**4. Left Regular and Right Regular Elements in** *S***(***X***,** *Y***)**

**Theorem 17** ([18])**.** *Let α* ∈ *S*(*X*,*Y*)*. Then, α is left regular if and only if Xα* = *Xα*<sup>2</sup> *and Yα* = *Yα*2*.*

**Theorem 18** ([18])**.** *Let α* ∈ *S*(*X*,*Y*)*. Then, α is right regular if and only if π*(*α*) = *π*(*α*2) *and σ*(*α*) = *σ*(*α*2) *where σ*(*α*) = {*yα*−<sup>1</sup> : *y* ∈ *Xα* ∩ *Y*}*.*

Although the left and right regular elements of *S*(*X*,*Y*) were characterized in [18], in this section we obtain the different results; see the following theorems.

**Theorem 19.** *Let α* ∈ *S*(*X*,*Y*)*. Then, α is a left regular element if and only if for every P* ∈ *π*(*α*)*, P* ∩ *Xα* = ∅ *and for every P* ∈ *πY*(*α*)*, P* ∩ *Yα* = ∅*.*

**Proof.** Assume that *α* is a left regular element. Thus, *α* = *βα*<sup>2</sup> for some *β* ∈ *S*(*X*,*Y*). Let *P* ∈ *π*(*α*) and let *x* ∈ *P*. Then, *P* = (*xα*)*α*−<sup>1</sup> and *xα* = *xβα*<sup>2</sup> = [(*xβ*)*α*]*α*. We see that (*xβ*)*α* ∈ (*xα*)*α*−<sup>1</sup> = *P* and (*xβ*)*α* ∈ *Xα*. Therefore, *P* ∩ *Xα* = ∅. Let *P* ∈ *πY*(*α*) and let *y* ∈ *P* ∩*Y*. Then, *P* = (*yα*)*α*<sup>−</sup>1, *yβ* ∈ *Y* and *yα* = *yβα*<sup>2</sup> = [(*yβ*)*α*]*α*. We note that (*yβ*)*α* ∈ *Yα* and (*yβ*)*α* ∈ (*yα*)*α*−<sup>1</sup> = *P*. Hence, *P* ∩ *Yα* = ∅.

Conversely, suppose the conditions hold. Let *x* ∈ *X*. Since *π*(*α*) is a partition of *X*, there is a unique *Px* ∈ *π*(*α*) satisfying *x* ∈ *Px*. If *Px* ∩ *Y* = ∅, then *Px* ∈ *πY*(*α*) and *Px* ∩ *Yα* = ∅ by our assumption. So, we choose *x* ∈ *Y* satisfying *x α* ∈ *Px*. If *Px* ∩ *Y* = ∅, then since *Px* ∩ *Xα* = ∅, we choose *x* ∈ *X* satisfying *x α* ∈ *Px*. Define *β* : *X* → *X* by *xβ* = *x* for all *x* ∈ *X*. Then, *β* is well-defined and *xβα*<sup>2</sup> = *x α*<sup>2</sup> = (*x α*)*α* = *xα*. Let *y* ∈ *Y*. Then, there is a unique *Py* ∈ *π*(*α*) such that *y* ∈ *Py*. Thus, *Py* ∩ *Y* = ∅. Therefore, *yβ* = *y* ∈ *Y*; that is, *Yβ* ⊆ *Y*. So *α* is left regular.

**Theorem 20.** *Let α* ∈ *S*(*X*,*Y*)*. Then, α is a right regular element if and only if α*|*X<sup>α</sup> is injective and* (*Xα* \ *Y*)*α* ⊆ *X* \ *Y.*

**Proof.** Assume that *α* is right regular. Thus, *α* is also a right regular element in *T*(*X*). From Corollary 2, we have *α*|*X<sup>α</sup>* is injective. Next, we will show that (*Xα* \ *Y*)*α* ⊆ *X* \ *Y*. Let *z* ∈ *Xα* \ *Y*. Then, *z* = *z α* for some *z* ∈ *X*. Thus, *z* = *z α*2*β* for some *β* ∈ *S*(*X*,*Y*) since *α* is a right regular element in *S*(*X*,*Y*). If *z α*<sup>2</sup> ∈ *Y*, then *z* = (*z α*2)*β* ∈ *Yβ* ⊆ *Y*, which is a contradiction. Hence, *zα* = *z α*<sup>2</sup> ∈/ *Y* and so (*Xα* \ *Y*)*α* ⊆ *X* \ *Y*.

