8 Regular Systems

In the previous chapter, we have described the construction of a regular groupoid starting with a symmetric presheaf of groups

G

over a biordered set

E

with respect to a regular equivalence relation

δ

E

and a regular family of matrices

P

for

G

. Here we describe the construction of regular semigroups from regular groupoids.

Throughout this chapter,

S

will denote a regular groupoid,

E

, the set of idempotents of

S

and

δ

, the equivalence relation on

E

induced by the distinct Rees groupoids of

S

. Since inverses of elements of

S

can be defined as in the case of regular semigroups, notations used for regular semigroups in (1.14) and (1.15) can be carried over to regular groupoids also. In particular,

i

will denote the inverse relation (cf. (1.14)) of

S

and

i (x)

the set of inverse of

x \in S

A partial transformation

t

S

is said to be idempotent preserving if and only if for all

X \subseteq S

(E (X)) t \subseteq E

Definition 8.1. Let $S$ be a regular groupiod. For $e^{'} ω^{r} e [e^{'} ω^{l} e]$ let $σ^{r} (e, e^{'}) [σ^{l} (e, e^{'})]$ denote a mapping of $R_{e} [L_{e}]$ into $R_{e^{'}} [L_{e^{'}}]$ and let

σ^{r} = {σ^{r} (e, e^{'}) | e^{'} ω^{r} e}, σ^{l} = {σ^{l} (e, e^{'}) | e^{'} ω^{l} e} .

(8.1)

Then $ℛ = (S, σ^{r}, σ^{l})$ is a regular system if the following axioms (and their duals) are satisfied. In these statements products shown are products in $S$ .

For all $e \in E, σ^{r} (e, e)$ is the identity map of $R_{e}$ and for all $e^{'} ω e, e σ^{r} (e, e^{'}) = e^{'}$ .
Let $e^{'}, e^{''} \in ω^{r} (e)$ and $e_{1} \overset{r}{\approx} e$ . Then
$σ^{r} (e, e^{'}) = σ^{r} (e, e^{''})$

if either $e^{''} = e^{'}$ or for some $e_{1} \overset{r}{\approx} e, e^{''} = e^{'} τ^{r} (e_{1})$ .
Let $x \in R_{e}, x^{'} \in i (x; L_{e})$ and $e^{'}, e^{''} \in ω (e)$ . If $e^{'} \overset{r}{\approx} e^{''}$ , then $x σ^{l} (e, e^{'}), x^{'} σ^{l} (e, e^{''})$ are inverses of $x σ^{r} (e, e^{'})$ .
Let $x \in R_{e}$ and $x^{'} \in i (x; L_{e})$ . For $e^{'} \in ω (e), x^{'} σ^{l} (e, e^{'})$ is an inverse of $x σ^{r} (e, e^{'})$ and the mapping $a (x, x^{'}) : ω (e) \to ω (x^{'} x)$ defined by
$e^{'} a (x, x^{'}) = (x^{'} σ^{l} (e, e^{'})) \cdot (x σ^{r} (e, e^{'}))$

is an $ω$ -isomorphism.

Let

x \in R_{e}, x^{'} \in i (x; L_{e})

and

e^{'} \in ω (e)

. If

x^{'} x = f

and

e^{'} a (x, x^{'}) = f^{'}

, then

x σ^{r} (e, e^{'}) = x σ^{l} (f, f^{'}) .

Let $x \in H_{e_{1}, e}, y \in H_{f, f_{1}}, h \in S (e, f), x^{'} = i (x; e, e_{1})$ and $g \in ω^{l} (e) \cap ω^{r} (f)$ . If $h_{1} = (h τ^{'} (e)) a (x^{'}, x)$ and $g_{1} = (g τ^{l} (e)) a (x^{'}, x)$ , then
$[(x σ^{l} (e, h)) (y σ^{r} (f, h))] σ^{r} (h_{1}, g_{1}) = (x σ^{l} (e, g)) (y σ^{r} (f, g)) .$

(8.1)

a : (x, x^{'}) \to a (x, x^{'})

is a mapping of

i_{S}

into

T (E, δ)

such that for all

e \in E, a (e, e)

is the identity map of

ω (e)

Proof. Since by (R3) , $x^{'} σ^{l} (e, e^{'})$ is an inverse of $x σ^{r} (e, e^{'})$ (where $x \in R_{e}, x^{'} \in i (x; L_{e})$ and $e^{'} ω e$ ), $(x^{'} σ^{l} (e, e^{'})) (x σ^{r} (e, e^{'}))$ is an idempotent in the Rees groupoid $D_{e^{'}}$ of $S$ and hence $e^{'} δ e^{'} a (x, x^{'})$ . By (R4), $a$ is a mapping of $i_{S}$ into $\tilde{E_{1}}$ and the statement above implies that $a (x, x^{'}) \in T (E, δ)$ for all $(x, x^{'}) \in i_{S}$ .

