On the representation of ordered semigroups by transformations of ordered sets

Abstract

A transformation of an ordered set M is an isotone mapping of M into M. By a representation of an ordered semigroup S by transformations of an ordered set M we mean a homomorphism of S into the set of transformations of M, i.e. (since the set of transformations of M is an ordered semigroup) an isotone mapping from S into the set of transformations of M preserving the operations. We prove that this type of representation leads to an “action” of S on M and so we introduce the notion of a left operand of M over S. Also we introduce the notions of a left operator pseudoorder on a left operand over S and a left operator homomorphism between left operands over S. We show that the concept of left operator pseudoorders on left operands over S plays an important role in the study of left operator homomorphisms of left operands over S. In the case of right operands over S dually definitions and results hold.

1 Introduction

By a representation of a semigroup S by transformations of a set M we mean (see [2] and [1]) a homomorphism of S into ${{\mathcal {T}}_{M}}$, where ${{\mathcal {T}}_{M}}$ is the full transformation semigroup on M (i.e. the set of all mappings of M into itself). This type of representation of S leads to an “action” of S on M. The aim of the paper is to study similar concepts in the case of ordered semigroups and ordered sets. Since the equality relation is clearly an order relation (and hence each set can be regarded as an ordered set with the equality relation as its order relation) then, in the case of ordered sets, it is natural to consider any isotone mapping of an ordered set into itself as a transformation of the ordered set. Also from [2] we see that congruences are very important in the study of operator homomorphisms between operands over semigroups. In the case of ordered semigroups, the notion of congruence is usually replaced by the notion of pseudoorder (in many papers — most of them joint with Kehayopulu — [3,4,5,6,7,8,9,10,11,12,13] we have shown the very essential role of pseudoorders in Ordered Semigroup Theory). Following this, we introduce and study the concept of operator pseudoorder on operands over ordered semigroups and, from the results obtained, we can understand how fundamental the notion of operator pseudoorder is in the study of operator homomorphisms between operands over ordered semigroups.

2 Representation of ordered semigroups by transformations of ordered sets: operands of ordered sets over ordered semigroups

An ordered semigroup $\left( T,\centerdot ,{{\le }_{T}}\right)$ is a semigroup $\left( T,\centerdot \right)$ with an order relation “${{\le }_{T}}$” which is compatible with the operation “$\centerdot$” (i.e. for $\alpha ,b,c\in T$, $\alpha \ {{\le }_{T}}\ b$ implies $c\centerdot \alpha \ {{\le }_{T}}\ c\centerdot b$ and $\alpha \centerdot c\ {{\le }_{T}}\ b\centerdot c$). Let $\left( M,{{\le }_{M}}\right)$ be an ordered set. A mapping $\alpha :\left( M,{{\le }_{M}} \right) \rightarrow \left( M,{{\le }_{M}} \right)$ is called a transformation of M if $x\ {{\le }_{M}}\ y$ implies $\alpha \left( x\right) \ {{\le }_{M}}\ \alpha \left( y\right)$ for $x,y\in M$ (i.e. a transformation of M is simply an isotone mapping of $\left( M,{{\le }_{M}}\right)$). We denote by ${{\mathcal {T}}_{\left( M,{{\le }_{M}}\right) }}$ the set of all transformations of $\left( M,{{\le }_{M}}\right)$. If no confusion arises, we shall write ${{\mathcal {T}}_{M}}$ instead of ${{\mathcal {T}}_{\left( M,{{\le }_{M}}\right) }}$. For $\alpha ,\beta \in {{\mathcal {T}}_{M}}$ we define $\alpha \ {{{\underline{\prec }}}_{{{\mathcal {T}}_{M}}}}\beta$ if $\alpha \left( x\right) \ {{\le }_{M}}\ \beta \left( x\right)$ for each $x\in M$. It is a matter of routine to prove that $\left( {{\mathcal {T}}_{M}},\circ ,{{{{\underline{\prec }}}}_{{{\mathcal {T}}_{M}}}}\right)$ is an ordered semigroup where “$\circ$” is the composition of transformations of $\left( M,{{\le }_{M}}\right)$ (i.e. if $\alpha ,\beta \in {{\mathcal {T}}_{M}}$ then $\left( \alpha \circ \beta \right) \left( x\right) :=\alpha \left( \beta \left( x\right) \right)$ for all $x\in M$). Consider now an ordered semigroup $\left( S,\cdot ,{{\le }_{S}}\right)$. By a representation of $\left( S,\cdot ,{{\le }_{S}}\right)$ by transformations of $\left( M,{{\le }_{M}}\right)$ we mean a homomorphism of $\left( S,\cdot ,{{\le }_{S}}\right)$ into $\left( {{\mathcal {T}}_{M}},\circ ,{{{{\underline{\prec }}}}_{{{\mathcal {T}}_{M}}}}\right)$, that is, a mapping $\varphi :\left( S,\cdot ,{{\le }_{S}}\right) \rightarrow \left( {{\mathcal {T}}_{M}},\circ ,{{{\underline{\prec }}}_{{{\mathcal {T}}_{M}}}}\right)$ such that

• i) $\varphi \left( \alpha \cdot b\right) =\varphi \left( \alpha \right) \circ \varphi \left( b\right)$ for all $\alpha ,b\in S$, and

• ii) for $\alpha ,b\in S$, $\alpha \ {{\le }_{S}}\ b$ implies $\varphi \left( \alpha \right) \ {{{\underline{\prec }}}_{{{\mathcal {T}}_{M}}}}\varphi \left( b \right)$

Dually, by an anti-representation of $\left( S,\cdot ,{{\le }_{S}}\right)$ by transformations of $\left( M,{{\le }_{M}}\right)$ we mean an anti-homomorphism of $\left( S,\cdot ,{{\le }_{S}}\right)$ into $\left( {{\mathcal {T}}_{M}},\circ ,{{{{\underline{\prec }}}}_{{{\mathcal {T}}_{M}}}}\right)$, that is, a mapping $\psi :\left( S,\cdot ,{{\le }_{S}}\right) \rightarrow \left( {{\mathcal {T}}_{M}},\circ ,{{{{\underline{\prec }}}}_{{{\mathcal {T}}_{M}}}}\right)$ such that

• $\psi \left( \alpha \cdot b\right) =\psi \left( b\right) \circ \psi \left( \alpha \right)$ for all $\alpha ,b\in S$, and

• for $\alpha ,b\in S$, $\alpha \ {{\le }_{S}}\ b$ implies $\psi \left( \alpha \right) \ {{{\underline{\prec }}}_{{{\mathcal {T}}_{M}}}}\psi \left( b \right)$.

Now if

$$\begin{aligned} _{S}{{*}_{M}}:&S\times M\rightarrow M \\&\left( \alpha ,x\right) \ \rightarrow \alpha \ _{S}{{*}_{M}}\ x \end{aligned}$$

is a mapping satisfying the following two conditions:

• I) $\left( \alpha \cdot b\right) \ _{S}{{*}_{M}}\ x=\alpha \ _{S}{{*}_{M}}\left( b\ _{S}{{*}_{M}}\ x\right)$ for all $\alpha ,b\in S$, $x\in M$,

• II) for $\alpha ,b\in S$ and $x,y\in M$ such that $\alpha \ {{\le }_{S}}\ b$ and $x\ {{\le }_{M}}\ y$, it follows $\alpha \ _{S}{{*}_{M}}\ x\ {{\le }_{M}}\ b\ _{S}{{*}_{M}}\ y$,

then $\left[ \left( M,{{\le }_{M}}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{S}{{*}_{M}}\right]$ is called a left operand of $\left( M,{{\le }_{M}}\right)$ over $\left( S,\cdot , {{\le }_{S}}\right)$ and is denoted $_{S}M$.

Dually, if

$$\begin{aligned} _{M}{{*}_{S}}:&M\times S\rightarrow M \\&\left( x,\alpha \right) \ \rightarrow x\ _{M}{{*}_{S}}\ \alpha \end{aligned}$$

is a mapping satisfying the next two conditions:

• $x\ _{M}{{*}_{S}}\ \left( \alpha \cdot b\right) =\left( x\ _{M}{{*}_{S}}\ \alpha \right) \ _{M}{{*}_{S}}\ b$ for all $\alpha ,b\in S$, $x\in M$,

• for $x,y\in M$ and $\alpha ,b\in S$ such that $x\ {{\le }_{M}}\ y$ and $\alpha \ {{\le }_{S}}\ b$, it follows $x\ _{M}{{*}_{S}}\ \alpha \ {{\le }_{M}}\ y\ _{M}{{*}_{S}}\ b$,

then $\left[ \left( M,{{\le }_{M}}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{M}{{*}_{S}}\right]$ is called a right operand of $\left( M,{{\le }_{M}}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ and is denoted by ${{M}_{S}}$. If the context ensures that no confusion will arise, we abbreviate $\alpha \ _{S}{{*}_{M}}\ x$ to $\alpha x$ and $x\ _{M}{{*}_{S}}\ \alpha$ to $x\alpha$.

It is straightforward to prove the next

Proposition 2.1

Let $_{S}M$ be a left operand of M over S. Then the mapping

$$\begin{aligned} \varphi :&S\rightarrow {{\mathcal {T}}_{M}}\\&\alpha \rightarrow \varphi \left( \alpha \right) :M\rightarrow M \\&\quad \quad \quad \quad \quad \ x\ \rightarrow \varphi \left( \alpha \right) \left( x\right) :=\alpha x \end{aligned}$$

is a representation of S by transformations of M.

The following (dual) Proposition also holds.

Proposition 2.2

Let ${{M}_{S}}$ be a right operand of M over S. Then the mapping

$$\begin{aligned} \psi :&S\rightarrow {{\mathcal {T}}_{M}} \\&\alpha \rightarrow \psi \left( \alpha \right) :M\rightarrow M \\&\quad \quad \quad \quad \quad \ x\ \rightarrow \psi \left( \alpha \right) \left( x \right) :=x\alpha \end{aligned}$$

is an anti-representation of S by transformations of M.

Proposition 2.3

Let $\varphi :\left( S,\cdot ,{{\le }_{S}}\right) \rightarrow \left( {{\mathcal {T}}_{M}},\circ ,{{{{\underline{\prec }}}}_{{{\mathcal {T}}_{M}}}}\right)$ be a representation of S by transformations of M and

$$\begin{aligned} _{S}{{*}_{M}}:&S\times M\rightarrow M \\ {}&\left( \alpha ,x\right) \ \rightarrow \alpha \ _{S}{{*}_{M}}\ x:=\varphi \left( \alpha \right) \left( x \right) \end{aligned}$$

Then $\left[ \left( M,{{\le }_{M}}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{S}{{*}_{M}}\right]$ is a left operand of M over S.

