1 Introduction
Let
S
be a semigroup and
E
(
S
)
be the set of idempotents in
S
. Then an element
a
in
S
is called regular if there exists
x
∈
S
such that
a
=
a
x
a
. Moreover, if every element in
S
is regular, then we call
S
regular. It is well known that Green’s relations
ℒ
,
ℛ
,
ℋ
,
D
,
J
play an important role in the theory of semigroups (see [1]). Recall that
a
ℒ
b
(resp.
a
ℛ
b
) if and only if
S
1
a
=
S
1
b
(resp.
a
S
1
=
b
S
1
), and
ℋ
=
ℒ
∩
ℛ
. A semigroup
S
is regular if and only if every
ℒ
-class of
S
contains idempotents (if and only if every
ℛ
-class of
S
contains idempotents). A regular semigroup
S
is called orthodox (resp. completely regular) if the set of idempotents forms a subsemigroup of
S
(resp. if every
ℋ
-class of
S
contains an idempotent).
As generalizations of Green’s relations, Green’s
∗
-relations
ℒ
∗
,
ℛ
∗
,
ℋ
∗
,
D
∗
,
J
∗
were introduced and investigated in the 1970s (see [2,3]). Recall that the relations
ℒ
∗
and
ℛ
∗
on a semigroup
S
are defined as follows:
ℒ
∗
=
{
(
a
,
b
)
∈
S
×
S
∣
a
x
=
a
y
if and only if
b
x
=
b
y
for all
x
,
y
∈
S
1
}
,
ℛ
∗
=
{
(
a
,
b
)
∈
S
×
S
∣
x
a
=
y
a
if and only if
x
b
=
y
b
for all
x
,
y
∈
S
1
}
.
Obviously,
ℒ
∗
and
ℛ
∗
are equivalences on
S
. According to Fountain [3], an element
a
in
S
is said to be right abundant (resp. left abundant) if the
ℒ
∗
-class (resp.
ℛ
∗
-class) containing
a
has an idempotent. Moreover, an abundant element of
S
is an element which is both left and right abundant. Furthermore, if every element of a semigroup
S
is left abundant (resp. right abundant, abundant), then we call
S
left abundant (resp. right abundant, abundant).
In 1991, Lawson [4] introduced a new set of Green’s relations. Let
S
be a semigroup and
U
be a non-empty subset of
E
(
S
)
. Define the relations
ℒ
˜
U
and
ℛ
˜
U
on
S
as follows:
ℒ
˜
U
=
{
(
a
,
b
)
∈
S
×
S
∣
a
u
=
a
if and only if
b
u
=
b
for all
u
∈
U
}
,
ℛ
˜
U
=
{
(
a
,
b
)
∈
S
×
S
∣
u
a
=
a
if and only if
u
b
=
b
for all
u
∈
U
}
.
Then
ℒ
˜
U
and
ℛ
˜
U
are also equivalences on
S
. In particular, we write
ℒ
˜
E
(
S
)
and
ℛ
˜
E
(
S
)
as
ℒ
˜
and
ℛ
˜
, respectively. Moreover, we call
ℒ
˜
and
ℛ
˜
Green’s
∼
-relations of
S
. By Lawson [4],
(1.1)
ℒ
⊆
ℒ
∗
⊆
ℒ
˜
⊆
ℒ
˜
U
,
ℛ
⊆
ℛ
∗
⊆
ℛ
˜
⊆
ℛ
˜
U
for every non-empty subset
U
of
E
(
S
)
. For more information on Green’s relations and their generalizations, the reader can see the literature [5]. According to Lawson [4], an element
a
in
S
is left semiabundant (resp. right semiabundant) if the
ℛ
˜
-class (resp.
ℒ
˜
-class) containing
a
has an idempotent. Moreover, a semiabundant element of
S
is an element which is both left and right semiabundant. Furthermore, if every element of a semigroup
S
is left semiabundant (resp. right semiabundant, semiabundant), then we call
S
left semiabundant (resp. right semiabundant, semiabundant). By (1.1), for a semigroup
S
, regular elements are always abundant, and left abundant (resp. right abundant, abundant) elements are always left semiabundant (resp. right semiabundant, semiabundant).
On the other hand, Lawson [4] also introduced left and right Ehresmann semigroups by using
ℛ
˜
U
and
ℒ
˜
U
, respectively. In 2010, Gould [6] characterized left and right Ehresmann semigroups from the view of variety. A unary semigroup
(
S
,
⋅
,
+
)
is called left Ehresmann if for all
x
,
y
∈
S
,
(1.2)
x
+
x
=
x
,
(
x
y
)
+
=
(
x
y
+
)
+
,
(
x
+
y
+
)
+
=
x
+
y
+
=
y
+
x
+
,
x
+
(
x
y
)
+
=
(
x
y
)
+
.
