**3. Left Regular and Right Regular Elements in** *T***(***X***,** *Y***)**

Now, we start with the characterizations of left regular and right regular elements in *T*(*X*,*Y*). Moreover, we determine whenever *T*(*X*,*Y*) becomes a left regular semigroup and a right regular semigroup, respectively.

**Theorem 9.** *Let α* ∈ *T*(*X*,*Y*)*. Then, the following statements are equivalent.*

*(1) α is left regular.*

*(2) Yα*<sup>2</sup> = *Xα. (3) for any P* ∈ *π*(*α*)*, Yα* ∩ *P* = ∅*.*

**Proof.** (1) ⇒ (2). Suppose that *α* is left regular. Thus, there exists *β* ∈ *T*(*X*,*Y*) satisfying *α* = *βα*2. This implies that *Yα*<sup>2</sup> ⊆ *Xα* = *Xβα*<sup>2</sup> ⊆ *Yα*2. Thus, *Yα*<sup>2</sup> = *Xα*.

(2) ⇒ (3). Suppose that *Yα*<sup>2</sup> = *Xα* and let *P* ∈ *π*(*α*). Thus, there is *y* ∈ *Xα* satisfying *yα*−<sup>1</sup> = *P*. By assumption, we have *y* ∈ *Yα*2. Thus, there exists *z* ∈ *Y* such that *y* = *zα*2; that is, *zα* ∈ *yα*−<sup>1</sup> = *P*. It follows that *zα* ∈ *Yα* ∩ *P*. Therefore, *Yα* ∩ *P* = ∅.

(3) ⇒ (1). Assume that (3) holds. For *x* ∈ *X*, there is a unique *Px* ∈ *π*(*α*) satisfying *x* ∈ *Px*. From assumption, we have *Yα* ∩ *Px* = ∅. So there exists *x* ∈ *Y* such that *x α* ∈ *Px* = (*xα*)*α*<sup>−</sup>1. Define *β* : *X* → *X* by *xβ* = *x* for all *x* ∈ *X*. It is obvious that *β* ∈ *T*(*X*,*Y*). Let *x* ∈ *X*. This implies that *xβα*<sup>2</sup> = *x α*<sup>2</sup> = (*x α*)*α* = *xα*. Hence, *α* = *βα*<sup>2</sup> and so *α* is left regular.

If we replace *Y* with *X* in Theorem 9, we have the following corollary.

**Corollary 1.** *Let α* ∈ *T*(*X*)*. Then, α is left regular if and only if for each P* ∈ *π*(*α*), *Xα* ∩ *P* = ∅*.*

**Proof.** By taking *X* = *Y*, we obtain *T*(*X*,*Y*) = *T*(*X*, *X*) = *T*(*X*) and *Xα* = *Yα*. By Theorem 9(3), we obtain that *α* is regular if and only if for each *P* ∈ *π*(*α*), *Xα* ∩ *P* = *Yα* ∩ *P* = ∅.

**Theorem 10.** *Let α* ∈ *T*(*X*,*Y*)*. Then, α is right regular if and only if α*|*X<sup>α</sup> is injective.*

**Proof.** Assume that *α* is right regular. Then, there exists *β* ∈ *T*(*X*,*Y*) such that *α* = *α*2*β*. We will show that *α*|*X<sup>α</sup>* is injective. Let *x*, *y* ∈ *Xα* be such that *xα* = *yα*. Thus, there exist *x* , *y* ∈ *X* such that *x* = *x α* and *y* = *y α*. We obtain that

$$\mathfrak{a} = \mathfrak{x}'\mathfrak{a} = \mathfrak{x}'\mathfrak{a}^2\mathfrak{z} = (\mathfrak{x}'\mathfrak{a})\mathfrak{a}\mathfrak{z} = \mathfrak{x}\mathfrak{a}\mathfrak{z} = \mathfrak{y}\mathfrak{a}\mathfrak{z} = \mathfrak{y}'\mathfrak{a}^2\mathfrak{z} = \mathfrak{y}'\mathfrak{a} = \mathfrak{y}.$$

Therefore, *α*|*X<sup>α</sup>* is injective.