Proof

It is clear that “$_{S}{{*}_{M}}$” is well defined.

• i) Let $\alpha ,b\in S$ and $x\in M$. Then (since $\varphi$ is a representation of S by transformations of M) we have

  $$\begin{aligned} \left( \alpha \cdot b\right) \ _{S}{{*}_{M}}\ x&=\varphi \left( \alpha \cdot b\right) \left( x \right) =\left( \varphi \left( \alpha \right) \circ \varphi \left( b\right) \right) \left( x\right) =\varphi \left( \alpha \right) \left( \varphi \left( b\right) \left( x\right) \right) \\ {}&=\varphi \left( \alpha \right) \left( b\ _{S}{{*}_{M}}\ x \right) =\alpha \ _{S}{{*}_{M}}\ \left( b\ _{S}{{*}_{M}}\ x\right) \end{aligned}$$

• ii) Let $\alpha ,b\in S$ and $x,y\in M$ such that $\alpha \ {{\le }_{S}}\ b$ and $x\ {{\le }_{M}}\ y$. Since $\varphi$ is a representation of S by transformations of M, we have $\varphi \left( \alpha \right) \ {{{\underline{\prec }}}_{{{\mathcal {T}}_{M}}}}\varphi \left( b\right)$ and hence $\varphi \left( \alpha \right) \left( y\right) \ {{\le }_{M}}\ \varphi \left( b\right) \left( y\right)$. Also (since $\varphi \left( \alpha \right)$ is a transformation of M) we have $\varphi \left( \alpha \right) \left( x\right) \ {{\le }_{M}}\ \varphi \left( \alpha \right) \left( y\right)$. Thus $\varphi \left( \alpha \right) \left( x\right) \ {{\le }_{M}}\ \varphi \left( b\right) \left( y\right)$, that is,

  $$\begin{aligned} \alpha \ _{S}{{*}_{M}}\ x\ {{\le }_{M}}\ b\ _{S}{{*}_{M}}\ y \end{aligned}$$

$\square$

In a similar way we prove the next (dual) Proposition.

Proposition 2.4

Let $\psi :\left( S,\cdot ,{{\le }_{S}}\right) \rightarrow \left( {{\mathcal {T}}_{M}},\circ ,{{{{\underline{\prec }}}}_{{{\mathcal {T}}_{M}}}}\right)$ be anti-representation of S by transformations of M and

$$\begin{aligned} _{M}{{*}_{S}}:&M\times S\rightarrow M \\ {}&\left( x,\alpha \right) \ \rightarrow x\ _{M}{{*}_{S}}\ \alpha :=\psi \left( \alpha \right) \left( x \right) \end{aligned}$$

Then $\left[ \left( M,{{\le }_{M}}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{M}{{*}_{S}}\right]$ is a right operand of M over S.

In the next we shall deal only with left operands. By Propositions 2.1, 2.3 we immediately have the following.

Theorem 2.5

There exists a left operand of $\left( M,{{\le }_{M}}\right)$ over $\left( S,\cdot ,{{\le }_{S}} \right)$ if and only if there exists a representation of $\left( S,\cdot ,{{\le }_{S}}\right)$ by transformations of $\left( M,{{\le }_{M}}\right)$.

3 Operator pseudoorders of operands

Let $_{S}M$ be a left operand of M over S. A relation $\sigma$ on M is called left operator pseudoorder of $_{S}M$ if

• A) ${{\le }_{M}}\subseteq \sigma$,

• B) $\sigma$ is transitive,

• C) for $\alpha \in S$ and $x,y\in M$, $\left( x,y\right) \in \sigma$ implies $\left( \alpha x,\alpha y\right) \in \sigma$.

Proposition 3.1

Let $_{S}M$ be a left operand of M over S and $\sigma$ be a left operator pseudoorder of $_{S}M$. Also let $\alpha ,b\in S$ and $x,y\in M$ such that $\alpha \ {{\le }_{S}}\ b$ and $\left( x,y\right) \in \sigma$. Then $\left( \alpha x,by\right) \in \sigma$.

Proof

Since $_{S}M$ is a left operand of M over S, $\alpha \ {{\le }_{S}}\ b$ and $x\ {{\le }_{M}}\ x$, then $\alpha x\ {{\le }_{M}}\ bx$ and hence (since ${{\le }_{M}}\subseteq \sigma$) it follows that $\left( \alpha x,bx\right) \in \sigma$. Also (since $\sigma$ is a left operator pseudoorder of $_{S}M$ and $\left( x,y\right) \in \sigma$, $b\in S$) we have $\left( bx,by\right) \in \sigma$. Then, from the transitivity of $\sigma$, it follows immediately that $\left( \alpha x,by\right) \in \sigma$. $\square$

Let now $\sigma$ be a left operator pseudoorder of $_{S}M$. We define ${\bar{\sigma }}:=\sigma \cap {{\sigma }^{-1}}$ where ${\sigma }^{-1}=\left\{ \left( x,y\right) \in M\times M:\left( y,x\right) \in \sigma \right\}$. By a left operator congruence of $_{S}M$ we mean an equivalence relation $\rho$ on M having the property: for $x,y\in M$ and $\alpha \in S$, $\left( x,y\right) \in \rho$ implies $\left( \alpha x,\alpha y\right) \in \rho$.

If $\rho$ is a left operator congruence of $_{S}M$, we define $^{M}/_{\rho }$ as the set of all $\rho$-classes of M (i.e. $^{M}/_{\rho }=\left\{ {{\left( x\right) }_{\rho }}:x\in M\right\}$ where ${{\left( x\right) }_{\rho }}=\left\{ y\in M:\left( x,y\right) \in \rho \right\}$, $x\in M$).

Proposition 3.2

Let $\sigma$ be a left operator pseudoorder of $_{S}M$. Then ${\bar{\sigma }}$ is a left operator congruence of $_{S}M$.

Proof

It is easy to prove that ${\bar{\sigma }}$ is an equivalence relation of $_{S}M$. Take now $\alpha \in S$ and $x,y\in M$ such that $\left( x,y\right) \in {\bar{\sigma }}$. Thus $\left( x,y\right) ,\left( y,x\right) \in \sigma$ and since $\sigma$ is a left operator pseudoorder of $_{S}M$, it follows that $\left( \alpha x,\alpha y\right) ,\left( \alpha y,\alpha x \right) \in \sigma$ which means $\left( \alpha x,\alpha y\right) \in {\bar{\sigma }}$. $\square$

If $\sigma$ is a left operator pseudoorder of $_{S}M$, we define a relation “${{\le }_{\sigma }}$” on $^{M }/_{{{\bar{\sigma }}}}$ by the rule that

$$\begin{aligned} {{\le }_{\sigma }}:=\left\{ \left( A,B\right) \in \ ^{M}/_{{{\bar{\sigma }}}}\times \ ^{M}/{}_{{{\bar{\sigma }}}}:\left( \exists \ x\in A,y\in B \right) \ \left( x,y\right) \in \sigma \right\} \end{aligned}$$

Lemma 3.3

Let $\sigma$ be a left operator pseudoorder of $_{S}M$ and $x,y\in M$. Then ${\left( x\right) }_{{{\bar{\sigma }}}}\ {\le }_{\sigma }\ {\left( y\right) }_{{{\bar{\sigma }}}}$ if and only if $\left( x,y\right) \in \sigma$.

Proof

If $\left( x,y\right) \in \sigma$ then clearly $\left( x\right) _{{\bar{\sigma }}}\ {\le }_{\sigma }\ \left( y\right) _{{\bar{\sigma }}}$. Let now ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$. By definition of ${{\le }_{\sigma }}$ there exist $z\in {{\left( x\right) }_{{{\bar{\sigma }}}}}$ and $w\in {{\left( y\right) }_{{{\bar{\sigma }}}}}$ such that $\left( z,w\right) \in \sigma$. Since $z\in {{\left( x\right) }_{{{\bar{\sigma }}}}}$ then $\left( x,z\right) \in \sigma$. Similarly $\left( w,y\right) \in \sigma$. By the transitivity of $\sigma$ it follows immediately that $\left( x,y\right) \in \sigma$. $\square$

As in the case of ordered semigroups (cf. [4, Lemma 1]), we can similarly prove the following

Proposition 3.4

Let $\sigma$ be a left operator pseudoorder of $_{S}M$ (resp. ${{M}_{S}}$). Then the relation ${{\le }_{\sigma }}$ is an order relation on $^{M}/_{{{\bar{\sigma }}}}$.

Proposition 3.5

Let $\sigma$ be a left operator pseudoorder of $_{S}M$. Also let $x,y\in M$ and $\alpha ,b\in S$ such that $\left( x\right) _{{\bar{\sigma }}}\ {\le }_{\sigma } \ \left( y\right) _{{\bar{\sigma }}}$ and $\alpha \ {\le }_{S}\ b$. Then ${{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( by\right) }_{{{\bar{\sigma }}}}}$.

Proof

Since ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$ then, by Lemma 3.3, $\left( x,y\right) \in \sigma$. Therefore, by Proposition 3.1 (since $\alpha \ {{\le }_{S}}\ b$), we have $\left( \alpha x,by\right) \in \sigma$ and hence (from the definition of ${{\le }_{\sigma }}$) ${{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( by\right) }_{{{\bar{\sigma }}}}}$. $\square$

Let $\rho$ be a left operator congruence of $_{S}M$. We define

$$\begin{aligned} _{S}{{*}_{^{M}/{}_{\rho }}}:&\ S\times ^{M}/_{\rho }\rightarrow \ ^{M}/{}_{\rho } \\&\left( \alpha ,{{\left( x\right) }_{\rho }}\right) \rightarrow \alpha \ _{S}{{*}_{^{M}/_{\rho }}}\ {{\left( x \right) }_{\rho }}:={{\left( \alpha x\right) }_{\rho }} \end{aligned}$$

As we have already mentioned, if there is no confusion, we usually write $\alpha {{\left( x\right) }_{\rho }}$, instead of $\alpha \ _{S}{{*}_{{}^{M}/{}_{\rho }}}\ {{\left( x\right) }_{{{\bar{\sigma }}}}}$ and hence $\alpha {{\left( x\right) }_{\rho }}={{\left( \alpha x\right) }_{\rho }}$.