A left Ehresmann semigroup
(
S
,
⋅
,
+
)
is called a left restriction semigroup if
(
x
y
)
+
x
=
x
y
+
for all
x
,
y
∈
S
. Dually, a unary semigroup
(
S
,
⋅
,
∗
)
is called right Ehresmann if for all
x
,
y
∈
S
,
(1.3)
x
x
∗
=
x
,
(
x
y
)
∗
=
(
x
∗
y
)
∗
,
(
x
∗
y
∗
)
∗
=
x
∗
y
∗
=
y
∗
x
∗
,
(
x
y
)
∗
y
∗
=
(
x
y
)
∗
.
A right Ehresmann semigroup
(
S
,
⋅
,
∗
)
is called a right restriction semigroup if
x
∗
y
=
y
(
x
y
)
∗
for all
x
,
y
∈
S
.
Let
X
be a non-empty set and
Y
a non-empty subset of
X
. Denote the full transformation semigroup on
X
by
T
(
X
)
and use
∣
X
∣
to represent the cardinality of
X
. For all
U
⊆
X
and
f
∈
T
(
X
)
, write
ker
f
=
{
(
x
,
y
)
∈
X
×
X
∣
f
(
x
)
=
f
(
y
)
}
,
f
(
U
)
=
{
f
(
x
)
∣
x
∈
U
}
and
X
⧹
U
=
{
x
∈
X
∣
x
∉
U
}
. Obviously, the set
T
(
X
,
Y
)
=
{
f
∈
T
(
X
)
∣
f
(
X
)
⊆
Y
}
is a subsemigroup of
T
(
X
)
. In [7,8], the authors investigated Green’s relations and the regularity, and Green’s
∗
-relations and the abundance of
T
(
X
,
Y
)
, respectively. In particular, Sanwong and Sommanee proved in [7] that the set
R
T
(
X
,
Y
)
of all regular elements of
T
(
X
,
Y
)
in fact a regular subsemigroup of
T
(
X
,
Y
)
.
The aim of this paper is to study
T
(
X
,
Y
)
along this direction. We first give some characterizations of Green’s
∼
-relations
ℒ
˜
and
ℛ
˜
on
T
(
X
,
Y
)
and obtain a necessary and sufficient condition under which
ℒ
=
ℒ
∗
(resp.
ℛ
=
ℛ
∗
,
ℒ
∗
=
ℒ
˜
and
ℛ
˜
=
ℛ
∗
). Next, we prove that
T
(
X
,
Y
)
is always a left Ehresmann semigroup but it is neither a left restriction semigroup nor a right Ehresmann semigroup unless
∣
Y
∣
=
1
or
Y
=
X
. Finally, we also consider the orthodoxy and complete regularity of
R
T
(
X
,
Y
)
. In the continuation of this section, we recall some known results from [5] and [7,8].
Lemma 1.1
(Theorem 4.2.2 in [5] and its dual) Let
S
be a semigroup and
a
,
b
∈
S
. If
a
and
b
are regular (resp. right abundant; left abundant), then
a
ℒ
b
⇔
a
ℒ
∗
b
⇔
a
ℒ
˜
b
,
a
ℛ
b
⇔
a
ℛ
∗
b
⇔
a
ℛ
˜
b
(
resp.
a
ℒ
∗
b
⇔
a
ℒ
˜
b
;
a
ℛ
∗
b
⇔
a
ℛ
˜
b
)
.
As consequences, if
S
is regular (resp. right abundant; left abundant), then
ℒ
=
ℒ
∗
=
ℒ
˜
,
ℛ
=
ℛ
∗
=
ℛ
˜
(resp.
ℒ
∗
=
ℒ
˜
;
ℛ
∗
=
ℛ
˜
).
Lemma 1.2
(Theorems 2.4, 3.2 and 3.3 in [7]) Let
f
,
g
∈
T
(
X
,
Y
)
. Then the following statements hold:
f
is regular in
T
(
X
,
Y
)
if and only if
f
(
X
)
=
f
(
Y
)
.
f
ℛ
g
if and only if
f
=
g
or (
f
,
g
are regular and
f
(
X
)
=
g
(
X
)
).
f
ℒ
g
if and only if
ker
f
=
ker
g
.
Lemma 1.3
(Lemma 2, Corollary 1 and Lemma 3 in [8]) Let
f
,
g
∈
T
(
X
,
Y
)
. Then the following statements hold:
(
f
,
g
)
∈
ℛ
∗
if and only if
f
(
X
)
=
g
(
X
)
.
T
(
X
,
Y
)
is left abundant.
(
f
,
g
)
∈
ℒ
∗
if and only if one of the following conditions holds:
f
(
X
)
=
f
(
Y
)
,
g
(
X
)
=
g
(
Y
)
and
ker
f
=
ker
g
;
f
(
X
)
≠
f
(
Y
)
,
g
(
X
)
≠
g
(
Y
)
and
ker
f
∩
(
Y
×
Y
)
=
ker
g
∩
(
Y
×
Y
)
.
