X/ρ=1,3,5,…,2,4,6,…. Let α be a function in EFSX,ρ defined by, for all x∈X,(2)xα=1,ifxis odd,2,ifxis even.
By Lemma2, the function α is the only function in EFSX,ρ. So EFSX,ρ has no nonempty proper subset, and hence, there exists no magnifying element in EFSX,ρ.
For the rest of this section, we focus on nontrivial cases, i.e.,ρ≠ΔX and ρ≠X×X, and establish the existence of right magnifying elements in EFSX,ρ. We thus assume now that EFSX,ρ≠∅.Lemma 3.
Suppose thatX is countably infinite. If there are infinite xρ∈X/ρ such that S∩xρ=∅, then there is a surjective function in EFSX,ρ. Consequently, if α∈EFSX,ρ is a right magnifying element, then α is onto.Proof.
Suppose that there are infinitexρ∈X/ρ such that S∩xρ=∅. Hence, X\S is infinite. Let A=xρ∣S∩xρ=∅. By assumption, A is infinite. Then there is a bijective function σ from A to X\S. For each x∈X∖S such that S∩xρ≠∅, there exists a unique sx∈S such that x,sx∈ρ, by Lemma 1. Define a function η by(3)xη=xρσ,ifx∈X\Sandxρ∈A,sx,ifx∈X\Sandxρ∉A,x,ifx∈S.
Clearly,η∈EFSX,ρ. Moreover, η is onto. Let α be a right magnifying element in EFSX,ρ. There is a proper subset M of EFSX,ρ such that Mα=EFSX,ρ. Then βα=η for some β∈M. This implies that α is onto.Suppose thatX is uncountably infinite and A=xρ∣S∩xρ=∅. It is easy to prove that if X and A have the same cardinality, then X∖S and A have the same cardinality as well. The proof of our next result is similar to the proof of Lemma 3 and so will be omitted.Lemma 4.
Suppose thatX is uncountably infinite and A=xρ∣S∩xρ=∅. If X and A have the same cardinality, then there is a surjective function in EFSX,ρ. Consequently, if α∈EFSX,ρ is a right magnifying elements, then α is onto.Theorem 5.
Suppose thatρ≠ΔX and ρ≠X×X. If S∩xρ≤1 for all x∈X and there are infinite xρ∈X/ρ such that S∩xρ=∅, then a function α∈EFSX,ρ is a right magnifying element if and only if α is onto.Proof.
By Lemmas3 and 4, the necessity is clear. Conversely, assume that α∈EFSX,ρ is onto. Let M=β∈EFSX,ρ∣βis not onto. Clearly, M is a proper subset of EFSX,ρ since α∉M. Let γ be a function in EFSX,ρ. Since α is onto, for each x∈X, there exists an element yx∈X such that yxα=xγ. For each xρ∈X/ρ, if S∩xρ≠∅, then there is a unique element sx∈S such that x,sx∈ρ; and if S∩xρ=∅, we choose only one yx to define the function β as(4)aβ=sx,ifa∈xρandS∩xρ≠∅,yx,ifa∈xρandS∩xρ=∅,for all a∈X. It is readily seen that β fixes every element in S. Next, we let a,b∈X such that a,b∈ρ. Then a,b∈xρ for some xρ∈X/ρ. If S∩xρ≠∅, then aβ=sx=bβ. If S∩xρ=∅, then aβ=yx=bβ. Therefore, β∈EFSX,ρ. Since ρ≠ΔX and α∈EFSX,ρ, the function α is not one-to-one. Then there are distinct elements ya,yb∈X such that yaα=ybα. So at least one of ya and yb does not belong to ranβ, and hence, β is not onto. So β∈M. For all x∈X, xβα=xγ. This shows that βα=γ. Hence, Mα=EFSX,ρ, which implies that α is a right magnifying element.Example 2.
ConsiderX=Z and S=0. Let ρ be an equivalence relation on X such that X/ρ=0,−1,−2,−3,…,1,2,3,…. Assume that α is a function in EFSX,ρ defined by, for all x∈X,(5)xα=x2,ifx∈Z+and2∣x,−n,ifx∈Z+andx=2n−1for somen∈N,0,otherwise.
