Regularity and Green's Relations on a Semigroup of Transformations with Restricted Range

Abstract

Let 𝑇(𝑋) be the full transformation semigroup on the set 𝑋 and let 𝑇(𝑋,𝑌)={𝛼∈𝑇(𝑋)∶𝑋𝛼⊆𝑌}. Then 𝑇(𝑋,𝑌) is a sub-semigroup of 𝑇(𝑋) determined by a nonempty subset 𝑌 of 𝑋. In this paper, we give a necessary and sufficient condition for 𝑇(𝑋,𝑌) to be regular. In the case that 𝑇(𝑋,𝑌) is not regular, the largest regular sub-semigroup is obtained and this sub-semigroup is shown to determine the Green's relations on 𝑇(𝑋,𝑌). Also, a class of maximal inverse sub-semigroups of 𝑇(𝑋,𝑌) is obtained.

1. Introduction

Let 𝑌 be a nonempty subset of 𝑋 and let 𝑇(𝑋) denote the semigroup of transformations from 𝑋 into itself. We consider the sub-semigroup of 𝑇(𝑋) defined by 𝑇(𝑋,𝑌)={𝛼∈𝑇(𝑋)∶𝑋𝛼⊆𝑌}, (1.1) when 𝑋𝛼 denotes the range of 𝛼. In fact, if |𝑌|=1, then 𝑇(𝑋,𝑌) contains exactly one element (namely, the constant map with range 𝑌) and if 𝑌=𝑋, then 𝑇(𝑋,𝑌)=𝑇(𝑋).

In 1975, Symons [1] described all the automorphisms of 𝑇(𝑋,𝑌) and found that the most difficult case occurs when |𝑌|=2. He also determined when 𝑇(𝑋₁,𝑌₁) is isomorphic to 𝑇(𝑋₂,𝑌₂) and, surprisingly, the answer depends on the cardinals |𝑋_𝑖| and |𝑋_𝑖⧵𝑌_𝑖|, not on |𝑌_𝑖| for 𝑖=1,2. Here, we study other algebraic properties of this semigroup. Recall that an element 𝑎 of a semigroup 𝑆 is called regular if 𝑎=𝑎𝑥𝑎 for some 𝑥 in 𝑆. A semigroup 𝑆 is regular if every element of 𝑆 is regular. It is already known that 𝑇(𝑋) is a regular semigroup (see [[2], page 33]). But 𝑇(𝑋,𝑌) is not regular in general. So, in Section 2, we prove that 𝑇(𝑋,𝑌) is regular if and only if |𝑌|=1 or 𝑌=𝑋. We also prove that if 𝑇(𝑋,𝑌) is not regular, the set 𝐹={𝛼∈𝑇(𝑋,𝑌)∶𝑋𝛼⊆𝑌𝛼} is the largest regular sub-semigroup of 𝑇(𝑋,𝑌). In Section 3, we characterize the Green's relations on 𝑇(𝑋,𝑌) and find that its 𝒟 and 𝒥 relations are surprising, but they reduce to those on 𝑇(𝑋) when 𝑌=𝑋. And in Section 4, we give a class of maximal inverse sub-semigroups of 𝑇(𝑋,𝑌), of the form 𝐹_𝑎={𝛼∈𝐹∶𝑎𝛼=𝑎 and 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹}. When 𝑌=𝑋, the set 𝐹_𝑎={𝛼∈𝑇(𝑋)∶𝑎𝛼=𝑎 and 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹} is a class of maximal inverse sub-semigroups of 𝑇(𝑋) given in [4].

Note that throughout the paper, we write functions on the right; in particular, this means that for a composition 𝛼𝛽, 𝛼 is applied first.

2. Regularity of 𝑇(𝑋,𝑌)

To give a necessary and sufficient condition for the semigroup 𝑇(𝑋,𝑌) to be regular, we first note the following.

(1)If |𝑌|=1, say 𝑌={𝑎}, then 𝑇(𝑋,𝑌) contains exactly one element (namely, the constant map 𝑋_𝑎 with range {𝑎}), so 𝑇(𝑋,𝑌) is regular.(2)If 𝑌=𝑋, then 𝑇(𝑋,𝑌)=𝑇(𝑋) which is a regular semigroup.(3)If |𝑋|≤2, then |𝑌|=1 or 𝑌=𝑋, and 𝑇(𝑋,𝑌) is regular by (1) and (2). Now, we need some notation. We adopt the convention introduced in ([[3], page 241]), namely, if 𝛼∈𝑇(𝑋,𝑌), then we write 𝑋𝛼=^𝑖𝑎_𝑖, (2.1) and take as understood that the subscript 𝑖 belongs to some (unmentioned) index set 𝐼, the abbreviation {𝑎_𝑖} denotes {𝑎_𝑖∶𝑖∈𝐼}, and that 𝑋𝛼 = {𝑎_𝑖} and 𝑎_𝑖𝛼⁻¹=𝑋_𝑖.

Theorem 2.1. 𝑇(𝑋,𝑌) is a regular semigroup if and only if |𝑌|=1 or 𝑌=𝑋.

Proof. Assume that |𝑌|≠1 and 𝑌≠𝑋. Let 𝑎,𝑏∈𝑌 be such that 𝑎≠𝑏 and choose 𝑐∈𝑋⧵𝑌. Let 𝑋𝛼=^𝑖𝑦_𝑖 (2.2) be any element in 𝑇(𝑋,𝑌), where 𝑋𝛼={𝑦_𝑖}⊆𝑌 and 𝑋_𝑖=𝑦_𝑖𝛼⁻¹.
We define 𝛽=(^{𝑐𝑋⧵{𝑐}}_𝑎𝑏), and it is clear that 𝛽∈𝑇(𝑋,𝑌).
Since 𝑐∉𝑌, so 𝑦_𝑖≠𝑐 for all 𝑖, and 𝑋𝛼𝛽=^𝑖𝑦_𝑖=𝑋𝑏𝑐𝑋⧵{𝑐}𝑎𝑏≠𝛽. (2.3) So, we conclude that 𝛼𝛽≠𝛽 for all 𝛼∈𝑇(𝑋,𝑌), this implies that 𝛽 is a nonregular element in 𝑇(𝑋,𝑌). Therefore, 𝑇(𝑋,𝑌) is not a regular semigroup. The converse is clear by the previous note.