Assume that *α*|*X<sup>α</sup>* is injective and (*Xα* \ *Y*)*α* ⊆ *X* \ *Y*. Let *β* be defined in the converse part of Theorem 10; we note that *α* = *α*2*β*. It is enough to verify that *β* ∈ *S*(*X*,*Y*). Let *x* ∈ *Y*. If *x* ∈/ *Xα*2, then by the definition of *β*, we have *xβ* ∈ *Y*. Assume that *x* ∈ *Xα*2. There is a unique *x* ∈ *Xα* satisfying *x α* = *x*. If *x* ∈/ *Y*, then *x* ∈ *Xα* \ *Y*. By assumption, we have *x α* ∈ *X* \ *Y* which is a contradiction. This means that *xβ* = *x* ∈ *Y*. Hence, *Yβ* ⊆ *Y* and so *β* ∈ *S*(*X*,*Y*).

**Example 9.** *Let X* = N *and Y* = 2N*. Define α* : *X* → *X by*

$$\mathfrak{x}\mathfrak{a} = \begin{cases} 2 & \text{if } \mathfrak{x} \in \mathcal{Y}\_{\prime} \\ -4 & \text{if } \mathfrak{x} = 1\_{\prime} \\ \mathfrak{x} - 2 & \text{otherwise} \end{cases}$$

*Then, Xα* = {2, 4}∪{2*n* − 1 : *n* ∈ N} *and Yα* = {2} ⊆ *Y. So α* ∈ *S*(*X*,*Y*)*. Moreover, we obtain π*(*α*) = {*Y*} ∪ {{2*n* − 1} : *n* ∈ N} *and πY*(*α*) = {*Y*}*. It is clear that P* ∩ *Xα* = ∅ *for every P* ∈ *π*(*α*) *and P* ∩ *Yα* = ∅ *for all P* ∈ *πY*(*α*)*. From Theorem 19, α is left regular. Note that Xα* ∩ *Y* = {2, 4} = *Yα. By Theorem 3, α is also not regular.*

**Example 10.** *Let α be defined in Example 2. Then, Yα* ⊆ *Y and also Xα* ∩ *Y* = *Yα. Thus, α* ∈ *S*(*X*,*Y*) *and α is regular. Note that* 4*α*−<sup>1</sup> = {1, 2} *and Xα* ∩ 4*α*−<sup>1</sup> = ∅*. Hence, α is not a left regular element of S*(*X*,*Y*)*.*

**Example 11.** *Let α be defined in Example 5. Then, Yα* ⊆ *Y and so α* ∈ *S*(*X*,*Y*)*. Consider Y* ∩ *Xα* = 2N \ {2} = *Yα and α*|*X<sup>α</sup> is not injective. Hence, α is regular but not right regular in S*(*X*,*Y*)*.*

**Example 12.** *Recall α from Example 4. Then, Yα* ⊆ *Y and so α* ∈ *S*(*X*,*Y*)*. We see that α*|*X<sup>α</sup> is injective and* (*Xα* \ *Y*)*α* = ∅ ⊆ *X* \ *Y. From Theorem 20, we obtain α is right regular. Consider Xα* ∩ *Y* = 2N *and Yα* = 2N \ {2}*. Hence, Xα* ∩ *Y* = *Yα. From Theorem 3, we obtain α is not regular.*

**Theorem 21.** *The following statements are equivalent.*

*(1) S*(*X*,*Y*) = *LReg*(*S*(*X*,*Y*))*. (2) S*(*X*,*Y*) = *RReg*(*S*(*X*,*Y*))*. (3)* |*X*| ≤ 2*.*

**Proof.** (1) ⇔ (3). Assume that *S*(*X*,*Y*) = *LReg*(*S*(*X*,*Y*)). We will show that |*X*| ≤ 2. Suppose that |*X*| ≥ 3. Let *a*, *b*, *c* ∈ *X* be distinct elements and *a* ∈ *Y*. Define *α* : *X* → *X* by

$$\mathfrak{x}\mathfrak{a} = \begin{cases} b & \text{if } \mathfrak{x} = c, \\ a & \text{if } \mathfrak{x} \neq c. \end{cases}$$