That is,(6)α=⋯−2−1012345⋯⋯000−11−22−3⋯.
Clearly, the functionα is onto. By Theorem 5, the function α is a right magnifying element. Let M=β∈EFSX,ρ∣βis not onto. Then there is a proper subset M of EFSX,ρ such that Mα=EFSX,ρ. Let γ be a function in EFSX,ρ defined by, for all x∈X,(7)xγ=−x,ifx∈Z+,0,ifx∈Z−∪0.
That is,(8)γ=⋯−2−1012345⋯⋯000−1−2−3−4−5⋯.
Then there is a functionβ∈M such that βα=γ. For more details, define a function β∈EFSX,ρ by, for all x∈X,(9)xβ=2x−1,ifx∈Z+,0,ifx∈Z−∪0.
That is,(10)β=⋯−2−1012345⋯⋯00013579⋯.
We obtain thatβ∈M and βα=γ, as required.Corollary 1.
The following statements hold:(1)
Suppose thatρ=ΔX and S≠X. Then α∈EFSX,ρ is a right magnifying element if and only if α is one-to-one but not onto.(2)
Suppose thatρ≠ΔX and ρ≠X×X. Let X be infinite and A=xρ∣S∩xρ=∅. If X and A have the same cardinality, and for all x∈X, S∩xρ≤1, then α∈EFSX,ρ is a right magnifying element if and only if α is onto.


### 4.2. Left Magnifying Elements

By Proposition2, we obtain the next theorem.Theorem 6 (see [14]).
Suppose thatρ=ΔX, S≠X,and there are infinite xρ∈X/ρ such that S∩xρ=∅. The function α is a left magnifying element in EFSX,ρ if and only if α is one-to-one but not onto.Next, we will consider the case in which the equivalence relationρ≠ΔX and ρ≠X×X.Lemma 5.
Suppose thatρ≠ΔX and ρ≠X×X. Let α be a function in EFSX,ρ. Then α is a left magnifying element if the following conditions hold:(1)
For eachxρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.Proof.
Letα be a function in EFSX,ρ satisfying(1)
For eachxρ∈X/ρ with xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.
Sinceyx∈ranα, there is a unique x′ρ∈X/ρ such that x′α=yx. Fix an element y0∈X and let M=β∈EFSX,ρ∣xβ=y0for allx∈Xsuch thatxρ∩ranα=∅. Clearly, M is a proper subset of EFSX,ρ. Let γ be a function in EFSX,ρ. We then define a function β by, for all x∈X,(11)xβ=x′γ,ifxρ∩ranα≠∅,y0,ifxρ∩ranα=∅.
Clearly,β∈M, and for all x∈X, xαβ=xγ. Hence, αβ=γ, and so αM=EFSX,ρ. Therefore, α is a left magnifying element.Lemma 6.
Suppose thatρ≠ΔX and ρ≠X×X. If α∈EFSX,ρ is a left magnifying element, then the following conditions hold:(1)
For eachxρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.Proof.
Sinceρ≠ΔX and ρ≠X×X, X>2. Let α∈EFSX,ρ be a left magnifying element. Then there exists a proper subset M of EFSX,ρ such that αM=EFSX,ρ. Suppose that there exists xρ∈X/ρ such that there are two distinct elements y1,y2∈xρ∩ranα. Since y1,y2∈ranα and y1≠y2, there are two distinct equivalence classes x1ρ,x2ρ∈X/ρ such that xα=y1 for all x∈x1ρ and xα=y2 for all x∈x2ρ.
Case 1 (S∩x1ρ≠∅ and S∩x2ρ≠∅): Then y1,y2∈S. This implies y1,y2∈S∩xρ, which is impossible, by Lemma 1.
Case 2 (S∩x1ρ=∅ and S∩x2ρ=∅): Note that, for any x∈X, by Lemma 1, if S∩xρ≠∅, then there exists a unique sx∈X such that sx∈S∩xρ. Let a,b,c∈X be distinct elements and let γ be a function in EFSX,ρ defined by, for all x∈X,(12)xγ=a,ifx∈x1ρ,b,ifx∈x2ρ,sx,ifS∩xρ≠∅,c,otherwise.
