(inherited from $\alpha$), while $\alpha\eta$ does not.

For more on monoids generated by idempotents and one-sided units, see [34,
35].

Example 4.27 involved a (strong) action pair $(U,S)$ for which $U$ and $S$
were both submonoids of $M$; in such cases, the monoid $US$ contains both $U$
and $S$. The next example, which is one of our original sources of motivation,
provides a contrast.

######  Example 4.30.

Let $n$ be a positive integer, and write ${\bf n}=\\{1,\ldots,n\\}$. In
addition to the monoids $\mathcal{P}\mathcal{T}_{n}$, $\mathcal{T}_{n}$, and
so on, let $\mathcal{E}_{n}=\\{{\mathrm{id}_{A}}:{A\subseteq{\bf n}}\\}$ be
the semilattice of partial identities, and write
$\operatorname{Sing}(\mathcal{E}_{n})=\mathcal{E}_{n}\setminus\\{\mathrm{id}_{\bf
n}\\}$. Then the following are all strong action pairs in
$\mathcal{P}\mathcal{T}_{n}$:

$(\mathcal{E}_{n},\mathcal{T}_{n}),\qquad(\mathcal{E}_{n},\operatorname{Sing}(\mathcal{T}_{n})),\qquad(\operatorname{Sing}(\mathcal{E}_{n}),\mathcal{T}_{n}),\qquad(\operatorname{Sing}(\mathcal{E}_{n}),\operatorname{Sing}(\mathcal{T}_{n})).$

Indeed, conditions (SA1) and (SA2) follow from the laws

$\alpha\cdot\mathrm{id}_{A}=\mathrm{id}_{A\alpha^{-1}}\cdot\alpha\qquad\text{and}\qquad\mathrm{id}_{A}\cdot\alpha=\alpha\mathord{\restriction}_{A}.$
(4.31)

One can check (though it is not entirely obvious, and in any case follows from
results of later chapters; cf. Remark 8.64) that the subsemigroups $US$ of
$\mathcal{P}\mathcal{T}_{n}$ corresponding to the above pairs $(U,S)$ are:

  * •

$\mathcal{E}_{n}\cdot\mathcal{T}_{n}=\mathcal{P}\mathcal{T}_{n}$,

  * •

$\mathcal{E}_{n}\cdot\operatorname{Sing}(\mathcal{T}_{n})=\operatorname{Sing}(\mathcal{P}\mathcal{T}_{n})$,
and

  * •

$\operatorname{Sing}(\mathcal{E}_{n})\cdot\mathcal{T}_{n}=\operatorname{Sing}(\mathcal{E}_{n})\cdot\operatorname{Sing}(\mathcal{T}_{n})=\mathcal{P}\mathcal{T}_{n}\setminus\mathcal{T}_{n}$.

It is also worth noting that
$\mathcal{E}_{n}\cdot\operatorname{Sing}(\mathcal{T}_{n})=\operatorname{Sing}(\mathcal{P}\mathcal{T}_{n})$
contains $\operatorname{Sing}(\mathcal{T}_{n})$ but not $\mathcal{E}_{n}$; it
does however contain
$\operatorname{Sing}(\mathcal{E}_{n})=\mathcal{E}_{n}\setminus\\{\mathrm{id}_{\bf
n}\\}$, a crucial property that features heavily in Chapter 6. Presentations
for the semigroups $\operatorname{Sing}(\mathcal{T}_{n})$,
$\operatorname{Sing}(\mathcal{P}\mathcal{T}_{n})$ and
$\mathcal{P}\mathcal{T}_{n}\setminus\mathcal{T}_{n}$ can be found in [29, 31,
30, 32]. Also, while
$\operatorname{Sing}(\mathcal{E}_{n})\cdot\mathcal{T}_{n}=\operatorname{Sing}(\mathcal{E}_{n})\cdot\operatorname{Sing}(\mathcal{T}_{n})=\mathcal{P}\mathcal{T}_{n}\setminus\mathcal{T}_{n}$
contains $\operatorname{Sing}(\mathcal{E}_{n})$, it is disjoint from both
$\mathcal{T}_{n}$ and $\operatorname{Sing}(\mathcal{T}_{n})$.

We also observe that if
$(U,S)=(\operatorname{Sing}(\mathcal{E}_{n}),\mathcal{T}_{n})$ or
$(\operatorname{Sing}(\mathcal{E}_{n}),\operatorname{Sing}(\mathcal{T}_{n}))$,
then the condition $\alpha U^{1}\subseteq U^{1}\alpha$ in (SA1) cannot be
replaced by $\alpha U\subseteq U\alpha$ (cf. Remark 4.6). Indeed, if
$\alpha\in\mathcal{T}_{n}$ is any map with image $A\subsetneq{\bf n}$, then
$\alpha\cdot\mathrm{id}_{A}=\alpha\not=\mathrm{id}_{B}\cdot\alpha$ for any
$B\subsetneq{\bf n}$. For the same reason, the action of $S$ on
$U^{1}=\mathcal{E}_{n}$ does not restrict to an action of $S$ on
$U=\operatorname{Sing}(\mathcal{E}_{n})$, as
${}^{\alpha}\mathrm{id}_{A}=\mathrm{id}_{\bf n}$ when
$\operatorname{im}(\alpha)=A\subsetneq{\bf n}$.

Similar considerations show that none of the above pairs satisfy the right-
handed version $U^{1}\alpha\subseteq\alpha U^{1}$ of (SA1) when $n\geq 2$.
Indeed, let $\alpha\in\mathcal{T}_{n}$ be an arbitrary constant map, and fix
some $\varnothing\subsetneq A\subsetneq{\bf n}$. Then for any $B\subseteq{\bf
n}$, $\alpha\cdot\mathrm{id}_{B}$ is equal to either $\alpha$ or else the
empty map, and so is never equal to $\mathrm{id}_{A}\cdot\alpha$. This all
shows that $\mathrm{id}_{A}\cdot\alpha\in U\alpha\setminus\alpha U^{1}$, and
so $U^{1}\alpha\not\subseteq\alpha U^{1}$.

We also have strong action pairs $(\mathcal{E}_{n},\mathcal{G}_{n})$ and
$(\operatorname{Sing}(\mathcal{E}_{n}),\mathcal{G}_{n})$ in
$\mathcal{P}\mathcal{T}_{n}$. These lead to the (almost-)factorisable inverse
semigroups

$\mathcal{E}_{n}\cdot\mathcal{G}_{n}=\mathcal{I}_{n}\qquad\text{and}\qquad\operatorname{Sing}(\mathcal{E}_{n})\cdot\mathcal{G}_{n}=\operatorname{Sing}(\mathcal{I}_{n}),$

which are special cases of Example 4.27.

Generalisations of Example 4.30 will be discussed in Chapters 8 and 9.

######  Example 4.32.

Extending the last example, consider again the full and partial transformation
semigroups $\mathcal{T}_{X}$ and $\mathcal{P}\mathcal{T}_{X}$ over an
arbitrary set $X$. Consider also the power set
${{\mathcal{P}}(X)=\\{{A}:{A\subseteq X}\\}}$, as a semilattice under
intersection. Note that $\mathcal{T}_{X}$ acts on ${\mathcal{P}}(X)$ via

${}^{\alpha}\\!A=A\alpha^{-1}\qquad\text{for $A\in{\mathcal{P}}(X)$ and
$\alpha\in\mathcal{T}_{X}$.}$

Now let $S$ be a subsemigroup of $\mathcal{T}_{X}$, let $F$ be a
subsemilattice of ${\mathcal{P}}(X)$ such that $F^{1}=F\cup\\{X\\}$ is closed
under the action of $S$, and set $U=\\{{\mathrm{id}_{A}}:{A\in F}\\}$. Then
$(U,S)$ is a strong action pair in $M=\mathcal{P}\mathcal{T}_{X}$, as again
follows from (4.31). So we have the subsemigroup

$US=\\{{\mathrm{id}_{A}\cdot\alpha}:{\alpha\in S,\ A\in
F}\\}=\\{{\alpha\mathord{\restriction}_{A}}:{\alpha\in S,\ A\in
F}\\}\leq\mathcal{P}\mathcal{T}_{X},$

with composition given by

$\alpha\mathord{\restriction}_{A}\cdot\beta\mathord{\restriction}_{B}=(\alpha\beta)\mathord{\restriction}_{A\cap
B\alpha^{-1}}\qquad\text{for $\alpha,\beta\in S$ and $A,B\in F$.}$

When $S=\mathcal{T}_{X}$ and $F={\mathcal{P}}(X)$, we of course obtain
$US=\mathcal{P}\mathcal{T}_{X}$, but varying $S$ and $F$ gives rise to other
interesting cases.

  1. (i)

Consider the case in which $X=\mathbb{N}=\\{0,1,2,\ldots\\}$. Define
transformations $\gamma_{1},\gamma_{2}\in\mathcal{T}_{\mathbb{N}}$ by

$\gamma_{1}=\left(\begin{smallmatrix}0&1&2&3&\cdots\\\
1&2&3&4&\cdots\end{smallmatrix}\right)\qquad\text{and}\qquad\gamma_{2}=\left(\begin{smallmatrix}0&1&2&3&\cdots\\\
0&2&3&4&\cdots\end{smallmatrix}\right),$

and define the (infinite cyclic) subsemigroups
$S_{1}=\langle\gamma_{1}\rangle$ and $S_{2}=\langle\gamma_{2}\rangle$ of
$\mathcal{P}\mathcal{T}_{\mathbb{N}}$. Let $F$ be the set of all subsets of
$\mathbb{N}$ of the form $x+\mathbb{N}=\\{x,x+1,x+2,\ldots\\}$ with $x\geq 1$.
So $F$ is a subsemilattice of ${\mathcal{P}}(\mathbb{N})$, and
$F\cup\\{\mathbb{N}\\}$ is closed under the actions of both $S_{1}$ and
$S_{2}$ ($F$ is closed under the action of $S_{2}$, but not of $S_{1}$). So
$(U,S_{1})$ and $(U,S_{2})$ are both strong action pairs in
$\mathcal{P}\mathcal{T}_{\mathbb{N}}$, and in fact we have $US_{1}=US_{2}$.
This shows that the ‘$S$ component’ of a (strong) action pair $(U,S)$ is not
uniquely determined by the semigroup $US$ itself. Note also that
$US_{1}=US_{2}$ contains no total maps, and also no idempotents. So in
general, a semigroup $US$ can be disjoint from both $U$ and $S$. Clearly
$US_{1}=US_{2}$ contains no isomorphic copy of $U$ (as $U$ consists entirely
of idempotents), but any element of $US_{1}=US_{2}$ generates an infinite
cyclic subsemigroup, which is isomorphic to both $S_{1}$ and $S_{2}$.