Now, we consider the set 𝐹={𝛼∈𝑇(𝑋,𝑌)∶𝑋𝛼⊆𝑌𝛼}. (2.4) It is easy to see that 𝐹={𝛼∈𝑇(𝑋,𝑌)∶(𝑋⧵𝑌)𝛼⊆𝑌𝛼}={𝛼∈𝑇(𝑋,𝑌)∶𝑋𝛼=𝑌𝛼}. Since 𝑌≠∅, there exists 𝑎∈𝑌 and we see that the constant map 𝑋_𝑎 with range {𝑎} satisfies the condition in 𝐹, therefore, 𝑋_𝑎∈𝐹 and so 𝐹≠∅. And for each 𝛼∈𝐹 and 𝛽∈𝑇(𝑋,𝑌), we have 𝑋𝛼𝛽=(𝑋𝛼)𝛽⊆(𝑌𝛼)𝛽=𝑌𝛼𝛽, and thus 𝛼𝛽∈𝐹. This proves the following.

Lemma 2.2. 𝐹 is a right ideal of 𝑇(𝑋,𝑌).

In general, 𝐹 is not a left ideal of 𝑇(𝑋,𝑌) as shown in the following example.

Example 2.3. Let 𝑋=ℕ denote the set of positive integers, let 𝑌 denote the set of all positive even integers, and define 𝑛𝛼=2𝑛,𝛽=2𝑛𝑋⧵𝑌2𝑛2. (2.5) Then 𝛼∈𝑇(𝑋,𝑌)⧵𝐹 and 𝛽∈𝐹, but 𝛼𝛽=𝛼∉𝐹. Thus 𝐹 is not a left ideal of 𝑇(𝑋,𝑌).

Theorem 2.4. 𝐹 is the largest regular sub-semigroup of 𝑇(𝑋,𝑌).

Proof. From Lemma 2.2, we see that 𝐹 is a sub-semigroup of 𝑇(𝑋,𝑌). Let 𝛼∈𝐹 and write 𝑥𝛼=^𝑖𝛼⁻¹𝑥_𝑖, (2.6) where ⋃_𝑖∈𝐼𝑥_𝑖𝛼⁻¹=𝑋 and 𝑋𝛼={𝑥_𝑖∶𝑖∈𝐼}=𝑌𝛼. For each 𝑥∈𝑌𝛼, choose 𝑑_𝑥∈𝑥𝛼⁻¹⋂𝑌, so 𝑑_𝑥𝛼=𝑥, and 𝑑_𝑦≠𝑑_𝑧, for all 𝑦,𝑧∈𝑌𝛼 such that 𝑦≠𝑧. Choose 𝑘∈𝐼 and let 𝐽=𝐼⧵{𝑘}. Define 𝑥𝛽=^𝑗𝑋⧵{𝑥_𝑗}𝑑_𝑥𝑗𝑑_𝑥𝑘, (2.7) where {𝑥_𝑗∶𝑗∈𝐽}=𝑌𝛼⧵{𝑥_𝑘}. Then 𝛽∈𝑇(𝑋,𝑌) and 𝛼𝛽𝛼=𝛼. Since 𝑋𝛽={𝑑_𝑥𝑖∶𝑖∈𝐼}⊆(𝑌𝛼)𝛽⊆𝑌𝛽, we have 𝛽∈𝐹. Hence 𝐹 is a regular sub-semigroup of 𝑇(𝑋,𝑌). Now, let 𝛼 be any regular element in 𝑇(𝑋,𝑌). Then 𝛼𝛽𝛼=𝛼, for some 𝛽∈𝑇(𝑋,𝑌), so 𝑋𝛼=𝑋𝛼𝛽𝛼=(𝑋𝛼𝛽)𝛼⊆𝑌𝛼, and thus 𝛼∈𝐹. Therefore, 𝐹 is the largest regular sub-semigroup of 𝑇(𝑋,𝑌) as required.

Note that if 𝑌={𝑎}, then for each 𝛼∈𝑇(𝑋,𝑌), we have 𝑋𝛼={𝑎}=𝑌𝛼 which implies that   𝐹=𝑇(𝑋,𝑌) consists of only one element and so is a regular semigroup, and if 𝑋=𝑌, then 𝐹={𝛼∈𝑇(𝑋,𝑌)∶𝑋𝛼⊆𝑌𝛼}={𝛼∈𝑇(𝑋)∶𝑋𝛼⊆𝑋𝛼}=𝑇(𝑋) which is also a regular semigroup.

3. Green's Relations on 𝑇(𝑋,𝑌)

Let 𝑆 be a semigroup. Then we define 𝑆¹ to be a semigroup of adding an identity to 𝑆 if 𝑆 does not already have an identity element in it and 𝑆¹=𝑆 if 𝑆 contains an identity. The following definitions are due to J. A. Green. For any 𝑎,𝑏∈𝑆, we define 𝑎ℒ𝑏iff𝑆¹𝑎=𝑆¹𝑏, (3.1) or equivalently; 𝑎ℒ𝑏 if and only if 𝑎=𝑥𝑏,𝑏=𝑦𝑎 for some 𝑥,𝑦∈𝑆¹.

Dually, we define 𝑎ℛ𝑏tomeans𝑎𝑆¹=𝑏𝑆¹, (3.2) or equivalently; 𝑎ℛ𝑏 if and only if 𝑎=𝑏𝑥,𝑏=𝑎𝑦 for some 𝑥,𝑦∈𝑆¹.

And we define 𝑎𝒥𝑏tomeans𝑆¹𝑎𝑆¹=𝑆¹𝑏𝑆¹, (3.3) or equivalently; 𝑎𝒥𝑏 if and only if 𝑎=𝑥𝑏𝑦,𝑏=𝑢𝑎𝑣 for some 𝑥,𝑦,𝑢,𝑣∈𝑆¹.

Finally, we define ℋ=ℒ∩ℛ and 𝒟=ℒ∘ℛ.

In [2, 3], Clifford and Preston characterized Green's relations on the full transformation semigroup 𝑇(𝑋), where 𝑋 is an arbitrary set. They proved that 𝛼ℒ𝛽iff𝑋𝛼=𝑋𝛽,𝛼ℛ𝛽iff𝜋_𝛼=𝜋_𝛽. (3.4) Here, we do the same for the semigroup 𝑇(𝑋,𝑌) and we obtain results which generalize the same results on 𝑇(𝑋).