Then there is a functionβ∈M such that αβ=γ. Hence, a=x1γ=x1αβ=y1β=y2β=x2αβ=x2γ=b, which is a contradiction.
Case 3 (WLOG, S∩x1ρ≠∅ and S∩x2ρ=∅): By Lemma 1, if S∩x1ρ≠∅, then there exists a unique sx1∈X such that sx1∈S∩xρ, and hence, y1=sx1. Let a∈X such that sx1,a∉ρ, and let γ be a function in EFSX,ρ defined by, for all x∈X,(13)xγ=sx,ifS∩xρ≠∅,a,ifS∩xρ=∅.
Then there is a functionβ∈M such that αβ=γ. Since y1,y2∈xρ and β∈EFSX,ρ, we obtain sx1=x1γ=x1αβ=y1β=y2β=x2αβ=x2γ=a, which is a contradiction. Therefore, for each xρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα.
Lety∈ranα. Suppose that there are two distinct equivalence classes aρ,bρ∈X/ρ such that xα=y for all x∈aρ∪bρ.
Case 1 (S∩aρ≠∅ and S∩bρ≠∅): Then we have y∈S∩aρ and y∈S∩bρ. Hence, y∈aρ and y∈bρ, which is a contradiction.
Case 2 (S∩aρ=∅ and S∩bρ=∅): For any x∈X, by Lemma 1, if S∩xρ≠∅, then there exists a unique sx∈X such that sx∈S∩xρ. Let c∈X and define a function γ in EFSX,ρ by(14)xγ=a,ifx∈aρ,b,ifx∈bρ,sx,ifS∩xρ≠∅,c,otherwise,for all x∈X. Then there is a function β∈M such that αβ=γ. Hence, a=aγ=aαβ=yβ=bαβ=bγ=b, which is a contradiction.
Case 3 (WLOG, S∩aρ≠∅ and S∩bρ=∅): By Lemma 1, if S∩xρ≠∅, then there exists a unique sx∈X such that sx∈S∩xρ. Since S∩aρ≠∅ and xα=y for all x∈aρ, there exists a unique element sa∈S∩aρ such that sa=saα=y. Define a function γ in EFSX,ρ by(15)xγ=sx,ifS∩xρ≠∅,b,otherwise,for all x∈X. Then there is a function β∈M such that αβ=γ. Hence, y=sa=aγ=aαβ=yβ=bαβ=bγ=b. Then y∈bρ, and hence S∩bρ≠∅, which is a contradiction. All the cases show that there is a unique zρ∈X/ρ such that xα=y for all x∈zρ. Therefore, the proof is complete.By Lemmas5 and 6, we obtain the following theorem.Theorem 7.
Suppose thatρ≠ΔX and ρ≠X×X. A function α is a left magnifying element in EFSX,ρ if and only if the following conditions hold:(1)
For eachxρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.Example 3.
ConsiderX=N and S=1,5,9,13,…. Let ρ be an equivalence relation on X such that X/ρ=1,2,3,4,5,6,7,8,…. Assume that α is a function in EFSX,ρ defined by, for all x∈X,(16)xα=x,ifx≡1mod4,x−1,ifx≡2mod4,x+4,ifx≡3mod4,x+3,ifx≡0mod4.
That is,(17)α=123456789⋯11775511119⋯.
Clearly,X/ρ=1,2,3,4,5,6,7,8,…=1ρ,3ρ,5ρ,7ρ,… and ranα=1,5,9,13,…∪7,11,15,19,…. We can see that 1∈1ρ∩ranα, 5∈5ρ∩ranα, 7∈7ρ∩ranα, 9∈9ρ∩ranα, and 11∈11ρ∩ranα,…. So for each xρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα. Moreover, for all y∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ. By Theorem 7, α is a left magnifying element. Let M=β∈EFSX,ρ∣xβ=1for allx∈Xsuch thatxρ∩ranα=∅. Then there is a proper subset M of EFSX,ρ such that αM=EFSX,ρ. Let γ be a function in EFSX,ρ defined by, for all x∈X,(18)xγ=x,ifx≡1mod4,x−1,ifx≡2,3mod4,x−2,ifx≡0mod4.
That is,(19)γ=123456789⋯112255669⋯.