  2. (ii)

Now let $X={\mathbb{Z}}=\\{0,\pm 1,\pm 2,\ldots\\}$, and define

$\gamma_{3}=\left(\begin{smallmatrix}\cdots&0&1&2&3&\cdots\\\
\cdots&1&2&3&4&\cdots\end{smallmatrix}\right)\in\mathcal{T}_{\mathbb{Z}}.$

Let $S_{3}=\langle\gamma_{3}\rangle$, and let $F$ be as in (i), but now
considered as a subsemilattice of ${\mathcal{P}}({\mathbb{Z}})$. Note that
$F\cup\\{{\mathbb{Z}}\\}$ is not closed under the action of $S_{3}$, and in
fact $(U,S_{3})$ is not a weak action pair at all, as even (SA1) fails. For
example, with $\mathbb{P}=\\{1,2,3,\ldots\\}\in F$ and
$\mathbb{N}={}^{\gamma_{3}}\mathbb{P}=\\{0,1,2,\ldots\\}\not\in
F\cup\\{{\mathbb{Z}}\\}$, we have
$\gamma_{3}\cdot\mathrm{id}_{\mathbb{P}}\in\gamma_{3}U^{1}$; on the other
hand,
$\operatorname{dom}(\gamma_{3}\cdot\mathrm{id}_{\mathbb{P}})=\mathbb{N}$, yet
every element $\delta$ of $U^{1}\gamma_{3}$ has
$\operatorname{dom}(\delta)={\mathbb{Z}}$ or else
$\operatorname{dom}(\delta)\subseteq\mathbb{P}$. However, we do have

$A\cap B\alpha^{-1}\in F\qquad\text{for all $A,B\in F$ and $\alpha\in
S_{3}$,}$

and it follows from this and the rule
$\alpha\mathord{\restriction}_{A}\cdot\beta\mathord{\restriction}_{B}=(\alpha\beta)\mathord{\restriction}_{A\cap
B\alpha^{-1}}$ that $US_{3}$ is still a semigroup. In fact, by identifying
$\mathcal{P}\mathcal{T}_{\mathbb{N}}$ as a subset of
$\mathcal{P}\mathcal{T}_{\mathbb{Z}}$ (consisting of all partial maps with
domain and range contained in $\mathbb{N}$), we have ${US_{1}=US_{2}=US_{3}}$,
where $S_{1}$ and $S_{2}$ are as in (i) above. This time, note that $S_{3}$
consists of units of $\mathcal{T}_{\mathbb{Z}}$.

  3. (iii)

Again consider $X=\mathbb{N}$, and let $S=\langle\gamma_{1}\rangle$, with
$\gamma_{1}\in\mathcal{T}_{\mathbb{N}}$ as in (i). This time we let $F$
consist of all finite subsets of $\mathbb{N}$, which is again a subsemilattice
of ${\mathcal{P}}(\mathbb{N})$ closed under the action of $S$. The resulting
strong action pair $(U,S)$ leads to the subsemigroup $US$ of
$\mathcal{P}\mathcal{T}_{\mathbb{N}}$, as usual. This time $US$ contains a
single idempotent, namely the empty map $\varnothing$ (which is of course a
zero of $US$), and we note that $US$ is a _nilsemigroup_ , in the sense that
every element of $US$ has a power equal to $\varnothing$. It follows that $US$
does not contain any subsemigroup isomorphic to $U$ or to $S$.

Among other things, the above examples show that if one is given a semigroup
known to be of the form $US$ for some (strong) action pair $(U,S)$, it is not
necessarily obvious what $U$ and/or $S$ might be.

It is also worth noting that the _reflection monoids_ of Everitt and Fountain
[39, 40] are also special cases of the construction in Example 4.32, or indeed
of Example 4.27.

The next example features a _weak_ action pair $(U,S)$ in partial
transformation monoids, and here we no longer assume that $U$ consists of
partial identities.

######  Example 4.33.

Let $Y$ and $Z$ be disjoint non-empty sets, let $X=Y\cup Z$, and let
$M=\mathcal{P}\mathcal{T}_{X}$. Let $S\leq\mathcal{P}\mathcal{T}_{Y}$ and
$V\leq\mathcal{P}\mathcal{T}_{Z}$ be arbitrary subsemigroups. We may also
regard $S$ as a subsemigroup of $\mathcal{P}\mathcal{T}_{X}$ (as any partial
transformation of $Y$ is of course a partial transformation of $X$), and we
additionally define $U=\\{{\mathrm{id}_{Y}\cup\alpha}:{\alpha\in V}\\}$. So
$U$ is a subsemigroup of $\mathcal{P}\mathcal{T}_{X}$, and clearly $U\cong V$.
It is easy to check that

$\alpha\beta=\beta\alpha=\alpha\qquad\text{for all $\alpha\in S$ and $\beta\in
U^{1}=U\cup\\{\mathrm{id}_{X}\\}$.}$ (4.34)

It immediately follows that condition (SA1) holds for the pair $(U,S)$.
However, condition (SA2) only holds if $|U|=1$, as
$\alpha\beta=\alpha^{\prime}\beta$ for all $\alpha,\alpha^{\prime}\in U$ and
$\beta\in S$. By contrast, the right-handed version of (SA2) holds:
$\alpha\beta=\alpha^{\prime}\beta^{\prime}\ \Rightarrow\ \beta=\beta^{\prime}$
for all $\alpha,\alpha^{\prime}\in U$ and $\beta,\beta^{\prime}\in S$.

It also follows from (4.34) that condition (A1) holds with respect to the
constant action: ${}^{\alpha}\beta=\mathrm{id}_{X}$ for all $\alpha\in S$ and
$\beta\in U^{1}$. Condition (A2) is equivalent to (SA2), so still fails
(unless $|U|=1$). So the weak action pair $(U,S)$ is not an action pair. In
this case we have $US=S(=SU)$.

In Example 4.27 we considered factorisable inverse monoids, which came from
strong action pairs present in any inverse monoid. As a foreshadowing of the
next section, we now consider another action pair present in any inverse
semigroup, though these are generally not strong.

######  Example 4.35.

Let $S$ be an inverse semigroup with semilattice of idempotents $E=E(S)$.
Define the map $S\to E^{1}:s\mapsto s^{+}=ss^{-1}$. It is easy to check that
(SA1) holds for the pair $(E,S)$, and that the unary operation $s\mapsto
s^{+}$ satisfies ($+$1)–($+$4). For example, if $s,t\in S$, then

$st^{+}=stt^{-1}=ss^{-1}stt^{-1}=stt^{-1}s^{-1}s=(st)(st)^{-1}s=(st)^{+}s,$

giving ($+$2). For ($+$4), suppose $es=ft$ for some $e,f\in E$ and $s,t\in S$.
Then certainly $(es)^{+}=(ft)^{+}$. But

$(es)^{+}=(es)(es)^{-1}=ess^{-1}e^{-1}=ess^{-1}e=eess^{-1}=es^{+}\qquad\text{and
similarly}\qquad(ft)^{+}=ft^{+},$

so that $es^{+}=ft^{+}$, as required. It follows from Proposition 4.7 that
$(E,S)$ is an action pair in the monoid $S^{1}$. By Lemma 4.15(ii), this pair
is strong if and only if $ss^{-1}=1$ for all $s\in S$, which occurs precisely
when $S$ is a group.

To understand the pair $(E,S)$ via the original Definition 4.2, we need to
understand the action of $S$ on $E$. For this, let $s\in S$ and $e\in E$. Then
$se=ss^{-1}se=ses^{-1}s=ses^{-1}\cdot s$. It follows from the proof of
Proposition 4.7 (see (4.8)) that ${}^{s}e=ses^{-1}\cdot s^{+}=ses^{-1}\cdot
ss^{-1}=ses^{-1}$. This all shows that the action of $S$ on $E$ is given by
conjugation, as in Example 3.3(ii).

As usual, the pair $(E,S)$ leads to the subsemigroup $ES$ of $S$, and of
course we have $ES=S$. Although this is not an interesting/new subsemigroup,
it leads to a number of other interesting subsemigroups, as $(E,T)$ is an
action pair for any $T\leq S$, by Lemma 4.18(i). We will explore this in more
detail in the next section, where we extend our scope to left restriction
monoids.

####  4.3 Left restriction monoids

As just noted, another family of natural examples comes from the class of left
restriction monoids, as defined in Section 2.3. For the duration of this
section we fix some such monoid $M$ with identity $1$. So $M$ has a (basic)
unary operation $s\mapsto s^{+}$ satisfying

  1.   2. (L1)

$x^{+}x=x$,

  3. (L2)

$x^{+}y^{+}=y^{+}x^{+}$,

  4. (L3)

$(x^{+}y)^{+}=x^{+}y^{+}$,

  5. (L4)

$xy^{+}=(xy)^{+}x$,

  6. (L5)

$x^{+}x^{+}=x^{+}$,

  7. (L6)

$(x^{+})^{+}=x^{+}$.

(Recall that the defining identities are (L1)–(L4), while (L5) and (L6) are
consequences.) It follows quickly from (L1) that $1^{+}=1$. As before, in the
calculations to follow we write $=_{1}$ to indicate an application of (L1),
and similarly for $=_{2}$, and so on.

The following subsets of $M$ will play an important role in all that follows:

$\displaystyle E=E(M)$ $\displaystyle=\\{{s\in M}:{s=s^{2}}\\},$
$\displaystyle L=L(M)$ $\displaystyle=\\{{s\in M}:{1\in Ms}\\},$
$\displaystyle P=P(M)$ $\displaystyle=\\{{s^{+}}:{s\in M}\\},$ $\displaystyle
R=R(M)$ $\displaystyle=\\{{s\in M}:{1\in sM}\\},$ $\displaystyle T=T(M)$
$\displaystyle=\\{{s\in M}:{s^{+}=1}\\},$ $\displaystyle G=G(M)$
$\displaystyle=L\cap R.$

The elements of $P$ are called _projections_. By (L2), (L3) and (L5), $P$ is a
(monoid) semilattice. By (L6), $P$ is fixed pointwise by the + operation,
meaning that $u=u^{+}$ for all $u\in P$. If $M$ is inverse (with
$s^{+}=ss^{-1}$), then $P=E$ is precisely the semilattice of (all) idempotents
of $M$. However, this need not be the case for an arbitrary left restriction
semigroup, as for example with $\mathcal{P}\mathcal{T}_{X}$ (with
$\alpha^{+}=\mathrm{id}_{\operatorname{dom}(\alpha)}$ and
$P=\\{{\mathrm{id}_{A}}:{A\subseteq X}\\}$), although we always of course have
${P\subseteq E}$. In fact, $E$ need not be a subsemigroup at all, let alone a
semilattice; again consider $\mathcal{P}\mathcal{T}_{X}$. The other subsets
defined above _are_ submonoids, however. The submonoids $L$ and $R$ consist of
all left and right units of $M$, respectively, and $G$ is the group of (two-
sided) units. We have $L=R=G$ if $M$ is finite (see for example [34, Lemma
2.3]), but this need not hold for infinite $M$. We will discuss $T$ shortly,
but our first goal is to show that $(P,M)$ is an action pair; see Proposition
4.37. For this, and for later use, it is convenient to prove a simple fact,
which we have already seen in Example 4.35 for the special case of inverse
semigroups.

######  Lemma 4.36.

If $S$ is a left restriction semigroup with semilattice of projections
$P=P(S)$, then $(us)^{+}=us^{+}$ for any $u\in P^{1}$ and $s\in S$.