2 Main results
Throughout the rest of the paper, let
X
be a fixed non-empty set and
Y
a fixed non-empty subset of
X
. Denote the set of all idempotents in
T
(
X
,
Y
)
by
E
(
T
)
. We first characterize Green’s
∼
-relations
ℒ
˜
and
ℛ
˜
on
T
(
X
,
Y
)
. By Lemmas 1.1 and 1.3 (2), we have
ℛ
∗
=
ℛ
˜
, and so
(
f
,
g
)
∈
ℛ
˜
if and only if
f
(
X
)
=
g
(
X
)
for all
f
,
g
∈
T
(
X
,
Y
)
. To give a characterization of
ℒ
˜
, we need the following lemma.
Lemma 2.1
Let
f
∈
T
(
X
,
Y
)
. Then
f
(
X
)
=
f
(
Y
)
if and only if there exists
e
∈
E
(
T
)
such that
f
e
=
f
.
Proof
If
f
(
X
)
=
f
(
Y
)
, then by Lemma 1.2,
f
is regular in
T
(
X
,
Y
)
, and so there is
h
∈
T
(
X
,
Y
)
such that
f
=
f
h
f
. Take
e
=
h
f
. Then
f
=
f
e
and
e
∈
E
(
T
)
. Conversely, if there exists
e
∈
E
(
T
)
such that
f
e
=
f
, then
e
is regular, and so
e
(
X
)
=
e
(
Y
)
by Lemma 1.2. Therefore,
f
(
X
)
=
f
e
(
X
)
=
f
e
(
Y
)
=
f
(
Y
)
.□
Theorem 2.2
Let
f
,
g
∈
T
(
X
,
Y
)
. Then
(
f
,
g
)
∈
ℒ
˜
if and only if one of the following conditions holds:
f
(
X
)
=
f
(
Y
)
,
g
(
X
)
=
g
(
Y
)
,
ker
f
=
ker
g
;
f
(
X
)
≠
f
(
Y
)
,
g
(
X
)
≠
g
(
Y
)
.
Proof
Let
(
f
,
g
)
∈
ℒ
˜
. If
f
(
X
)
=
f
(
Y
)
, then by Lemma 2.1 there exists
e
∈
E
(
T
)
such that
f
=
f
e
. Since
(
f
,
g
)
∈
ℒ
˜
, it follows that
g
=
g
e
. Using Lemma 2.1 again, we have
g
(
X
)
=
g
(
Y
)
. Similarly, it follows that
g
(
X
)
=
g
(
Y
)
implies
f
(
X
)
=
f
(
Y
)
. Thus,
f
(
X
)
=
f
(
Y
)
if and only if
g
(
X
)
=
g
(
Y
)
. If this is the case, there exists
h
∈
T
(
X
,
Y
)
such that
f
=
f
h
f
by Lemma 1.2. Since
(
f
,
g
)
∈
ℒ
˜
and
h
f
∈
E
(
T
)
, it follows that
g
=
g
h
f
. If
(
x
,
y
)
∈
ker
f
, then
f
(
x
)
=
f
(
y
)
, and so
g
(
x
)
=
g
h
f
(
x
)
=
g
h
f
(
y
)
=
g
(
y
)
.
This implies that
ker
f
⊆
ker
g
. Dually,
ker
g
⊆
ker
f
, and hence
ker
f
=
ker
g
. In other words, we have shown that one of the above two conditions must be true if
(
f
,
g
)
∈
ℒ
˜
.
Conversely, if (1) is true, by Lemma 1.3 we have
(
f
,
g
)
∈
ℒ
∗
, and so
(
f
,
g
)
∈
ℒ
˜
by (1.1). If (2) holds, then by Lemma 2.1 we have
f
e
≠
f
and
g
e
≠
g
for all
e
∈
E
(
T
)
. This also gives that
(
f
,
g
)
∈
ℒ
˜
.□
The following corollary gives the semiabundance of
T
(
X
,
Y
)
.
Corollary 2.3
If
∣
Y
∣
≠
1
and
Y
≠
X
, then
T
(
X
,
Y
)
is left semiabundant but not right semiabundant.
Proof
By Lemma 1.3 (2),
T
(
X
,
Y
)
is left abundant, and so it is left semiabundant. We next show that
T
(
X
,
Y
)
is not right semiabundant. Since
∣
Y
∣
≥
2
, we can take distinct elements
a
,
b
in
Y
. Consider the mapping
f
:
X
→
X
defined by
f
(
x
)
=
a
,
x
∈
Y
,
b
,
otherwise
.
Then
f
∈
T
(
X
,
Y
)
and
{
a
,
b
}
=
f
(
X
)
≠
f
(
Y
)
=
{
a
}
(note that
Y
≠
X
). Suppose that the
ℒ
˜
-class containing
f
has an idempotent
e
. By Theorem 2.2, we have
e
(
X
)
≠
e
(
Y
)
. However, by Lemma 2.1,
e
(
X
)
=
e
(
Y
)
, a contradiction. This implies that the
ℒ
˜
-class containing
f
has no idempotent. Thus,
T
(
X
,
Y
)
is not right semiabundant.□
Remark 2.4
Corollary 2.3 shows that the semigroup
T
(
X
,
Y
)
with
∣
Y
∣
≠
1
and
Y
≠
X
provides an explicit example of a left semiabundant semigroup, which is not right semiabundant. Certainly, its dual semigroup gives an explicit example of a right semiabundant semigroup, which is not left semiabundant.