Proposition 3.6

Let $\rho$ be a left operator congruence of $_{S}M$. Then

• i) $_{S}{{*}_{{}^{M}/{}_{\rho }}}$ is well defined (i.e. $_{S}{{*}_{{}^{M}/{}_{\rho }}}$ is a single valued mapping),

• ii) for each $\alpha ,b\in S$ and $x\in M$ we have that

  $$\begin{aligned} \left( \alpha \cdot b\right) \ _{S}{{*}_{^{M}/_{\rho }}}\ {{\left( x\right) }_{{{\bar{\sigma }}}}}=\alpha \ _{S}{{*}_{^{M}/_{\rho }}}\left( b\ _{S}{{*}_{^{M}/{}_{\rho }}}\ {{\left( x\right) }_{\rho }}\right) \end{aligned}$$

Proof

i) Let $\alpha \in S$ and $x,y\in M$ such that ${{\left( x\right) }_{\rho }}={{\left( y\right) }_{\rho }}$. Then $\left( x,y\right) \in \rho$. Since $\rho$ is a left operator congruence of $_{S}M$ we have $\left( \alpha x,\alpha y\right) \in \rho$. Therefore ${{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}={{\left( \alpha y\right) }_{{{\bar{\sigma }}}}}$ and thus $\alpha \ _{S}{{*}_{^{M}/_{\rho }}}\ {{\left( x\right) }_{\rho }}=\alpha \ _{S}{{*}_{^{M}/_{\rho }}}\ {{\left( y\right) }_{\rho }}$.

ii) Let $\alpha ,b\in S$ and $x\in M$. Then $\left( \alpha \cdot b\right) \ _{S}{{*}_{^{M}/_{\rho }}}\ {{\left( x\right) }_{\rho }}={{\left( \left( \alpha \cdot b\right) x\right) }_{\rho }}={{\left( \alpha \left( bx\right) \right) }_{\rho }}=\alpha \ _{S}{{*}_{^{M}/_{\rho }}}\ {{\left( bx\right) }_{\rho }}=\alpha \ _{S}{{*}_{^{M}/_{\rho }}}\left( b\ _{S}{{*}_{^{M}/_{\rho }}}\ {{\left( x\right) }_{\rho }}\right)$. $\square$

Proposition 3.7

Let $\sigma$ be a left operator pseudoorder of $_{S}M$. Also let $\alpha ,b\in S$ and $x,y\in M$ such that $\alpha \ {{\le }_{S}}\ b$ and ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$. Then

$$\begin{aligned} \alpha \ _{S}{{*}_{^{M}/_{{{\bar{\sigma }}}}}}\ {{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ b\ _{S}{{*}_{^{M}/_{{{\bar{\sigma }}}}}}\ {{\left( y\right) }_{{{\bar{\sigma }}}}} \end{aligned}$$

Proof

Since ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$ then (by Lemma 3.3) $\left( x,y\right) \in \sigma$ and thus (by Proposition 3.1) $\left( \alpha x,by\right) \in \sigma$ which means (by Lemma 3.3) ${{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( by\right) }_{{{\bar{\sigma }}}}}$. Consequently $\alpha \ _{S}{{*}_{^{M}/_{{{\bar{\sigma }}}}}}\ {{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ b\ _{S}{{*}_{^{M}/_{{{\bar{\sigma }}}}}}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$. $\square$

Remark 3.8

If $\sigma$ is a left operator pseudoorder of ${}_{S}M$ then, by Propositions 3.4,  3.2, 3.6 and 3.7 we immediately have that $\left[ \left( ^{M}/_{{{\bar{\sigma }}}},{{\le }_{\sigma }}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;_{S}{{*}_{^{M}/_{{{\bar{\sigma }}}}}}\right]$ is a left operand of $\left( ^{M}/_{{{\bar{\sigma }}}},{{\le }_{\sigma }}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ denoted by $^{_{S}M}/_{{{\bar{\sigma }}}}$. We call $^{_{S}M}/_{{{\bar{\sigma }}}}$ left factor operand of $_{S}M$ by $\sigma$.

4 Operator homomorphisms of operands

Let $_{S}M$, $_{S}{M}'$ be left operands of ordered sets $\left( M,{{\le }_{M}}\right)$ and $\left( {M}',{{\le }_{{{M}'}}}\right)$ respectively over $\left( S,\cdot ,{{\le }_{S}}\right)$. A mapping $\theta :M\rightarrow {M}'$ (we usually write $\theta :_{S}M\rightarrow {}_{S}{M}'$) is called

• i) a left operator homomorphism if

  • i1) for all $\alpha \in S$ and $x\in M$, we have that $\theta \left( \alpha x \right) =\alpha \ \theta \left( x\right)$,

  • i2) for $x,y\in M$, $x\ {{\le }_{M}}\ y$ implies $\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right)$.

• ii) a left operator isomorphism if

  • ii1) it is a left operator homomorphism,

  • ii2) for $x,y\in M$, $\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right)$ implies $x\ {{\le }_{M}}\ y$ (i.e. $\theta$ is reverse isotone and hence 1-1),

  • ii3) it is onto.

If $\theta$ is a left operator isomorphism, we write $_{S}M\simeq {}_{S}{M}'$.

Now let $_{S}M$ be a left operand of M over S and $\sigma$ be a left operator pseudoorder of $_{S}M$. We define $_{S}{{\sigma }^{\#}}:{}_{S}M\rightarrow {}^{_{S}M }/_{{{\bar{\sigma }}}},\ _{S}{{\sigma }^{\#}}\left( x\right) :={{\left( x\right) }_{{{\bar{\sigma }}}}}$

Proposition 4.1

Let $_{S}M$ be a left operand of M over S and $\sigma$ be a left operator pseudoorder of $_{S}M$. Then $_{S}{{\sigma }^{\#}}$ is a left operator homomorphism and onto.

Proof

It is clear that $_{S}{{\sigma }^{\#}}$ is onto.

i1) Let $\alpha \in S$ and $x\in M$. Then

$$\begin{aligned} _{S}{{\sigma }^{\#}}\left( \alpha x\right) ={{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}=\alpha {{\left( x \right) }_{{{\bar{\sigma }}}}}=\alpha \ _{S}{{\sigma }^{\#}}\left( x\right) \end{aligned}$$

i2) Let $x,y\in M$ such that $x\ {{\le }_{M}}\ y$. Since $\sigma$ is a left operator pseudoorder of $_{S}M$ we have ${{\le }_{M}}\subseteq \sigma$ and hence $\left( x,y\right) \in \sigma$. Then (by definition of ${{\le }_{\sigma }}$) ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$, that is, $_{S}{{\sigma }^{\#}}\left( x\right) \ {{\le }_{\sigma }}\ _{S}{{\sigma }^{\#}}\left( y\right)$. $\square$

A left operator congruence $\rho$ of ${}_{S}M$ is said to be regular if there exists an order relation “${{{\underline{\prec }}}_{\rho }}$” on $^{M}/_{\rho }$ such that

• a) $\left[ \left( ^{M}/_{\rho },{{{{\underline{\prec }}}}_{\rho }}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{S}{{*}_{^{M}/_{\rho }}}\right]$ is a left operand of $\left( ^{M}/_{\rho },{{\underline{{\prec }}}_{\rho }}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$, say $^{{}_{S}M}/_{\rho }$,

• b) the mapping $_{S}\pi :{}_{S}M \rightarrow {}^{{}_{S}M }/_{\rho }$, $_{S}\pi \left( x\right) :={{\left( x\right) }_{\rho }}$ is a left operator homomorphism.

In general, any left operator congruence is not regular. We prove it in the following

Example

Let $\left( S,\cdot ,{{\le }_{S}} \right)$ be an arbitrary ordered semigroup. We consider the set $M=\left\{ x,y,z,w \right\}$ and the relation

$$\begin{aligned} {{\le }_{M}}:=\left\{ \left( x,x \right) ,\left( x,z \right) ,\left( y,y \right) ,\left( y,w \right) ,\left( z,z \right) ,\left( w,w \right) \right\} \end{aligned}$$

Clearly $\left( M,{{\le }_{M}} \right)$ is an ordered set. For $\alpha \in S$ and $u\in M$ we define

$$\begin{aligned} \alpha {}_{S}{{*}_{M}}u:=u \end{aligned}$$

It is straightforward to prove that $\left[ \left( M,{{\le }_{M}} \right) ,\left( S,\cdot ,{{\le }_{S}} \right) ;{}_{S}{{*}_M} \right]$ is a left operand of $\left( M,{{\le }_{M}} \right)$ over $\left( S,\cdot ,{{\le }_{S}} \right)$. Now let

$$\begin{aligned} \rho :=\left\{ \left( x,x \right) ,\left( x,w \right) ,\left( y,y \right) ,\left( y,z \right) ,\left( z,y \right) ,\left( z,z \right) ,\left( w,x \right) ,\left( w,w \right) \right\} \end{aligned}$$

It follows immediately that $\rho$ is a left operator congruence of ${}_{S}M$ with ${{\left( x \right) }_{\rho }}=\left\{ x,w \right\} ={{\left( w \right) }_{\rho }}$ and ${{\left( y \right) }_{\rho }}=\left\{ y,z \right\} ={{\left( z \right) }_{\rho }}$. Assume that $\rho$ is regular. Then there exists an order relation “${{{\underline{\prec }}}_{\rho }}$” on ${}^{M}/{}_{\rho }$ such that the mapping ${}_{S}\pi$ is a left operator homomorphism of ${}_{S}M$ into ${}^{{}_{S}M}/{}_{\rho }$ and hence for $u,v\in M$, $u\ {{\le }_{M}}\ v$ implies ${{\left( u \right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( v \right) }_{\rho }}$. So, since $x\ {{\le }_{M}}\ z$ and $y\ {{\le }_{M}}\ w$, we have ${{\left( x \right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( z \right) }_{\rho }}$ and ${{\left( y \right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( w \right) }_{\rho }}$, that is, ${{\left( x \right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( y \right) }_{\rho }}$ and ${{\left( y \right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( x \right) }_{\rho }}$. Thus ${{\left( x \right) }_{\rho }}={{\left( y \right) }_{\rho }}$ which obviously does not hold. Therefore $\rho$ is not regular.

Theorem 4.2

Let $_{S}M$ be a left operand of $\left( M,{{\le }_{M}}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ and $\rho$ be a left operator congruence of $_{S}M$. The following are equivalent:

• i) $\rho$ is a regular.