Conversely, suppose *α*|*X<sup>α</sup>* is injective. Let *z* ∈ *Xα*2. Then, there exists a unique *z* ∈ *Xα* such that *z α* = *z*. We choose *y* ∈ *Y*. Define *β* : *X* → *X* by

$$z\beta = \begin{cases} z' & \text{if } z \in Xa^2, \\ y & \text{otherwise.} \end{cases}$$

From *Xα* ⊆ *Y*, we obtain that *Xβ* ⊆ *Xα* ∪ {*y*} ⊆ *Y*. Let *x* ∈ *X*. Note that *xα*<sup>2</sup> ∈ *Xα*<sup>2</sup> and (*xα*2)*β* = (*xα*2) where (*xα*2) *α* = *xα*<sup>2</sup> = *xαα*. Since *α*|*X<sup>α</sup>* is injective, we have (*xα*2) = *xα*. So *xα*2*β* = (*xα*2) = *xα*. Hence, *α*2*β* = *α* and so *α* is right regular.

**Corollary 2.** *Let α* ∈ *T*(*X*)*. Then, α is right regular if and only if α*|*X<sup>α</sup> is injective.*

From Theorems 4 and 10, we obtain the following corollary immediately.

**Corollary 3.** *For X* = *Y, every left magnifying element of a semigroup T*(*X*,*Y*) *is right regular.*

**Corollary 4.** *Every right magnifying element of a semigroup T*(*X*,*Y*) *is left regular.*

**Proof.** Let *α* be a right magnifying element. By Remark 1, we have *Yα* = *Y*. It follows that *Xα* ⊆ *Y* = *Yα* = (*Yα*)*α* = *Yα*<sup>2</sup> ⊆ *Xα*. Hence, *Yα*<sup>2</sup> = *Xα* and so *α* is left regular.

**Example 4.** *Let X* = N *and Y* = 2N*. Define α* : *X* → *Y by*

$$\mathfrak{X}\mathfrak{a} = \begin{cases} \mathfrak{x} + 2 & \text{if } \mathfrak{x} \in 2\mathbb{N}\_{\prime} \\ \mathfrak{x} + 1 & \text{otherwise} \end{cases}$$

*Then, α* ∈ *T*(*X*,*Y*)*. Clearly, α is not injective. We see that Xα* = 2N*. This means that α*|*X<sup>α</sup>* : *Xα* → *Xα is an injection. This means that α is right regular but it is not left magnifying.*

**Example 5.** *Let X* = N *and Y* = 2N*. Define α* : *X* → *Y by*

$$\mathfrak{X}\mathfrak{A} = \begin{cases} 4 & \text{if } \mathfrak{x} \in \{1, 2, 3, 4\}, \\ \mathfrak{x} - 2 & \text{if } \mathfrak{x} \in 2\mathbb{N} \backslash \{2, 4\}, \\ \mathfrak{x} + 1 & \text{otherwise}. \end{cases}$$

*Clearly, α* ∈ *T*(*X*,*Y*)*. We see that Xα* = 2N \ {2} = *Yα*<sup>2</sup> *and Yα* = 2N \ {2} = *Y. Hence, α is a left regular element but it is not right magnifying.*

Notice that for |*X*| ≤ 2, we obtain that *T*(*X*,*Y*) is left and right regular. Now, we consider the other case.

**Theorem 11.** *Let X* = *Y be such that* |*X*| ≥ 3*. Then, T*(*X*,*Y*) *is a right regular semigroup if and only if* |*Y*| = 1*.*

**Proof.** If |*Y*| = 1, then |*T*(*X*,*Y*)| = 1 and *T*(*X*,*Y*) is a right regular semigroup. Assume that *T*(*X*,*Y*) is a right regular semigroup and suppose |*Y*| = 1. Let *a*, *b*, *c* ∈ *X* be distinct elements and *a*, *b* ∈ *Y*. Define *α* : *X* → *X* by

$$\mathfrak{X} \mathfrak{A} = \begin{cases} a & \text{if } \mathfrak{x} \in \{a, b\}, \\ b & \text{otherwise}. \end{cases}$$

Then, *α* ∈ *T*(*X*,*Y*). However, *α*|*X<sup>α</sup>* is not injective. Thus, *α* is not right regular, which is a contradiction. Hence, |*Y*| = 1.