Then there is a functionβ∈M such that αβ=γ. Define a function β∈EFSX,ρ by, for all x∈X,(20)xβ=x,ifx≡1mod4,x−1,ifx≡2mod4,x−5,ifx≡3mod4,andx≠3,x−6,ifx≡0mod4,andx≠4,1,ifx=3,4.
That is,(21)β=123456789⋯111155229⋯.
Soβ∈M, and hence,(22)αβ=123456789⋯11775511119⋯123456789⋯111155229⋯=123456789⋯112255669⋯=γ.


## 4.1. Right Magnifying Elements

By Proposition2, we obtain the next theorem.Theorem 4 (see [14]).
Suppose thatρ=ΔX and S≠X. A function α is right magnifying in EFSX,ρ if and only if α is onto but not one-to-one.Our next two results are related to the existence of magnifying elements inEFSX,ρ.Lemma 1.
IfS∩xρ>1 for some x∈X, then EFSX,ρ is empty.Proof.
Letx∈X. Assume that S∩xρ>1. Suppose that there is an element α belonging to EFSX,ρ. By assumption, there are two distinct elements s1,s2∈S∩xρ. Then s1,s2∈ρ. So s1=s1α=s2α=s2, which is a contradiction.Lemma 2.
IfS∩xρ=1 for all x∈X, then EFSX,ρ=1.Proof.
Ifρ=ΔX and S∩xρ=1 for all x∈X, then X=S, and hence, EFSX,ρ=idX. So EFSX,ρ=1. Next, assume that ρ≠ΔX and S∩xρ=1 for all x∈X. Then S≠X, and hence for each x∈X∖S, there is a unique element sx∈S such that x,sx∈ρ. Define a function α by(1)xα=x,ifx∈S,sx,ifx∈X∖S,for all x∈X. Clearly, α∈EFSX,ρ. Next, we will show that α is the only element in EFSX,ρ. Let β be a function in EFSX,ρ. For all s∈S, sα=s=sβ. By assumption, for each x∈X∖S, there is a unique sx∈S such that x,sx∈ρ, and hence, xβ=sxβ=sx=xα. Therefore, α=β and EFSX,ρ=1.By Lemmas1 and 2, if ρ≠ΔX and either S∩xρ=1 for all x∈X or S∩xρ>1 for some x∈X, then there exists no magnifying element in EFSX,ρ.Example 1.
ConsiderX=N, S=1,2, and x,y∈ρ if and only if x≡ymod2.
Clearly,ρ is an equivalence relation on X and X/ρ=1,3,5,…,2,4,6,…. Let α be a function in EFSX,ρ defined by, for all x∈X,(2)xα=1,ifxis odd,2,ifxis even.
By Lemma2, the function α is the only function in EFSX,ρ. So EFSX,ρ has no nonempty proper subset, and hence, there exists no magnifying element in EFSX,ρ.
For the rest of this section, we focus on nontrivial cases, i.e.,ρ≠ΔX and ρ≠X×X, and establish the existence of right magnifying elements in EFSX,ρ. We thus assume now that EFSX,ρ≠∅.Lemma 3.
Suppose thatX is countably infinite. If there are infinite xρ∈X/ρ such that S∩xρ=∅, then there is a surjective function in EFSX,ρ. Consequently, if α∈EFSX,ρ is a right magnifying element, then α is onto.Proof.
Suppose that there are infinitexρ∈X/ρ such that S∩xρ=∅. Hence, X\S is infinite. Let A=xρ∣S∩xρ=∅. By assumption, A is infinite. Then there is a bijective function σ from A to X\S. For each x∈X∖S such that S∩xρ≠∅, there exists a unique sx∈S such that x,sx∈ρ, by Lemma 1. Define a function η by(3)xη=xρσ,ifx∈X\Sandxρ∈A,sx,ifx∈X\Sandxρ∉A,x,ifx∈S.
Clearly,η∈EFSX,ρ. Moreover, η is onto. Let α be a right magnifying element in EFSX,ρ. There is a proper subset M of EFSX,ρ such that Mα=EFSX,ρ. Then βα=η for some β∈M. This implies that α is onto.Suppose thatX is uncountably infinite and A=xρ∣S∩xρ=∅. It is easy to prove that if X and A have the same cardinality, then X∖S and A have the same cardinality as well. The proof of our next result is similar to the proof of Lemma 3 and so will be omitted.Lemma 4.