If $e \in E$ , and $e^{'} \in ω (e)$ , by (R1) and ${(R 1)}^{*} e σ^{r} (e, e^{'}) = e^{'} = e σ^{l} (e, e^{'})$ . Thus $e^{'} a (e, e) = e^{'}$ . □

(8.2) For all

e^{'} ω^{r} e

and

e_{1} \overset{r}{\approx} e, e_{1} σ^{r} (e, e^{'}) = e^{'} τ^{r} (e_{1})

Proof. By axioms (R2) and (R1),

For every $e \in E, a (e, e)$ and $σ^{r} (e, e)$ are identity maps of $ω (e)$ and $R_{e}$ respectively.

e^{'} ω^{r} (e)

and

x \in R_{e}

, then

x σ^{r} (e, e^{'}) \in H_{e^{'}, e^{'} a^{r} {(x, x^{'})}^{'}}

for every $x^{'} \in i (x)$ .

Let $e^{'} ω^{r} (e)$ and either $e^{'} = e^{''}$ or for some $e_{2} \overset{r}{\approx} e, e^{''} = e^{'} τ^{r} (e_{2})$ . If $e \overset{r}{\approx} e_{1}$ , then
$σ^{r} (e, e^{'}) = σ^{r} (e_{1}, e^{''}) .$
Let $e^{'}, e^{''} \in ω^{r} (e)$ and $e^{'} \overset{l}{\approx} e^{''}$ . Then for every $x \in R_{e}$ ,
$x σ^{r} (e, e^{''}) = e^{''} (x σ^{r} (e, e^{'}))$

where $e^{''} (σ^{r} (e, e^{'}))$ denotes the product of $e^{''}$ and $x σ^{r} (e, e^{'})$ in $S$ .
Let $e^{'} ω^{r} e$ and $x \in R_{e}$ . If for some $x^{'} \in i (x)$ such that $x x^{'} = e \cdot f^{'} = e^{'} a^{r} (x, x^{'})$ and if $f = x^{'} x$ then
$x σ^{r} (e, e^{'}) = x σ^{l} (f, f^{'}) and f^{'} a^{l} (x, x^{'}) = e^{'} τ^{r} (e) .$
Let $x \in H_{e_{1}, e}, y \in H_{f, f_{1}}, h \in S (e, f), x^{'} = i (x, e, e_{1})$
$e_{1} σ^{r} (e, e^{'}) = e_{1} σ^{r} (e_{1}, e^{'} τ^{r} (e_{1})) = e^{'} τ^{r} (e_{1}) .$

□

(8.3) Let

x \in R_{e}

and

e^{'} ω^{r} e

. If

a^{r} (x, x^{'}) = τ^{r} (x, x^{'}) \cdot a (x, x^{'})

. Then

for all

x^{'} \in i (x)

. In particular

e^{'} a^{r} (x, x^{'}) \overset{l}{\approx} e^{'} a^{r} (x, x^{''})

for

x^{'}, x^{''} \in i (x)

Proof. Suppose that $x^{'} \in i (x)$ . Let $x x^{'} = e_{1}$ . Then $e \overset{r}{\approx} e_{1}$ and hence, $e_{1}^{'} = e^{'} τ^{r} (e_{1})$ , and by axiom (R2), $σ^{r} (e, e^{'}) = σ^{r} (e_{1}, e_{1}^{'})$ . Now by (R3), $x^{'} σ^{l} (e_{1}, e_{1}^{'})$ is an inverse of $x σ^{r} (e_{1}, e_{1}^{'})$ and hence $x σ^{r} (e_{1}, e_{1}^{'})$ is in the $ℒ$ -class of $(x^{'} σ^{l} (e_{1}, e_{1}^{'})) (x σ^{r} (e_{1}, e_{1}^{'})) = e_{1}^{'} a (x, x^{'})$ . Since by definitions $x σ^{r} (e, e^{'}) = x σ^{r} (e, e_{1}^{'}) \in R_{e^{'}}$ , we have the required result.

It follows from the remarks above that $H_{e^{'}, e^{'} a^{r} (x, x^{'})} = H_{e^{'}, e^{'} a^{r} (x, x^{''})}$ and hence $e^{'} a^{r} (x, x^{'})$ is $ℛ$ -equivalent to $e^{'} a^{r} (x, x^{''})$ □

Proof. Let $e^{'} ω e$ where $e = x x^{'}$ . If $f = x^{'} x$ and $f^{'} = e^{'} a (x, x^{'})$ , then

f^{'} a (x^{'} x) = (x σ^{l} (f, f^{'})) (x^{'} σ^{r} (f, f^{'})) .