By (1.1), Lemmas 1.2 (1) and 2.1, the following corollary is clear.
Corollary 2.5
Let
f
∈
T
(
X
,
Y
)
. Then
f
i
s
r
e
g
u
l
a
r
⇔
f
i
s
a
b
u
n
d
a
n
t
⇔
f
i
s
s
e
m
i
a
b
u
n
d
a
n
t
⇔
f
i
s
r
i
g
h
t
a
b
u
n
d
a
n
t
⇔
f
i
s
r
i
g
h
t
s
e
m
i
a
b
u
n
d
a
n
t
.
We have already observed that
ℛ
∗
=
ℛ
˜
in
T
(
X
,
Y
)
. We now consider the conditions under which the equality
ℒ
=
ℒ
∗
(resp.
ℛ
=
ℛ
∗
,
ℒ
∗
=
ℒ
˜
) holds in
T
(
X
,
Y
)
. We start with the following simple result: if
∣
Y
∣
=
1
or
Y
=
X
, then
ℒ
=
ℒ
∗
=
ℒ
˜
and
ℛ
=
ℛ
∗
=
ℛ
˜
. In fact, in this case, for all
f
∈
T
(
X
,
Y
)
, we have
f
(
X
)
⊆
Y
with
∣
Y
∣
=
1
or
Y
=
X
, which implies that
f
(
X
)
=
f
(
Y
)
. So
T
(
X
,
Y
)
is regular by Lemma 1.2 (1). In view of Lemma 1.1 the desired result holds. Moreover, we can prove the following results.
Theorem 2.6
For the semigroup
T
(
X
,
Y
)
in which
∣
Y
∣
≠
1
and
Y
≠
X
, the following statements hold:
ℒ
=
ℒ
∗
if and only if
∣
X
⧹
Y
∣
=
1
.
ℒ
∗
=
ℒ
˜
if and only if
∣
Y
∣
=
2
.
Proof
(1) By hypothesis,
∣
Y
∣
≥
2
and
∣
X
⧹
Y
∣
≥
1
. Take different elements
a
,
b
in
Y
and define
f
,
g
∈
T
(
X
,
Y
)
by
f
=
Y
X
\
Y
a
b
,
g
=
Y
X
\
Y
b
a
.
Then
f
(
X
)
=
g
(
X
)
. By Lemma 1.3 (1), we have
(
f
,
g
)
∈
ℛ
∗
. However, by (1) and (2) of Lemma 1.2, neither
f
nor
g
is regular, and so
(
f
,
g
)
∉
ℛ
. Thus,
ℛ
≠
ℛ
∗
.
(2) Necessity. By hypothesis,
∣
Y
∣
≥
2
and
∣
X
⧹
Y
∣
≥
1
. If
∣
X
⧹
Y
∣
≥
2
, then we can take different elements
a
,
b
in
Y
and different elements
c
,
d
in
X
⧹
Y
. Define
f
,
g
∈
T
(
X
,
Y
)
by
f
=
a
b
c
d
X
\
{
a
,
b
,
c
,
d
}
a
a
b
b
a
,
g
=
a
b
c
d
X
\
{
a
,
b
,
c
,
d
}
b
b
b
a
b
.
Obviously,
ker
f
≠
ker
g
,
f
(
Y
)
=
{
a
}
≠
{
a
,
b
}
=
f
(
X
)
and
g
(
X
)
=
{
a
,
b
}
≠
{
b
}
=
g
(
Y
)
, and
ker
f
∩
(
Y
×
Y
)
=
Y
×
Y
=
ker
g
∩
(
Y
×
Y
)
.
This yields that
(
f
,
g
)
∈
ℒ
∗
and
(
f
,
g
)
∉
ℒ
by Lemmas 1.3 (3) and 1.2 (3).
Sufficiency. Let
∣
X
⧹
Y
∣
=
1
,
f
,
g
∈
T
(
X
,
Y
)
and
(
f
,
g
)
∈
ℒ
∗
. In view of Lemma 1.3 (3), we have one of the following two cases:
Case 1.
f
(
X
)
=
f
(
Y
)
,
g
(
X
)
=
g
(
Y
)
and
ker
f
=
ker
g
. In this case, we have
(
f
,
g
)
∈
ℒ
by Lemma 1.2 (3).