Suppose thatX is uncountably infinite and A=xρ∣S∩xρ=∅. If X and A have the same cardinality, then there is a surjective function in EFSX,ρ. Consequently, if α∈EFSX,ρ is a right magnifying elements, then α is onto.Theorem 5.
Suppose thatρ≠ΔX and ρ≠X×X. If S∩xρ≤1 for all x∈X and there are infinite xρ∈X/ρ such that S∩xρ=∅, then a function α∈EFSX,ρ is a right magnifying element if and only if α is onto.Proof.
By Lemmas3 and 4, the necessity is clear. Conversely, assume that α∈EFSX,ρ is onto. Let M=β∈EFSX,ρ∣βis not onto. Clearly, M is a proper subset of EFSX,ρ since α∉M. Let γ be a function in EFSX,ρ. Since α is onto, for each x∈X, there exists an element yx∈X such that yxα=xγ. For each xρ∈X/ρ, if S∩xρ≠∅, then there is a unique element sx∈S such that x,sx∈ρ; and if S∩xρ=∅, we choose only one yx to define the function β as(4)aβ=sx,ifa∈xρandS∩xρ≠∅,yx,ifa∈xρandS∩xρ=∅,for all a∈X. It is readily seen that β fixes every element in S. Next, we let a,b∈X such that a,b∈ρ. Then a,b∈xρ for some xρ∈X/ρ. If S∩xρ≠∅, then aβ=sx=bβ. If S∩xρ=∅, then aβ=yx=bβ. Therefore, β∈EFSX,ρ. Since ρ≠ΔX and α∈EFSX,ρ, the function α is not one-to-one. Then there are distinct elements ya,yb∈X such that yaα=ybα. So at least one of ya and yb does not belong to ranβ, and hence, β is not onto. So β∈M. For all x∈X, xβα=xγ. This shows that βα=γ. Hence, Mα=EFSX,ρ, which implies that α is a right magnifying element.Example 2.
ConsiderX=Z and S=0. Let ρ be an equivalence relation on X such that X/ρ=0,−1,−2,−3,…,1,2,3,…. Assume that α is a function in EFSX,ρ defined by, for all x∈X,(5)xα=x2,ifx∈Z+and2∣x,−n,ifx∈Z+andx=2n−1for somen∈N,0,otherwise.
That is,(6)α=⋯−2−1012345⋯⋯000−11−22−3⋯.
Clearly, the functionα is onto. By Theorem 5, the function α is a right magnifying element. Let M=β∈EFSX,ρ∣βis not onto. Then there is a proper subset M of EFSX,ρ such that Mα=EFSX,ρ. Let γ be a function in EFSX,ρ defined by, for all x∈X,(7)xγ=−x,ifx∈Z+,0,ifx∈Z−∪0.
That is,(8)γ=⋯−2−1012345⋯⋯000−1−2−3−4−5⋯.
Then there is a functionβ∈M such that βα=γ. For more details, define a function β∈EFSX,ρ by, for all x∈X,(9)xβ=2x−1,ifx∈Z+,0,ifx∈Z−∪0.
That is,(10)β=⋯−2−1012345⋯⋯00013579⋯.
We obtain thatβ∈M and βα=γ, as required.Corollary 1.
The following statements hold:(1)
Suppose thatρ=ΔX and S≠X. Then α∈EFSX,ρ is a right magnifying element if and only if α is one-to-one but not onto.(2)
Suppose thatρ≠ΔX and ρ≠X×X. Let X be infinite and A=xρ∣S∩xρ=∅. If X and A have the same cardinality, and for all x∈X, S∩xρ≤1, then α∈EFSX,ρ is a right magnifying element if and only if α is onto.


## 4.2. Left Magnifying Elements

By Proposition2, we obtain the next theorem.Theorem 6 (see [14]).
Suppose thatρ=ΔX, S≠X,and there are infinite xρ∈X/ρ such that S∩xρ=∅. The function α is a left magnifying element in EFSX,ρ if and only if α is one-to-one but not onto.Next, we will consider the case in which the equivalence relationρ≠ΔX and ρ≠X×X.Lemma 5.