###### Proof.

This is obvious for $u=1$; otherwise,
$(us)^{+}=_{6}(u^{+}s)^{+}=_{3}u^{+}s^{+}=_{6}us^{+}$. ∎

######  Proposition 4.37.

Let $M$ be a left restriction monoid, and write $P=P(M)$. Then $(P,M)$ is an
action pair.

###### Proof.

We use Proposition 4.7. For (SA1), let $s\in M$. Then for any $u\in P$ we have

$su=_{6}su^{+}=_{4}(su)^{+}s,$ (4.38)

which gives $sP\subseteq Ps$.

It remains to check that the map $M\to P:s\mapsto s^{+}$ (i.e., the +
operation itself) satisfies conditions ($+$1)–($+$4). Now, conditions ($+$1)
and ($+$2) are simply (L1) and (L4), respectively. For ($+$3), let $s,t\in M$.
Then

$(st)^{+}=_{1}((s^{+}s)t)^{+}=(s^{+}(st))^{+}=_{3}s^{+}(st)^{+}=_{2}(st)^{+}s^{+}.$

Finally, for ($+$4), suppose $u,v\in P$ and $s,t\in M$ are such that $us=vt$.
Then certainly $(us)^{+}=(vt)^{+}$, and it follows from Lemma 4.36 that
$us^{+}=vt^{+}$. ∎

######  Remark 4.39.

If $S$ is a left restriction _semigroup_ , then $(P,S)$ is an action pair in
the monoid $S^{1}$, where again we write $P=P(S)=\\{{s^{+}}:{s\in S}\\}$.
Indeed, this follows from the same proof as above.

Alternatively, when $S$ is a left restriction semigroup, $S^{1}$ becomes a
left restriction _monoid_ in a unique way by putting $1^{+}=1$. Proposition
4.37 says that $(P^{1},S^{1})$ is an action pair in $S^{1}$, and we then apply
(both parts of) Lemma 4.18.

######  Remark 4.40.

Given any action pair $(U,S)$ we have the associated semigroup $US$. In the
case of the pair $(P,M)$, this subsemigroup is precisely $PM=M$ itself. Thus,
the pair $(P,M)$ does not lead to an ‘interesting’ subsemigroup of $M$.
Similarly, the results of Chapter 6 will be of no use in describing
presentations for $M$ by means of the pair $(P,M)$, since they assume we
already know a presentation for $M$ (and $P$). Nevertheless, knowing that
$(P,M)$ is an action pair is useful in describing many more such pairs below
(using Lemma 4.18). The pair $(P,M)$ will also be extremely important in
Chapter 5.

The proof of Proposition 4.37 utilised the alternative formulation of action
pairs from Proposition 4.7, but it is convenient to understand the associated
action of $M$ on $P(=P^{1})$. To do so, fix some $u\in P$ and $s\in M$. Then
$su=(su)^{+}s$, by (4.38). By the proof of Proposition 4.7 (see (4.8)), we
therefore have ${}^{s}u=(su)^{+}s^{+}=(su)^{+}$, where we used ($+$3) in the
last step. To summarise, the action of $M$ on $P$ is given by

${}^{s}u=(su)^{+}\qquad\text{for $s\in M$ and $u\in P$.}$ (4.41)

If $M$ is inverse (with $s^{+}=ss^{-1}$), then this action has the simple form
${{}^{s}u=su(su)^{-1}=sus^{-1}}$; cf. Examples 3.3(ii) and 4.35. It is worth
noting that the action (4.41) goes back to work of Fountain on the narrower
class of _adequate semigroups_ [44]; see also [53] where it is used in the
broader class of _Ehresmann semigroups_.

Next we wish to prove a result concerning action pairs of the form $(Q,S)$,
where $Q\leq P$ and $S\leq M$. The set $T$ introduced above will play an
important role in this, specifically in determining which of these pairs are
strong.

######  Lemma 4.42.

Let $M$ be a left restriction monoid, and write $T=T(M)$ and $R=R(M)$. Then

  1. (i)

$T$ is a submonoid of $M$,

  2. (ii)

$R\subseteq T$, with equality if $M$ is inverse,

  3. (iii)

$M\setminus T=\\{{s\in M}:{s^{+}\not=1}\\}$ is a right ideal (and hence a
subsemigroup) of $M$.

###### Proof.

(i). Define the relation $\sigma=\big{\\{}{(s,t)\in M\times
M}:{s^{+}=t^{+}}\big{\\}}$. Clearly $\sigma$ is an equivalence (it is the
kernel of the map $s\mapsto s^{+}$), and it follows quickly from Lemma 3.1(ii)
that $\sigma$ is a left congruence. Lemma 2.1 then tells us that $T=\\{{s\in
M}:{s\mathrel{\sigma}1}\\}$ is a submonoid.

(ii). Lemma 4.23(i) gives $R\subseteq T$. If $M$ is inverse, and if $s\in T$,
then $1=s^{+}=ss^{-1}$, so $s\in R$.

(iii). Let $s\in M\setminus T$ and $t\in M$. So $s^{+}\not=1$, and we must
show that $(st)^{+}\not=1$. But if $(st)^{+}=1$, then $s^{+}=s^{+}\cdot
1=s^{+}(st)^{+}=_{3}(s^{+}\cdot st)^{+}=_{1}(st)^{+}=1$, a contradiction. ∎

######  Remark 4.43.

The equality $T=R$ need not hold if $M$ is not inverse. For example, in
$\mathcal{P}\mathcal{T}_{X}$ we have

$T(\mathcal{P}\mathcal{T}_{X})=\mathcal{T}_{X}\qquad\text{and}\qquad
R(\mathcal{P}\mathcal{T}_{X})=\\{{\alpha\in\mathcal{T}_{X}}:{\alpha\text{ is
injective}}\\}.$

The same example also shows that the right ideal $M\setminus T$ (which is
$\mathcal{P}\mathcal{T}_{X}\setminus\mathcal{T}_{X}$ in this case) is not
always a left ideal.

The next result follows immediately from Proposition 4.37, and Lemmas 4.15 and
4.18.

######  Proposition 4.44.

Let $M$ be a left restriction monoid, and write $P=P(M)$ and $T=T(M)$. Then

  1. (i)

$(Q,S)$ is an action pair for any $Q\leq P$ and $S\leq M$ such that $Q^{1}$ is
closed under the action of $S$ given in (4.41),

  2. (ii)

such a pair $(Q,S)$ is strong if and only if $S\subseteq T$. ∎

######  Remark 4.45.

In particular, $(P,S)$ and $(P^{\flat},S)$ are both action pairs for any
$S\leq M$, where we write $P^{\flat}=P\setminus\\{1\\}\leq P$. Such pairs are
strong if and only if $S\subseteq T$. Thus, the following action pairs are all
strong:

$(P,T),\qquad(P,R),\qquad(P,G),\qquad(P^{\flat},T),\qquad(P^{\flat},R)\qquad\text{and}\qquad(P^{\flat},G).$

As usual, these lead to the subsemigroups $PT$, $PR$ and so on. We first note
that $PG(=GP)$ is in fact a factorisable inverse monoid. Indeed, this follows
from the easily checked facts that:

  * •

$E(PG)=P$,

  * •

$g^{-1}u={}^{g^{-1}}ug^{-1}\in PG$ is an inverse of $ug$ for any $u\in P$ and
$g\in G$.

The inverse semigroup $P^{\flat}G(=GP^{\flat})=PG\setminus G$ is _almost-
factorisable_. The submonoid $PT$ and the subsemigroup $P^{\flat}T$ are _left-
factorisable_ and _almost-left-factorisable_ in the terminology of [114]. Note
that $PG=GP$ has a left-right duality, where the corresponding right action is
given by ${e^{g}=g^{-1}eg}$, making it a restriction monoid; similar comments
apply to the semigroup ${P^{\flat}G=GP^{\flat}}$.

(As we will see in Chapter 6, the very fact that $P^{\flat}=P\setminus\\{1\\}$
is a subsemigroup of $P$ is important in certain applications; see for example
Assumption 6.32 and Remark 6.33.)

The action pair $(P,L)$ is strong precisely when $L\subseteq T$. When $M$ is
inverse, this is equivalent to $L\subseteq R$ (cf. Lemma 4.42(ii)), and hence
to $L=R=G$.

Finally, the action pair $(P,M)$ itself is strong if and only if $M=T$,
meaning that the left restriction structure of $M$ is trivial ($s^{+}=1$ for
all $s\in M$). When $M$ is inverse, this is equivalent to $M$ being a group
($ss^{-1}=1$ for all $s\in M$), as we observed in Example 4.35. For the action
pair $(P^{\flat},M)$, one can check that

$P^{\flat}M=M\setminus T=\\{{s\in M}:{s^{+}\not=1}\\}.$

We showed in Lemma 4.42(iii) that this subsemigroup is in fact a right ideal
of $M$.

####  4.4 The first structure theorem

Having now discussed an extensive collection of examples, including the class
of left restriction semigroups, we return to the general theory. The main goal
of this section is to describe the structure of a semigroup $US$ arising from
an action pair $(U,S)$, showing how such a semigroup can be built from a
semidirect product $U\rtimes S$ and a special congruence. We do this in
Theorem 4.53 below, but we begin with a simple result that holds more
generally for a _weak_ action pair $(U,S)$. For such a pair, the action from
(A1) allows for the formation of the (external) semidirect product $U\rtimes
S$, as in Definition 3.4.

######  Proposition 4.46.

If $(U,S)$ is a weak action pair, then the map $\pi:U\rtimes S\to US$ given by
$(u,s)\pi=us$ is a surmorphism with kernel

$\theta=\big{\\{}{((u,s),(v,t))}:{u,v\in U,\ s,t\in S,\ us=vt}\big{\\}}.$

Consequently, $\theta$ is a congruence on $U\rtimes S$, and $US\cong(U\rtimes
S)/\theta$.

###### Proof.

If $u,v\in U$ and $s,t\in S$, then

$(u,s)\pi\cdot(v,t)\pi=us\cdot vt=u\cdot{}^{s}v\cdot
st=(u\cdot{}^{s}v,st)\pi=((u,s)\cdot(v,t))\pi,$

so that $\pi$ is a morphism. Surjectivity is clear, as is the description of
the kernel. The final assertion follows from the First Isomorphism Theorem. ∎

######  Remark 4.47.

Recall from Proposition 3.32 that

$M=\big{\\{}{(u,s)\in U\rtimes S}:{u=us^{+}}\big{\\}}$

is a subsemigroup of $U\rtimes S$, and that the map

$F:U\rtimes S\to M:(u,s)\mapsto(us^{+},s)$

is a retraction. (Note that to apply Proposition 3.32 we need to keep Remark
4.4 in mind, which says that if $S$ is a monoid then it acts monoidally on
$U^{1}$.) Because of the identity $s=s^{+}s$ (cf. Remark 4.4), the following
diagram commutes:

${M}$${U\rtimes
S}$${US}$$\scriptstyle{\pi\mathord{\restriction}_{M}}$$\scriptstyle{F}$$\scriptstyle{\pi}$
(4.48)

In particular, $\pi\mathord{\restriction}_{M}:M\to US$ is still a surmorphism,
so we also have

$US\cong
M/\vartheta\qquad\text{where}\qquad\vartheta=\ker(\pi\mathord{\restriction}_{M})=\theta\mathord{\restriction}_{M}.$

Now consider an action pair $(U,S)$. Of course Proposition 4.46 holds for this
pair, and we keep the meaning of the surmorphism $\pi:U\rtimes S\to US$, and
the congruence $\theta=\ker(\pi)$. For any $u\in U^{1}$, we define the
relation

$\theta_{u}=\big{\\{}{(s,t)\in S\times S}:{us=ut}\big{\\}}.$ (4.49)

We also continue to write $s^{+}={}^{s}1$ for $s\in S$.