Case 2.
f
(
X
)
≠
f
(
Y
)
,
g
(
X
)
≠
g
(
Y
)
and
ker
f
∩
(
Y
×
Y
)
=
ker
g
∩
(
Y
×
Y
)
. Let
x
1
,
x
2
∈
X
and
(
x
1
,
x
2
)
∈
ker
f
. Then
f
(
x
1
)
=
f
(
x
2
)
. If both
x
1
and
x
2
are in
Y
, then
(
x
1
,
x
2
)
∈
ker
f
∩
(
Y
×
Y
)
=
ker
g
∩
(
Y
×
Y
)
, and so
(
x
1
,
x
2
)
∈
ker
g
. If neither
x
1
nor
x
2
is in
Y
, then
x
1
=
x
2
by the fact
∣
X
⧹
Y
∣
=
1
, and certainly
(
x
1
,
x
2
)
∈
ker
g
. If
x
1
∈
Y
and
x
2
∉
Y
, then we have
X
=
Y
∪
{
x
2
}
by the fact
∣
X
⧹
Y
∣
=
1
. This implies that
f
(
X
)
=
f
(
Y
)
since
f
(
x
1
)
=
f
(
x
2
)
, which leads to a contradiction. Dually, it is also impossible that
x
1
∉
Y
and
x
2
∈
Y
. Thus,
ker
f
⊆
ker
g
in this case. This together with its dual give that
ker
f
=
ker
g
. By Lemma 1.2 (3),
(
f
,
g
)
∈
ℒ
.
We have shown that
ℒ
∗
⊆
ℒ
in both cases. By (1.1), we have
ℒ
=
ℒ
∗
.
(3) Let
∣
Y
∣
=
2
,
f
,
g
∈
T
(
X
,
Y
)
and
(
f
,
g
)
∈
ℒ
˜
. Then
f
(
X
)
=
f
(
Y
)
,
g
(
X
)
=
g
(
Y
)
,
ker
f
=
ker
g
or
f
(
X
)
≠
f
(
Y
)
,
g
(
X
)
≠
g
(
Y
)
by Theorem 2.2. If the former case holds, then
(
f
,
g
)
∈
ℒ
∗
by Lemma 1.3 (3). Now assume that
f
(
X
)
≠
f
(
Y
)
and
g
(
X
)
≠
g
(
Y
)
. Since
f
(
Y
)
⊆
f
(
X
)
⊆
Y
and
∣
Y
∣
=
2
, we have
∣
f
(
X
)
∣
=
2
and
∣
f
(
Y
)
∣
=
1
. Similarly,
∣
g
(
X
)
∣
=
2
and
∣
g
(
Y
)
∣
=
1
. Thus,
ker
f
∩
(
Y
×
Y
)
=
Y
×
Y
=
ker
g
∩
(
Y
×
Y
)
.
By Lemma 1.3 (3),
(
f
,
g
)
∈
ℒ
∗
. Hence, we have
ℒ
˜
⊆
ℒ
∗
in both cases. By (1.1),
ℒ
˜
=
ℒ
∗
.
Conversely, we need to prove that
ℒ
∗
≠
ℒ
˜
when
∣
Y
∣
≥
3
as
∣
Y
∣
≠
1
and
Y
≠
X
by hypothesis. Take different elements
a
,
b
,
c
in
Y
. Define
f
,
g
∈
T
(
X
,
Y
)
by
f
=
Y
X
\
Y
a
b
,
g
=
{
a
,
b
}
Y
\
{
a
,
b
}
X
\
Y
a
b
c
.
Then
f
(
X
)
≠
f
(
Y
)
,
g
(
X
)
≠
g
(
Y
)
. By Theorem 2.2, we have
(
f
,
g
)
∈
ℒ
˜
. Since
(
a
,
c
)
∈
Y
×
Y
,
(
a
,
c
)
∈
ker
f
and
(
a
,
c
)
∉
ker
g
, it follows that
ker
f
∩
(
Y
×
Y
)
≠
ker
g
∩
(
Y
×
Y
)
.
This implies that
(
f
,
g
)
∉
ℒ
∗
by Lemma 1.3. Therefore,
ℒ
∗
≠
ℒ
˜
.□
The next result shows that
T
(
X
,
Y
)
is always a left Ehresmann semigroup but it is neither a left restriction semigroup nor a right Ehresmann semigroup unless
∣
Y
∣
=
1
or
Y
=
X
.
Theorem 2.7
T
(
X
,
Y
)
is a left Ehresmann semigroup. However, if
∣
Y
∣
≠
1
and
Y
≠
X
,
then
T
(
X
,
Y
)
is neither right Ehresmann nor left restriction.
Proof
Fix an element
a
∈
Y
. Define
e
:
X
→
X
by
e
(
x
)
=
x
,
x
∈
Y
,
a
,
otherwise
.
Then
e
is a left identity of
T
(
X
,
Y
)
. For all
f
∈
T
(
X
,
Y
)
, let
f
+
=
e
. Then the identities in (1.2) are all satisfied, and so
(
T
(
X
,
Y
)
,
⋅
,
+
)
is left Ehresmann. Assume that
∣
Y
∣
≠
1
,
Y
≠
X
and
(
T
(
X
,
Y
)
,
⋅
,
∗
)
is right Ehresmann. Then for all
f
∈
T
(
X
,
Y
)
, we have
f
f
∗
=
f
, and so
f
∗
=
(
f
f
∗
)
∗
=
(
f
∗
f
∗
)
∗
=
f
∗
f
∗
by (1.3). In view of Lemma 2.1, we obtain that
f
(
X
)
=
f
(
Y
)
. This implies that
f
(
X
)
=
f
(
Y
)
for all
f
∈
T
(
X
,
Y
)
. However, this is impossible. In fact, fix an element
a
∈
X
⧹
Y
and two distinct elements
b
,
c
∈
Y
and define
g
:
X
→
X
by
g
(
x
)
=
b
,
x
=
a
,
c
,
otherwise
.