• ii) There exists a left operator pseudoorder of $_{S}M$, say $\sigma$, such that $\rho ={\bar{\sigma }}$.

Proof

i) $\Rightarrow$ ii) Since $\rho$ is regular then there exists an order relation “${{{\underline{\prec }}}_{\rho }}$” on $^{M}/{}_{\rho }$ such that

$\circ$:

$\left[ \left( ^{M}/{}_{\rho },{{{\underline{\prec }}}_{\rho }}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{S}{{*}_{{}^{M}/{}_{\rho }}}\right]$ is a left operand of $\left( ^{M}/{}_{\rho },{{{{\underline{\prec }}}}_{\rho }}\right)$ over $\left( S,\cdot , {{\le }_{S}}\right)$, say $^{{}_{S}M}/_{\rho }$, and

$\circ$:

the mapping $_{S}\pi :{}_{S}M \rightarrow {}^{_{S}M}/_{\rho }$, $_{S}\pi \left( x\right) :={{\left( x\right) }_{\rho }}$ is a left operator homomorphism.

Consider $\sigma :=\left\{ \left( x,y\right) \in M\times M:{{\left( x\right) }_{\rho }}\ {{{{\underline{\prec }}}}_{\rho }}{{\left( y \right) }_{\rho }}\right\}$.

A) Let $x,y\in M$ such that $x\ {{\le }_{M}}\ y$. Since $_{S}\pi$ is a left operator homomorphism, we have $_{S}\pi \left( x\right) \ {{{\underline{\prec }}}_{\rho }}\ _{S}\pi \left( y\right)$, that is, ${{\left( x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}{{\left( y\right) }_{\rho }}$ and hence $\left( x,y\right) \in \sigma$. Therefore ${{\le }_{M}}\subseteq \sigma$.

B) Obviously $\sigma$ is transitive.

C) Let $\alpha \in S$ and $x,y\in M$ such that $\left( x,y\right) \in \sigma$. Since $\left( x,y\right) \in \sigma$, then ${{\left( x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}{{\left( y\right) }_{\rho }}$. Thus, since $\left[ \left( ^{M}/{}_{\rho },{{{{\underline{\prec }}}}_{\rho }}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{S}{{*}_{{}^{M}/{}_{\rho }}}\right]$ is a left operand of $\left( {}^{M}/{}_{\rho },{{{{\underline{\prec }}}}_{\rho }}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ and $\alpha \ {{\le }_{S}}\ \alpha$, it follows that $\alpha {{\left( x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ \alpha {{\left( y\right) }_{\rho }}$, that is, ${{\left( \alpha x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( \alpha y\right) }_{\rho }}$ and hence $\left( \alpha x,\alpha y\right) \in \sigma$.

From A) – C) we immediately obtain that $\sigma$ is a left operator pseudoorder of $_{S}M$. Now let $x,y\in M$. Then

$$\begin{aligned} \left( x,y\right) \in {\bar{\sigma }}\Leftrightarrow \left( x,y\right) \in \sigma \cap {{\sigma }^{-1}}\Leftrightarrow \left( x,y\right) ,\left( y,x\right) \in \sigma \Leftrightarrow \\ \Leftrightarrow {{\left( x\right) }_{\rho }}\ {{{{\underline{\prec }}}}_{\rho }}{{\left( y\right) }_{\rho }},{{\left( y \right) }_{\rho }}\ {{{{\underline{\prec }}}}_{\rho }}{{\left( x\right) }_{\rho }}\Leftrightarrow {{\left( x\right) }_{\rho }}={{\left( y\right) }_{\rho }}\Leftrightarrow \left( x,y\right) \in \rho . \end{aligned}$$

Consequently $\rho ={\bar{\sigma }}$.

ii) $\Rightarrow$ i) Consider ${{{\underline{\prec }}}_{\rho }}:={{\le }_{\sigma }}$. Then clearly $_{S}\pi ={}_{S}{{\sigma }^{\#}}$ and so, by Proposition 4.1, $_{S}\pi$ is a left operator homomorphism. $\square$

Proposition 4.3

Let $_{S}M$ be a left operand of $\left( M,{{\le }_{M}}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ and $\left\{ {{\rho }_{i}}:i\in I\right\}$ be a nonempty family of regular left operator congruences of $_{S}M$. Then $\bigcap \nolimits _{i\in I}{{{\rho }_{i}}}$ is a regular left operator congruence of $_{S}M$.

Proof

Since $\left\{ {{\rho }_{i}}:i\in I\right\}$ is a family of regular left operator congruences of $_{S}M$, then for each $i\in I$ there exists an order relation “${{{\underline{\prec }}}_{{{\rho }_{i}}}}$” on $^{M}/{}_{{{\rho }_{i}}}$ such that

• $\left[ \left( {}^{M}/{}_{{{\rho }_{i}}},{{{{\underline{\prec }}}}_{{{\rho }_{i}}}} \right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{S}{{*}_{^{M}/_{{{\rho }_{i}}}}}\right]$ is a left operand of $\left( ^{M}/{}_{{{\rho }_{i}}},{{{{\underline{\prec }}}}_{{{\rho }_{i}}}}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$, say $^{{}_{S}M}/_{{{\rho }_{i}}}$, and

• the mapping $_{S}{{\pi }_{i}}:{}_{S}M\rightarrow {}^{{}_{S}M}/_{{{\rho }_{i}}}$, $_{S}{{\pi }_{i}}\left( x\right) :={{\left( x\right) }_{{{\rho }_{i}}}}$ is a left operator homomorphism.

Let $\rho =\bigcap \nolimits _{i\in I}{{{\rho }_{i}}}$. It is a matter of routine to prove that $\rho$ is a left operator congruence of $_{S}M$. Consider

$$\begin{aligned} {{{\underline{\prec }}}_{\rho }}:=\left\{ \left( A,B \right) \in {}^{M}/{}_{\rho }\times {}^{M}/{}_{\rho }:\left( \exists \ x\in A,y\in B \right) \left( \forall \ i\in I \right) {{\left( x\right) }_{{{\rho }_{i}}}}\ {{{{\underline{\prec }}}}_{{{\rho }_{i}}}}\ {{\left( y\right) }_{{{\rho }_{i}}}}\right\} \end{aligned}$$

One can easily prove that for $x,y\in M$ it holds

$$\begin{aligned} {{\left( x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( y\right) }_{\rho }}\Leftrightarrow {{\left( x\right) }_{{{\rho }_{i}}}}\ {{\underline{{\prec }}}_{{{\rho }_{i}}}}{{\left( y\right) }_{{{\rho }_{i}}}}, \ i\in I \end{aligned}$$

(4.1)

Using (4.1) we immediately have that ${{{\underline{\prec }}}_{\rho }}$ is an order relation on $^{M}/_{\rho }$. Thus $\left( ^{M}/{}_{\rho },{{{\underline{{\prec }}}_{\rho }}}\right)$ is an ordered set. Now let $\alpha ,b\in S$ and $x,y\in M$ such that $\alpha \ {{\le }_{S}}\ b$ and ${{\left( x\right) }_{\rho }}\ {{\le }_{\rho }}\ {{\left( y\right) }_{\rho }}$. Then, by (4.1), ${{\left( x\right) }_{{{\rho }_{i}}}}\ {{{\underline{\prec }}}_{{{\rho }_{i}}}}{{\left( y\right) }_{{{\rho }_{i}}}}$, $i\in I$ and hence $\alpha \ _{S}{{*}_{^{M}/_{{{\rho }_{i}}}}}{{\left( x\right) }_{{{\rho }_{i}}}}\ {{{\underline{\prec }}}_{{{\rho }_{i}}}}b\ _{S}{{*}_{^{M}/_{{{\rho }_{i}}}}}{{\left( y\right) }_{{{\rho }_{i}}}}$ for all $i\in I$, that is, ${{\left( \alpha x\right) }_{{{\rho }_{i}}}}\ {{{\underline{\prec }}}_{{{\rho }_{i}}}}{{\left( by\right) }_{{{\rho }_{i}}}}$, $i\in I$ Therefore (see (4.1)) ${{\left( \alpha x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ {{\left( by\right) }_{\rho }}$, i.e.

$$\begin{aligned} \alpha \ _{S}{{*}_{^{M}/{\rho }}}\ {{\left( x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}\ b\ _{S}{{*}_{^{M}/{}_{\rho }}}\ {{\left( y\right) }_{\rho }}. \end{aligned}$$

(4.2)

From (4.2) and Proposition 3.6ii) we immediately have that

$$\begin{aligned} \left[ \left( ^{M}/{}_{\rho },{{{{\underline{\prec }}}}_{\rho }}\right) ,\left( S,\cdot ,{{\le }_{S}}\right) ;{}_{S}{{*}_{^{M}/{}_{\rho }}}\right] \end{aligned}$$

is a left operand of $\left( ^{M}/_{\rho },{{{{\underline{\prec }}}}_{\rho }}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$, say $^{_{S}M}/_{\rho }$. To complete the proof, we need to show that the mapping

$$\begin{aligned} _{S}\pi :{}_{S}M\rightarrow {}^{_{S}M}/{}_{\rho },\ _{S}\pi \left( x\right) :={{\left( x\right) }_{\rho }} \end{aligned}$$

is a left operator homomorphism. It is clear that $_{S}\pi \left( \alpha x\right) =\alpha \ _{S}\pi \left( x\right)$ for all $\alpha \in S$, $x\in M$. Now let $x,y\in M$ such that $x\ {{\le }_{M}}\ y$. Since $_{S}{{\pi }_{i}}$ is a left operator homomorphism for each $i\in I$, we have $_{S}{{\pi }_{i}}\left( x\right) \ {{{\underline{\prec }}}_{{{\rho }_{i}}}}{}_{S}{{\pi }_{i}}\left( y\right)$, $i\in I$, that is,

$$\begin{aligned} {\left( x\right) }_{{{\rho }_{i}}}\ {{{\underline{\prec }}}_{{{\rho }_{i}}}}{{\left( y\right) }_{{{\rho }_{i}}}}, i\in I. \end{aligned}$$

Then, by (4.1), it follows that ${{\left( x\right) }_{\rho }}\ {{{\underline{\prec }}}_{\rho }}{{\left( y\right) }_{\rho }}$, i.e. $_{S}\pi \left( x\right) \ {{{\underline{\prec }}}_{\rho }}\ {}_{S}\pi \left( y\right)$. Thus $_{S}\pi$ is a left operator homomorphism. Therefore $\rho =\bigcap \nolimits _{i\in I}{{{\rho }_{i}}}$ is a regular left operator congruence of $_{S}M$. $\square$

Proposition 4.4

Let $_{S}M$ be a left operand of $\left( M,{{\le }_{M}}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ and $\rho$ be a left operator congruence of $_{S}M$. Then there exists a regular left operator congruence of $_{S}M$, say $\tau$, containing $\rho$ such that if $\nu$ is a regular left operator congruence of $_{S}M$ containing $\rho$ then $\tau \subseteq \nu$.

