Báo cáo toán học: Regularity and Isomorphism Theorems of Generalized Order - Preserving Transformation Semigroups

Trình tự bảo quản đầy đủ chuyển đổi nửa nhóm OT (X) trên X poset từ lâu đã được nghiên cứu. Trong bài báo này, chúng ta nghiên cứu các-nửa nhóm (OT (X, Y), θ) trong đó X và Y là chuỗi, OT (X, Y) là tập hợp của tất cả các bản đồ để bảo quản từ X vào Y, θ ∈ OT (Y , X) và hoạt động...
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Báo cáo toán học: "Regularity and Isomorphism Theorems of Generalized Order - Preserving Transformation Semigroups" 9LHWQDP -RXUQDOVietnam Journal of Mathematics 33:3 (2005) 253–260 RI 0$7+(0$7,&6 ‹ 9$67 Regularity and Isomorphism Theorems of Generalized Order - Preserving Transformation Semigroups Yupaporn Kemprasit1 and Sawian Jaidee2 1 Department of Mathematics, Faculty of Science Chulalongkorn University, Bangkok 10330, Thailand 2 Department of Mathematics, Faculty of Science, Khon Kean University Khon Kean 40002, Thailand Received April 15, 2003 Revised June 6, 2005Abstract. The full order-preserving transformation semigroup OT (X ) on a posetX has long been studied. In this paper, we study the semigroup (OT (X, Y ), θ)where X and Y are chains, OT (X, Y ) is the set of all order-preserving maps fromX into Y , θ ∈ OT (Y, X ) and the operation ∗ is defined by α ∗ β = αθβ for allα, β ∈ OT (X, Y ). We characterize when (OT (X, Y ), θ) is regular, (OT (X, Y ), θ)∼ OT (X ) and (OT (X, Y ), θ) ∼ OT (Y ).= =1. IntroductionThe full transformation semigroup on a set X is denoted by T (X ) and for α ∈T (X ), let ran α denote the range of α. It is well-known that T (X ) is regular forany set X , that is, for every α ∈ T (X ), α = αβα for some β ∈ T (X ). Next, let X and Y be posets. A map α : X → Y is said to be order-preservingif for x1 , x2 ∈ X, x1 ≤ x2 implies x1 α ≤ x2 α. A bijection ϕ : X → Y is calledan order-isomorphism if ϕ and ϕ−1 are order-preserving. It is clear that if bothX and Y are chains and ϕ : X → Y is an order-preserving bijection, then ϕ isan order-isomorphism. We say that X and Y are order-isomorphic if there is anorder-isomorphism from X onto Y . Naturally, X and Y are said to be anti-order-isomorphic if there exists a bijection ϕ : X → Y such that for x1 , x2 ∈ X, x1 ≤x2 if and only if x2 ϕ ≤ x1 ϕ. Let OT (X ) denote the subsemigroup of T (X )254 Yupaporn Kemprasit and Sawian Jaideeconsisting of all order-preserving transformations α : X → X . The semigroupOT (X ) may be called the full order-preserving transformation semigroup onX (see [6]). The full order-preserving transformation semigroup on a poset haslong been studied. For examples, see [5, Theorem V.8.9], [2, (Exercise 6.1.7),3, 7, 1, 4]. Theorem V.8.9 of [5] gives an interesting isomorphism theorem asfollows: For posets X and Y , OT (X ) ∼ OT (Y ) if and only if X and Y are order- =isomorphic or anti- order-isomorphic. Let Z and R denote the set of integersand the set of real numbers, respectively. In [3], Kemprasit and Changphascharacterized when OT (X ) is regular where X is a nonempty subset of Z or Xis a nonempty interval of R with their natural order as follows:Theorem 1.1 [4]. For any nonempty subset X of Z, OT (X ) is regular.Theorem 1.2 [4]. For a nonempty interval X of R, OT (X ) is regular if andonly if X is closed and bounded. In this paper, the semigroup OT (X ) is replaced by the semigroup (OT (X, Y ),θ) where OT (X, Y ) is the set of all order-preserving maps α : X → Y , θ ∈OT (Y, X ) and the operation ∗ is defined by α ∗ β = αθβ for all α, β ∈ OT (X, Y ).Note that OT (X ) = (OT (X, X ), 1X ) where 1X is the identity map on X .We confine our attention to study the semigroup (OT (X, Y ), θ) when X andY are chains. In this paper we characterize the regularity of the semigroup(OT (X, Y ), θ). Further we provide necessary and sufficient conditions for thissemigroup to be isomorphic to OT (X ) and, respectively, isomorphic to OT (Y ).Our main results are Theorems 3.1, 3.5 and 3.6. From now on we assume thatX and Y are chains and θ ∈ OT (Y, X ).2. LemmasThe following series of the lemmas is required to obtain our main results.Lemma 2.1. Let a,b ∈ X and c, d ∈ Y be such that a < b, c < d and cθ = dθ.If α : X → Y is defined by c if x < b, xα = if x ≥ b, dthen α ∈ OT (X, Y ), |ran α| = 2 and |ran(αθ)| = 1.Proof. It is clear that α ∈ OT (X, Y ), ran α = {c, d} and ran(αθ) = (ran α)θ ={cθ, dθ} = {cθ}.Lemma 2.2. If |X | > 1 and (OT (X, Y ), θ) is regular, then θ is one-to-one.Proof. Let |X | > 1 and assume that θ is not one-to-one. Then there are a, b ∈ Xand c, d ∈ Y such that a < b, c < d and cθ = dθ. Define α : X → Y as in Lemma ...