Then
g
∈
T
(
X
,
Y
)
. Obviously,
g
(
X
)
≠
g
(
Y
)
. Thus,
T
(
X
,
Y
)
is not right Ehresmann.
Let
∣
Y
∣
≠
1
,
Y
≠
X
and assume that
(
T
(
X
,
Y
)
,
⋅
,
+
)
is a left restriction semigroup. Then by (1.2) and the “left restriction” condition, for all
u
,
v
∈
T
(
X
,
Y
)
,
(2.1)
u
+
u
=
u
,
u
+
v
+
=
v
+
u
+
,
(
u
+
v
)
+
=
u
+
v
+
,
u
v
+
=
(
u
v
)
+
u
.
Fix two distinct elements
a
,
b
∈
Y
and define
f
:
X
→
X
,
g
:
X
→
X
by
f
(
x
)
=
x
,
x
∈
Y
,
a
,
otherwise
,
g
(
x
)
=
x
,
x
∈
Y
,
b
,
otherwise
.
Then
f
,
g
∈
T
(
X
,
Y
)
and
f
g
=
g
,
g
f
=
f
. By (2.1), for all
x
∈
Y
,
(2.2)
f
+
(
x
)
=
f
+
(
f
(
x
)
)
=
(
f
+
f
)
(
x
)
=
f
(
x
)
=
x
=
g
(
x
)
=
(
g
+
g
)
(
x
)
=
g
+
(
x
)
.
Since
g
(
x
)
∈
Y
for all
x
∈
X
,
(
f
+
g
)
(
x
)
=
f
+
(
g
(
x
)
)
=
f
(
g
(
x
)
)
=
g
(
x
)
. This gives that
f
+
g
=
g
. Dually,
g
+
f
=
f
. By (2.1),
g
+
=
(
f
+
g
)
+
=
f
+
g
+
=
g
+
f
+
=
(
g
+
f
)
+
=
f
+
.
On the other hand, let
x
0
∈
X
⧹
Y
. Then by (2.1), the fact
f
g
=
g
, the definition of
f
and (2.2), we have
f
(
g
+
(
x
0
)
)
=
f
g
+
(
x
0
)
=
(
(
f
g
)
+
f
)
(
x
0
)
=
(
g
+
f
)
(
x
0
)
=
g
+
(
f
(
x
0
)
)
=
g
+
(
a
)
=
a
.
Observe that the definition of
f
and the fact
g
+
∈
T
(
X
,
Y
)
, it follows that
g
+
(
x
0
)
=
a
. Dually,
f
+
(
x
0
)
=
b
. This implies that
f
+
≠
g
+
, a contradiction. Thus,
T
(
X
,
Y
)
is not left restriction.□
Remark 2.8
Theorem 2.7 shows that the semigroup
T
(
X
,
Y
)
with
∣
Y
∣
≠
1
and
Y
≠
X
provides an explicit example of a left Ehresmann semigroup which is neither right Ehresmann nor left restriction. Certainly, its dual semigroup gives an explicit example of a right Ehresmann semigroup which is neither left Ehresmann nor right restriction.
Finally, we consider the orthodoxy and complete regularity of
R
T
(
X
,
Y
)
(the set of all regular elements of
T
(
X
,
Y
)
). For this purpose, we need the lemma below.
Lemma 2.9
Let
e
∈
T
(
X
,
Y
)
. Then
e
∈
E
(
T
)
if and only if
e
(
X
)
=
e
(
Y
)
and
e
(
a
)
=
a
for all
a
∈
e
(
Y
)
.
Proof
If
e
2
=
e
, then by Lemma 2.1, we have
e
(
X
)
=
e
(
Y
)
. Let
a
∈
e
(
Y
)
=
e
(
X
)
. Then there exists
x
0
∈
X
such that
a
=
e
(
x
0
)
. Thus,
e
(
a
)
=
e
e
(
x
0
)
=
e
(
x
0
)
=
a
. Conversely, since
e
(
X
)
=
e
(
Y
)
, it follows that for every
x
∈
X
, there is
y
∈
Y
such that
e
(
x
)
=
e
(
y
)
∈
e
(
Y
)
, and so
e
e
(
x
)
=
e
e
(
y
)
=
e
(
y
)
=
e
(
x
)
. This implies that
e
2
≐
e
.□
Theorem 2.10
For the semigroup
T
(
X
,
Y
)
,
the following statements are equivalent:
R
T
(
X
,
Y
)
is orthodox.