Lemma 3.1. Let 𝛼,𝛽∈𝑇(𝑋,𝑌). If 𝛽∈𝐹, then 𝑋𝛼⊆𝑋𝛽 if and only if 𝛼=𝛾𝛽 for some 𝛾∈𝑇(𝑋,𝑌).

Proof. Let 𝛽 be an element of 𝐹. It is clear that if 𝛼=𝛾𝛽 for some 𝛾∈𝑇(𝑋,𝑌), then 𝑋𝛼⊆𝑋𝛽. Now, we assume that 𝑋𝛼⊆𝑋𝛽 and write 𝑎𝛼=^𝑖𝛼⁻¹𝑎_𝑖𝑏,𝛽=^𝑗𝛽⁻¹𝑏_𝑗, (3.5) where {𝑎_𝑖}⊆{𝑏_𝑗}. For each 𝑎∈𝑋𝛼⊆𝑋𝛽⊆𝑌𝛽 (since 𝛽∈𝐹), we get 𝑎=𝑦𝛽 for some 𝑦∈𝑌 which implies 𝑦∈𝑎𝛽⁻¹ and thus 𝑦∈𝑌∩𝑎𝛽⁻¹≠∅. Choose 𝑑_𝑎∈𝑌∩𝑎𝛽⁻¹, so 𝑑_𝑎∈𝑌 and 𝑑_𝑎𝛽=𝑎. Since 𝑋=⋃_{𝑎𝑖∈𝑋𝛼}𝑎_𝑖𝛼⁻¹ is the disjoint union of the 𝑎_𝑖𝛼⁻¹, we can define 𝑎𝛾=^𝑖𝛼⁻¹𝑑_𝑎𝑖. (3.6) Then 𝛾∈𝑇(𝑋,𝑌) and 𝛾𝛽=𝛼.

From now on, the notations 𝐿_𝛼(𝑅_𝛼,𝐻_𝛼,𝐷_𝛼) denote the set of all elements of 𝑇(𝑋,𝑌) which are ℒ-related (ℛ-related, ℋ-related, 𝒟-related) to 𝛼, where 𝛼∈𝑇(𝑋,𝑌).

Theorem 3.2. For 𝛼∈𝑇(𝑋,𝑌), the following statements hold.
(1)If 𝛼∈𝐹, then 𝐿_𝛼={𝛽∈𝐹∶𝑋𝛼=𝑋𝛽}.(2)If 𝛼∈𝑇(𝑋,𝑌)⧵𝐹, then 𝐿_𝛼={𝛼}.

Proof. Let 𝛼 be any element in 𝑇(𝑋,𝑌) and let 𝛽∈𝐿_𝛼. Then 𝛼ℒ𝛽 which implies that 𝛼=𝛼^′𝛽 and 𝛽=𝛽^′𝛼, for some 𝛼^′,𝛽^′∈𝑇(𝑋,𝑌)¹.
(1) Assume that 𝛼∈𝐹. If 𝛽=𝛼, then 𝛽∈𝐹 and 𝑋𝛼=𝑋𝛽. If 𝛽≠𝛼, then 𝛼^′ and 𝛽^′ both belong to 𝑇(𝑋,𝑌). Thus 𝑋𝛽=(𝑋𝛽^′𝛼^′)𝛽⊆𝑌𝛽, and hence 𝛽∈𝐹. From 𝛼∈𝐹 and 𝛽=𝛽^′𝛼, we get 𝑋𝛽⊆𝑋𝛼 by Lemma 3.1. Similarly, from 𝛽∈𝐹 and 𝛼=𝛼^′𝛽, we get 𝑋𝛼⊆𝑋𝛽. Therefore, 𝑋𝛼=𝑋𝛽. Now, if 𝛾∈𝐹 and 𝑋𝛼=𝑋𝛾, then it is clear by Lemma 3.1 that 𝛾∈𝐿_𝛼.
(2) Assume that 𝛼∈𝑇(𝑋,𝑌)⧵𝐹. If 𝛼^′,𝛽^′∈𝑇(𝑋,𝑌), then 𝑋𝛼=𝑋(𝛼^′𝛽)=𝑋(𝛼^′(𝛽^′𝛼))=(𝑋𝛼^′𝛽^′)𝛼⊆𝑌𝛼. Thus 𝛼∈𝐹 which is a contradiction, so 𝛼^′=1 or 𝛽^′=1 and 𝛽=𝛼.

We note that for any 𝛼∈𝑇(𝑋,𝑌), 𝜋_𝛼={(𝑎,𝑏)∈𝑋×𝑋∶𝑎𝛼=𝑏𝛼} is an equivalence on 𝑋 and |𝑋/𝜋_𝛼|=|𝑋𝛼|. The relation 𝜋_𝛼 is usually called the kernel of 𝛼.

Theorem 3.3. Let 𝛼,𝛽∈𝑇(𝑋,𝑌). Then 𝜋_𝛽⊆𝜋_𝛼 if and only if 𝛼=𝛽𝛾 for some 𝛾∈𝑇(𝑋,𝑌). Hence 𝛼ℛ𝛽 if and only if 𝜋_𝛼=𝜋_𝛽.

Proof. It is clear that if 𝛼=𝛽𝛾 for some 𝛾∈𝑇(𝑋,𝑌), then 𝜋_𝛽⊆𝜋_𝛼. Now, suppose that 𝜋_𝛽⊆𝜋_𝛼. If 𝑥∈𝑋𝛽, then 𝑥=𝑧𝛽 for some 𝑧∈𝑋, so we define 𝛾∶𝑋→𝑋 by 𝑥𝛾=𝑧𝛼,if𝑥∈𝑋𝛽,𝑥𝛽,if𝑥∈𝑋⧵𝑋𝛽. (3.7) Then 𝛾 is well defined (since 𝜋_𝛽⊆𝜋_𝛼) and 𝛾∈𝑇(𝑋,𝑌). For each 𝑥∈𝑋, let 𝑦=𝑥𝛽∈𝑋𝛽, so 𝑥𝛽𝛾=(𝑥𝛽)𝛾=𝑦𝛾=𝑥𝛼 by the definition of 𝛾. Thus 𝛼=𝛽𝛾 as required, and the remaining assertion is clear.

Lemma 3.4. Let 𝛼,𝛽∈𝑇(𝑋,𝑌). If 𝜋_𝛼=𝜋_𝛽 then either both 𝛼 and 𝛽 are in 𝐹, or neither is in 𝐹.