######  Lemma 4.50.

Let $(U,S)$ be an action pair, and let $u,v\in U$ and $s,t\in S$.

  1. (i)

We have $(u,s)\mathrel{\theta}(v,t)\ \Leftrightarrow\ us^{+}=vt^{+}$ and
$(s,t)\in\theta_{us^{+}}$.

  2. (ii)

If $(U,S)$ is strong, then $(u,s)\mathrel{\theta}(v,t)\ \Leftrightarrow\ u=v$
and $(s,t)\in\theta_{u}$.

###### Proof.

The second part follows from the first, given Lemma 4.15(ii). For the first
part, suppose $(u,s)\mathrel{\theta}(v,t)$: i.e., $us=vt$. Then
$us^{+}=vt^{+}$ follows from ($+$4). Together with ($+$1), it follows that
$us^{+}s=us=vt=vt^{+}t=us^{+}t$, so $(s,t)\in\theta_{us^{+}}$. The converse is
similar. ∎

Proposition 4.46 showed that, given an action pair $(U,S)$, the semigroup $US$
can be viewed abstractly as a homomorphic image of a semidirect product
$U\rtimes S$. Our next goal is to establish a converse of this fact; namely,
we wish to describe _all_ homomorphic images of semidirect products that
correspond to action pairs in this way. To do so, we first establish some
special properties of the above congruence $\theta$, and the associated
relations $\theta_{u}$. Here and elsewhere, if $\sigma$ is a relation on a
semigroup $T$, and if $x\in T$, we write
$x\cdot\sigma=\big{\\{}{(xs,xt)}:{(s,t)\in\sigma}\big{\\}}$. The
diagonal/equality relation on any set $X$ is denoted
$\Delta_{X}=\big{\\{}{(x,x)}:{x\in X}\big{\\}}$.

######  Lemma 4.51.

For an action pair $(U,S)$, and with the above notation, the following all
hold:

  1. (i)

For all $u\in U$ and $s\in S$, we have $(u,s)\mathrel{\theta}(us^{+},s)$.

  2. (ii)

For all $s,t\in S$, we have $(s^{+},s)\mathrel{\theta}(t^{+},t)\ \Rightarrow\
s=t$.

  3. (iii)

For all $u,v\in U$ and $s,t\in S$, we have $(u,s)\mathrel{\theta}(v,t)\
\Rightarrow\ us^{+}=vt^{+}$.

  4. (iv)

We have $\theta_{1}=\Delta_{S}$.

  5. (v)

For all $u\in U$, the relation $\theta_{u}$ is a right congruence on $S$.

  6. (vi)

For all $u,w\in U$, we have $\theta_{u}\subseteq\theta_{wu}$.

  7. (vii)

For all $u\in U$ and $x\in S$, we have
$x\cdot\theta_{u}\subseteq\theta_{{}^{x}u}$.

  8. (viii)

For all $u,w\in U$ and $(s,t)\in\theta_{u}$, we have
$u\cdot{}^{s}w=u\cdot{}^{t}w$ and $(s,t)\in\theta_{u\cdot{}^{s}w}$.

###### Proof.

These are easily checked, using the definition of $\theta$ and $\theta_{u}$.
We additionally use ($+$1) for (i) and (ii), and Lemma 4.50 for (iii). We just
give the details for (vii) and (viii).

(vii). We have

$(s,t)\in\theta_{u}\ \Rightarrow\ us=ut\ \Rightarrow\ xus=xut\ \Rightarrow\
{}^{x}u\cdot xs={}^{x}u\cdot xt\ \Rightarrow\ (xs,xt)\in\theta_{{}^{x}u}.$

(viii). We begin by noting that

$(s,t)\in\theta_{u}\ \Rightarrow\ us=ut\ \Rightarrow\
u\cdot{}^{s}w=u\cdot{}^{t}w,$

by Lemma 4.25. Combining this with (A1) and $us=ut$, it also follows that

$u\cdot{}^{s}w\cdot s=usw=utw=u\cdot{}^{t}w\cdot t=u\cdot{}^{s}w\cdot t,$

so that $(s,t)\in\theta_{u\cdot{}^{s}w}$. ∎

We now use these properties of $\theta$ (and the $\theta_{u}$) to construct a
set of axioms to abstractly characterise the quotients of semidirect products
giving rise to action pairs. To keep the ‘concrete’ pairs $(U,S)$ separate
from the abstract, we will use the notation $V$ and $T$ for the latter.

######  Definition 4.52.

Suppose $V$ and $T$ are semigroups such that $T$ has a left action on $V^{1}$
(the monoid completion of $V$) by semigroup morphisms, written
$(s,u)\mapsto{}^{s}u$, for $s\in T$ and $u\in V^{1}$. For $s\in T$, we write
$s^{+}={}^{s}1\in V^{1}$. Suppose $\sigma$ is a congruence on the semidirect
product $V\rtimes T$, and for each $u\in V$ define the relation

$\sigma_{u}=\big{\\{}{(s,t)\in T\times
T}:{(u,s)\mathrel{\sigma}(u,t)}\big{\\}}.$

If $V$ is not a monoid (so $1\not\in V$), we additionally define
$\sigma_{1}=\Delta_{T}$. We say $\sigma$ is _special_ if the following all
hold:

  1. (S1)

For all $u\in V$ and $s\in T$, we have $(u,s)\mathrel{\sigma}(us^{+},s)$.

  2. (S2)

For all $s,t\in T$, we have $(s^{+},s)\mathrel{\sigma}(t^{+},t)\ \Rightarrow\
s=t$.

  3. (S3)

For all $u,v\in V$ and $s,t\in T$, we have $(u,s)\mathrel{\sigma}(v,t)\
\Rightarrow\ us^{+}=vt^{+}$.

  4. (S4)

We have $\sigma_{1}=\Delta_{T}$.

  5. (S5)

For all $u\in V$, $\sigma_{u}$ is a right congruence on $T$.

  6. (S6)

For all $u,w\in V$, we have $\sigma_{u}\subseteq\sigma_{wu}$.

  7. (S7)

For all $u\in V$ and $x\in T$, we have
$x\cdot\sigma_{u}\subseteq\sigma_{{}^{x}u}$.

  8. (S8)

For all $u,w\in V$ and $(s,t)\in\sigma_{u}$, we have
$u\cdot{}^{s}w=u\cdot{}^{t}w$ and $(s,t)\in\sigma_{u\cdot{}^{s}w}$.

If $T$ acts by monoid morphisms (i.e., if $s^{+}=1$ for all $s\in T$), then
several simplifications arise in (S1)–(S8). For example, (S1) is trivially
true, and (S3) says $(u,s)\mathrel{\sigma}(v,t)\ \Rightarrow\ u=v$.

Here is the main structural result of this chapter.

######  Theorem 4.53.

Suppose $V$ and $T$ are semigroups such that $T$ has a left action on $V^{1}$
by semigroup morphisms. Suppose also that $\sigma$ is a special congruence on
the semidirect product $V\rtimes T$. Then there is an action pair $(U,S)$ in
some monoid $M$ such that

$U\cong V,\qquad S\cong T\qquad\text{and}\qquad US\cong(V\rtimes T)/\sigma.$

If $T$ acts on $V^{1}$ by monoid morphisms, then $(U,S)$ is strong.

Conversely, given any action pair $(U,S)$, we have $US\cong(U\rtimes
S)/\theta$ for some special congruence $\theta$ on the semidirect product
$U\rtimes S$.

###### Proof.

The last assertion follows from Proposition 4.46 and Lemma 4.51. The remainder
of the proof concerns the first two assertions, for which we fix $V$, $T$ and
$\sigma$ with the specified properties. Without loss of generality, we may
assume that $T$ and $V^{1}$ are disjoint. With a slight abuse of notation, we
let $T^{1}=T\cup\\{1\\}$ be the monoid obtained from $T$ by adjoining $1$ (the
identity of $V^{1}$) as an identity, _whether or not $T$ was already a
monoid_. The reason for doing this is that $T$ might be a monoid acting non-
monoidally on $V$, and we will soon construct an action of $T^{1}$ on $V^{1}$
that is required to be monoidal. (But we stress that $V=V^{1}$ if $V$ is a
monoid.)

We denote the action of $T$ on $V^{1}$ by $(s,u)\mapsto{}^{s}u$ ($s\in T$,
$u\in V^{1}$), and we extend this to a monoidal action of $T^{1}$ on $V^{1}$
in the usual way, by further defining ${}^{1}u=u$ for all $u\in V^{1}$. We
then have the semidirect product $V^{1}\rtimes T^{1}$, which contains
$V\rtimes T$ as a proper subsemigroup. (Note that $V^{1}\rtimes T^{1}$ is not
necessarily a monoid, although $(1,1)$ is always a left identity, as the
action of $T^{1}$ on $V^{1}$ is monoidal; cf. Lemma 3.11.) Throughout the
proof, we write $s^{+}={}^{s}1$ for $s\in T^{1}$; we will make frequent use of
the properties of these elements listed in Lemma 3.1, often without explicit
reference. Since the action is monoidal, we have $1^{+}=1$.

The main effort in the proof goes into establishing the following:

######  Lemma 4.54.

There is a congruence $\Sigma$ on $V^{1}\rtimes T^{1}$ such that the following
all hold:

  1. ($\Sigma$1)

$\sigma=\Sigma\mathord{\restriction}_{V\rtimes T}$,

  2. ($\Sigma$2)

For all $u,v\in V^{1}$, we have $(u,1)\mathrel{\Sigma}(v,1)\ \Leftrightarrow\
u=v$.

  3. ($\Sigma$3)

For all $u\in V^{1}$ and $s\in T^{1}$, we have
$(u,s)\mathrel{\Sigma}(us^{+},s)$.

  4. ($\Sigma$4)

For all $u,v\in V^{1}$ and $s,t\in T^{1}$, we have
$(u,s)\mathrel{\Sigma}(v,t)\ \Rightarrow\ us^{+}=vt^{+}$.

###### Proof.

We define $\Sigma$ by specifying its equivalence classes. Each $\sigma$-class
in $V\rtimes T$ will determine a unique $\Sigma$-class, so consider some such
$\sigma$-class $K$. By (S2), $K$ contains at most one element of the form
$(s^{+},s)$ with $s\in T$.

  1. (K1)

If $K$ does contain such an element $(s^{+},s)$, then
$K\cup\big{\\{}(1,s)\big{\\}}$ is a $\Sigma$-class. (Note that this is just
$K$ itself if $s^{+}=1\in V$. If $s^{+}=1\not\in V$, then no $\sigma$-class
contains $(s^{+},s)$.)