Proof

Let $\mathcal {A}$ be the set of all regular left operator congruences of $_{S}M$ containing $\rho$. $\mathcal {A}$ is nonempty since $S\times S\in \mathcal {A}$. Set $\tau :=\bigcap \nolimits _{\mu \in \mathcal {A}}{\mu }$. Then, by Proposition 4.3, we immediately have that $\nu$ is a regular left operator congruence of $_{S}M$. Clearly $\rho \subseteq \tau$ and $\nu$ is contained in every regular left operator congruence of $_{S}M$ containing $\rho$. $\square$

Remark 4.5

From Proposition 4.4 it follows that for every $\rho$ left operator congruence of $_{S}M$ there exists a least (under the inclusion relation) regular left operator congruence of $_{S}M$ containing $\rho$.

Now let $_{S}M$, $_{S}{M}'$ be respectively left operands of ordered sets $\left( M,{{\le }_{M}}\right)$ and $\left( {M}',{{\le }_{{{M}'}}}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ and $\theta :{}_{S}M\rightarrow {}_{S}{M}'$ be a left operator homomorphism. We define

$$\begin{aligned} \underset{\scriptstyle \thicksim }{\theta }:=\left\{ \left( x,y\right) \in M\times M:\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right) \right\} \end{aligned}$$

Proposition 4.6

$\underset{\scriptstyle \thicksim }{\theta }$ is a left operator pseudoorder of $_{S}M$.

Proof

A) Let $x,y\in M$ such that $x\ {{\le }_{M}}\ y$. Then (since $\theta$ is a left operator homomorphism) $\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right)$. Therefore $\left( x,y\right) \in \underset{\scriptstyle \thicksim }{\theta }$ and hence ${{\le }_{M}}\subseteq \underset{\scriptstyle \thicksim }{\theta }$.

B) Clearly $\theta$ is transitive.

C) Let $\alpha \in S$ and $x,y\in M$ such that $\left( x,y\right) \in \underset{\scriptstyle \thicksim }{\theta }$. Then (by definition of $\underset{\scriptstyle \thicksim }{\theta }$) $\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right)$. Therefore, since $_{S}{M}'$ is a left operand of M over S, $\alpha \ \theta \left( x\right) \ {{\le }_{{{M}'}}}\ \alpha \ \theta \left( y\right)$ and hence (since $\theta$ is a left operator homomorphism) $\theta \left( \alpha x\right) \ {{\le }_{{{M}'}}}\ \theta \left( \alpha y\right)$, that is, $\left( \alpha x,\alpha y\right) \in \underset{\scriptstyle \thicksim }{\theta }$. $\square$

Since, by Proposition 4.6, $\underset{\scriptstyle \thicksim }{\theta }$ is a left operator pseudoorder of $_{S}M$, then

• by Proposition 3.2, $\bar{\underset{\scriptstyle \thicksim }{\theta }}$ is a left operator congruence of $_{S}M$ (usually denoted by ker$\theta$). Obviously for $x,y\in M$ we have

  $$\begin{aligned} \left( x,y\right) \in \underset{\scriptstyle \thicksim }{{\bar{\theta }}}\Leftrightarrow \theta \left( x\right) =\theta \left( y\right) , \end{aligned}$$

• from Remark 3.8 we can consider the left factor operand $^{{}_{S}M}/{}_{{\bar{\underset{\scriptstyle \thicksim }{\theta }}}}$ of $_{S}M$ by $\underset{\scriptstyle \thicksim }{\theta }$.

Proposition 4.7

Let $\sigma$ be a left operator pseudoorder of $_{S}M$. The following are equivalent:

• i) $\sigma \subseteq \underset{\scriptstyle \thicksim }{\theta }$.

• ii) There exists a (unique) left operator homomorphism $f:{}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}\rightarrow {}_{S}{M}'$ such that $f\circ {}_{S}{{\sigma }^{\#}}=\theta$.

Proof

i) $\Rightarrow$ ii) Consider

$$\begin{aligned} f:{}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}\rightarrow {}_{S}{M}', \ f\left( {{\left( x\right) }_{{{\bar{\sigma }}}}}\right) :=\theta \left( x\right) \end{aligned}$$

$\alpha$) f is well defined: Let $x,y\in M$ such that ${{\left( x\right) }_{{{\bar{\sigma }}}}}={{\left( y\right) }_{{{\bar{\sigma }}}}}$. Since ${{\left( x\right) }_{{{\bar{\sigma }}}}}={{\left( y\right) }_{{{\bar{\sigma }}}}}$ then $\left( x,y\right) \in {\bar{\sigma }}$. Also, since $\sigma \subseteq \underset{\scriptstyle \thicksim }{\theta }$ then ${\bar{\sigma }}\subseteq \bar{\underset{\scriptstyle \thicksim }{\theta }}$ and hence $\left( x,y\right) \in \bar{\underset{\scriptstyle \thicksim }{\theta }}$ which means that $\theta \left( x\right) =\theta \left( y\right)$. Therefore $f\left( {{\left( x\right) }_{{{\bar{\sigma }}}}} \right) =f\left( {{\left( y\right) }_{{{\bar{\sigma }}}}} \right)$.

$\beta$) Let $\alpha \in S$ and $x\in M$. We shall prove that $f\left( \alpha {{\left( x \right) }_{{{\bar{\sigma }}}}} \right) =\alpha f\left( {{\left( x \right) }_{{{\bar{\sigma }}}}} \right)$. Since $\alpha {{\left( x\right) }_{{{\bar{\sigma }}}}}={{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}$ then

$$\begin{aligned} f\left( \alpha {{\left( x\right) }_{{{\bar{\sigma }}}}} \right)&=f\left( {{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}\right) =\theta \left( \alpha x\right) {}&{} (\theta \text{ is } \text{ a } \text{ left } \text{ operator } \text{ homomorphism})\\ {}&=\alpha \ \theta \left( x\right) =\alpha \ f\left( {{\left( x\right) }_{{{\bar{\sigma }}}}}\right) \end{aligned}$$

$\gamma$) Let $x,y\in M$ such that ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$. We will show that

$$\begin{aligned} f\left( {{\left( x\right) }_{{{\bar{\sigma }}}}}\right) \ \ {{\le }_{{{M}'}}}\ f\left( {{\left( y\right) }_{{{\bar{\sigma }}}}}\right) \end{aligned}$$

Since ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$ then $\left( x,y\right) \in \sigma$ and hence (since $\sigma \subseteq \underset{\scriptstyle \thicksim }{\theta }$) $\left( x,y\right) \in \underset{\scriptstyle \thicksim }{\theta }$, i.e. $\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right)$. Consequently $f\left( {{\left( x\right) }_{{{\bar{\sigma }}}}}\right) \ \ {{\le }_{{{M}'}}}\ f\left( {{\left( y\right) }_{{{\bar{\sigma }}}}}\right)$.

From $\alpha$) – $\gamma$) we immediately have that f is a left operator homomorphism. It is clear that $f\circ {}_{S}{{\sigma }^{\#}}=\theta$. Regarding the uniqueness of f, consider a left operator homomorphism $\varphi :{}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}\rightarrow {}_{S}{M}'$ such that $\varphi \circ {}_{S}{{\sigma }^{\#}}=\theta$. Then for every $x\in M$ we have

$$\begin{aligned} f\left( {}_{S}{{\sigma }^{\#}}\left( x\right) \right) \ \ {{\le }_{{{M}'}}}\ f\left( {}_{S}{{\sigma }^{\#}}\left( y\right) \right) \Rightarrow \theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right) \end{aligned}$$

ii) $\Rightarrow$ i) Let $f:{}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}\rightarrow {}_{S}{M}'$ be a left operator homomorphism such that $f\circ {}_{S}{{\sigma }^{\#}}=\theta$. Take $x,y\in M$ such that $\left( x,y\right) \in \sigma$. We shall prove that $\left( x,y\right) \in \underset{\scriptstyle \thicksim }{\theta }$. Since $\left( x,y\right) \in \sigma$ then ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$ and hence (f is a left operator homomorphism) $f\left( {{\left( x\right) }_{{{\bar{\sigma }}}}}\right) \ \ {{\le }_{{{M}'}}}\ f\left( {{\left( y\right) }_{{{\bar{\sigma }}}}}\right)$, that is,

$$\begin{aligned} f\left( {}_{S}{{\sigma }^{\#}}\left( x\right) \right) \ \ {{\le }_{{{M}'}}}\ f\left( {}_{S}{{\sigma }^{\#}}\left( y\right) \right) \Rightarrow \theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right) \end{aligned}$$

Thus $\left( x,y\right) \in \underset{\scriptstyle \thicksim }{\theta }$. $\square$

Remark 4.8

If the statement i) of the previous Proposition holds, then the unique left operator homomorphism $f:{}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}\rightarrow {}_{S}{M}'$ such that $f\circ {}_{S}{{\sigma }^{\#}}=\theta$ is defined by the rule that

$$\begin{aligned} f\left( {{\left( x\right) }_{{{\bar{\sigma }}}}} \right) :=\theta \left( x\right) , \ x\in M \end{aligned}$$

It is straightforward to prove the next proposition.

Proposition 4.9

The mapping

$$\begin{aligned} {\theta }':{}^{{}_{S}M}/{}_{{\bar{\underset{\scriptstyle \thicksim }{\theta }}}}\rightarrow {}_{S}{M}',\ {\theta }'\left( {{\left( x\right) }_{{\bar{\underset{\scriptstyle \thicksim }{\theta }}}}}\right) :=\theta \left( x\right) \end{aligned}$$

is a reverse isotone left operator homomorphism such that ${\theta }'\circ {}_{S}{{\sigma }^{\#}}=\theta$.