R
T
(
X
,
Y
)
is completely regular.
Proof
(
1
)
⇒
(
2
)
. Suppose that
∣
Y
∣
≥
3
and fix distinct elements
a
,
b
,
c
∈
Y
. Define
e
,
f
∈
T
(
X
,
Y
)
by
e
=
c
X
⧹
{
c
}
c
a
,
f
=
b
X
⧹
{
b
}
b
c
.
Then
e
and
f
are idempotents, and so
e
,
f
∈
R
T
(
X
,
Y
)
. However,
e
f
=
b
X
⧹
{
b
}
a
c
is not an idempotent, and this implies that
R
T
(
X
,
Y
)
is not orthodox.
(
2
)
⇒
(
1
)
, (3). If
∣
Y
∣
=
1
, then
T
(
X
,
Y
)
contains only one element, and
R
T
(
X
,
Y
)
is certainly orthodox and completely regular. Now let
∣
Y
∣
=
2
and write
Y
=
{
a
,
b
}
. By Lemma 2.9, the set of idempotents of
R
T
(
X
,
Y
)
is
E
(
R
T
(
X
,
Y
)
)
=
{
f
∈
T
(
X
,
Y
)
∣
f
(
a
)
=
a
,
f
(
b
)
=
b
}
∪
{
e
1
,
e
2
}
,
where
e
1
=
X
a
,
e
2
=
X
b
.
Obviously,
E
(
R
T
(
X
,
Y
)
)
is a subsemigroup of
R
T
(
X
,
Y
)
. So
R
T
(
X
,
Y
)
is orthodox.
On the other hand, in view of Lemma 1.2 (1), we have
R
T
(
X
,
Y
)
=
E
(
R
T
(
X
,
Y
)
)
∪
{
f
∈
T
(
X
,
Y
)
∣
f
(
a
)
=
b
,
f
(
b
)
=
a
}
.
For a given
f
∈
R
T
(
X
,
Y
)
satisfying
f
(
a
)
=
b
and
f
(
b
)
=
a
, define
e
:
X
→
X
by
e
(
x
)
=
x
,
x
∈
Y
,
b
,
x
∈
X
⧹
Y
and
f
(
x
)
=
a
,
a
,
x
∈
X
⧹
Y
and
f
(
x
)
=
b
.
Then
e
∈
E
(
R
T
(
X
,
Y
)
)
and
e
(
X
)
=
f
(
X
)
=
{
a
,
b
}
,
ker
e
=
ker
f
. This implies that
(
e
,
f
)
∈
ℋ
in
R
T
(
X
,
Y
)
by (2) and (3) of Lemma 1.2 and Proposition 4.1.1 in [1], and so
f
is completely regular. Thus,
R
T
(
X
,
Y
)
is completely regular.
(
3
)
⇒
(
2
)
. We need to prove that
R
T
(
X
,
Y
)
is not completely regular when
∣
Y
∣
≥
3
. Let
∣
Y
∣
≥
3
and take different elements
a
,
b
,
c
in
Y
. Define
f
:
X
→
X
,
f
=
c
X
⧹
{
c
}
b
a
,
then
f
∈
R
T
(
X
,
Y
)
by Lemma 1.2 (1). Moreover, in view of (2) and (3) of Lemma 1.2, the
ℋ
-class containing
f
just has the following two elements:
c
X
⧹
{
c
}
b
a
,
c
X
⧹
{
c
}
a
b
.
Obviously, the above two elements are not idempotents. Thus,
R
T
(
X
,
Y
)
is not completely regular.□
To end the paper, we give a concrete example to illustrate the above results.
Example 2.11
Let
X
=
{
1
,
2
,
3
,
4
}
and
Y
=
{
1
,
2
,
3
}
. Then the semigroup
T
(
X
,
Y
)
contains 81 elements. For the sake of simplicity, we denote a typical element
1
2
3
4
i
j
k
l
in
T
(
X
,
Y
)
by
(
i
,
j
,
k
,
l
)
, where
i
,
j
,
k
,
l
∈
Y
. With the above notation, by Lemma 1.2 (3), the 14
ℒ
-classes of
T
(
X
,
Y
)
can be described as follows:
L
1
=
{
(
i
,
i
,
i
,
i
)
∣
i
∈
Y
}
,
L
2
=
{
(
j
,
i
,
i
,
i
)
∣
i
,
j
∈
Y
,
i
≠
j
}
,
L
3
=
{
(
i
,
j
,
i
,
i
)
∣
i
,
j
∈
Y
,
i
≠
j
}
,
L
4
=
{
(
i
,
i
,
j
,
i
)
∣
i
,
j
∈
Y
,
i
≠
j
}
,
L
5
=
{
(
i
,
j
,
j
,
i
)
∣
i
,
j
∈
Y
,
i
≠
j
}
,
L
6
=
{
(
i
,
j
,
i
,
j
)
∣
i
,
j
∈
Y
,
i
≠
j
}
,
L
7
=
{
(
i
,
i
,
j
,
j
)
∣
i
,
j
∈
Y
,
i
≠
j
}
,
L
8
=
{
(
i
,
i
,
i
,
j
)
∣
i
,
j
∈
Y
,
i
≠
j
}
,
L
9
=
{
(
i
,
i
,
j
,
k
)
∣
{
i
,
j
,
k
}
=
Y
}
,
L
10
=
{
(
i
,
j
,
i
,
k
)
∣
{
i
,
j
,
k
}
=
Y
}
,
L
11
=
{
(
j
,
i
,
i
,
k
)
∣
{
i
,
j
,
k
}
=
Y
}
,
L
12
=
{
(
i
,
j
,
k
,
i
)
∣
{
i
,
j
,
k
}
=
Y
}
,
L
13
=
{
(
j
,
i
,
k
,
i
)
∣
{
i
,
j
,
k
}
=
Y
}
,
L
14
=
{
(
j
,
k
,
i
,
i
)
∣
{
i
,
j
,
k
}
=
Y
}
.