  2. (K2)

Otherwise, $K$ is a $\Sigma$-class.

All elements of $V^{1}\rtimes T^{1}$ not yet assigned to $\Sigma$-classes are
now assigned to singleton $\Sigma$-classes. These elements are precisely:

  * •

$(u,1)$, for each $u\in V^{1}$, and

  * •

$(1,s)$, for each $s\in T$ for which $s^{+}=1\not\in V$ (if any). (Note that
such an element is still of the form $(1,s)=(s^{+},s)$.)

It is clear from the construction that $\Sigma$ is an equivalence on
$V^{1}\rtimes T^{1}$, and that ($\Sigma$1) and ($\Sigma$2) both hold. So it
remains to show that $\Sigma$ is compatible, and establish ($\Sigma$3) and
($\Sigma$4). We begin with these two conditions.

($\Sigma$3). This follows from (S1) when $u\in V$ and $s\in T$, and is trivial
when $s=1$. When $u=1$, ($\Sigma$3) says $(1,s)\mathrel{\Sigma}(s^{+},s)$;
this is clear if $s^{+}=1$, and is true by definition of $\Sigma$ otherwise;
cf. (K1).

($\Sigma$4). Suppose $(u,s)\mathrel{\Sigma}(v,t)$, where $u,v\in V^{1}$ and
$s,t\in T^{1}$. We must show that $us^{+}=vt^{+}$. This is obvious if
$(u,s)=(v,t)$, or follows from (S3) if $(u,s)\mathrel{\sigma}(v,t)$, so we
assume otherwise. By construction (and by symmetry), it follows that
$(u,s)=(1,s)$ and $(v,t)$ belongs to the $\sigma$-class of $(s^{+},s)$, with
$s^{+}\in V$; cf. (K1). But then $(v,t)\mathrel{\sigma}(s^{+},s)$, and it
follows from (S3) and Lemma 3.1(i) that

$vt^{+}=s^{+}s^{+}=s^{+}=1\cdot s^{+}=us^{+},$

as required.

As noted above, it remains (for the proof of the lemma) to establish the
compatibility of $\Sigma$. To do so, let $({\bf a},{\bf b})\in\Sigma$ and
${\bf c}\in V^{1}\rtimes T^{1}$. We must show that

${\bf c}\cdot({\bf a},{\bf b})=({\bf c}{\bf a},{\bf c}{\bf
b})\in\Sigma\qquad\text{and}\qquad({\bf a},{\bf b})\cdot{\bf c}=({\bf a}{\bf
c},{\bf b}{\bf c})\in\Sigma.$ (4.55)

Clearly (4.55) holds if ${\bf a}={\bf b}$, so we now assume that ${\bf
a}\not={\bf b}$. Since ${\bf a}\mathrel{\Sigma}{\bf b}$, it follows that ${\bf
a}$ and ${\bf b}$ belong to a $\Sigma$-class of type (K1) or (K2). By the form
of these classes, we may write

${\bf a}=(u,s),\ \ \ {\bf b}=(v,t)\quad\text{and}\quad{\bf
c}=(w,x),\qquad\text{where $u,v,w\in V^{1}$, $s,t\in T$ and $x\in T^{1}$,}$

and we have

${\bf c}\cdot({\bf a},{\bf
b})=((w\cdot{}^{x}u,xs),(w\cdot{}^{x}v,xt))\qquad\text{and}\qquad({\bf a},{\bf
b})\cdot{\bf c}=((u\cdot{}^{s}w,sx),(v\cdot{}^{t}w,tx)).$ (4.56)

Case 1. Suppose first that $({\bf a},{\bf b})\in\sigma$; in particular, we
have $u,v\in V$. If ${\bf c}\in V\rtimes T$, then (4.55) follows from the fact
that $\sigma$ is compatible and $\sigma\subseteq\Sigma$. So we are left to
consider the cases in which $w=1$ and/or $x=1$. We will soon consider these
separately, but we first note that (S3) gives $us^{+}=vt^{+}$. Combining this
with (S1), we have

$(us^{+},s)\mathrel{\sigma}(u,s)\mathrel{\sigma}(v,t)\mathrel{\sigma}(vt^{+},t)=(us^{+},t)\
\ \Rightarrow\ \ (s,t)\in\sigma_{us^{+}}.$ (4.57)

Case 1.1. Suppose $w=1$. Examining (4.56), we must show that

  1.   2. (a)

$({}^{x}u,xs)\mathrel{\Sigma}({}^{x}v,xt)$, and

  3. (b)

$(us^{+},sx)\mathrel{\Sigma}(vt^{+},tx)$.

For (b), we note that (4.57) and (S5) give $(sx,tx)\in\sigma_{us^{+}}$, and so

$(us^{+},sx)\mathrel{\sigma}(us^{+},tx)=(vt^{+},tx).$

For (a), we note that (4.57) and (S7) give
$(xs,xt)\in\sigma_{{}^{x}(us^{+})}$. There is now a subtle point concerning
the possibility that ${}^{x}(us^{+})$ might equal $1$.

  * •

If ${}^{x}(us^{+})=1$, then it follows from (S4) that $xs=xt$, and so
$({}^{x}(us^{+}),xs)=({}^{x}(us^{+}),xt)$.

  * •

If ${}^{x}(us^{+})\not=1$, then ${}^{x}(us^{+})\in V$ and
$({}^{x}(us^{+}),xs)\mathrel{\sigma}({}^{x}(us^{+}),xt)$, by definition of
$\sigma_{{}^{x}(us^{+})}$, since $(xs,xt)\in\sigma_{{}^{x}(us^{+})}$.

Either way, we have $({}^{x}(us^{+}),xs)\mathrel{\Sigma}({}^{x}(us^{+}),xt)$.
Now,

${}^{x}(us^{+})={}^{x}u\cdot{}^{x}(s^{+})={}^{x}u(xs)^{+}\qquad\text{and
similarly}\qquad{}^{x}(us^{+})={}^{x}(vt^{+})={}^{x}v(xt)^{+}.$

Combining all the above with ($\Sigma$3), we conclude that

$({}^{x}u,xs)\mathrel{\Sigma}({}^{x}u(xs)^{+},xs)=({}^{x}(us^{+}),xs)\mathrel{\Sigma}({}^{x}(us^{+}),xt)=({}^{x}v(xt)^{+},xt)\mathrel{\Sigma}({}^{x}v,xt),$

completing the proof of (a).

Case 1.2. Now suppose $x=1$. Since we have already dealt with the $w=1$ case,
we will also assume here that $w\not=1$. Examining (4.56) again, this time we
must show that

  1.   2. (c)

$(wu,s)\mathrel{\Sigma}(wv,t)$, and

  3. (d)

$(u\cdot{}^{s}w,s)\mathrel{\Sigma}(v\cdot{}^{t}w,t)$.

For (c), we note that (4.57) and (S6) give $(s,t)\in\sigma_{wus^{+}}$. Since
$wus^{+}\in V$ (as $u\in V$), it follows that
$(wus^{+},s)\mathrel{\sigma}(wus^{+},t)$. Combining this with (S1) and
$us^{+}=vt^{+}$, we have

$(wu,s)\mathrel{\sigma}(wus^{+},s)\mathrel{\sigma}(wus^{+},t)=(wvt^{+},t)\mathrel{\sigma}(wv,t).$

For (d), and keeping $w\not=1$ in mind, we note that (4.57) and (S8) give

$\displaystyle us^{+}\cdot{}^{s}w=us^{+}\cdot{}^{t}w$
$\displaystyle\qquad\text{and}\qquad(s,t)\in\sigma_{us^{+}\cdot{}^{s}w}.$ By
Lemma 3.1(iv) we have $us^{+}\cdot{}^{s}w=u\cdot{}^{s}w$ and
$us^{+}\cdot{}^{t}w=vt^{+}\cdot{}^{t}w=v\cdot{}^{t}w$. So in fact,
$\displaystyle u\cdot{}^{s}w=v\cdot{}^{t}w$
$\displaystyle\qquad\text{and}\qquad(s,t)\in\sigma_{u\cdot{}^{s}w},$

which gives
$(u\cdot{}^{s}w,s)\mathrel{\sigma}(u\cdot{}^{s}w,t)=(v\cdot{}^{t}w,t)$,
completing the proof of (d).

Case 2. We now assume that $({\bf a},{\bf b})\notin\sigma$ (but we still of
course have $({\bf a},{\bf b})\in\Sigma$). By definition of $\Sigma$, and
swapping ${\bf a}$ and ${\bf b}$ if necessary, we may assume that ${\bf
a}=(1,s)$, and that ${\bf b}=(v,t)$ belongs to the $\sigma$-class of ${\bf
d}=(s^{+},s)$, with $s^{+}\in V$. Since $({\bf b},{\bf d})\in\sigma$, it
follows from Case 2 that ${\bf c}\cdot({\bf b},{\bf d})\in\Sigma$ and $({\bf
b},{\bf d})\cdot{\bf c}\in\Sigma$. So by transitivity, we can complete the
proof of (4.55), and hence of the lemma, by showing that

$\displaystyle{\bf c}\cdot({\bf a},{\bf d})\in\Sigma$
$\displaystyle\qquad\text{and}\qquad({\bf a},{\bf d})\cdot{\bf c}\in\Sigma.$
For this, we first note that $\displaystyle{\bf c}\cdot({\bf a},{\bf
d})=((wx^{+},xs),(w\cdot{}^{x}(s^{+}),xs)$
$\displaystyle\qquad\text{and}\qquad({\bf a},{\bf d})\cdot{\bf
c}=(({}^{s}w,sx),(s^{+}\cdot{}^{s}w,sx)).$

By ($\Sigma$3), we have

${\bf c}{\bf
a}=(wx^{+},xs)\mathrel{\Sigma}(wx^{+}(xs)^{+},xs)=(w(xs)^{+},xs)=(w\cdot{}^{x}(s^{+}),xs)={\bf
c}{\bf d}.$

On the other hand, we have ${\bf a}{\bf c}={\bf d}{\bf c}$, as
${}^{s}w=s^{+}\cdot{}^{s}w$. We have now completed the proof of Lemma 4.54. ∎

Returning now to the proof of the theorem, we fix a congruence $\Sigma$ on
$V^{1}\rtimes T^{1}$ as in Lemma 4.54, and we define $M=(V^{1}\rtimes
T^{1})/\Sigma$. We denote the $\Sigma$-class of $(u,s)\in V^{1}\rtimes T^{1}$
by $[u,s]$, so that

$M=\big{\\{}{[u,s]}:{u\in V^{1},\ s\in T^{1}}\big{\\}}\qquad\text{with
operation}\qquad[u,s]\cdot[v,t]=[u\cdot{}^{s}v,st].$

We already observed that $(1,1)$ is a left identity for $V^{1}\rtimes T^{1}$,
so of course $[1,1]$ is a left identity for $M$. It follows from ($\Sigma$3)
that

$[u,s]\cdot[1,1]=[us^{+},s]=[u,s]\qquad\text{for all $u\in V^{1}$ and $s\in
T^{1}$,}$

so that $[1,1]$ is a right identity for $M$. Consequently, $M$ is a monoid
with identity $[1,1]$.