By (R5) and ${(R 5)}^{*}$ , $x σ^{l} (f, f^{'}) = x σ^{r} (e, e^{'})$ and $x^{'} σ^{r} (f, f^{'}) = x^{'} σ^{l} (e e^{'})$ . Hence

\begin{array}{rcl} f^{'} a (x^{'}, x) & = & (x σ^{r} (e, e^{'})) (x^{'} σ^{l} (e, e^{'})) \\ = & (x σ^{r} (e, e^{'})) f^{'} (x^{'} σ^{l} (e, e^{'})) \\ = & (x σ^{r} (e, e^{'})) (x^{'} σ^{l} (e, e^{'})) (x σ^{r} (e, e^{'})) (x^{'} σ^{l} (e, e^{'})) \\ = & e^{'} \cdot e^{'} = e^{'} . \end{array}

This proves that $a (x, x^{'}) \circ a (x^{'}, x)$ is the identity map of $ω (e)$ . Similarly it can be shown that $a (x^{'}, x) \circ a (x, x^{'})$ is the identity map on $ω (f)$ . Hence

a (x^{'}, x) = {(a (x, x^{'}))}^{- 1} .

□

(8.5) Let

x \in R_{e}

, and

e^{'}, e^{''} \in ω^{r} (e)

. If

e^{'} \overset{l}{\approx} e^{''}

then

Proof. Choose $x^{'} \in i (x; L_{e})$ and let $e_{1}^{'} = e^{'} τ^{l} (e)$ and $e_{1}^{''} = e^{''} τ^{r} (e)$ . Thus by (R2)

x σ^{r} (e, e^{'}) = x σ^{r} (e, e_{1}^{'}), x σ^{r} (e, e^{''}) = x σ^{r} (e, e_{1}^{''}) .

By axioms ${(R 3)}^{*}$ , $x σ^{r} (e, e_{1}^{'})$ and $x σ^{r} (e, e_{1}^{''})$ are $ℒ$ -equivalent inverses of $x^{'} σ^{l} (e, e_{1}^{'})$ . Thus the required result follows from the fact that $e^{''} (x σ^{r} (e, e_{1}^{'}))$ is also an inverse $x^{'} σ^{l} (e, e_{1}^{'})$ in $H_{e^{''}, e^{''} a^{r} (x, x^{'})}$ , the product $e^{''} (x σ^{r} (e, e_{1}^{'}))$ being defined in $S$ . □

Proof. Since $h \in S (e, f)$ and $g \in ω^{l} (e) \cap ω^{r} (f)$ , we have $g τ^{l} (e) ω^{r} h τ^{l} (e)$ . Since $a (x^{'}, x)$ is an $ω$ -isomorphism we have

g_{1} = (g τ^{l} (e)) a (x^{'}, x) \neq ω^{r} (h τ^{l} (e)) a (x^{'}, x) = h_{1} .

□

Proof. Since $h \in S (e, f)$ and $g \in ω^{l} (e) \cap ω^{r} (f)$ ,

g τ^{l} (e) ω^{r} h τ^{l} (e) .

Hence

\begin{array}{rcl} g_{1} & = & (g) a^{l} (x, x^{'}) \\ = & (g τ^{l} (e)) a (x^{'}, x) ω^{r} (h τ^{l} (e)) a (x^{'}, x) \\ = & (h) a^{l} (x, x^{'}) = h_{1} . \end{array}

□

By the definition of

σ^{l} (e, h), x σ^{l} (e, h) \in L_{h}

and hence

h \in L_{x σ^{l} (e, h)}

. Similarly it can be seen that

h \in R_{y σ^{r} (f, h)}

. Consequently

Proof. Since $σ^{r} (e, e^{'})$ is idempotent preserving $e σ^{r} (e, e^{'})$ is an idempotent of $R_{e^{'}}$ . Choose $(e, e) \in i$ .

Then

a^{r} (e, e) = τ^{r} (e) \circ a (e, e) = τ^{r} (e)

by axiom (R1). Hence for $e^{'} ω^{r} e$

e^{'} a (e, e) = e^{'} τ^{r} (e) .

Again by axiom (R2), since $e^{'} \overset{r}{\approx} e^{'} τ^{r} (e)$ we have

e σ^{r} (e, e^{'}) \in H_{e^{'}, e^{'} τ^{r} (e)} = H_{e^{'} τ^{r} (e), e^{'} τ^{r} (e)} .