Proof. Assume that 𝜋_𝛼=𝜋_𝛽 and suppose that 𝛼,𝛽∈𝐹 is false. So one of 𝛼 or 𝛽 is not in 𝐹, we suppose that 𝛼∉𝐹. Thus (𝑋⧵𝑌)𝛼⊈𝑌𝛼, so there is 𝑥₀∈𝑋⧵𝑌 such that 𝑥₀𝛼≠𝑦𝛼 for all 𝑦∈𝑌. Thus (𝑥₀,𝑦)∉𝜋_𝛼 for all 𝑦∈𝑌. If 𝛽∈𝐹, then 𝑋𝛽=𝑌𝛽, so (𝑥₀,𝑦)∈𝜋_𝛽 for some 𝑦∈𝑌 which contradicts 𝜋_𝛼=𝜋_𝛽. Therefore, 𝛽∉𝐹.

Using Theorem 3.3 and Lemma 3.4, we have the following corollary.

Corollary 3.5. For 𝛼∈𝑇(𝑋,𝑌), the following statements hold.
(1)If 𝛼∈𝐹, then 𝑅_𝛼={𝛽∈𝐹∶𝜋_𝛼=𝜋_𝛽}.(2)If 𝛼∈𝑇(𝑋,𝑌)⧵𝐹, then 𝑅_𝛼={𝛽∈𝑇(𝑋,𝑌)⧵𝐹∶𝜋_𝛼=𝜋_𝛽}.

As a direct consequence of Theorems 3.2 and 3.3, we have the following.

Theorem 3.6. For 𝛼∈𝑇(𝑋,𝑌), the following statements hold.
(1)If 𝛼∈𝐹, then 𝐻_𝛼={𝛽∈𝐹∶𝑋𝛼=𝑋𝛽and𝜋_𝛼=𝜋_𝛽}.(2)If 𝛼∈𝑇(𝑋,𝑌)⧵𝐹, then 𝐻_𝛼={𝛼}.

In [2, 3], volume 1, Clifford and Preston proved that two elements of 𝑇(𝑋) are 𝒟-related if and only if they have the same rank (i.e., the ranges of the two elements have the same cardinality). But for 𝑇(𝑋,𝑌), we have the following.

Theorem 3.7. For 𝛼∈𝑇(𝑋,𝑌), the following statements hold.
(1)If 𝛼∈𝐹, then 𝐷_𝛼={𝛽∈𝐹∶|𝑋𝛼|=|𝑋𝛽|}.(2)If 𝛼∈𝑇(𝑋,𝑌)⧵𝐹, then 𝐷_𝛼={𝛽∈𝑇(𝑋,𝑌)⧵𝐹∶𝜋_𝛼=𝜋_𝛽}.

Proof. Let 𝛼 be any element in 𝑇(𝑋,𝑌) and let 𝛽∈𝐷_𝛼. Then 𝛼ℒ𝛾 and 𝛾ℛ𝛽 for some 𝛾∈𝑇(𝑋,𝑌).
(1) If 𝛼∈𝐹, then since 𝛼ℒ𝛾, we must have 𝛾∈𝐹 and 𝑋𝛼=𝑋𝛾. From 𝛾ℛ𝛽, we get 𝜋_𝛾=𝜋_𝛽 and 𝛽=𝛾𝜆 for some 𝜆∈𝑇(𝑋,𝑌)¹. Since 𝐹 is a right ideal of 𝑇(𝑋,𝑌), so 𝛽∈𝐹. And |𝑋𝛼|=|𝑋𝛾|=|𝑋/𝜋_𝛾|=|𝑋/𝜋_𝛽|=|𝑋𝛽|. Conversely, assume that 𝜆∈𝐹 and |𝑋𝛼|=|𝑋𝜆|. Then there is a bijection 𝜃∶𝑋𝜆→𝑋𝛼. We let 𝜇=𝜆𝜃, then 𝜇∈𝑇(𝑋,𝑌) and 𝑋𝜇=𝑋𝜆𝜃=(𝑋𝜆)𝜃=𝑋𝛼. Since 𝜆∈𝐹 implies 𝑋𝜆⊆𝑌𝜆, so 𝑋𝜇=𝑋𝜆𝜃⊆𝑌𝜆𝜃=𝑌𝜇, hence 𝜇∈𝐹. Since 𝛼,𝜇∈𝐹 and 𝑋𝛼=𝑋𝜇, so 𝛼ℒ𝜇 by Theorem 3.2. Now, since 𝜇=𝜆𝜃 and 𝜃 is injective on 𝑋𝜆, we get 𝜋_𝜇=𝜋_𝜆, so 𝜇ℛ𝜆. Therefore, 𝛼 and 𝜆 are 𝒟-related and 𝜆∈𝐷_𝛼.
(2) If 𝛼∈𝑇(𝑋,𝑌)⧵𝐹, then 𝛾=𝛼 (since 𝛼ℒ𝛾) and thus 𝛼ℛ𝛽 which implies that 𝜋_𝛼=𝜋_𝛽. So by Lemma 3.4 we must have 𝛽∈𝑇(𝑋,𝑌)⧵𝐹. The other containment is clear since ℛ⊆𝒟.

In order to characterize the 𝒥-relation on 𝑇(𝑋,𝑌), the following lemma is needed.

Lemma 3.8. Let 𝛼,𝛽∈𝑇(𝑋,𝑌). If 𝛼=𝜆𝛽𝜇 for some 𝜆∈𝑇(𝑋,𝑌) and 𝜇∈𝑇(𝑋,𝑌)¹, then |𝑋𝛼|≤|𝑌𝛽|.

Proof. If 𝛼=𝜆𝛽𝜇 for some 𝜆∈𝑇(𝑋,𝑌) and 𝜇∈𝑇(𝑋,𝑌)¹. Then 𝑋𝜆⊆𝑌, which implies that (𝑋𝜆)𝛽⊆𝑌𝛽 and so |𝑋𝜆𝛽|≤|𝑌𝛽|. If 𝜇=1, then 𝛼=𝜆𝛽 and so |𝑋𝛼|=|𝑋𝜆𝛽|≤|𝑌𝛽|. If 𝜇∈𝑇(𝑋,𝑌), then |𝑋𝛼|=|𝑋(𝜆𝛽𝜇)|=|(𝑋𝜆𝛽)𝜇|≤|𝑋𝜆𝛽|≤|𝑌𝛽|. Thus |𝑋𝛼|≤|𝑌𝛽| as required.