Claim that *α* ∈ *S*(*X*,*Y*). If *c* ∈/ *Y*, then *Yα* = {*a*} ⊆ *Y*. Moreover, if *b*, *c* ∈ *Y*, then *Yα* = {*a*, *b*} ⊆ *Y*. Thus, *α* ∈ *S*(*X*,*Y*) when *c* ∈/ *Y* or *b*, *c* ∈ *Y*. Note that *π*(*α*) = {{*c*}, *X* \ {*c*}} and *Xα* = {*a*, *b*}. Clearly, *Xα* ∩ {*c*} = ∅. Hence, *α* is not left regular. For the case *b* ∈/ *Y* and *c* ∈ *Y*, we define *β* : *X* → *X* by

$$\mathfrak{x}\mathfrak{z} = \begin{cases} \mathfrak{c} & \text{if } \mathfrak{x} = b\_{\mathfrak{z}} \\ a & \text{if } \mathfrak{x} \neq b\_{\mathfrak{z}} \end{cases}$$

Then, *Yβ* = {*a*} ⊆ *Y*, *π*(*β*) = {{*b*}, *X* \ {*b*}} and *Xβ* = {*a*, *c*}. It follows that *β* ∈ *S*(*X*,*Y*) but *β* ∈/ *LReg*(*S*(*X*,*Y*)). We conclude that *S*(*X*,*Y*) = *LReg*(*S*(*X*,*Y*)), which is a contradiction. So |*X*| ≤ 2.

Conversely, assume that |*X*| ≤ 2. Then, it is easy to verify that *S*(*X*,*Y*) is a left regular semigroup. Hence, *S*(*X*,*Y*) = *LReg*(*S*(*X*,*Y*)).

(2) ⇔ (3). Assume that *S*(*X*,*Y*) = *RReg*(*S*(*X*,*Y*)). We will show that |*X*| ≤ 2. Suppose that |*X*| ≥ 3. We consider *α*, *β* from condition (1) ⇔ (3). Then *α*, *β* ∈ *S*(*X*,*Y*). Since *Xα* = {*a*, *b*} and *α*|*X<sup>α</sup>* is not injective, we have *α* is not right regular. Similarly, *Xβ* = {*a*, *c*} and *β*|*X<sup>β</sup>* is not injective. So *β* is not right regular, which is a contradition. Hence, |*X*| ≤ 2.

Conversely, suppose that |*X*| ≤ 2. Then, it is clear that *S*(*X*,*Y*) is a right regular semigroup. Hence, *S*(*X*,*Y*) = *RReg*(*S*(*X*,*Y*)).

**Theorem 22.** *The following statements are equivalent.*


**Proof.** (1) ⇔ (3). Suppose that |*X*| ≥ 3. Let *a*, *b*, *c* ∈ *X* be distinct elements and *a* ∈ *Y*. It is enough to consider only two cases.

Case 1: {*b*, *c*} ⊆ *Y*. Recall *α*, *β* from Theorem 14, we have *α*, *β* ∈ *S*(*X*,*Y*). Note that *π*(*α*) = *πY*(*α*) = {{*c*}, *X* \ {*c*}} and *Xα* = {*b*, *c*}. Clearly, *Yα* ∩ {*c*} = ∅ and *Yα* ∩ *X* \ {*c*} = ∅. Thus, *α* is left regular; that is, *α* ∈ *LReg*(*S*(*X*,*Y*)). Similarly, *β* ∈ *LReg*(*S*(*X*,*Y*)). Consider {*c*} ∈ *π*(*αβ*) and *Xαβ* = {*a*, *b*}. Therefore, *Xαβ* ∩ {*c*} = ∅ and so *αβ* is not left regular; that is, *αβ* ∈/ *LReg*(*S*(*X*,*Y*)). Hence, *LReg*(*S*(*X*,*Y*)) is not a semigroup.