Thus $e σ^{r} (e, e^{'})$ is the idempotent of $H_{e^{'} τ^{r} (e), e^{'} τ^{r} (e)}$ . Consequently

e σ^{r} (e, e^{'}) = e^{'} τ^{r} (e) .

The bracketed statement can be proved dually. □

Proof. For $x, y \in S$ define

x \cdot y = (x σ^{l} (e, h)) (y σ^{r} (f, h))

(8.2)

where $e \in E (L_{x}), f \in E (R_{y}), h \in S (x, y)$ and $(x σ^{l} (e, h)) (y σ^{r} (f, h))$ denotes the product in $S$ . This product is defined in $S$ since $h \in H_{y σ^{r} (f, h), x σ^{l} (e, h)}$ .

If $x^{'} \in i (x), y^{'} \in i (y), h_{1} = h a^{l} (x, x^{'})$ and $h_{2} = h a^{r} (y, y^{'})$ , then by axiom (R2),

x \cdot y \in H_{h_{1}, h_{2}} .

(8.3)

By axioms (R3), (R3)* and Lemma 2.6, the right hand side of (8.2) does not depend on the choice of $e \in E (L_{x})$ and $f \in E (R_{y})$ . Hence to prove that (8.2) defines a binary operation on $S$ , it is necessary to show that for $h, h^{'} \in S (e, f)$ ,

(x σ^{l} (e, h)) (y σ^{r} (f, h)) = (x σ^{l} (e, h^{'})) (y σ^{r} (f, h^{'})) .

(8.4)

First suppose that $h \overset{l}{\approx} h^{'}$ . Then $h τ^{l} (e) = h^{'} τ^{l} (e)$ , and hence by (R3)*,

\begin{array}{rcl} x σ^{l} (e, h) & = & x σ^{l} (e, h τ^{l} (e)) \\ = & x σ^{l} (e, h^{'} τ^{l} (e)) \\ = & x σ^{l} (e, h^{'}) . \end{array}

Further by axiom (R4),

h^{'} (y σ^{r} (f, h)) = y σ^{r} (f, h^{'}) .

Thus

(x σ^{l} (e, h^{'})) (y σ^{r} (f, h^{'})) = (x σ^{l} (e, h) (h^{'} (y σ^{r} (f, h)))) .

Now $x σ^{l} (e, h) \in L_{h} = L_{h^{'}}$ and so, the products $(x σ^{l} (e, h)) h^{'}$ and $h^{'} (y σ^{r} (f, h))$ are defined in $S$ . Since $S$ is associative and since $h^{'}$ is a right identity of $x σ^{l} (e, h)$ , we see that (8.4) holds.

If $h \overset{r}{\approx} h^{'}$ , dually we can prove that (8.4) holds. If $h$ and $h^{'}$ are any two members of $S (e, f)$ by Lemma 2.7, there exists $h^{''} \in S (e, f)$ such that $h \overset{l}{\approx} h^{''} \overset{r}{\approx} h^{'}$ . Hence combining the two cases we see that (8.4) holds in general.

For $x, y \in S$ , if the product $x y$ is defined in $S$ then $H_{y, x}$ contains an idempotent $e$ (say) and by Lemma 2.4, $S (e, e) = {e}$ . Therefore by axiom (R1) and (R1)*

x \cdot y = (x σ^{l} (e, e)) (y σ^{r} (e, e)) = x y .

Thus the operation in $S$ , whenever it is defined, coincides with that given by (8.2).

If $e^{'} ω^{r} e$ , by Lemma 2.5, $e^{'} \in S (e^{'}, e)$ . Hence by axiom (R1)* and (R2) we obtain

\begin{array}{rcl} e^{'} \cdot x & = & (e^{'} σ^{r} (e^{'}, e^{'})) x σ^{r} (e, e^{'}) & (8.5) \\ = & e^{'} (x σ^{r} (e, e^{'})) \\ = & x σ^{r} (e, e^{'}) . \end{array}

Dually if $e^{'} ω^{l} e$

x \cdot e^{'} = x σ^{l} (e, e^{'}) .

(8.5*)

Thus (8.2), (8.5) and (8.5*) together give

x \cdot y = (x \cdot h) (h \cdot y) .

We shall now prove that the operation defined by (8.2) is associative. Let $x, y, z \in S$ , $e \in E (L_{x}) y \in H_{f, f^{'}}$ and $g \in E (R_{z})$ . Choose $h_{1} \in S (e, f)$ , $h_{2} \in S (f^{'}, g)$ and let $y^{'} = i (y; f^{'}, f)$ . Then $a (y, y^{'})$ is an $ω$ -isomorphism of $ω (f)$ onto $ω (f^{'})$ and by axiom (R5)

a {(y, y^{'})}^{- 1} = a (y^{'}, y) .