**Theorem 12.** *Let X* = *Y be such that* |*X*| ≥ 3*. Then, T*(*X*,*Y*) *is a left regular semigroup if and only if* |*Y*| = 1*.*

**Proof.** If |*Y*| = 1, then |*T*(*X*,*Y*)| = 1 and *T*(*X*,*Y*) is a left regular semigroup. Assume that *T*(*X*,*Y*) is a left regular semigroup and |*Y*| = 1. Let *a*, *b*, *c* ∈ *X* be distinct elements and *a*, *b* ∈ *Y*. Define *α* : *X* → *X* by

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

Then, *α* ∈ *T*(*X*,*Y*). Note that *a*, *b* ∈/ *bα*−<sup>1</sup> and *Yα* ⊆ {*a*, *b*}. So *Yα* ∩ *bα*−<sup>1</sup> = ∅. Therefore, *α* is not left regular. This is a contradiction. Hence, |*Y*| = 1.

**Corollary 5.** *Every left regular element of a semigroup T*(*X*,*Y*) *is regular.*

**Proof.** We first note that *F*(*X*,*Y*) = {*α* ∈ *T*(*X*,*Y*) : *Xα* = *Yα*} is the largest regular subsemigroup of *T*(*X*,*Y*). Let *α* be a left regular element of *T*(*X*,*Y*). It follows from Theorem 9(2) that *Xα* = *Yα*<sup>2</sup> = (*Yα*)*α* ⊆ *Yα* ⊆ *Xα*. Thus, *Xα* = *Yα* and so *α* ∈ *F*(*X*,*Y*). Therefore, *α* is a regular element of *T*(*X*,*Y*).

**Example 6.** *Let X* = N *and Y* = 2N*. Define α* : *X* → *Y by*

$$\mathfrak{X}\mathfrak{a} = \begin{cases} \mathfrak{x} + 2 & \text{if } \mathfrak{x} \in 2\mathbb{N}\_{\prime} \\ \mathfrak{x} + 3 & \text{otherwise} \end{cases}$$

*Clearly, α* ∈ *T*(*X*,*Y*)*. Consider Xα* = *Yα* = 2N \ {2} *and Yα*<sup>2</sup> = 2N \ {2, 4}*. Then, α is regular but it is not left regular.*

**Example 7.** *Let X* = N *and Y* = 2N + 1*. Define α* : *X* → *Y by*

*xα* = *x* + 1 *if x* ∈ 2N, *x* + 2 *otherwise.*

*It is verifiable that α* ∈ *T*(*X*,*Y*)*. We obtain that α*|*X<sup>α</sup> is injective and Xα* = *Yα. Thus, α is right regular but it is not regular.*

Finally, we consider the set of all left regular elements *LReg*(*T*(*X*,*Y*)) and the set of all right regular elements *RReg*(*T*(*X*,*Y*)) of *T*(*X*,*Y*). We begin with the following example.

**Example 8.** *Let X* = {1, 2, 3, 4, 5} *and Y* = {1, 2, 3, 4}*. We consider the mappings α* = 12345 33343! *and β* = 12345 21121! *. We note that α*, *β* ∈ *T*(*X*,*Y*)*. Moreover, α*|*X<sup>α</sup> and β*|*X<sup>β</sup> are injective; that is α*, *β* ∈ *RReg*(*T*(*X*,*Y*))*. Clearly, αβ* = 12345 11121! *. This implies that αβ*|*Xαβ is not injective and so αβ* ∈/ *RReg*(*T*(*X*,*Y*))*. In this case, we obtain that RReg*(*T*(*X*,*Y*)) *is not a semigroup.*

**Theorem 13.** *Let* |*X*| ≥ 3*. Then, RReg*(*T*(*X*,*Y*)) *is a semigroup if and only if* |*Y*| ≤ 2*.*

**Proof.** Suppose that |*Y*| ≥ 3. Let *a*, *b*, *c* ∈ *Y* be distict elements. Define *α* : *X* → *Y* by

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

We see that *Xα* = {*a*, *b*, *c*} and *α*|*X<sup>α</sup>* is injective. Define *β* : *X* → *Y* by

*xβ* = *a* if *x* = *a*, *c* otherwise.