Theorem 3.9. Let 𝛼,𝛽∈𝑇(𝑋,𝑌). Then 𝛼𝒥𝛽𝑖𝑓𝑓𝜋_𝛼=𝜋_𝛽𝑜𝑟|𝑋𝛼|=|𝑌𝛼|=|𝑌𝛽|=|𝑋𝛽|. (3.8)

Proof. First, assume that 𝛼𝒥𝛽. Then 𝛼=𝛾𝛽𝜆 and 𝛽=𝛾^′𝛼𝜆^′ for some 𝛾,𝜆,𝛾^′,𝜆^′∈𝑇(𝑋,𝑌)¹. If 𝛾=1=𝛾^′, then 𝛼=𝛽𝜆 and 𝛽=𝛼𝜆^′ which imply 𝛼ℛ𝛽 and thus 𝜋_𝛼=𝜋_𝛽. If 𝛾∈𝑇(𝑋,𝑌) or 𝛾^′∈𝑇(𝑋,𝑌), then we conclude that 𝛼=𝜎𝛽𝛿 and 𝛽=𝜎^′𝛼𝛿^′ for some 𝜎,𝜎^′∈𝑇(𝑋,𝑌) and 𝛿,𝛿^′∈𝑇(𝑋,𝑌)¹. For example, if 𝛾=1 and 𝛾^′∈𝑇(𝑋,𝑌), then 𝛼=𝛽𝜆 and 𝛽=𝛾^′𝛼𝜆^′ imply 𝛼=𝛽𝜆=(𝛾^′𝛼𝜆^′)𝜆=𝛾^′𝛼(𝜆^′𝜆)=𝛾^′(𝛽𝜆)𝜆^′𝜆=𝛾^′𝛽(𝜆𝜆^′𝜆). By using Lemma 3.8, we get that |𝑌𝛽|≥|𝑋𝛼|≥|𝑌𝛼|≥|𝑋𝛽|≥|𝑌𝛽|, so it follows that |𝑋𝛼|=|𝑌𝛼|=|𝑌𝛽|=|𝑋𝛽|.
Conversely, if 𝜋_𝛼=𝜋_𝛽, then 𝛼ℛ𝛽 which implies that 𝛼𝒥𝛽 since ℛ⊆𝒥. If |𝑋𝛼|=|𝑌𝛼|=|𝑌𝛽|=|𝑋𝛽|, then by applying Lemma 2.7 in [2, 3], to |𝑋/𝜋_𝛼|=|𝑋𝛼|=|𝑌𝛽| and |𝑋/𝜋_𝛽|=|𝑋𝛽|=|𝑌𝛼|, we get there are 𝛾,𝜆∈𝑇(𝑋,𝑌) such that 𝜋_𝛾=𝜋_𝛼, 𝑋𝛾=𝑌𝛽; and 𝜋_𝜆=𝜋_𝛽, 𝑋𝜆=𝑌𝛼. From 𝜋_𝛾=𝜋_𝛼 and 𝜋_𝜆=𝜋_𝛽, we get 𝛾ℛ𝛼 and 𝜆ℛ𝛽, so 𝛼=𝛾𝛾^′ and 𝛽=𝜆𝜆^′ for some 𝛾^′,𝜆^′∈𝑇(𝑋,𝑌)¹. And from 𝑋𝛾=𝑌𝛽, we write 𝑌𝛽={𝑦_𝑖∶𝑖∈𝐼}, so 𝑦𝛾=^𝑖𝛾⁻¹𝑦_𝑖. (3.9) For each 𝑖∈𝐼, choose 𝑎_𝑖∈𝑦_𝑖𝛽⁻¹∩𝑌 and define 𝛽^′∶𝑋→𝑋 by 𝛽^′=𝑦_𝑖𝛾⁻¹𝑎_𝑖. (3.10) Then 𝛽^′∈𝑇(𝑋,𝑌) and 𝛾=𝛽^′𝛽. Similarly, from 𝑋𝜆=𝑌𝛼, we can prove that 𝜆=𝛼^′𝛼 for some 𝛼^′∈𝑇(𝑋,𝑌).
Therefore, 𝛼=𝛾𝛾^′=𝛽^′𝛽𝛾^′ and 𝛽=𝜆𝜆^′=𝛼^′𝛼𝜆^′ which implies that 𝛼𝒥𝛽 as required.

Recall that 𝒟⊆𝒥 on any semigroup and 𝒟=𝒥 on 𝑇(𝑋); but in 𝑇(𝑋,𝑌), this is not always true, as shown in the following example.

Example 3.10. Let 𝑋=ℕ denote the set of positive integers and let 𝑌 denote the set of all positive even integers. Then we define 𝑛𝛼=2𝑛,𝛽=2𝑛𝑋⧵𝑌4𝑛2. (3.11) Hence 𝛼,𝛽∈𝑇(𝑋,𝑌)⧵𝐹 and |𝑋𝛼|=|𝑌𝛼|=ℵ₀=|𝑌𝛽|=|𝑋𝛽|, so 𝛼𝒥𝛽. Since 𝜋_𝛼≠𝜋_𝛽, we have 𝛼 and 𝛽 are not 𝒟-related on 𝑇(𝑋,𝑌).
As a consequence of Theorems 3.7 and 3.9, we see that 𝒟=𝒥 on the sub-semigroup 𝐹 of 𝑇(𝑋,𝑌).

Corollary 3.11. If 𝛼,𝛽∈𝐹, then 𝛼𝒥𝛽 on 𝑇(𝑋,𝑌) if and only if 𝛼𝒟𝛽 on 𝑇(𝑋,𝑌).

Proof. In general, we have 𝒟⊆𝒥. Let 𝛼,𝛽∈𝐹 and 𝛼𝒥𝛽 on 𝑇(𝑋,𝑌). Then 𝜋_𝛼=𝜋_𝛽 or |𝑋𝛼|=|𝑌𝛼|=|𝑌𝛽|=|𝑋𝛽|. If 𝜋_𝛼=𝜋_𝛽, then |𝑋𝛼|=|𝑋/𝜋_𝛼|=|𝑋/𝜋_𝛽|=|𝑋𝛽|. Thus, both cases imply |𝑋𝛼|=|𝑋𝛽| and 𝛼𝒟𝛽 on 𝑇(𝑋,𝑌) by Theorem 3.7.

If we replace 𝑌 with 𝑋 in the above corollary, we then get 𝒟=𝒥 on 𝑇(𝑋,𝑌)=𝑇(𝑋). Next, we will consider the case when 𝑌 is a finite subset of 𝑋.