Hence if $h_{1}^{'} = h_{1} a^{r} (y, y^{'})$ and $h_{2}^{'} = h_{2} a^{l} (y, y^{'})$ , then by Lemma 3.9, there exist $h \in S (e, h_{2}^{'})$ and $h^{'} \in S (h_{1}^{'}, g)$ such that

h a^{r} (y, y^{'}) = h^{'} τ^{l} (f^{'}) .

Now by axioms (R3)* and (R5) we have

y σ^{r} (f, h) = y σ^{l} (f^{'}, h^{'} τ^{l} (f^{'})) = y σ^{l} (f^{'}, h^{'}) .

Therefore by (8.5) and (8.5*)

h \cdot y = y \cdot h^{'} .

(8.6)

Since

\begin{array}{rcl} h_{2}^{'} ω f & = & f a^{l} (y, y^{'}) \\ = & f^{'} a {(y, y^{'})}^{- 1} \end{array}

and $h_{1} \in S (e, f)$ , we have

h \in ω^{l} (e) \cap ω^{r} (h_{2}^{'}) \subseteq ω^{l} (e) \cap ω^{r} (f) .

Hence by axioms (R3), (R6) and (8.2)

\begin{array}{rcl} (x y) σ^{r} (h_{1}, h a^{r} (y, y^{'})) & = & (x y) σ^{r} (h_{1}, h^{'} τ^{l} (f^{'})) \\ = & (x y) σ^{r} (h_{1}, h^{'}) \\ = & (x σ^{l} (e, h)) (y σ^{r} (f, h)) . \end{array}

Therefore by (8.5) and (8.5*) we get

(x \cdot y) \cdot h^{'} = (x \cdot h) (h \cdot y) .

Dually we have

h \cdot (y \cdot z) = (y \cdot h^{'}) (h^{'} \cdot z)

and so

\begin{array}{rcl} (x \cdot y) \cdot z & = & ((x \cdot y) \cdot h^{'}) (h^{'} \cdot z) \\ = & ((x \cdot h) (h \cdot y)) (h^{'} \cdot z) \\ = & ((x \cdot h) (y \cdot h^{'})) (h^{'} \cdot z) . \end{array}

Now the products $(x \cdot h) (h \cdot y), ((x \cdot h) (h \cdot y)) (h^{'} \cdot z)$ and $(y \cdot h^{'}) (h^{'} \cdot z)$ are defined in $S$ . Thus

\begin{array}{rcl} ((x \cdot h) (y \cdot h^{'})) (h^{'} \cdot z) & = & (x \cdot h) (y \cdot h^{'}) (h^{'} \cdot z) \\ = & (x \cdot h) (h \cdot (y \cdot z)) \\ = & x \cdot (y \cdot z) . \end{array}

We have proved that $S$ with the operation defined by (8.2) is a semigroup. Let this semigroup be denoted by $S (ℛ)$ . Evidently two elements $x, y \in S$ that are $ℒ (ℛ)$ -equivalent in $S$ are also $ℒ (ℛ)$ -equivalent in $S (ℛ)$ . An idempotent in $S$ is also an idempotent in $S (ℛ)$ . This implies that $S (ℛ)$ is a regular semigroup.

Conversely let $S$ be a regular semigroup and $\bar{S}$ be its regular groupoid. For $(x, x^{'}) \in i$ , we define $a_{s} (x, x^{'})$ as the $ω$ -isomorphism given by Lemma 4.11. Then it is clear that $a_{s}$ is a mapping of $i$ into $T (E, δ)$ where in this case $E = E (S), δ = δ (S)$ . Further for $e^{'} ω^{r} e (e^{'} ω^{l} e)$ and $x \in R_{e} (x \in L_{e})$ define

x σ_{s}^{r} (e, e^{'}) = e^{'} x (x σ_{s}^{l} (e, e^{'}) = x e^{'}) .

(8.7)

σ_{s}^{r} = {σ_{s}^{r} (e, e^{'}) ∣ e^{'} ω^{r} e}, σ_{s}^{l} = {σ_{s}^{l} (e, e^{'}) ∣ e^{'} ω^{l} e}

we then show that

ℛ_{s} = (\bar{S}, a_{s}, σ_{s}^{r}, σ_{s}^{l})

(8.8)

is a regular system.

Axioms (R1) and (R1)* are clearly satisfied. Also if $e^{'} ω^{r} e (e^{'} ω^{l} e), x \in R_{e} (x \in L_{e})$ and if $x^{'} \in i (x)$ then by Lemma 4.11.

e^{'} a_{s}^{r} (x, x^{'}) = (e^{'} τ^{r} (e)) a_{s} (x, x^{'}) = x^{'} (e^{'} e) x = x^{'} e^{'} x,

and similarly

e^{'} a_{s}^{l} (x, x^{'}) = x e^{'} x^{'} .