Now we define

$U=\big{\\{}{[u,1]}:{u\in V}\big{\\}}\qquad\text{and}\qquad
S=\big{\\{}{[s^{+},s]}:{s\in T}\big{\\}}.$

Clearly $U\leq M$, and it follows from ($\Sigma$2) that $U\cong V$. We have
$S\leq M$ because of the rule $[s^{+},s]\cdot[t^{+},t]=[(st)^{+},st]$, which
itself follows from $s^{+}\cdot{}^{s}(t^{+})={}^{s}(t^{+})=(st)^{+}$. We have
$S\cong T$ because of (S2).

Next we verify that $(U,S)$ is an action pair in $M$ by using Proposition 4.7.
Property (SA1) follows from

$[s^{+},s]\cdot[u,1]=[{}^{s}u,s]=[{}^{s}u,1]\cdot[s^{+},s].$

We define the map $S\to U^{1}:[s^{+},s]\mapsto[s^{+},s]^{+}=[s^{+},1]$. When
checking ($+$1)–($+$3) we make extensive use of Lemma 3.1. For ($+$4), we must
show that

$[u,1]\cdot[s^{+},s]=[v,1]\cdot[t^{+},t]\ \Rightarrow\
[u,1]\cdot[s^{+},s]^{+}=[v,1]\cdot[t^{+},t]^{+}\qquad\text{for all $u,v\in
V^{1}$ and $s,t\in T$.}$

Expanding the products, this amounts to showing that

$[us^{+},s]=[vt^{+},t]\ \Rightarrow\ [us^{+},1]=[vt^{+},1]$

for all such $u,v,s,t$. For this we use ($\Sigma$3) and ($\Sigma$4):

$\displaystyle[us^{+},s]=[vt^{+},t]\ \Rightarrow\ [u,s]=[v,t]\ \Rightarrow\
us^{+}=vt^{+}\ \Rightarrow\ [us^{+},1]=[vt^{+},1].$

This completes the proof that $(U,S)$ is an action pair in $M$.

If $T$ acts on $V^{1}$ by monoid morphisms, then $s^{+}={}^{s}1=1$ for all
$s\in T$, and it follows that $[s^{+},s]^{+}=[1,1]$ for all $s$, so that
$(U,S)$ is strong by Lemma 4.15(ii).

Next we note that

$US=\big{\\{}{[u,1][s^{+},s]}:{u\in V,\ s\in
T}\big{\\}}=\big{\\{}{[u,s]}:{u\in V,\ s\in T}\big{\\}},$

where in the last equality we use the fact that
$[u,1][s^{+},s]=[us^{+},s]=[u,s]$ for $u\in V$ and $s\in T$, which itself
follows from ($\Sigma$3). Finally, we define a map

$\Psi:V\rtimes T\to US\qquad\text{by}\qquad(u,s)\Psi=[u,s].$

It is clear that $\Psi$ is a surmorphism (it is the restriction to $V\rtimes
T$ of the natural projection $V^{1}\rtimes T^{1}\to(V^{1}\rtimes
T^{1})/\Sigma=M$). If $u,v\in V$ and $s,t\in T$, then by the definition of
$\Psi$, and by ($\Sigma$1), we also have

$(u,s)\Psi=(v,t)\Psi\ \Leftrightarrow\ [u,s]=[v,t]\ \Leftrightarrow\
(u,s)\mathrel{\Sigma}(v,t)\ \Leftrightarrow\ (u,s)\mathrel{\sigma}(v,t),$

so that $\ker(\Psi)=\sigma$. Thus, $US\cong(V\rtimes T)/\sigma$ by the First
Isomorphism Theorem. ∎

######  Remark 4.58.

Recall from Remark 4.47 that the surmorphism $\pi:U\rtimes S\to US$ factors
through the subsemigroup

$M=\big{\\{}{(u,s)\in U\rtimes S}:{u=us^{+}}\big{\\}}\leq U\rtimes S.$

It is possible to use Theorem 4.53 to classify the congruences $\sigma$ on
semigroups of the form $M$ such that $M/\sigma$ comes from an action pair. We
have chosen not to do this, however, because semidirect products are arguably
the more natural/direct construction. Nevertheless, the semigroup $M$ will
play a very important role in Chapter 5.

####  4.5 Congruence conditions

We conclude Chapter 4 with a sequence of results concerning generating sets
for the congruence $\theta$ on $U\rtimes S$ defined in Proposition 4.46, where
$(U,S)$ is an action pair in a monoid $M$. For convenience, let us recall that

$\theta=\big{\\{}{((u,s),(v,t))}:{u,v\in U,\ s,t\in S,\ us=vt}\big{\\}}.$

The reason we are interested in describing such generating sets is that they
feed into presentations for the semigroup $US\cong(U\rtimes S)/\theta$ in
Chapter 6.

The first result is the most general, and applies to any action pair $(U,S)$.
Roughly speaking, it shows that $\theta$ is generated by pairs of two kinds:

  * •

pairs with fixed $U$-coordinate, coming from the right congruences
$\theta_{u}$ in (4.49), and

  * •

pairs with fixed $S$-coordinate, coming from the $s\mapsto s^{+}$ map in
Proposition 4.7.

######  Lemma 4.59.

If $(U,S)$ is an action pair, then $\theta=\Omega^{\sharp}$, where

$\Omega=\big{\\{}{((u,s),(u,t))}:{u\in U,\
(s,t)\in\theta_{u}}\big{\\}}\cup\big{\\{}{((u,s),(us^{+},s))}:{u\in U,\ s\in
S}\big{\\}}.$

###### Proof.

Write $\theta^{\prime}=\Omega^{\sharp}$. It is easy to check that
$\Omega\subseteq\theta$, so $\theta^{\prime}\subseteq\theta$. It remains to
show that $\theta\subseteq\theta^{\prime}$, so suppose
$(u,s)\mathrel{\theta}(v,t)$. By Lemma 4.50(i), we have $us^{+}=vt^{+}$ and
$(s,t)\in\theta_{us^{+}}$. But then

$(u,s)\mathrel{\theta}^{\prime}(us^{+},s)\mathrel{\theta}^{\prime}(us^{+},t)=(vt^{+},t)\mathrel{\theta}^{\prime}(v,t).\qed$

If the pair $(U,S)$ is strong, then the previous result is essentially
vacuous; cf. Lemma 4.50(ii). The generating set $\Omega$ from Lemma 4.59 can
be reduced when $U$ and $S$ are both submonoids of the over-monoid $M$, as we
now show. For the statement, recall from Remark 4.4 that $s=s^{+}s$, which
tells us that $(1,s)\mathrel{\theta}(s^{+},s)$.

######  Lemma 4.60.

If $(U,S)$ is an action pair, and if $U$ and $S$ are both submonoids, then
$\theta=\Omega^{\sharp}$, where

$\Omega=\big{\\{}{((u,s),(u,t))}:{u\in U,\
(s,t)\in\theta_{u}}\big{\\}}\cup\big{\\{}{((1,s),(s^{+},s))}:{s\in
S}\big{\\}}.$

###### Proof.

Write $\theta^{\prime}=\Omega^{\sharp}$. It again suffices to show that
$\theta\subseteq\theta^{\prime}$. Given Lemma 4.59, it suffices to show that
$(u,s)\mathrel{\theta}^{\prime}(us^{+},s)$ for all $u\in U$ and $s\in S$. For
this, we have

$(u,s)=(u,1)(1,s)\mathrel{\theta}^{\prime}(u,1)(s^{+},s)=(us^{+},s).\qed$

Further simplifications are available when the pair $(U,S)$ is strong. We
begin by discussing certain conditions that might be satisfied by such a pair.

######  Lemma 4.61.

Suppose $(U,S)$ is a strong action pair, and consider the following
conditions:

  1.   2. (i)

$(\exists u\in U)\ (\forall s\in S)\ s=su$,

  3. (ii)

$(\forall s\in S)\ (\exists u\in U)\ s=su$,

  4. (iii)

$(\exists u\in U)\ (\forall s\in S)\ s=us$,

  5. (iv)

$(\forall s\in S)\ (\exists u\in U)\ s=us$,

  6. (v)

$(\forall s\in S)\ (\exists u\in U)\ (\exists t\in S)\ s=ut$,

  7. (vi)

$(\exists s,t\in S)\ (\exists u\in U)\ s=ut$,

  8. (vii)

$(\exists v\in U)\ (\forall u\in U)\ (\forall s\in S)\ us=usv$,

  9. (viii)

$(\forall u\in U)\ (\forall s\in S)\ (\exists v\in U)\ us=usv$,

  10. (ix)

$(\exists v\in U)\ (\forall u\in U)\ (\forall s\in S)\ u=u\cdot{}^{s}v$,

  11. (x)

$(\forall u\in U)\ (\forall s\in S)\ (\exists v\in U)\ u=u\cdot{}^{s}v$.

The following implications hold:

(iii)(iv)(v)(vi)(vii)(viii)(i)(ii)(ix)(x)$\Rightarrow$$\Rightarrow$$\Rightarrow$$\Rightarrow$$\Rightarrow$$\Rightarrow$$\Rightarrow$$\Downarrow$$\Downarrow$$\Updownarrow$$\Updownarrow$

###### Proof.

All implications other than (vi)$\ \Rightarrow\ $(vii), (vii)$\
\Leftrightarrow\ $(ix) and (viii)$\ \Leftrightarrow\ $(x) are obvious, and in
fact hold for non-strong action pairs as well. We treat the other implications
now.

(vi)$\ \Rightarrow\ $(vii). If (vi) holds, then we have $x=vy$ for some
$x,y\in S$ and $v\in U$. To show that (vii) holds (with respect to this $v$),
let $u\in U$ and $s\in S$ be arbitrary. Then $u\cdot
sx=usvy=u\cdot{}^{s}v\cdot sy$, so (SA2) gives $u=u\cdot{}^{s}v$. But then
$us=u\cdot{}^{s}v\cdot s=usv$, as required.

(vii)$\ \Leftrightarrow\ $(ix) and (viii)$\ \Leftrightarrow\ $(x). Using
$usv=u\cdot{}^{s}v\cdot s$ and (SA2), we have ${us=usv\ \Leftrightarrow\
u=u\cdot{}^{s}v}$. ∎

######  Remark 4.62.

The conditions listed in Lemma 4.61 are quite natural. Indeed, items
(i)–(viii) have the equivalent formulations:

  1. (i)

Some element of $U$ is a right identity for $S$.

  2. (ii)

Every element of $S$ has a right identity from $U$.

  3. (iii)

Some element of $U$ is a left identity for $S$.

  4. (iv)

Every element of $S$ has a left identity from $U$.

  5. (v)

$S\subseteq US$.

  6. (vi)

$S\cap US\not=\varnothing$.

  7. (vii)

Some element of $U$ is a right identity for $US$.

  8. (viii)

Every element of $US$ has a right identity from $U$.

While (ix) and (x) may not appear to be quite as natural, we have shown that
they are equivalent (when $(U,S)$ is strong) to (vii) and (viii), and these
alternative formulations will be useful in the next proof.