Then, *Xβ* = {*a*, *c*} and *β*|*X<sup>β</sup>* is injective. Since *bαβ* = *a* and *cαβ* = *c*, we obtain *a*, *c* ∈ *Xαβ*. From *aαβ* = *c* = *cαβ*, we conclude that *αβ*|*Xαβ* is not injective. Therefore, *αβ* ∈/ *RReg*(*T*(*X*,*Y*)) and *RReg*(*T*(*X*,*Y*)) is not closed.

Assume that |*Y*| ≤ 2. If |*Y*| = 1, then *T*(*X*,*Y*) is a right regular semigroup. Therefore, *RReg*(*T*(*X*,*Y*)) is a semigroup. Suppose that |*Y*| = 2. Let *Y* = {*a*, *b*} and *α*, *β* ∈ *RReg*(*T*(*X*,*Y*)). We will show that *αβ*|*Xαβ* is injective. Let *x*, *y* ∈ *Xαβ* be such that *xαβ* = *yαβ*. If *α* or *β* is a constant mapping, then *αβ* is a constant mapping. Thus, *αβ*|*Xαβ* is injective. Suppose that *α* and *β* are not constant mappings. Then, *Xα* = *Y* = *Xβ*. We observe that *x*, *y* ∈ *Xαβ* ⊆ *Xβ* = *Xα*. If *x* = *y*, then *xα* = *yα* since *α*|*X<sup>α</sup>* is injective. Note that *xα*, *yα* ∈ *Y* = *Xβ*. Then, (*xα*)*β* = (*yα*)*β* since *β*|*X<sup>β</sup>* is injective. This is a contradition. Hence, *x* = *y*. So *αβ*|*Xαβ* is injective. Therefore, *RReg*(*T*(*X*,*Y*)) is closed.

**Theorem 14.** *Let* |*X*| ≥ 3*. Then, LReg*(*T*(*X*,*Y*)) *is a semigroup if and only if* |*Y*| ≤ 2*.*

**Proof.** Suppose that |*Y*| ≥ 3. Let *a*, *b*, *c* ∈ *Y* be distinct elements. Define *α* : *X* → *Y* by

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

And we define *β* : *X* → *Y* by

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

Then, *Yα*<sup>2</sup> = (*Yα*)*α* = {*b*, *c*}*α* = {*b*, *c*} = *Xα* and *Yβ*<sup>2</sup> = (*Yβ*)*β* = {*a*, *b*}*β* = {*a*, *b*} = *Xβ*. Thus, *α*, *β* ∈ *LReg*(*T*(*X*,*Y*)). It is easy to verify that

*xαβ* = *a* if *x* = *c*, *b* otherwise

such that *π*(*αβ*) = {{*c*}, *X* \ {*c*}} and *Yαβ* = {*a*, *b*}. Clearly, *Yαβ* ∩ {*c*} = ∅. Hence, *αβ* ∈/ *LReg*(*T*(*X*,*Y*)). So *αβ* is not left regular.