By Lemma 4.1 (Lemma 4.1*) $x$ translates $L_{e^{'}} (R_{e^{'}})$ to an $ℒ (ℛ)$ class of $S$ and therefore $e^{'} x \in R_{e^{'}} (x e^{'} \in L_{e^{'}})$ . Also $e^{'} x \in L_{x^{'} e^{'} x} (x e^{'} \in R_{x e^{'} x})$ . Hence axioms (R2) and (R2)* hold. Axioms (R3), (R3)*, (R4) and (R4)* are easily verified. Axioms (R5) and (R5)* follow from the fact that for every $(x, x^{'}) \in i, a_{s} (x, x^{'})$ and $a_{s} (x^{'}, x)$ are mutually inverse $ω$ -isomorphisms (cf. Lemma 4.11) and the equality

e^{'} x = x (x^{'} e^{'} x) (x e^{'} = (x e^{'} x^{'}) x)

where $e^{'} ω^{r} e (e^{'} ω^{l} e)$ .

To verify axiom (R6), assume that $x \in H_{e, e}, y \in H_{f, f_{1}}, x^{'} = i (x; e, e_{1}), h \in S (e, f) = S (x, y)$ and $g \in ω^{l} (e) \cap ω^{r} (f)$ . Then

h_{1} = h a_{s}^{l} (x, x^{'}) = x h x^{'}

and

g_{1} = g a_{s}^{l} (x, x^{'}) = x g x^{'} .

Therefore by (4.8) and (8.7) we have

\begin{array}{rcl} ((x σ^{l} (e, h)) (y σ^{r} (f, h))) σ^{r} (h_{1}, g_{1}) & = & g_{1} ((x \cdot h) (h y)) \\ = & (x g^{'} x^{'}) (x y) \\ = & x g y \\ = & (x g) (g y) \\ = & (x σ^{l} (e, g)) (y σ^{r} (f, g)) . \end{array}

Axiom (R6)* can be verified similarly.

We have now proved that $ℛ_{s}$ is a regular system. With respect to this system the product $x \cdot y$ of $x, y \in S$ is given by

x \cdot y = (x σ^{l} (e, h)) (y σ^{r} (f, h))

where $e \in E (L_{x}), f \in E (R_{y})$ and $h \in S (x, y)$ .

Then by (8.7) and (4.8)

x \cdot y = (x \cdot h) (h \cdot y) = x y

where the product on the right is that of $S$ . Therefore the new operation coincides with the operation in $S$ .

Hence the theorem. □

Proof. Let $S = S (ℛ)$ , $E_{s} = E (S (ℛ))$ and $E = E (S)$ . Denote by $ω^{r}, ω^{l}$ and $τ^{r}, τ^{l}$ etc. the quasi-orders and partial translations of $E$ and by $ω_{s}^{r}, ω_{s}^{l}, τ_{s}^{r}, τ_{s}^{l}$ etc. those of $E_{s}$ . Since $E_{s}$ is the biordered set of idempotents of the regular semigroup $S, ω_{s}^{r}, ω_{s}^{l}, τ_{s}^{r}, τ_{s}^{l}$ etc. are quasi-orders and partial translations defined by (1.10), (1.11) and (4.1). For $x, y \in S$ the product of $x$ and $y$ in $S$ , if it is defined, will be denoted by $x y$ . The product of the same elements in $S$ is denoted by $x \cdot y$ .

Since the operation of $S$ is the extension of the partial operation in $S$ , it is evident that

E \subseteq E_{s} .

On the other hand suppose that $x \in E_{s}$ . Choose $e \in E (L_{x}), f \in E (R_{x})$ and $h \in S (e, f)$ . Then using the fact that $x$ is an idempotent in $S$ , (8.5) and (8.5*), we obtain

x = (x σ^{l} (e, h)) (x σ^{r} (f, h)) = (x \cdot h) (h \cdot x) = x \cdot h \cdot x .

By the definition of product in $S$ , we obtain from above that

h \cdot x ℒ x ℒ e in S,

and therefore

(h \cdot x) \cdot h = (h \cdot x) σ^{r} (e, h) \in L_{h} .

