【摘要】 设ＴＸ 表示集合Ｘ 上的全变换半群，Ｃｏｎ（Ｓ）表示半群Ｓ上的同余格． 对Ｘ 上任一非平凡等价关 系Ｅ，令ＴＥ（Ｘ）＝ ｛ｆ∈Ｔｘ：（ａ，ｂ）∈Ｅ，（ｆ（ａ），ｆ（ｂ））∈Ｅ｝． 据［４］，ＴＥ（Ｘ）构成一个α半群，且Ｃｏｎ（ＴＥ（Ｘ））可以分解为三个互不相交的完全子格，其中的一个为［Ｃ（Ｅ），Ｃａ（Ｅ）］．本文报道了ＴＥ（Ｘ）的同余τ，并证明了当Ｅ为单等价关系时，τ是［Ｃ（Ｅ），Ｃａ（Ｅ）］中的唯一原子．更多还原

【Abstract】 T X denotes the full transformation semigroup on a set X and Con(S) denotes the complete lattice of all congruences on a semigroup S. For a nontrivial equivalence E on X, let T E(X)={f∈T X:(a,b) ∈E,(f(a),f(b)∈E}. As we have seen in [4],T E(X) forms an α semigroup and Con(T E(X)) can be decomposed into three disjoint complete sublattices, one of which is [C(E),C α(E)].In this paper we find out a congruence τ on T E(X) which is contained in [C(E),C α(E)] and we show that whenever E is a simple equivalence relation, τ is a unique atom of the complete sublattice [C(E),C α(E)] of Con(T E(X)).更多还原