Suppose thatρ≠ΔX and ρ≠X×X. Let α be a function in EFSX,ρ. Then α is a left magnifying element if the following conditions hold:(1)
For eachxρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.Proof.
Letα be a function in EFSX,ρ satisfying(1)
For eachxρ∈X/ρ with xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.
Sinceyx∈ranα, there is a unique x′ρ∈X/ρ such that x′α=yx. Fix an element y0∈X and let M=β∈EFSX,ρ∣xβ=y0for allx∈Xsuch thatxρ∩ranα=∅. Clearly, M is a proper subset of EFSX,ρ. Let γ be a function in EFSX,ρ. We then define a function β by, for all x∈X,(11)xβ=x′γ,ifxρ∩ranα≠∅,y0,ifxρ∩ranα=∅.
Clearly,β∈M, and for all x∈X, xαβ=xγ. Hence, αβ=γ, and so αM=EFSX,ρ. Therefore, α is a left magnifying element.Lemma 6.
Suppose thatρ≠ΔX and ρ≠X×X. If α∈EFSX,ρ is a left magnifying element, then the following conditions hold:(1)
For eachxρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.Proof.
Sinceρ≠ΔX and ρ≠X×X, X>2. Let α∈EFSX,ρ be a left magnifying element. Then there exists a proper subset M of EFSX,ρ such that αM=EFSX,ρ. Suppose that there exists xρ∈X/ρ such that there are two distinct elements y1,y2∈xρ∩ranα. Since y1,y2∈ranα and y1≠y2, there are two distinct equivalence classes x1ρ,x2ρ∈X/ρ such that xα=y1 for all x∈x1ρ and xα=y2 for all x∈x2ρ.
Case 1 (S∩x1ρ≠∅ and S∩x2ρ≠∅): Then y1,y2∈S. This implies y1,y2∈S∩xρ, which is impossible, by Lemma 1.
Case 2 (S∩x1ρ=∅ and S∩x2ρ=∅): Note that, for any x∈X, by Lemma 1, if S∩xρ≠∅, then there exists a unique sx∈X such that sx∈S∩xρ. Let a,b,c∈X be distinct elements and let γ be a function in EFSX,ρ defined by, for all x∈X,(12)xγ=a,ifx∈x1ρ,b,ifx∈x2ρ,sx,ifS∩xρ≠∅,c,otherwise.
Then there is a functionβ∈M such that αβ=γ. Hence, a=x1γ=x1αβ=y1β=y2β=x2αβ=x2γ=b, which is a contradiction.
Case 3 (WLOG, S∩x1ρ≠∅ and S∩x2ρ=∅): By Lemma 1, if S∩x1ρ≠∅, then there exists a unique sx1∈X such that sx1∈S∩xρ, and hence, y1=sx1. Let a∈X such that sx1,a∉ρ, and let γ be a function in EFSX,ρ defined by, for all x∈X,(13)xγ=sx,ifS∩xρ≠∅,a,ifS∩xρ=∅.
Then there is a functionβ∈M such that αβ=γ. Since y1,y2∈xρ and β∈EFSX,ρ, we obtain sx1=x1γ=x1αβ=y1β=y2β=x2αβ=x2γ=a, which is a contradiction. Therefore, for each xρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα.
Lety∈ranα. Suppose that there are two distinct equivalence classes aρ,bρ∈X/ρ such that xα=y for all x∈aρ∪bρ.
Case 1 (S∩aρ≠∅ and S∩bρ≠∅): Then we have y∈S∩aρ and y∈S∩bρ. Hence, y∈aρ and y∈bρ, which is a contradiction.
Case 2 (S∩aρ=∅ and S∩bρ=∅): For any x∈X, by Lemma 1, if S∩xρ≠∅, then there exists a unique sx∈X such that sx∈S∩xρ. Let c∈X and define a function γ in EFSX,ρ by(14)xγ=a,ifx∈aρ,b,ifx∈bρ,sx,ifS∩xρ≠∅,c,otherwise,for all x∈X. Then there is a function β∈M such that αβ=γ. Hence, a=aγ=aαβ=yβ=bαβ=bγ=b, which is a contradiction.