Dually $h \cdot (x \cdot h) \in R_{h}$ and hence $h \cdot x \cdot h \in H_{h, h}$ . Further, since $H_{h, h}$ contains an idempotent, $(h \cdot x \cdot h) (h \cdot x \cdot h) \in H_{h, h}$ . But $(h \cdot x \cdot h) (h \cdot x \cdot h) = h \cdot (x \cdot h \cdot x) \cdot h = h \cdot x \cdot h$ . Hence $h = h \cdot x \cdot h$ , and so $h \cdot x$ is an idempotent in $S$ . This implies that $e$ and $h \cdot x$ are $ℒ$ equivalent idempotents in $S$ and therefore

\begin{array}{rcl} e = e \cdot (h \cdot x) & = & (e σ^{l} (e, h)) ((h \cdot x) σ^{r} (h, h)) \\ = & (h τ^{l} (e)) (h \cdot x) (by Remark 8.3 and axiom (R1)) . \end{array}

Therefore $e \overset{r}{\approx} h τ^{l} (e)$ and hence $e = h τ^{l} (e)$ . This implies that $h ℒ x$ and dually we can show that $h ℛ x$ . Hence $x \in H_{h, h}$ and therefore $x = h$ . Thus $x \in E$ .

This proves that $E = E_{s}$ .

Let $e, f \in E, e ω^{r} f$ . Then by Lemma 2.5 $e^{'} = e τ^{r} (f) \in S (f, e)$ and by (8.2),

f \cdot e = (f σ^{l} (f, e^{'})) (e σ^{r} (e, e^{'})) .

Now by Remark 8.3 $f σ^{l} (f, e^{'}) = e^{'}$ . Since $e \overset{r}{\approx} e^{'}$ in $E$ , by axioms (R1) and (R3),

σ^{r} (e, e^{'}) = σ^{r} (e^{'}, e^{'}) = I_{R_{e^{'}}} = I_{R_{e}} .

Thus $e σ^{r} (e, e^{'}) = e$ . Hence

f \cdot e = e^{'} e .

But $e^{'} e = e$ in $S$ and hence

f \cdot e = e .

This implies by (4.1) that $e ω_{s}^{r} f$ . Consequently $ω^{r} \subseteq ω_{s}^{r}$ and dually $ω^{l} \subseteq ω_{s}^{l}$ .

Further if $e ω^{r} f$ , by Remarks 8.3, 8.5 and (4.1),

e τ^{r} (f) = f σ^{r} (f, e) = e \cdot f = e τ_{s}^{r} (f) .

Hence

τ^{r} (f) = τ_{s}^{r} (f) | ω^{r} (f) .

Now for $e, f \in E$ suppose that $e ω_{s}^{r} f$ . Then for $h \in S (f, e)$ ,

e = f \cdot e = (f \cdot h) (h \cdot e) .

Since $e, f, h \in E$ ,

h \cdot e = h τ_{s}^{r} (e) = h τ^{r} (e),

and

f \cdot h = h τ_{s}^{l} (f) = h τ^{l} (f) .

Since $(f \cdot h) (h \cdot e)$ is the product in $S$ , $h \cdot e \overset{l}{\approx} e$ and $h \overset{r}{\approx} h \cdot e$ . But $h \cdot e ω e$ and hence $h \cdot e = e$ , i.e., $h \overset{r}{\approx} e$ . Since $(f \cdot h) e$ is defined in $S$ and $(f, h) e = e$ , $f \cdot h \overset{r}{\approx} e$ . This implies that

f \cdot h \overset{l}{\approx} h \overset{r}{\approx} e \overset{r}{\approx} f \cdot h,

and so $f \cdot h = h$ . This shows that

h τ^{l} (f) = h,

and hence $h ω f$ . Therefore $e ω^{r} f$ .

Thus $ω^{r} = ω_{s}^{r} \cap E^{2}$ and dually $ω^{l} = ω_{s}^{l} \cap E^{2}$ . Suppose that $h \in S (e, f)$ , where $e, f \in E$ . Since $h$ is an idempotent in $S$ ,

e \cdot h \cdot f = (e \cdot h) \cdot (h \cdot f) .

Now the product of $(e \cdot h)$ and $(h \cdot f)$ is defined in $S$ and hence by (8.2), (8.5) and (8.5*)

(e \cdot h) \cdot (h \cdot f) = (e \cdot h) (h \cdot f) = e \cdot f .

Therefore by Theorem 4.4, $h \in S_{s} (e, f)$ . Thus we have proved that $E$ is a biordered subset of $E_{s}$ .

Now let $x \in E_{s}$ and let $e \in L_{x} \cap E$ , $f \in R_{x} \cap E$ . Then by (8.2),

x = x \cdot x = (x \cdot h) (h \cdot x),

where $h \in S (e, f)$ .