Conversely, suppose |*Y*| ≤ 2. If |*Y*| = 1, then *T*(*X*,*Y*) is a left regular semigroup. We have *LReg*(*T*(*X*,*Y*)) is a semigroup. Assume that *Y* = {*a*, *b*}. Let *α*, *β* ∈ *LReg*(*T*(*X*,*Y*)). If |*Xα*| = 1 or |*Xβ*| = 1, then *αβ* is a constant mapping. Suppose that |*Xα*| = |*Xβ*| = 2. Then, *π*(*α*) = {*aα*<sup>−</sup>1, *bα*−1} and so it is a left regular element. From *α* being a left regular element, we obtain *Yα* ∩ *aα*−<sup>1</sup> = ∅ and *Yα* ∩ *bα*−<sup>1</sup> = ∅. This implies that |*aα*−<sup>1</sup> ∩ *Y*| = 1 = |*bα*−<sup>1</sup> ∩ *Y*| and *Xα* = *Yα* = {*a*, *b*}. Similarly, |*aβ*−<sup>1</sup> ∩ *Y*| = 1 = |*bβ*−<sup>1</sup> ∩ *Y*| and *Xβ* = *Yβ* = {*a*, *b*}. It is easy to verify that *π*(*αβ*) = *π*(*α*) and *Yαβ* = {*a*, *b*} = *Yα*. Hence, *Yαβ* ∩ *aα*−<sup>1</sup> = ∅ and *Yαβ* ∩ *bα*−<sup>1</sup> = ∅. Therefore, *αβ* is a left regular element.

**Theorem 15.** *If Y is finite, then RReg*(*T*(*X*,*Y*)) = *LReg*(*T*(*X*,*Y*))*.*

**Proof.** Assume that *Y* is finite. Let *α* be a left regular element of *T*(*X*,*Y*). Then, *Yα*<sup>2</sup> = *Xα*. It follows that *Xα* = *Yα*<sup>2</sup> ⊆ *Xα*<sup>2</sup> ⊆ *Xα*; that is, *Xα*<sup>2</sup> = *Xα*. From (*Xα*)*α* = *Xα*<sup>2</sup> = *Xα*, we obtain *α*|*X<sup>α</sup>* : *Xα* → *Xα* is surjective. Since *Xα* ⊆ *Y* and *Y* is a finite set, we obtain *α*|*X<sup>α</sup>* is an injection. So *α* is right regular.

Assume that *α* is right regular. Thus, *α*|*X<sup>α</sup>* : *Xα* → *Xα* is injective and also *α*|*X<sup>α</sup>* : *Xα* → *Xα* is surjective since *Xα* is finite. This means that (*Xα*)*α* = *Xα*. We see that *Xα* = (*Xα*)*α* ⊆ *Yα* ⊆ *Xα*. Hence, *Xα* = *Yα* and so *Yα*<sup>2</sup> = *Xα*<sup>2</sup> = *Xα*. Therefore, *α* is a left regular element of *T*(*X*,*Y*).

Next, the cardinality of right regular elements in the semigroup *T*(*X*,*Y*) are investigated when *X* is finite.

**Theorem 16.** *Let* |*X*| = *n and* |*Y*| = *r. Then,*

$$|L\text{Reg}(T(X,Y))| = |R\text{Reg}(T(X,Y))| = \sum\_{k=1}^{r} k! \binom{r}{k} k^{n-k}$$

*where* 1 ≤ *k* ≤ *r.*

**Proof.** By Theorem 15, we have *LReg*(*T*(*X*,*Y*)) = *RReg*(*T*(*X*,*Y*)). This implies that |*LReg*(*T*(*X*,*Y*))| = |*RReg*(*T*(*X*,*Y*))|. Let 1 ≤ *k* ≤ *r* and *Bk* = {*α* ∈ *RReg*(*T*(*X*,*Y*)) : |*Xα*| = *k*}. From *Y* being finite, we have *α*|*X<sup>α</sup>* : *Xα* → *Xα* is bijective for all *α* ∈ *Bk* by Theorem 10. Notice that the number of image sets in *Y* of cardinality *k* is equal to ( *r <sup>k</sup>*). Since there are *kn*−*<sup>k</sup>* ways of partitioning the remaining *n* − *k* elements into *k* subsets, we obtain |*Bk*| = ( *r <sup>k</sup>*)*k*! *<sup>k</sup>n*−*k*. Therefore,

$$\left| R \operatorname{Re} \mathcal{g} (T(X, \mathcal{Y})) \right| = \sum\_{k=1}^{r} \left| B\_k \right| = \sum\_{k=1}^{r} k! (\,^{r}\_{k}) k^{n-k}.$$