Theorem 3.12. If 𝑌 is a finite subset of 𝑋, then 𝒟=𝒥 on 𝑇(𝑋,𝑌).

Proof. Let 𝑌 be a finite subset of 𝑋 and let 𝛼,𝛽∈𝑇(𝑋,𝑌) be such that 𝛼 and 𝛽 are 𝒥-related. Then 𝜋_𝛼=𝜋_𝛽 or |𝑋𝛼|=|𝑌𝛼|=|𝑌𝛽|=|𝑋𝛽|. We note that if 𝛼∉𝐹, then |𝑋𝛼|>|𝑌𝛼|. For if |𝑋𝛼|=|𝑌𝛼|, then 𝑋𝛼 is a finite set (since 𝑌 is finite) which implies that 𝑋𝛼=𝑌𝛼 and thus 𝛼∈𝐹. Now, if 𝛼∈𝐹 and 𝛽∉𝐹, then |𝑋𝛼|=|𝑌𝛼| but |𝑌𝛽|<|𝑋𝛽|, so 𝜋_𝛼=𝜋_𝛽 which contradicts Lemma 3.4. Therefore, either both 𝛼 and 𝛽 are in 𝐹, or neither is in 𝐹. If 𝛼,𝛽∈𝐹, then 𝛼𝒟𝛽 by Corollary 3.11. If 𝛼,𝛽∉𝐹, then |𝑋𝛼|>|𝑌𝛼| which implies that 𝜋_𝛼=𝜋_𝛽 and thus 𝛼𝒟𝛽 by Theorem 3.7. Therefore, 𝒥⊆𝒟 and the other containment is clear.

4. Maximal Inverse Sub-Semigroups on 𝑇(𝑋,𝑌)

We first recall that a semigroup is said to be an inverse semigroup if it is regular and any two idempotents commute. In this section, we give one class of maximal inverse sub-semigroups on 𝑇(𝑋,𝑌). If |𝑌|=1, then there is only one element in 𝑇(𝑋,𝑌), the constant map. Hence, in this case, there is no maximal inverse sub-semigroup on 𝑇(𝑋,𝑌). Therefore, from now on, we assume that |𝑌|≥2.

In 1976, Nichols [4] gave a class of maximal inverse sub-semigroups of 𝑇(𝑋). Later in 1978, Reilly [5] generalized Nichols' result. Here, with some mild modifications of the proof given in [4], we get one class of maximal inverse sub-semigroups of 𝑇(𝑋,𝑌) which generalizes Nichols' result.

Let 𝑋 be a set and 𝑌 a nonempty subset of 𝑋. For each 𝑎∈𝑌, define 𝐹_𝑎={𝛼∈𝐹∶𝑎𝛼=𝑎and𝛼isinjectiveon𝑋⧵𝑎𝛼⁻¹}. (4.1) We see that 𝐹_𝑎≠∅, since the constant map 𝑋_𝑎∈𝐹_𝑎. To describe maximal inverse sub-semigroups on 𝑇(𝑋,𝑌), we first prove the following.

Theorem 4.1. Let 𝛼∈𝑇(𝑋,𝑌) and 𝑎∈𝑌. Then 𝛼∈𝐹_𝑎 if and only if {𝑎}∪(𝑋⧵𝑌)⊆𝑎𝛼⁻¹ and 𝛼 is injective on 𝑌⧵𝑎𝛼⁻¹.

Proof. If 𝑌=𝑋, then 𝑋⧵𝑌=∅ and 𝐹=𝑇(𝑋), thus 𝛼∈𝐹_𝑎 if and only if 𝛼∈𝑇(𝑋), {𝑎}⊆𝑎𝛼⁻¹ and 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹.
Now, we prove for the case 𝑌≠𝑋. Assume that 𝛼∈𝐹_𝑎. So 𝛼∈𝐹,𝑎𝛼=𝑎 and 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹. We show that (𝑋⧵𝑌)𝛼={𝑎}. Let 𝑏∈(𝑋⧵𝑌)𝛼, so there exists 𝑥∈𝑋⧵𝑌 such that 𝑥𝛼=𝑏. Thus 𝑏∈(𝑋⧵𝑌)𝛼⊆𝑌𝛼 since 𝛼∈𝐹. Hence 𝑏=𝑦𝛼 for some 𝑦∈𝑌. So, 𝑥,𝑦∈𝑏𝛼⁻¹ and 𝑥≠𝑦. By the definition of 𝐹_𝑎, we must have 𝑏=𝑎. Therefore, {𝑎}∪(𝑋⧵𝑌)⊆𝑎𝛼⁻¹ and 𝛼 is injective on 𝑌⧵𝑎𝛼⁻¹. Conversely, assume that the conditions hold. Since 𝑎∈𝑌 and {𝑎}⊆𝑎𝛼⁻¹, we get 𝑎=𝑎𝛼∈𝑌𝛼, and thus {𝑎}⊆𝑌𝛼. So (𝑋⧵𝑌)𝛼={𝑎}⊆𝑌𝛼 and therefore 𝛼∈𝐹. Since 𝛼 is injective on 𝑌⧵𝑎𝛼⁻¹ and 𝑌⧵𝑎𝛼⁻¹=𝑋⧵𝑎𝛼⁻¹, it follows that 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹, and so 𝛼∈𝐹_𝑎.

Recall that for each 𝛼∈𝑇(𝑋), 𝛼 is an idempotent in 𝑇(𝑋) if and only if 𝑥𝛼=𝑥 for all 𝑥∈𝑋𝛼. Since 𝑇(𝑋,𝑌) is a sub-semigroup of 𝑇(𝑋), we conclude that 𝛼 is an idempotent in 𝑇(𝑋,𝑌) if and only if 𝑥𝛼=𝑥 for all 𝑥∈𝑋𝛼. And by Theorem 4.1, if 𝛼 is an idempotent in 𝐹_𝑎, then 𝑥𝛼=𝑥 for all 𝑥∈𝑌⧵𝑎𝛼⁻¹.

Lemma 4.2. Let 𝐿 be a regular sub-semigroup of 𝑇(𝑋,𝑌) such that 𝐹_𝑎⊊𝐿 and suppose that 𝛼∈𝐿⧵𝐹_𝑎. Then
(i)𝛼ℒ𝛽 on 𝐿 for some idempotent 𝛽∈𝐹_𝑎.(ii)If 𝑎𝛼=𝑎, then 𝛼 is not ℛ-related on 𝐿 to any element in 𝐹_𝑎.