Now $(x \cdot h) (h \cdot x)$ is the product in $S$ and hence $x \cdot h \in H_{h_{1}, x}$ and $h \cdot x \in H_{x, h}$ . Since $x, e, f$ are idempotents in $S$ and $e \overset{l}{\approx} x \overset{r}{\approx} f$ , $e \cdot f \in H_{e f}$ and $e \cdot f = (e \cdot h) (h \cdot f)$ . Hence $e \cdot h \in H_{h, e \cdot f}$ . But since $e \cdot h = e σ^{r} (e, h)$ , $e \cdot h \in L_{h}$ . This implies that $e \cdot h = h$ and dually $h \cdot f = h$ . Consequently $e \cdot f = h$ and $e \overset{r}{\approx} h \overset{l}{\approx} f$ . But since $h ω^{l} e$ , $(h, e) \in ω^{l} \cap {(ω^{r})}^{- 1}$ and hence $h = e$ . Similarly $h = f$ . This implies that $h = x$ . Therefore $x \in E$ , and thus $E = E_{s}$ .

Now the sets underlying $S$ and $\bar{S (ℛ)}$ are the same. Also product of $x$ and $y$ is defined in $\bar{S (ℛ)}$ if it is defined in $S$ . Conversely if $x \cdot y$ is defined in $\bar{S (ℛ)}$ , $H_{y, x}$ contains an idempotent. Since this idempotent is an idempotent of $S$ , $x y$ is defined in $S$ also. Thus $S$ and $\bar{S (ℛ)}$ denote the same regular groupoids. □

Proof. Let $ℛ = (S, a, σ^{r}, σ^{l})$ . By (8.2), (8.5) and (8.7), if $e^{'} ω^{r} e$ and $x \in R_{e}$ , then

e^{'} \cdot x = x σ^{r} (e, e^{'})

and

x σ^{r} (e, e^{'}) = e^{'} \cdot x .

Hence $x σ^{r} (e, e^{'}) = x σ_{s}^{r} (e, e^{'})$ . □

Now let

(x, x^{'}) \in i, x x^{'} = e, x^{'} x = f

. By the definition of the mapping

S

, the domain of

S (x, x^{'})

ω (e)

which is the same as the domain of

S (x, x^{'})

(cf. Lemma 4.11). Now let

e^{'} \in ω (e)

. Then if

Proof. Let $x, y \in S$ , $e \in E (L_{x}), f \in E (R y)$ and $h \in S (e, f)$ . Then by (8.2)

x \cdot y = (x σ^{l} (e, h)) (y σ^{r} (f, h))

where the product on the right hand side is defined in $S$ hence if $ϕ$ is a homomorphism of $S$ into $S^{'}$ , then

(x \cdot y) ϕ = (x σ^{l} (e, h)) ϕ (y σ^{r} (f, h)) ϕ .

By the definition of homomorphisms of regular groupoids $ϕ | E (S)$ is a bimorphism and hence $h ϕ \in S (e ϕ, f ϕ)$ . $e ϕ \in E (L_{x ϕ})$ and $f ϕ \in E (R_{y ϕ})$ . Hence if satisfies (8.9), then

\begin{array}{rcl} (x \cdot y) ϕ & = & (x ϕ σ^{' l} (e ϕ, h ϕ)) (y ϕ σ^{' r} (f ϕ, h ϕ)) \\ = & x ϕ \cdot y ϕ \end{array}

i.e., $ϕ$ extends to a homomorphism.

Conversely if $ϕ$ is a homomorphism and if $e^{'} ω^{r} e$ , $x \in R_{e}$ , then $e^{'} ϕ ω^{r} e ϕ, x ϕ \in R_{e ϕ}$ and

\begin{array}{rcl} (x σ^{r} (e, e^{'})) ϕ & = & (e^{'} x) ϕ \\ = & e^{'} ϕ x ϕ \\ = & x ϕ σ^{r} (e ϕ, e^{'} ϕ) . \end{array}

Statement (8.9)* can be verified dually.

If $ϕ$ is an isomorphism it is clear that its extension is also an isomorphism. Hence the bracketed statement now follows. □

By Theorem 5.17 and the definiton of strictly regular groupoid we notice that the regular groupoid

\bar{S}

of a regular semigroup

S

is strictly regular if and only if

S

is strictly regular. Similarly

S

is an inverse semigroup if and only if

\bar{S}

is an inverse groupoid. Thus we have the following theorem:

If we define an inverse system

\sum = (S, S, σ^{r}, σ^{l})

as a regular system whose regular groupoid is an inverse groupoid, then we can state the axioms for inverse system as follows (here except for the changes introduced above, all other notations are those of definition (8.1)).

Theorems corresponding to Theorems 8.4 and 8.6 can also be proved for inverse systems in a similar manner.

Chapter 8Regular Systems

Chapter 8
Regular Systems