Case 3 (WLOG, S∩aρ≠∅ and S∩bρ=∅): By Lemma 1, if S∩xρ≠∅, then there exists a unique sx∈X such that sx∈S∩xρ. Since S∩aρ≠∅ and xα=y for all x∈aρ, there exists a unique element sa∈S∩aρ such that sa=saα=y. Define a function γ in EFSX,ρ by(15)xγ=sx,ifS∩xρ≠∅,b,otherwise,for all x∈X. Then there is a function β∈M such that αβ=γ. Hence, y=sa=aγ=aαβ=yβ=bαβ=bγ=b. Then y∈bρ, and hence S∩bρ≠∅, which is a contradiction. All the cases show that there is a unique zρ∈X/ρ such that xα=y for all x∈zρ. Therefore, the proof is complete.By Lemmas5 and 6, we obtain the following theorem.Theorem 7.
Suppose thatρ≠ΔX and ρ≠X×X. A function α is a left magnifying element in EFSX,ρ if and only if the following conditions hold:(1)
For eachxρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα,(2)
For ally∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ.Example 3.
ConsiderX=N and S=1,5,9,13,…. Let ρ be an equivalence relation on X such that X/ρ=1,2,3,4,5,6,7,8,…. Assume that α is a function in EFSX,ρ defined by, for all x∈X,(16)xα=x,ifx≡1mod4,x−1,ifx≡2mod4,x+4,ifx≡3mod4,x+3,ifx≡0mod4.
That is,(17)α=123456789⋯11775511119⋯.
Clearly,X/ρ=1,2,3,4,5,6,7,8,…=1ρ,3ρ,5ρ,7ρ,… and ranα=1,5,9,13,…∪7,11,15,19,…. We can see that 1∈1ρ∩ranα, 5∈5ρ∩ranα, 7∈7ρ∩ranα, 9∈9ρ∩ranα, and 11∈11ρ∩ranα,…. So for each xρ∈X/ρ such that xρ∩ranα≠∅, there exists a unique element yx∈xρ∩ranα. Moreover, for all y∈ranα, there is a unique zρ∈X/ρ such that xα=y for all x∈zρ. By Theorem 7, α is a left magnifying element. Let M=β∈EFSX,ρ∣xβ=1for allx∈Xsuch thatxρ∩ranα=∅. Then there is a proper subset M of EFSX,ρ such that αM=EFSX,ρ. Let γ be a function in EFSX,ρ defined by, for all x∈X,(18)xγ=x,ifx≡1mod4,x−1,ifx≡2,3mod4,x−2,ifx≡0mod4.
That is,(19)γ=123456789⋯112255669⋯.
Then there is a functionβ∈M such that αβ=γ. Define a function β∈EFSX,ρ by, for all x∈X,(20)xβ=x,ifx≡1mod4,x−1,ifx≡2mod4,x−5,ifx≡3mod4,andx≠3,x−6,ifx≡0mod4,andx≠4,1,ifx=3,4.
That is,(21)β=123456789⋯111155229⋯.
Soβ∈M, and hence,(22)αβ=123456789⋯11775511119⋯123456789⋯111155229⋯=123456789⋯112255669⋯=γ.


## 5. Conclusion

In this paper, some particular properties ofEFSX,ρ are investigated in Section 3. The main task of this paper is to establish the necessary and sufficient conditions for an elements α∈EFSX,ρ to be right or left magnifying elements, which are summarized as follows:Ifρ=ΔX and S≠X, the following statements are true:(1)
The functionα is a right magnifying element if and only if α is onto but not one-to-one.(2)
Suppose that there are infinite equivalence classesxρ such that S∩xρ=∅. Then α is a left magnifying element if and only if α is one-to-one but not onto.Ifρ≠ΔX and ρ≠X×X, the following statements are true:(1)
If the setX and the set of all equivalence classes xρ such that S∩xρ=∅ have the same cardinality, and for all x∈X, S∩xρ≤1, then α is a right magnifying element if and only if α is onto.(2)
The functionα is a left magnifying element if and only if(i)
Ifxρ∩ranα is nonempty, then it is singleton,(ii)
For ally∈ranα, there is a unique equivalence class xρ whose members are all mapped to y.



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