Since $\varnothing \ne \theta \left( M \right) \subseteq {M}'$ then $\left( \theta \left( M\right) ,{{\le }_{\theta \left( M\right) }}\right)$, where ${{\le }_{\theta \left( M\right) }}:={{\le }_{{{M}'}}}\cap \left( \theta \left( M\right) \times \theta \left( M\right) \right)$, is an ordered set. We define

$$\begin{aligned} _{S}{{*}_{\theta \left( M\right) }}:&S\times \theta \left( M\right) \rightarrow \theta \left( M\right) \\&\quad \left( \alpha ,\ y \right) \ \ \rightarrow \alpha \ _{S}{{*}_{\theta \left( M\right) }}\ y:=\alpha \ _{S}{{*}_{{{M}'}}}\ y . \end{aligned}$$

Proposition 4.10

$\left[ \left( \theta \left( M\right) ,{{\le }_{\theta \left( M \right) }}\right) ,\left( S,\cdot ,{{\le }_{S}} \right) ;{}_{S}{{*}_{\theta \left( M\right) }}\right]$ is a left operand of $\left( \theta \left( M\right) ,{{\le }_{\theta \left( M\right) }}\right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$.

Proof

i) $_{S}{{*}_{\theta \left( M\right) }}$ is well defined: Let $\alpha \in S$ and $y\in \theta \left( M\right)$. We shall prove that $\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\ y\in \theta \left( M\right)$. Since $y\in \theta \left( M\right)$, there exists $x\in M$ such that $y=\theta \left( x\right)$. Then

$$\begin{aligned} \alpha \ {}_{S}{{*}_{{{M}'}}}\ y=\alpha \ {}_{S}{{*}_{{{M}'}}}\ \theta \left( x\right) =\theta \left( \alpha \ {}_{S}{{*}_{M}}\ x\right) \in \theta \left( M\right) \end{aligned}$$

ii) It is evident that $\left( \alpha \cdot b\right) \ {}_{S}{{*}_{\theta \left( M\right) }}\ y=\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\left( b\ {}_{S}{{*}_{\theta \left( M\right) }}\ y\right)$ for all $\alpha ,b\in S$, $y\in \theta \left( M\right)$.

iii) Let $\alpha ,b\in S$ and $y,z\in \theta \left( M\right)$ such that $\alpha \ {{\le }_{S}}\ b$ and $y\ {{\le }_{\theta \left( M\right) }}\ z$. We will show that $\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\ y\ \ {{\le }_{\theta \left( M\right) }}\ \ b\ {}_{S}{{*}_{\theta \left( M\right) }}\ z$. Clearly $y,z\in \theta \left( M\right)$ satisfying $y\ {{\le }_{{{M}'}}}\ z$. Since $\alpha \ {{\le }_{S}}\ b$ and $y\ {{\le }_{{{M}'}}}\ z$, then $\alpha \ {}_{S}{{*}_{{{M}'}}}\ y\ \ {{\le }_{{{M}'}}}\ b\ {}_{S}{{*}_{{{M}'}}}\ z$, that is, $\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\ y\ \ {{\le }_{{{M}'}}}\ b\ {}_{S}{{*}_{\theta \left( M \right) }}\ z$. By ${\varvec{i)}}$ we have $\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\ y\,\ b\ {}_{S}{{*}_{\theta \left( M\right) }}\ z\in \theta \left( M\right)$ and thus $\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\ y\ \ {{\le }_{\theta \left( M\right) }}\ b\ {}_{S}{{*}_{\theta \left( M\right) }}\ z$. $\square$

$\left[ \left( \theta \left( M\right) ,{{\le }_{\theta \left( M\right) }}\right) ,\left( S,\cdot ,{{\le }_{S}} \right) ;{}_{S}{{*}_{\theta \left( M\right) }}\right]$ is denoted by ${}_{S}\theta \left( M\right)$.

Theorem 4.11

The mapping

$$\begin{aligned} \Theta :{}^{_{S}M}/{}_{\ker \theta }\rightarrow {}_{S}\theta \left( M\right) ,\ \Theta \left( {{\left( x\right) }_{\ker \theta }}\right) =\theta \left( x\right) \end{aligned}$$

is a left operator isomorphism.

Proof

Clearly $\theta$ is well defined and onto.

i) Let $\alpha \in S$ and $x\in M$. Then

$$\begin{aligned} \Theta \left( \alpha \ {}_{S}{{*}_{{}^{M}/{}_{\ker \theta }}}{{\left( x\right) }_{\ker \theta }} \right)&=\Theta \left( {{\left( \alpha \ {}_{S}{{*}_{M}}\ x\right) }_{\ker \theta }}\right) =\theta \left( \alpha \ {}_{S}{{*}_{M}}\ x\right) =\\ {}&=\alpha \ {}_{S}{{*}_{{{M}'}}}\ \theta \left( x\right) =\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\ \theta \left( x\right) =\\ {}&=\alpha \ {}_{S}{{*}_{\theta \left( M\right) }}\ \Theta \left( {{\left( x\right) }_{\ker \theta }} \right) . \end{aligned}$$

ii) Let $x,y\in M$ such that ${{\left( x\right) }_{\ker \theta }}\ {{\le }_{{\underset{\scriptstyle \thicksim }{\theta }}}}\ {{\left( y\right) }_{\ker \theta }}$. Then $\left( x,y\right) \in \underset{\scriptstyle \thicksim }{\theta }$ which means that $\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right)$ and hence $\theta \left( x\right) \ {{\le }_{\theta \left( M\right) }}\ \theta \left( y\right)$. Therefore

$$\begin{aligned} \Theta \left( {{\left( x\right) }_{\ker \theta }}\right) \ {{\le }_{\theta \left( M\right) }}\ \Theta \left( {{\left( y\right) }_{\ker \theta }}\right) \end{aligned}$$

(i.e. $\Theta$ is isotone).

iii) Let $x,y\in M$ such that $\Theta \left( {{\left( x\right) }_{\ker \theta }}\right) \ {{\le }_{\theta \left( M\right) }}\ \Theta \left( {{\left( y\right) }_{\ker \theta }}\right)$, that is, $\theta \left( x\right) \ {{\le }_{\theta \left( M\right) }}\ \theta \left( y\right)$. Then we have $\theta \left( x\right) \ {{\le }_{{{M}'}}}\ \theta \left( y\right)$ and thus $\left( x,y\right) \in \underset{\scriptstyle \thicksim }{\theta }$. By definition of $\underset{\scriptstyle \thicksim }{\theta }$ it immediately follows that ${{\left( x\right) }_{{\bar{\underset{\scriptstyle \thicksim }{\theta }}}}}\ {{\le }_{{\underset{\scriptstyle \thicksim }{\theta }}}}\ {{\left( y\right) }_{{\bar{\underset{\scriptstyle \thicksim }{\theta }}}}}$, that is,

$$\begin{aligned} {{\left( x\right) }_{\ker \theta }}\ {{\le }_{{\underset{\scriptstyle \thicksim }{\theta }}}}\ {{\left( y\right) }_{\ker \theta }} \end{aligned}$$

(i.e. $\Theta$ is reverse isotone). $\square$

Due to Theorem 4.11 we are allowed to write ${}^{_{S}M}/{}_{\ker \theta }\ \simeq {}_{S}\theta \left( M\right)$.

Let $\sigma$, $\tau$ be left operator pseudoorders of $_{S}M$ such that $\sigma \subseteq \tau$. On $^{M}/{}_{{{\bar{\sigma }}}}$ we define a relation “${}^{\tau }/{}_{\sigma }$” as following

$$\begin{aligned} {}^{\tau }/{}_{\sigma }:=\left\{ \left( A,B\right) \in {}^{M}/{}_{{{\bar{\sigma }}}}\times {}^{M}/{}_{{{\bar{\sigma }}}}:\left( \exists \ x\in A,y\in B\right) \left( x,y\right) \in \tau \right\} \end{aligned}$$

Lemma 4.12

Let $x,y\in M$. Then $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}}\right) \in {}^{\tau }/{}_{\sigma }$ if and only if $\left( x,y\right) \in \tau$.

Proof

If $\left( x,y\right) \in \tau$ then clearly $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$. Let now $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$. By definition of ${}^{\tau }/{}_{\sigma }$ there exist $z\in {{\left( x\right) }_{{{\bar{\sigma }}}}}$ and $w\in {{\left( y\right) }_{{{\bar{\sigma }}}}}$ such that $\left( z,w\right) \in \tau$. Since $z\in {{\left( x\right) }_{{{\bar{\sigma }}}}}$ then $\left( x,z\right) \in {\bar{\sigma }}$ and hence $\left( x,z\right) \in \sigma \subseteq \tau$. Similarly $\left( w,y\right) \in \tau$. By the transitivity of $\tau$ it follows immediately that $\left( x,y\right) \in \tau$. $\square$

Proposition 4.13

${}^{\tau }/{}_{\sigma }$ is a left operator pseudoorder of $^{{}_{S}M}/{}_{{{\bar{\sigma }}}}$.

Proof

i) ${\le }_{\sigma }\subseteq {}^{\tau }/{}_{\rho }$: Let $x,y\in M$ such that ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$. Then $\left( x,y\right) \in \sigma \subseteq \tau$ and thus (by definition of $^{\tau }/{}_{\sigma }$) $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$.

ii) ${}^{\tau }/{}_{\sigma }$ is transitive: Let $x,y,z\in M$ such that $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) , \left( {{\left( y\right) }_{{{\bar{\sigma }}}}},{{\left( z\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$. Then, by Lemma 4.12, we obtain $\left( x,y\right) \, \left( y,z\right) \in \tau$ and hence ($\tau$ is transitive) $\left( x,z\right) \in \tau$. Now it immediately follows that $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( z\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$.

iii) Let $\alpha \in S$ and $x,y\in M$ such that $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$. We shall prove that $\left( \alpha {{\left( x\right) }_{{{\bar{\sigma }}}}},\alpha {{\left( y\right) }_{{{\bar{\sigma }}}}}\right) \in {}^{\tau }/{}_{\sigma }$. Since $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$, we obtain $\left( x,y\right) \in \tau$ and hence (since $\tau$ is a left operator pseudoorder of $_{S}M$) we have $\left( \alpha x,\alpha y\right) \in \tau$. Therefore $\left( {{\left( \alpha x\right) }_{{{\bar{\sigma }}}}},{{\left( \alpha y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$, i.e. $\left( \alpha {{\left( x\right) }_{{{\bar{\sigma }}}}},\alpha {{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$. $\square$

Proposition 4.14

Let $\sigma$ be a left operator pseudoorder of $_{S}M$ and $\nu$ be a relation on $^{M}/{}_{{{\bar{\sigma }}}}$. The following are equivalent:

• i) $\nu$ is a left operator pseudoorder of ${}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}$.