One important special case in which all of conditions (i)–(x) obviously hold
is when $U$ is a submonoid of the over-monoid $M$; note that we only have to
check (i) and (iii), which clearly hold in this case.

######  Remark 4.63.

It is also worth noting that the conditions listed in Lemma 4.61 are not all
equivalent. For example, consider the partial transformation monoid
$M=\mathcal{P}\mathcal{T}_{n}$, where $n\geq 2$, and the subsemigroups
$U=\operatorname{Sing}(\mathcal{E}_{n})$, $S_{1}=\mathcal{T}_{n}$ and
$S_{2}=\operatorname{Sing}(\mathcal{T}_{n})$. As noted in Example 4.30, the
strong action pairs $(U,S_{1})$ and $(U,S_{2})$ give rise to the same
subsemigroup:

$US_{1}=US_{2}=\operatorname{Sing}(\mathcal{P}\mathcal{T}_{n}).$

One can check that:

  * •

$(U,S_{1})$ satisfies only (viii) and (x), while

  * •

$(U,S_{2})$ satisfies only (ii), (viii) and (x).

On the other hand, some of the conditions of Lemma 4.61 are equivalent under
further assumptions. For example, if every element of $U$ has a left identity
(which happens for example if $U$ is regular), then (iv)$\ \Leftrightarrow\
$(v). Indeed, suppose (v) holds, and let $s\in S$ be arbitrary. Then (v) gives
$s=vt$ for some $v\in U$ and $t\in S$. But then if $u\in U$ is a left identity
for $v$, it follows that $s=vt=uvt=us$, showing that (iv) holds.

A similar argument shows that if $U$ has a left identity, then (v)$\
\Rightarrow\ $(iii), so it follows that (iii)$\ \Leftrightarrow\ $(iv)$\
\Leftrightarrow\ $(v) in this case.

######  Remark 4.64.

Finally, we note that the implication (vi)$\ \Rightarrow\ $(vii) need not hold
if $(U,S)$ is not strong. For example, consider the symmetric inverse monoid
$S=\mathcal{I}_{X}$ where $|X|\geq 2$. Then $(U,S)$ is an action pair, where
$U=\operatorname{Sing}(\mathcal{E}_{X})=\\{{\mathrm{id}_{A}}:{A\subsetneq
X}\\}$. Indeed, this is easy to check directly, and it also follows from
Remark 4.45, as $\mathcal{I}_{X}$ is inverse, and hence left restriction, and
$U=P^{\flat}$ in the notation of that remark. Condition (vi) holds trivially
for this pair, as we can take $s=t=u=\mathrm{id}_{A}$ for any $A\subsetneq X$.
However, (vii) does not hold since

$US=\\{{\alpha\in\mathcal{I}_{X}}:{\operatorname{dom}(\alpha)\not=X}\\}$

does not have a right identity, let alone a right identity from $U$.

For the rest of this section, we assume that for each $u\in U$, the right
congruence $\theta_{u}$ is generated (as a right congruence) by some set of
pairs $\Omega_{u}\subseteq S\times S$. For any set $X$, and any set of pairs
$\Sigma\subseteq X\times X$, we write

$\Sigma^{-1}=\big{\\{}{(y,x)}:{(x,y)\in\Sigma}\big{\\}}.$

######  Lemma 4.65.

If $(U,S)$ is a strong action pair, and if any of the conditions listed in
Lemma 4.61 hold (e.g., if $U$ is a submonoid), then $\theta=\Omega^{\sharp}$,
where

$\Omega=\big{\\{}{((u,s),(u,t))}:{u\in U,\ (s,t)\in\Omega_{u}}\big{\\}}.$

###### Proof.

By Lemma 4.61 we may assume that (x) holds. As usual, it suffices to show that
$\theta\subseteq\theta^{\prime}$, where $\theta^{\prime}=\Omega^{\sharp}$. By
Lemma 4.50(ii), this amounts to showing that
$(u,s)\mathrel{\theta}^{\prime}(u,t)$ for all $u\in U$ and
$(s,t)\in\theta_{u}$, so fix some such $u,s,t$. Since $\theta_{u}$ is
generated (as a right congruence) by $\Omega_{u}$, there is a sequence

$s=s_{1}\to s_{2}\to\cdots\to s_{k}=t,$

where $s_{1},\ldots,s_{k}\in S$, and such that for each $1\leq i<k$,

$s_{i}=a_{i}c_{i}\quad\text{and}\quad s_{i+1}=b_{i}c_{i}\qquad\text{for some
$(a_{i},b_{i})\in\Omega_{u}\cup\Omega_{u}^{-1}$ and $c_{i}\in S^{1}$.}$

Since $s=s_{1}$ and $t=s_{k}$, we can show that
$(u,s)\mathrel{\theta}^{\prime}(u,t)$ by showing that
$(u,s_{i})\mathrel{\theta}^{\prime}(u,s_{i+1})$ for each $1\leq i<k$. This is
clear if $c_{i}=1$, since then

$((u,s_{i}),(u,s_{i+1}))=((u,a_{i}),(u,b_{i}))\in\Omega\cup\Omega^{-1}\subseteq\theta^{\prime}.$

So now we assume that $c_{i}\not=1$. By (x), there exists $v\in U$ such that
$u=u\cdot{}^{a_{i}}v$. First note that

$u\cdot{}^{b_{i}}v\cdot b_{i}=u\cdot b_{i}\cdot v=u\cdot a_{i}\cdot
v=u\cdot{}^{a_{i}}v\cdot a_{i}=u\cdot a_{i},$

where we used $(a_{i},b_{i})\in\theta_{u}$ (which implies $ua_{i}=ub_{i}$) in
the second step; (SA2) then gives $u\cdot{}^{b_{i}}v=u$. It then follows that

$(u,s_{i})=(u,a_{i})\cdot(v,c_{i})\mathrel{\theta}^{\prime}(u,b_{i})\cdot(v,c_{i})=(u,s_{i+1}),$

as required. ∎

######  Remark 4.66.

The above proof shows that $\theta$ is in fact generated by $\Omega$ as a
_right_ congruence.

The previous result showed that, under certain conditions, a generating set
for the congruence $\theta$ can be constructed by incorporating generating
sets for the right congruences $\theta_{u}$. We now aim to take this further,
by additionally incorporating a generating set for the semigroup $U$. This of
course necessitates stronger assumptions, and these include requiring that the
$\theta_{u}$ behave well with respect to joins. Before stating the result, we
first establish some notation, which will be of use throughout. For $u,v\in U$
we write

$\displaystyle v\preceq u\quad$ $\displaystyle\ \Leftrightarrow\ \quad u=wv$
for some $w\in U^{1}$. (4.67) (So $\preceq$ is Green’s
$\geq_{\mathrel{\mathscr{L}}}$-preorder on $U$.) It is easy to see that
$\displaystyle v\preceq u\quad$ $\displaystyle\ \Rightarrow\
\quad\theta_{v}\subseteq\theta_{u}$ for all $u,v\in U$. (4.68) It follows
immediately that for any $V\subseteq U$, $\displaystyle\bigvee_{v\in V,\atop
v\preceq u}\theta_{v}$ $\displaystyle\subseteq\theta_{u}$ for all $u\in U$.
(4.69) Here and elsewhere, the join of equivalence relations is taken in the
lattice of all equivalence relations (over the same set). In particular, the
join in (4.69) is the smallest equivalence relation on $S$ containing all of
the $\theta_{v}$ ($v\in V$, $v\preceq u$); note that this join could involve
infinitely many terms. If $U$ is commutative, it also follows from (4.68) that
$\displaystyle\theta_{u_{1}}\vee\cdots\vee\theta_{u_{k}}$
$\displaystyle\subseteq\theta_{u_{1}\cdots u_{k}}$ for all
$u_{1},\ldots,u_{k}\in U$, (4.70)

as $u_{i}\preceq u_{1}\cdots u_{k}$ for each $i$, by commutativity.

The next result concerns the cases in which we have equality in (4.69) or
(4.70). The remaining results of this section assume $U$ and $S$ are
submonoids of the over-monoid $M$; one could state them in greater generality,
similarly to Lemma 4.65, but the combinations of conditions become unweildy,
and we prefer the cleaner statements below.

######  Lemma 4.71.

Suppose $(U,S)$ is a strong action pair in a monoid $M$, and suppose
additionally that $U$ and $S$ are submonoids of $M$.

  1. (i)

If there exists a subset $V\subseteq U$ such that

$\theta_{u}=\bigvee_{v\in V,\atop v\preceq u}\theta_{v}\qquad\text{for all
$u\in U$,}$

then $\theta=\Omega^{\sharp}$, where

$\Omega=\big{\\{}{((v,s),(v,t))}:{v\in V,\ (s,t)\in\Omega_{v}}\big{\\}}.$

  2. (ii)

If $U$ is commutative, and if $\theta_{uv}=\theta_{u}\vee\theta_{v}$ for all
$u,v\in U$, then for any monoid generating set $V$ for $U$ we have
$\theta=\Omega^{\sharp}$, where

$\Omega=\big{\\{}{((v,s),(v,t))}:{v\in V,\ (s,t)\in\Omega_{v}}\big{\\}}.$

###### Proof.

(i). Suppose $V\subseteq U$ satisfies the stated assumption. As ever, it
suffices to show that $\theta\subseteq\theta^{\prime}$, where
$\theta^{\prime}=\Omega^{\sharp}$. By Lemma 4.50(ii), this amounts to showing
that $(u,s)\mathrel{\theta}^{\prime}(u,t)$ for any $u\in U$ and
$(s,t)\in\theta_{u}$, so fix some such $u,s,t$. By the assumption on
$\theta_{u}$, there is a sequence

$s=s_{1}\to s_{2}\to\cdots\to s_{k}=t$

such that each $(s_{i},s_{i+1})\in\bigcup_{v\in
V_{u}}(\Omega_{v}\cup\Omega_{v}^{-1})$, where $V_{u}=\\{{v\in V}:{v\preceq
u}\\}$. Since ${(u,s)=(u,s_{1})}$ and $(u,t)=(u,s_{k})$, it suffices to show
that $(u,s_{i})\mathrel{\theta}^{\prime}(u,s_{i+1})$ for each $1\leq i<k$, so
fix some such $i$. Then $(s_{i},s_{i+1})\in\Omega_{v}\cup\Omega_{v}^{-1}$ for
some $v\in V_{u}$, and so $((v,s_{i}),(v,s_{i+1}))\in\Omega\cup\Omega^{-1}$.
Since $v\preceq u$ we have $u=wv$ for some $w\in U(=U^{1})$, and then

$(u,s_{i})=(w,1)\cdot(v,s_{i})\mathrel{\theta}^{\prime}(w,1)\cdot(v,s_{i+1})=(u,s_{i+1}).$

(ii). Suppose the assumptions of (ii) hold, and let $V$ be an arbitrary
generating set for $U$. It suffices to show that the assumption of (i) holds
with respect to this $V$. By (4.69), this amounts to showing that

$\bigvee_{v\in V,\atop v\preceq u}\theta_{v}\supseteq\theta_{u}\qquad\text{for
all $u\in U$,}$

so fix some such $u$. Since $U=\langle V\rangle$, we have $u=v_{1}\cdots
v_{k}$ for some $v_{1},\ldots,v_{k}\in V$, and so

$\theta_{u}=\theta_{v_{1}\cdots
v_{k}}=\theta_{v_{1}}\vee\cdots\vee\theta_{v_{k}}\subseteq\bigvee_{v\in
V,\atop v\preceq u}\theta_{v},$

since each $v_{i}\preceq u$ by commutativity of $U$. ∎

The generating sets in Lemmas 4.65 and 4.71 can be further simplified in the
(very) special case that $U$ and $S$ are submonoids of $M$, with $S$ a
_group_. To deal with this case, we introduce another piece of notation.