Proof. (i) We write 𝛼=𝑎𝛼⁻¹𝐴_𝑖𝑎𝑎_𝑖, (4.2) where ⋃𝐴_𝑖=𝑋⧵𝑎𝛼⁻¹, and define 𝛽∈𝑇(𝑋,𝑌) by 𝑎𝛽=^𝑖𝑎𝑋⧵𝐵_𝑖𝑎, (4.3) where {𝑎_𝑖}=𝐵. Then 𝛽∶𝑋→𝑌, is an identity map on 𝐵 and 𝑋⧵𝐵=𝑎𝛽⁻¹. So 𝐵=𝑋⧵𝑎𝛽⁻¹ and 𝛽 is injective on 𝑋⧵𝑎𝛽⁻¹=𝐵. Since 𝑎∈𝑋⧵𝐵, so 𝑎=𝑎𝛽∈𝑌𝛽 and {𝑎}⊆𝑌𝛽. From 𝐵={𝑎_𝑖}⊆𝑌, we get (𝑋⧵𝑌)𝛽⊆(𝑋⧵𝐵)𝛽={𝑎}⊆𝑌𝛽, so 𝛽∈𝐹. Thus 𝛽∈𝐹_𝑎 and it is also an idempotent. From the fact that 𝐿⊆𝐹 is a regular sub-semigroup of 𝑇=𝑇(𝑋,𝑌), it follows from Hall's theorem that ℒ^𝐿=ℒ^𝑇∩(𝐿×𝐿)⊆ℒ^𝑇∩(𝐹×𝐹), where (𝑎,𝑏)∈ℒ^𝑆 means there exist 𝑠,𝑡 in 𝑆¹ such that 𝑎=𝑠𝑏,𝑏=𝑡𝑎. Since 𝑋𝛼=𝑋𝛽, we must have by Theorem 3.2 that 𝛼ℒ𝛽 on 𝐿.
(ii) Assume that 𝑎𝛼=𝑎 and suppose that 𝛼ℛ𝛽 on 𝐿 for some 𝛽∈𝐹_𝑎. Thus 𝜋_𝛼=𝜋_𝛽 by Theorem 3.3, and hence 𝑎𝛼⁻¹=𝑎𝛽⁻¹. Now, let 𝑥₁,𝑥₂∈𝑋⧵𝑎𝛼⁻¹ be such that 𝑥₁𝛼=𝑥₂𝛼. Then (𝑥₁,𝑥₂)∈𝜋_𝛼 implies (𝑥₁,𝑥₂)∈𝜋_𝛽 (since 𝜋_𝛼=𝜋_𝛽), so 𝑥₁𝛽=𝑥₂𝛽 and since 𝛽 is injective on 𝑋⧵𝑎𝛽⁻¹=𝑋⧵𝑎𝛼⁻¹, we get 𝑥₁=𝑥₂ which implies that 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹, which contradicts 𝛼∉𝐹_𝑎. Therefore, 𝛼 is not ℛ-related to any element in 𝐹_𝑎.

Theorem 4.3. 𝐹_𝑎 is a maximal inverse sub-semigroup of 𝑇(𝑋,𝑌).

Proof. First, we prove that 𝐹_𝑎 is a sub-semigroup of 𝑇(𝑋,𝑌).
Let 𝛼,𝛽 be elements in 𝐹_𝑎. Then 𝛼,𝛽∈𝐹,𝑎𝛼=𝑎=𝑎𝛽, and 𝛼,𝛽 are injective on 𝑋⧵𝑎𝛼⁻¹ and 𝑋⧵𝑎𝛽⁻¹, respectively. Since 𝐹 is a right ideal of 𝑇(𝑋,𝑌), it follows that 𝛼𝛽∈𝐹. Clearly 𝑎(𝛼𝛽)=𝑎, and 𝛼𝛽 is injective on 𝑋⧵𝑎(𝛼𝛽)⁻¹. Therefore, 𝛼𝛽∈𝐹_𝑎.
Next, we show that 𝐹_𝑎 is a regular sub-semigroup of 𝑇(𝑋,𝑌). For each 𝛼∈𝐹_𝑎, 𝑎𝛼=𝑎 and |𝑥𝛼⁻¹|=1 for all 𝑥∈𝑌𝛼⧵{𝑎} (see Theorem 4.1). Let {𝑥_𝑖}=𝑌𝛼⧵{𝑎} and write 𝑥_𝑖𝛼⁻¹=𝑦_𝑖 for all 𝑖, thus 𝑦𝛼=^𝑖𝑎𝛼⁻¹𝑥_𝑖𝑎, (4.4) where ⋃{𝑦_𝑖}=𝑌⧵𝑎𝛼⁻¹, 𝑋=𝑌⋃𝑎𝛼⁻¹ and define 𝛽∈𝑇(𝑋,𝑌) by 𝑥𝛽=^𝑖𝐴𝑦_𝑖𝑎, (4.5) where {𝑥_𝑖}=𝑌𝛼⧵{𝑎} and 𝐴=𝑋⧵{𝑥_𝑖}. Since 𝑎∈𝐴, we get 𝑎=𝑎𝛽∈𝑌𝛽, and so {𝑎}⊆𝑌𝛽. And for each 𝑥∈𝑋⧵𝑌, 𝑥≠𝑥_𝑖 for all 𝑖 since 𝑥_𝑖∈𝑌. Thus (𝑋⧵𝑌)𝛽={𝑎}⊆𝑌𝛽, and so 𝛽∈𝐹. Since 𝑎𝛽=𝑎 and 𝛽 is injective on {𝑥_𝑖}=𝑋⧵𝑎𝛽⁻¹, it follows that 𝛽∈𝐹_𝑎. And, it is clear that 𝛼=𝛼𝛽𝛼.
Now, we prove that any two idempotents in 𝐹_𝑎 commute, which is enough to show that 𝐹_𝑎 is an inverse semigroup. Assume that 𝛼,𝛽 are idempotents in 𝐹_𝑎. Then 𝑥𝛼=𝑥 for all 𝑥∈𝑋⧵𝑎𝛼⁻¹ and 𝑥𝛽=𝑥 for all 𝑥∈𝑋⧵𝑎𝛽⁻¹. Let 𝑥 be any element in 𝑋.
Case 1. 𝑥∈𝑋⧵𝑎𝛼⁻¹. Then 𝑥𝛼=𝑥. So, if 𝑥∈𝑋⧵𝑎𝛽⁻¹, we get 𝑥𝛽=𝑥 and 𝑥(𝛼𝛽)=(𝑥𝛼)𝛽=𝑥𝛽=𝑥=𝑥𝛼=(𝑥𝛽)𝛼=𝑥(𝛽𝛼). But if 𝑥∈𝑎𝛽⁻¹, then 𝑥𝛽=𝑎, and 𝑥(𝛼𝛽)=(𝑥𝛼)𝛽=𝑥𝛽=𝑎=𝑎𝛼=(𝑥𝛽)𝛼=𝑥(𝛽𝛼). Thus in this case 𝛼𝛽=𝛽𝛼.
Case 2. 𝑥∈𝑎𝛼⁻¹. Then 𝑥𝛼=𝑎. So by the same proof as given in Case 1, we get 𝛼𝛽=𝛽𝛼.
Therefore, 𝐹_𝑎 is an inverse sub-semigroup of 𝑇(𝑋,𝑌).
To prove the maximality, we suppose that 𝐹_𝑎 is properly contained in an inverse sub-semigroup 𝐿 of 𝑇(𝑋,𝑌), where 𝐿⊆𝐹⊆𝑇(𝑋,𝑌) and let 𝛼∈𝐿⧵𝐹_𝑎. Let 𝛽 be the constant map with range {𝑎}, so 𝛽 is an idempotent in 𝐹_𝑎, and thus 𝛽𝛼 is an idempotent in 𝐿. Since 𝛽𝛼𝛽=𝛽 and every two idempotents in 𝐿 commute, it follows that 𝛽=𝛽𝛼𝛽=𝛽𝛽𝛼=𝛽𝛼 and 𝑎𝛼=(𝑎𝛽)𝛼=𝑎(𝛽𝛼)=𝑎𝛽=𝑎. Since 𝐿 is regular, 𝛼=𝛼𝛼^′𝛼 for some 𝛼^′∈𝐿 and 𝛼ℛ𝛼𝛼^′ on 𝐿 such that 𝛼𝛼^′ is an idempotent in 𝐿. Let 𝛾=𝛼𝛼^′, then by Lemma 4.2 we must have 𝛾∈𝐿⧵𝐹_𝑎 and 𝛾ℒ𝜎 for some idempotent 𝜎∈𝐹_𝑎. Since every idempotent 𝑒 in a semigroup is a right identity for 𝐿_𝑒, we have 𝛾=𝛾𝜎=𝜎𝛾=𝜎 which is a contradiction since 𝛾∉𝐹_𝑎 but 𝜎∈𝐹_𝑎. Therefore, 𝐿=𝐹_𝑎 as required.