• ii) There exists a left operator pseudoorder $\sigma$ of $_{S}M$ containing $\sigma$ such that $\nu ={}^{\tau }/{}_{\sigma }$.

Proof

Since the implication ii) $\Rightarrow$ i) is obvious (by Proposition 4.13, ${}^{\tau }/{}_{\sigma }$ is a left operator pseudoorder of ${}^{{}_{S}M}/{}_{{\bar{\sigma }}}$), then we only need to prove the implication i) $\Rightarrow$ ii):

Consider

$$\begin{aligned} \tau :=\left\{ \left( x,y\right) \in M\times M:\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in \nu \right\} \end{aligned}$$

A) ${{\le _{M}}\subseteq \tau }$: Let $x,y\in M$ such that $x\ {{\le }_{M}}\ y$. Then (since ${{\le }_{M}}\subseteq \sigma$) we have $\left( x,y\right) \in \sigma$ and so ${{\left( x\right) }_{{{\bar{\sigma }}}}}\ {{\le }_{\sigma }}\ {{\left( y\right) }_{{{\bar{\sigma }}}}}$ whence (since $\nu$ is a left operator pseudoorder of ${}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}$) it follows that $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in \nu$. Therefore $\left( x,y\right) \in \tau$.

B) Based on the transitivity of $\nu$, we immediately obtain the transitivity of $\tau$.

C) Let $\alpha \in S$ and $x,y\in M$ such that $\left( x,y\right) \in \tau$. We will show that $\left( \alpha x,\alpha y\right) \in \tau$. Since $\left( x,y\right) \in \tau$ we have $\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in \nu$. Since $\nu$ is a left operator homomorphism of ${}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}$, it follows that $\left( \alpha {{\left( x\right) }_{{{\bar{\sigma }}}}}\,\ \alpha {{\left( y\right) }_{{{\bar{\sigma }}}}}\right) \in \nu$, i.e. $\left( {{\left( \alpha x\right) }_{{{\bar{\sigma }}}}}\,\ {{\left( \alpha y\right) }_{{{\bar{\sigma }}}}}\right) \in \nu$ which means that $\left( \alpha x,\alpha y\right) \in \tau$.

From A)–C) we immediately obtain that $\tau$ is a left operator pseudoorder of $_{S}M$. Because for $x,y\in M$ we have (since $\nu$ is a left operator pseudoorder of ${}^{{}_{S}M}/{}_{{\bar{\sigma }}}$)

$$\begin{aligned} \left( x,y\right) \in {\bar{\sigma }}\Leftrightarrow \left( x,y\right) \in \sigma \cap {{\sigma }^{-1}}\Leftrightarrow \left( x,y\right) ,\left( y,x\right) \in \sigma \Leftrightarrow \ \ \ \ \\ ({{\left( x\right) }_{\rho }}\ {{{{\underline{\prec }}}}_{\rho }}{{\left( y\right) }_{\rho }} \ and \ {{\left( y \right) }_{\rho }}\ {{{{\underline{\prec }}}}_{\rho }}{{\left( x\right) }_{\rho }})\Leftrightarrow {{\left( x\right) }_{\rho }}={{\left( y\right) }_{\rho }}\Leftrightarrow \left( x,y\right) \in \rho \end{aligned}$$

then it holds $\sigma \subseteq \tau$. Now let $x,y\in M$. Then

$$\begin{aligned} \left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}}\right) \in \nu \Leftrightarrow \left( x,y\right) \in \tau \underset{(\textrm{Lemma}\, 4.12 )}{\mathop {\Leftrightarrow }}\left( {{\left( x\right) }_{{{\bar{\sigma }}}}},{{\left( y\right) }_{{{\bar{\sigma }}}}}\right) \in {}^{\tau }/{}_{\sigma }. \end{aligned}$$

Therefore $\nu ={}^{\tau }/{}_{\sigma }$. $\square$

If $\sigma$, $\tau$ are left operator pseudoorders of $_{S}M$ such that $\sigma \subseteq \tau$ then, by Proposition 4.13, ${}^{\tau }/{}_{\sigma }$ is a left operator pseudoorder of $^{{}_{S}M}/{}_{{{\bar{\sigma }}}}$ and so we can consider the left (factor) operand $\left[ \left( {}^{{}^{M}/{}_{{{\bar{\sigma }}}}}/{}_{\overline{{}^{\tau }/{}_{\sigma }}},{{\le }_{{}^{\tau }/{}_{\sigma }}} \right) ,\left( S,\cdot ,{{\le }_{S}} \right) ;\bullet \right]$ of ordered set $\left( {}^{{}^{M}/{}_{{{\bar{\sigma }}}}}/{}_{\overline{{}^{\tau }/{}_{\sigma }}}\,{{\le }_{{}^{\tau }/{}_{\sigma }}} \right)$ over $\left( S,\cdot ,{{\le }_{S}}\right)$ where for $\alpha \in S$ and $x,y\in M$ we have

$$\begin{aligned} {{\left( {{\left( x\right) }_{{\bar{\sigma }}}}\right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}\ \ {{\le }_{{}^{\tau }/{}_{\sigma }}}\ {{\left( {{\left( y\right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}\ \Leftrightarrow \ \left( {{\left( x\right) }_{{\bar{\sigma }}}},{{\left( y\right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }\ \Leftrightarrow \ \left( x,y\right) \in \tau \end{aligned}$$

(4.3)

and

$$\begin{aligned} \alpha \bullet {{\left( {{\left( x \right) }_{{{\bar{\sigma }}}}}\right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}={{\left( \alpha \ {}_{S}{{*}_{{}^{M}/{}_{{{\bar{\sigma }}}}}}\ {{\left( x\right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}={{\left( {{\left( \alpha \ {}_{S}{{*}_{M}}\ x \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}. \end{aligned}$$

(4.4)

As we mentioned above, if there is no confusion, we usually simplify the notation (4.4) to

$$\begin{aligned} \alpha \bullet {{\left( {{\left( x \right) }_{{\bar{\sigma }}}}\right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}={{\left( \alpha {{\left( x \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}={{\left( {{\left( \alpha x \right) }_{{\bar{\sigma }}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}. \end{aligned}$$

(4.5)

Consider now the mapping

$$\begin{aligned} F:{}^{{}_{S}M}/{}_{{{\bar{\tau }}}}\rightarrow {}^{{}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}}/{}_{\overline{{}^{\tau }/{}_{\sigma }}}, \ F\left( {{\left( x\right) }_{{{\bar{\tau }}}}} \right) :={{\left( {{\left( x\right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}. \end{aligned}$$

Regarding F, we have the following

Theorem 4.15

F is a left operator isomorphism.

Proof

i) F is well defined: Let $x,y\in M$ such that ${{\left( x \right) }_{{{\bar{\tau }}}}}={{\left( y \right) }_{{{\bar{\tau }}}}}$. We shall prove that ${{\left( {{\left( x \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}={{\left( {{\left( y \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}$. Since ${{\left( x \right) }_{{\bar{\tau }}}}={{\left( y \right) }_{{{\bar{\tau }}}}}$ then $\left( x,y \right) \in {\bar{\tau }}$ which means that $\left( x,y \right) , \left( y,x \right) \in \tau$ and so (by Lemma 4.12 ) $\left( {{\left( x \right) }_{{{\bar{\sigma }}}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \,\left( {{\left( y \right) }_{{{\bar{\sigma }}}}},{{\left( x \right) }_{{{\bar{\sigma }}}}} \right) \in {}^{\tau }/{}_{\sigma }$. Consequently $\left( {{\left( x \right) }_{{{\bar{\sigma }}}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in \overline{{}^{\tau }/{}_{\sigma }}$, that is, ${{\left( {{\left( x \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}={{\left( {{\left( y \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}$.

ii) Let $\alpha \in S$ and $x\in M$. Then $F\left( \alpha {{\left( x \right) }_{{{\bar{\tau }}}}} \right) =\alpha \bullet F\left( {{\left( x \right) }_{{{\bar{\tau }}}}} \right)$. Indeed:

$$\begin{aligned} F\left( \alpha {{\left( x \right) }_{{{\bar{\tau }}}}} \right) =F\left( {{\left( \alpha x \right) }_{{{\bar{\tau }}}}} \right) ={{\left( {{\left( \alpha x \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}\ \underset{(4.5)}{\mathop {=}}\ \alpha \bullet {{\left( {{\left( x \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}=\alpha \bullet F\left( {{\left( x \right) }_{{{\bar{\tau }}}}} \right) . \end{aligned}$$

iii) Let $x,y\in M$. Then

$$\begin{aligned} {{\left( x \right) }_{{{\bar{\tau }}}}}{{\le }_{\tau }}{{\left( y \right) }_{{{\bar{\tau }}}}}&\underset{( \textrm{Lemma}\,3.3)}{\mathop { \Leftrightarrow }}\,\left( x,y \right) \in \tau \ \underset{(4.3)}{\mathop {\Leftrightarrow }}\ {{\left( {{\left( x \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}\ {{\le }_{{}^{\tau }/{}_{\sigma }}}\ {{\left( {{\left( y \right) }_{{{\bar{\sigma }}}}} \right) }_{\overline{{}^{\tau }/{}_{\sigma }}}}\ \Leftrightarrow \ \\ {}&\ \ \ \ \ \Leftrightarrow \ F\left( {{\left( x \right) }_{{{\bar{\tau }}}}} \right) \ {{\le }_{{}^{\tau }/{}_{\sigma }}}\ F\left( {{\left( y \right) }_{{{\bar{\tau }}}}} \right) . \end{aligned}$$

Thus F is isotone and reverse isotone.

From i) – iii) and since F is clearly onto, we immediately have that F is a left operator isomorphism. $\square$

Due to Theorem 4.15 we are allowed to write ${}^{{}_{S}M}/{}_{{{\bar{\tau }}}}\ \simeq \ {}^{{}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}}/{}_{\overline{{}^{\tau }/{}_{\sigma }}}$.

Proposition 4.16

Let $_{S}M$ be a left operand of $\left( M,{{\le }_{M}} \right)$ over $\left( S,\cdot ,{{\le }_{S}} \right)$ and ${\mathfrak {A}}$ be a family of left operator pseudoorders of ${}_{S}M$. Then there exists a left operator pseudoorder of ${}_{S}M$, say $\tau$, containing all $\rho \in {\mathfrak {A}}$ such that if $\nu$ is a left operator pseudoorder of ${}_{S}M$ containing all $\rho \in {\mathfrak {A}}$ then $\tau \subseteq \nu$.