Case 2: {*b*, *c*} ⊆ *Y*. Assume that *c* ∈/ *Y*. Define *α* : *X* → *X* by

$$\mathfrak{x}\mathfrak{a} = \begin{cases} \mathfrak{c} & \text{if } \mathfrak{x} = \mathfrak{c}, \\ a & \text{otherwise.} \end{cases}$$

Then, *Yα* = {*a*} ⊆ *Y*. So *α* ∈ *S*(*X*,*Y*). Note that *π*(*α*) = {{*c*}, *X* \ {*c*}}, *πY*(*α*) = {*X* \ {*c*}} and *Xα* = {*a*, *c*}. Clearly, *Xα* ∩ {*c*} = ∅, *Xα* ∩ (*X* \ {*c*}) = ∅ and *Yα* ∩ (*X* \ {*c*}) = ∅. Therefore, *α* is a left regular element of *S*(*X*,*Y*). Define *β* : *X* → *X* by

$$x \emptyset = \begin{cases} b & \text{if } x \in \{b, c\}, \\ a & \text{otherwise}. \end{cases}$$

If *b* ∈ *Y*, then *Yβ* = {*a*, *b*} ⊆ *Y*. Thus, *β* ∈ *S*(*X*,*Y*). If *b* ∈/ *Y*, then *Yβ* = {*a*} ⊆ *Y*. So *β* ∈ *S*(*X*,*Y*). We can show that *β* ∈ *LReg*(*S*(*X*,*Y*)). Then, we note that *π*(*αβ*) = {{*c*}, *X* \ {*c*}} and *Xαβ* = {*a*, *b*}. Clearly, *Xαβ* ∩ {*c*} = ∅. Hence, *αβ* ∈/ *LReg*(*S*(*X*,*Y*)) and so *LReg*(*S*(*X*,*Y*)) is not a semigroup.

Conversely, suppose |*X*| ≤ 2. Then, we have *LReg*(*S*(*X*,*Y*)) = *S*(*X*,*Y*) is a semigroup from Theorem 21(1).

(2) ⇔ (3). Suppose that |*X*| ≥ 3. Let *a*, *b*, *c* ∈ *X* be distinct elements and *a* ∈ *Y*. Recall *α* and *β* from the proof of (1) ⇔ (3). It is enough to show that *α* and *β* are right regular elements of *S*(*X*,*Y*). Clearly, *α*|*X<sup>α</sup>* and *β*|*X<sup>β</sup>* are injective. Consider (*Xα* \*Y*)*α* = {*c*} ⊆ *X* \*Y*

and (*Xβ* \ *Y*)*β* = ∅ if *b* ∈ *Y*, {*b*} if *<sup>b</sup>* <sup>∈</sup>/ *<sup>Y</sup>* <sup>⊆</sup> *<sup>X</sup>* \ *<sup>Y</sup>*.

Then, *α* and *β* are right regular elements of *S*(*X*,*Y*); that is, *α*, *β* ∈ *RReg*(*S*(*X*,*Y*)). Note that *αβ*|*Xαβ* is not injective. We conclude that *αβ* is not right regular. Thus, *αβ* ∈/ *RReg*(*S*(*X*,*Y*)) and so *RReg*(*S*(*X*,*Y*)) is not a semigroup.

Conversely, suppose |*X*| ≤ 2. Then, we have *RReg*(*S*(*X*,*Y*)) = *S*(*X*,*Y*) is a semigroup from Theorem 21(2).

**Author Contributions:** Conceptualization, E.L.; Methodology, K.S.; Investigation, W.S.; Writing original draft, K.S. and E.L.; Writing—review & editing, W.S. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Acknowledgments:** I would like to thank the referee for his/her valuable suggestions and comments which helped to improve the readability of this paper. Moreover, we thank The 25th Annual Meeting in Mathematics (AMM 2021) for a good comments about our presentation and inspiring to find the number of left and right regular elements of *T*(*X*,*Y*) as well as characterize left and right regular elements of *S*(*X*,*Y*).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


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