As an application of Theorem 4.3, we get the following corollary which first appeared in [4].

Corollary 4.4. 𝐹_𝑎={𝛼∈𝑇(𝑋)∶𝑎𝛼=𝑎 and 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹} is a maximal inverse sub-semigroup of 𝑇(𝑋).

Proof. By taking 𝑌=𝑋 in Theorem 4.3, we get 𝑇(𝑋,𝑌)=𝑇(𝑋)=𝐹 and 𝐹_𝑎={𝛼∈𝑇(𝑋)∶𝑎𝛼=𝑎 and 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹} which is a maximal inverse sub-semigroup of 𝑇(𝑋).

Recall that the number of combinations of 𝑛 distinct things taken 𝑟 at a time written (^𝑛_𝑟) is given by 𝑛𝑟=𝑛!(𝑛−𝑟)!𝑟!. (4.6) That is, (^𝑛_𝑟) is the number of ways that 𝑟 objects can be chosen from 𝑛 distinct objects.

In the next result, we use the above information to find the number of elements in 𝐹_𝑎 when 𝑌 is a finite subset of 𝑋.

Theorem 4.5. Suppose that 𝑋 is an arbitrary set and 𝑌 is a nonempty subset of 𝑋 such that |𝑌|=𝑛. Then for each 𝑎∈𝑌, |𝐹_𝑎|=∑^𝑛−1_𝑟=0𝑟!(_𝑟^𝑛−1)².

Proof. Let 𝑎∈𝑌 and 𝛼∈𝐹_𝑎. Then by Theorem 4.1 we see that 𝑌𝛼=(𝑋⧵𝑌)₁𝑌₂𝑎𝑌₃, (4.7) where 𝑌=𝑌₁⋃̇𝑌₂,𝑎∈𝑌₁,𝑌₃⊆𝑌⧵{𝑎}, and |𝑌₂|=|𝑌₃|. If 𝑌₂=∅, then 𝛼 can have only one form, the constant map 𝑋_𝑎. If 𝑌₂ has 𝑡 elements, where 1≤𝑡≤𝑛−1, then 𝑌₂ can have (_𝑡^𝑛−1) choices and for each choice of 𝑌₂,𝑌₃ can have (_𝑡^𝑛−1) choices, thus there are (_𝑡^𝑛−1)² ways to choose 𝑌₂ and 𝑌₃. Since the restriction of 𝛼 to 𝑌₂ is a permutation, for each choice of 𝑌₂ and 𝑌₃, the map 𝛼 has 𝑡! possible forms. Hence in this case 𝛼 can have 𝑡!(_𝑡^𝑛−1)² forms.
Therefore, |𝐹_𝑎|=1+∑^𝑛−1_𝑟=1𝑟!(_𝑟^𝑛−1)²=∑^𝑛−1_𝑟=0𝑟!(_𝑟^𝑛−1)² as required.

We observe that the number of elements in 𝐹_𝑎 depends only on the elements of 𝑌, and when we replace 𝑌 with 𝑋 in Theorem 4.5, we have the following corollary which first appeared in [4].

Corollary 4.6. If 𝑋 is a finite set with |𝑋|=𝑛, then the number of elements in 𝐹_𝑎={𝛼∈𝑇(𝑋)∶𝑎𝛼=𝑎 and 𝛼 is injective on 𝑋⧵𝑎𝛼⁻¹} equals ∑^𝑛−1_𝑟=0𝑟!(_𝑟^𝑛−1)².

Acknowledgments

Most of the work in this paper formed part of an MS thesis written by W. Sommanee under the supervision of J. Sanwong. W. Sommanee greatly appreciates the help of his supervisor in this work; and both authors are very grateful to the referee for his/her careful reading of the original manuscript and helpful comments.