Observe that
∣
L
1
∣
=
3
and
∣
L
i
∣
=
6
,
i
=
2
,
3
,
…
,
14
. By Theorem 2.6 (2) and the fact that
∣
X
⧹
Y
∣
=
1
, we have
ℒ
=
ℒ
∗
in
T
(
X
,
Y
)
, and so the above
ℒ
-classes of
T
(
X
,
Y
)
are also the whole
ℒ
∗
-classes of
T
(
X
,
Y
)
. Moreover, by Lemma 1.2 (1), the non-regular
ℒ
-classes (a non-regular
ℒ
-class is an
ℒ
-class in which every element is not regular) of
T
(
X
,
Y
)
are
L
8
,
L
9
,
L
10
and
L
11
. In view of Theorem 2.2 and Lemma 1.2 (1),
T
(
X
,
Y
)
has 11
ℒ
˜
-classes which are
L
1
,
L
2
,
L
3
,
L
4
,
L
5
,
L
6
,
L
7
,
⋃
i
=
8
11
L
i
,
L
12
,
L
13
,
L
14
.
On the other hand, by Lemma 1.3 (1), the
ℛ
∗
-classes of
T
(
X
,
Y
)
are as follows:
R
1
∗
=
{
(
1
,
1
,
1
,
1
)
}
,
R
2
∗
=
{
(
2
,
2
,
2
,
2
)
}
,
R
3
∗
=
{
(
3
,
3
,
3
,
3
)
}
,
R
4
∗
=
{
f
∈
T
(
X
,
Y
)
∣
f
(
X
)
=
{
1
,
2
}
}
,
R
5
∗
=
{
f
∈
T
(
X
,
Y
)
∣
f
(
X
)
=
{
1
,
3
}
}
,
R
6
∗
=
{
f
∈
T
(
X
,
Y
)
∣
f
(
X
)
=
{
2
,
3
}
}
,
R
7
∗
=
{
f
∈
T
(
X
,
Y
)
∣
f
(
X
)
=
Y
}
.
Note that
R
7
∗
is the union of
L
9
,
L
10
,
…
,
L
13
and
L
14
, and
∣
R
1
∗
∣
=
∣
R
2
∗
∣
=
∣
R
3
∗
∣
=
1
,
∣
R
4
∗
∣
=
∣
R
5
∗
∣
=
∣
R
6
∗
∣
=
14
,
∣
R
7
∗
∣
=
36
.
In view of the comments before Lemma 2.1, we have
ℛ
˜
=
ℛ
∗
, and so the above
ℛ
∗
-classes are also the whole
ℛ
˜
-classes of
T
(
X
,
Y
)
. Furthermore, observe by Lemma 1.2 (1) that the set of non-regular elements in
R
4
∗
(resp.
R
5
∗
,
R
6
∗
) is
{
(
1
,
1
,
1
,
2
)
,
(
2
,
2
,
2
,
1
)
}
(
resp
.
{
(
1
,
1
,
1
,
3
)
,
(
3
,
3
,
3
,
1
)
}
,
{
(
2
,
2
,
2
,
3
)
,
(
3
,
3
,
3
,
2
)
}
)
,
and the set of non-regular elements in
R
7
∗
is the union of
L
9
,
L
10
and
L
11
. Thus, by Lemma 1.2 (2), there are 31
ℛ
-classes in
T
(
X
,
Y
)
in which the whole non-single element
ℛ
-classes are as follows:
R
4
∗
⧹
{
(
1
,
1
,
1
,
2
)
,
(
2
,
2
,
2
,
1
)
}
,
R
5
∗
⧹
{
(
1
,
1
,
1
,
3
)
,
(
3
,
3
,
3
,
1
)
}
,
and
R
6
∗
⧹
{
(
2
,
2
,
2
,
3
)
,
(
3
,
3
,
3
,
2
)
}
,
R
7
∗
=
⋃
n
=
9
14
L
n
⋃
n
=
9
11
L
n
=
⋃
n
=
12
14
L
n
.