Proof

Let $\mathcal {A}$ be the set of all left operator pseudoorders of ${}_{S}M$ containing all $\rho \in {\mathfrak {A}}$. $\mathcal {A}$ is nonempty since $S\times S\in \mathcal {A}$. Set $\tau :=\bigcap \nolimits _{\mu \in \mathcal {A}}{\mu }$. Then we immediately have that $\tau$ is a left operator pseudoorder of ${}_{S}M$. Clearly $\rho \subseteq \tau$ for each $\rho \in {\mathfrak {A}}$ and $\tau$ is contained in every left operator pseudoorder of ${}_{S}M$ containing all $\rho \in {\mathfrak {A}}$. $\square$

Remark 4.17

By Proposition 4.16 it follows that for every family ${\mathfrak {A}}$ of left operator pseudoorders of ${}_{S}M$ there exists a least (under the inclusion relation) left operator pseudoorder of ${}_{S}M$ containing all $\rho \in {\mathfrak {A}}$ which will be denoted by $\bigvee \nolimits _{\rho \in {\mathfrak {A}}} \rho$.

An ordered set L is said to be a complete lattice if for every nonempty subset D of L there exist $\inf D$ and $\sup D$ in L.

Remark 4.18

It is clear that the set of all left operator pseudoorder of $_{S}M$ is a complete lattice with the inclusion relation being its order relation since for every (nonempty) family ${\mathfrak {A}}$ of left operator pseudoorders of $_{S}M$ we immediately have

$\circ$:

$\inf \left\{ \rho :\rho \in {\mathfrak {A}} \right\} =\bigcap \nolimits _{\rho \in {\mathfrak {A}}}{\rho }$, and

$\circ$:

$\sup \left\{ \rho :\rho \in {\mathfrak {A}} \right\} =\bigvee \nolimits _{\rho \in {\mathfrak {A}}}\rho$.

Take now a left operator pseudoorders of $_{S}M$, say $\sigma$. We denote

• PS $\left( {}_{S}M;\sigma \right)$ the set of all left operator pseudoorders of ${}_{S}M$ containing $\sigma$, and

• $PS\left( {}^{{}_{S}M}/{}_{{{\bar{\sigma }}}} \right)$ the set of all left operator pseudoorders of ${}^{{}_{S}M}/{}_{{{\bar{\sigma }}}}$.

Remark 4.19

• i) Since obviously Remark 4.18 is true for any left operand over S, then $\left( PS\left( {}^{{}_{S}M}/{}_{{{\bar{\sigma }}}} \right) ,\subseteq \right)$ is a complete lattice.

• ii) Using Remarks 4.17 and  4.18 it is straightforward to prove that $\left( PS\left( {}_{S}M;\sigma \right) ,\subseteq \right)$ is a complete lattice.

Proposition 4.20

For ${{\rho }_{1}},{{\rho }_{2}}\in PS\left( {}_{S}M;\sigma \right)$ we have

$$\begin{aligned} {{\rho }_{1}}\subseteq {{\rho }_{2}}\ \Leftrightarrow \ {}^{{{\rho }_{1}}}/{}_{\sigma }\subseteq {}^{{{\rho }_{2}}}/{}_{\sigma }. \end{aligned}$$

Proof

Since the implication “$\Rightarrow$” is obvious, it remains to prove only the implication “$\Leftarrow$”. So suppose ${}^{{{\rho }_{1}}}/{}_{\sigma }\subseteq {}^{{{\rho }_{2}}}/{}_{\sigma }$ and take $x,y\in M$. Then

$$\begin{aligned} \left( x,y \right) \in {{\rho }_{1}}\Leftrightarrow \left( {{\left( x \right) }_{{{\bar{\sigma }}}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in {}^{{{\rho }_{1}}}/{}_{\sigma }\Rightarrow \left( {{\left( x \right) }_{{{\bar{\sigma }}}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in {}^{{{\rho }_{2}}}/{}_{\sigma }\Leftrightarrow \left( x,y \right) \in {{\rho }_{2}}. \end{aligned}$$

$\square$

Theorem 4.21

Let $_{S}M$ be a left operand of $\left( M,{{\le }_{M}} \right)$ over $\left( S,\cdot ,{{\le }_{S}} \right)$, $\sigma$ be a left operator pseudoorder of ${}_{S}M$ and

$$\begin{aligned} G:PS\left( {}_{S}M;\sigma \right) \rightarrow PS\left( {}^{{}_{S}M}/{}_{{{\bar{\sigma }}}} \right) , \ G\left( \rho \right) :={}^{\rho }/{}_{\sigma } \end{aligned}$$

Then for any $\varnothing \ne {\mathfrak {A}}\subseteq PS\left( {}_{S}M;\sigma \right)$ we have

• i) $G\left( \bigcap \nolimits _{\rho \in {\mathfrak {A}}}{\rho } \right) =\bigcap \nolimits _{\rho \in {\mathfrak {A}}}{G\left( \rho \right) }$,

• ii) $G\left( \bigvee \nolimits _{\rho \in \mathfrak {{A}}}\,\rho \right) =\bigvee \nolimits _{\rho \in \mathfrak {{A}}}\,G\left( \rho \right)$.

Proof

Let $\varnothing \ne {\mathfrak {A}}\subseteq PS\left( {}_{S}M;\sigma \right)$.

i) We must show that $^{\bigcap \nolimits _{\rho \in {\mathfrak {A}}}{\rho }}/{}_{\sigma }=\bigcap \nolimits _{\rho \in \mathfrak {{A}}}{{}^{\rho }/{}_{\sigma }}$: Obviously $\bigcap \nolimits _{\rho \in {\mathfrak {A}}}{\rho }\in PS \left( {}_{S}M;\sigma \right)$. For $x,y\in M$ we have

$$\begin{aligned} \left( {{\left( x \right) }_{{{\bar{\sigma }}}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in \ {}^{\bigcap \limits _{\rho \in {\mathfrak {A}}}{\rho }}/{}_{\sigma }\ \Leftrightarrow \ \left( x,y \right) \in \bigcap \limits _{\rho \in {\mathfrak {A}}}{\rho }\ \Leftrightarrow \ \left( {{\left( x \right) }_{{\bar{\sigma }}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in \bigcap \limits _{\rho \in {\mathfrak {A}}}{{}^{\rho }/{}_{\sigma }} \end{aligned}$$

ii) We must show that $\ {}^{\bigvee \nolimits _{\rho \in {\mathfrak {A}}}\,\rho }/{}_{\sigma }\ =\ \bigvee \nolimits _{\rho \in {\mathfrak {A}}}\,{}^{\rho }/{}_{\sigma }$: Let

$$\begin{aligned} {{{\mathfrak {B}}}_{{\mathfrak {A}}}}:=\left\{ \tau \in PS\left( {}_{S}M;\sigma \right) :\rho \subseteq \tau \text { for all }\ \rho \in {\mathfrak {A}} \right\} \end{aligned}$$

and

$$\begin{aligned} {{{\mathfrak {C}}}_{{\mathfrak {A}}}}:=\left\{ \omega \in PS\left( {}^{{}_{S}M}/{}_{{{\bar{\sigma }}}} \right) :{}^{\rho }/{}_{\sigma }\subseteq \omega \text{ for } \text{ all } \ \rho \in {\mathfrak {A}} \right\} \end{aligned}$$

Since each left operator pseudoorder of ${}_{S}M$ containing all $\rho \in {\mathfrak {A}}$ clearly contains $\sigma$, then $\bigvee \nolimits _{\rho \in {\mathfrak {A}}}\ \rho =\bigcap \nolimits _{\tau \in {{{\mathfrak {B}}}_{{\mathfrak {A}}}}}{\tau }$. Obviously $\bigvee \nolimits _{\rho \in {\mathfrak {A}}}{}^{\rho }/{}_{\sigma }=\bigcap \nolimits _{\omega \in {{{\mathfrak {C}}}_{{\mathfrak {A}}}}}{\omega }$. By Propositions 4.14 and 4.20 it follows immediately that

$$\begin{aligned} {{{\mathfrak {C}}}_{{\mathfrak {A}}}}=\left\{ {}^{\tau }/{}_{\sigma }:\tau \in {{{\mathfrak {B}}}_{{\mathfrak {A}}}} \right\} \end{aligned}$$

and hence $\bigvee \nolimits _{\rho \in {\mathfrak {A}}}\,{}^{\rho }/{}_{\sigma }=\bigcap \nolimits _{\tau \in {{{\mathfrak {B}}}_{{\mathfrak {A}}}}}{{}^{\tau }/{}_{\sigma }}$. Then for $x,y\in M$ we have

$$\begin{aligned} \left( {{\left( x \right) }_{{{\bar{\sigma }}}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in \ ^{\underset{\rho \in {\mathfrak {A}}}{\bigvee }\,\rho }/{}_{\sigma }\ {}&\Leftrightarrow \ \left( x,y \right) \in \underset{\rho \in {\mathfrak {A}}}{\bigvee }\,\rho \ \Leftrightarrow \ \left( x,y \right) \in \bigcap \limits _{\tau \in {{{\mathfrak {B}}}_{{\mathfrak {A}}}}}{\tau }\ \\ {}&\Leftrightarrow \ \left( {{\left( x \right) }_{{\bar{\sigma }}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in \bigcap \limits _{\tau \in {{{\mathfrak {B}}}_{{\mathfrak {A}}}}}{{}^{\tau }/{}_{\sigma }}\ \\ {}&\Leftrightarrow \left( {{\left( x \right) }_{{{\bar{\sigma }}}}},{{\left( y \right) }_{{{\bar{\sigma }}}}} \right) \in \underset{\rho \in {\mathfrak {A}}}{\bigvee } {}^{\rho }/{}_{\sigma } \end{aligned}$$

$\square$

Remark 4.22

From Propositions 4.20 and 4.14 it is direct that G is isotone, reverse isotone and onto. Hence the complete lattices $\left( PS\left( {}_{S}M;\sigma \right) ,\subseteq \right)$ and $\left( PS\left( {}^{{}_{S}M}/{}_{{{\bar{\sigma }}}} \right) ,\subseteq \right)$ are isomorphic.

Remark 4.23

It is clear that in the case of right operands, dual results (based on dual definitions) are true after the necessary modifications have been made.