For an arbitrary weak action pair $(U,S)$, we define the sets

$S_{u}=\\{{s\in S}:{us=u}\\}\qquad\text{for each $u\in U$.}$ (4.72)

It is easy to see that each $S_{u}$ is a (possibly empty) subsemigroup of $S$.
When $S$ is a submonoid of $M$, each $S_{u}$ is a (non-empty) submonoid of
$S$; if $S$ is additionally a group, each $S_{u}$ is a subgroup. The next
result concerns this group case, when $(U,S)$ is an action pair (not just a
weak action pair). For the statement, we fix a _group_ generating set
$\Gamma_{u}$ for $S_{u}$, for each $u\in U$; so each $S_{u}$ is generated as a
monoid by $\Gamma_{u}\cup\Gamma_{u}^{-1}$. Note that in this case $S$ is
contained in the group of units of $M$, so the action pair $(U,S)$ is
automatically strong by Lemma 4.23(ii).

######  Lemma 4.73.

Suppose $(U,S)$ is a (strong) action pair in a monoid $M$, and suppose
additionally that $U$ and $S$ are submonoids of $M$, with $S$ a group. Then
$\theta=\Omega^{\sharp}$, where

$\Omega=\big{\\{}{((u,1),(u,s))}:{u\in U,\ s\in\Gamma_{u}}\big{\\}}.$

###### Proof.

Once again, it suffices to show that $(u,s)\mathrel{\theta}^{\prime}(u,t)$ for
all $u\in U$ and $(s,t)\in\theta_{u}$, where
$\theta^{\prime}=\Omega^{\sharp}$. First note that $(s,t)\in\theta_{u}\
\Rightarrow\ us=ut\ \Rightarrow\ ust^{-1}=u$, so that $st^{-1}\in S_{u}$. It
then follows that $st^{-1}=g_{1}\cdots g_{k}$ for some
${g_{1},\ldots,g_{k}\in\Gamma_{u}\cup\Gamma_{u}^{-1}}$. For each $1\leq i\leq
k+1$ write $s_{i}=g_{i}\cdots g_{k}t$. Since $s_{1}=s$ and $s_{k+1}=t$, it
suffices to show that $(u,s_{i})\mathrel{\theta}^{\prime}(u,s_{i+1})$ for each
$1\leq i\leq k$.

So fix some such $1\leq i\leq k$. If $g_{i}\in\Gamma_{u}$, then clearly
$(u,1)\mathrel{\theta}^{\prime}(u,g_{i})$. Otherwise,
${(u,1)\mathrel{\theta}^{\prime}(u,g_{i}^{-1})}$, so post-multiplying by
$(1,g_{i})$ gives $(u,g_{i})\mathrel{\theta}^{\prime}(u,1)$. Either way (and
keeping $s_{i}=g_{i}s_{i+1}$ in mind), it follows that

$(u,s_{i})=(u,g_{i})\cdot(1,s_{i+1})\mathrel{\theta}^{\prime}(u,1)\cdot(1,s_{i+1})=(u,s_{i+1}),$

as required. ∎

As with Lemma 4.71, the conclusion of Lemma 4.73 can be strengthened when the
subgroups $S_{u}$ behave well with respect to joins. (The join
$\bigvee_{i}H_{i}$ of a collection of subgroups is the smallest subgroup
containing the union $\bigcup_{i}H_{i}$.) The statement again uses the
relation $\preceq$ on $U$ given in (4.67).

######  Lemma 4.74.

Suppose $(U,S)$ is a (strong) action pair in a monoid $M$, and suppose
additionally that $U$ and $S$ are submonoids of $M$, with $S$ a group.

  1. (i)

If there exists a subset $V\subseteq U$ such that

$S_{u}=\bigvee_{v\in V,\atop v\preceq u}S_{v}\qquad\text{for all $u\in U$,}$

then $\theta=\Omega^{\sharp}$, where

$\Omega=\big{\\{}{((v,1),(v,s))}:{v\in V,\ s\in\Gamma_{v}}\big{\\}}.$

  2. (ii)

If $U$ is commutative, and if $S_{uv}=S_{u}\vee S_{v}$ for all $u,v\in U$,
then for any monoid generating set $V$ for $U$ we have
$\theta=\Omega^{\sharp}$, where

$\Omega=\big{\\{}{((v,1),(v,s))}:{v\in V,\ s\in\Gamma_{v}}\big{\\}}.$

###### Proof.

As in Lemma 4.71, part (ii) follows from part (i).

To prove (i), it suffices by Lemma 4.73 to show that
$(u,1)\mathrel{\theta}^{\prime}(u,s)$ for all $u\in U$ and $s\in\Gamma_{u}$,
where $\theta^{\prime}=\Omega^{\sharp}$, so fix some such $u,s$. Then $s\in
S_{u}=\bigvee_{v\in V_{u}}S_{v}$, where again we write $V_{u}=\\{{v\in
V}:{v\preceq u}\\}$, so $s=g_{1}\cdots g_{k}$ for some
${g_{1},\ldots,g_{k}\in\bigcup_{v\in V_{u}}(\Gamma_{v}\cup\Gamma_{v}^{-1})}$.
(Recall that $\Gamma_{v}$ is a group generating set for $S_{v}$.) Writing
$s_{i}=g_{i}\cdots g_{k}$ for each $1\leq i\leq k+1$, we have $s_{1}=s$ and
$s_{k+1}=1$, so it suffices to show that
$(u,s_{i})\mathrel{\theta}^{\prime}(u,s_{i+1})$ for each $1\leq i\leq k$. But
for any such $i$, we have $g_{i}\in\Gamma_{v}\cup\Gamma_{v}^{-1}$ for some
$v\in V_{u}$, and $u=wv$ for some $w\in U$. As in the proof of Lemma 4.73, we
have $(v,1)\mathrel{\theta}^{\prime}(v,g_{i})$, and (again remembering that
$s_{i}=g_{i}s_{i+1}$), it follows that

$(u,s_{i})=(w,1)\cdot(v,g_{i})\cdot(1,s_{i+1})\mathrel{\theta}^{\prime}(w,1)\cdot(v,1)\cdot(1,s_{i+1})=(u,s_{i+1}).\qed$

######  Remark 4.75.

As in (4.68), we have

$\displaystyle v\preceq u\ \Rightarrow\ S_{v}$ $\displaystyle\subseteq S_{u}$
for all $u,v\in U$. It follows that $\displaystyle\bigvee_{v\in V,\atop
v\preceq u}S_{v}$ $\displaystyle\subseteq S_{u}$ for all $u\in U$, and if $U$
is commutative, that $\displaystyle S_{u_{1}}\vee\cdots\vee S_{u_{k}}$
$\displaystyle\subseteq S_{u_{1}\cdots u_{k}}$ for all $u_{1},\ldots,u_{k}\in
U$.

Thus, in order to apply Lemma 4.74, we only need to check the forwards
inclusion in the relevant join equation.

###  5 Proper action pairs

Proposition 4.46 showed that the semigroup $US$ arising from a weak action
pair $(U,S)$ is a homomorphic image of a semidirect product $U\rtimes S$.
Theorem 4.53 took this further in the case that $(U,S)$ is an action pair (not
just a weak action pair), by additionally classifying the congruences on
semidirect products that lead to action pairs in this way. The purpose of the
current chapter is to prove a number of additional structure theorems
concerning semigroups arising from (weak) action pairs. These results are
inspired by important classical results on inverse semigroups, and wider
classes such as left restriction semigroups, each of which involve the notion
of a _proper_ semigroup from the relevant class.

We discuss these classical results in Section 5.1 (see Theorems 5.11 and
5.12), as well as some of the backstory behind their development. Section 5.2
shows that the mere existence of an action of $S$ on $U^{1}$ (for semigroups
$U$ and $S$) allows one to construct an action pair
$(\underline{U},\overline{S})$ in a suitable monoid $M$, with
$\underline{U}\cong U$ and $\overline{S}\cong S$, and with the respective
actions of $S$ and $\overline{S}$ on $U^{1}$ and $\underline{U}^{1}$ matching
up appropriately. Some aspects of this construction bear similarities with
ideas in the proof of Theorem 4.53. In Section 5.3 we introduce the key notion
of a _proper_ action pair, and show that a semigroup $US$ arising from an
arbitrary weak action pair $(U,S)$ has a ‘proper cover’. More specifically,
Theorem 5.39 shows that there exists a proper pair $(U^{\prime},S^{\prime})$
with $U^{\prime}\cong U$ and $S^{\prime}\cong S$, and a natural surmorphism
$U^{\prime}S^{\prime}\to US$. As an application of this, we show how to deduce
the classical Theorem 5.11 as a corollary. Sections 5.4–5.7 contain four
embedding theorems for proper _monoids_.

  * •

Theorem 5.60 (in Section 5.5) is the most general of these, and states that a
monoid $US$ arising from a proper action pair $(U,S)$ can be embedded in a
semidirect product $\mathcal{U}\rtimes(S/\sigma)$, where $\mathcal{U}$ is a
monoid containing $U$, and $\sigma$ is a certain natural congruence on $S$.
When $S$ is inverse, $\sigma$ is the least group congruence. The monoid
$\mathcal{U}$ constructed in the proof is defined by means of a presentation
that incorporates the structure of the monoids $U$ and $S/\sigma$, and the
action of $S$ on ${U(=U^{1})}$.

  * •

Theorem 5.47 (in Section 5.4) is actually a special case of Theorem 5.60,
stated and proved under the additional assumption that the elements
$s^{+}={}^{s}1$ ($s\in S$) are central in $U$. This allows us to construct a
very different monoid $\mathcal{U}$ during the proof, and the resulting
semidirect product is in fact an (unrestricted) wreath product. In the special
case that $S$ is left restriction, this $\mathcal{U}$ turns out to be a
semilattice (built in a natural way from the ideals of $U=P(S)$), and this
allows us to deduce the classical Theorem 5.12 as a corollary.

  * •

Theorems 5.82 and 5.102 (in Sections 5.6 and 5.7) are specialised versions of
Theorem 5.60. They show that when $U$ is assumed to have certain additional
structural properties (such as being a semilattice or a left-regular band),
then $M=US$ embeds in $\mathcal{U}\rtimes(S/\sigma)$, as above, where
$\mathcal{U}$ has the same structural property as $U$.

####  5.1 Background

In the early years of the development of semigroup theory, three directions of
attack emerged to understand the structure and behaviour of an inverse
semigroup $S$. These all had reference to the semilattice $E=E(S)=\\{{x\in
S}:{x=x^{2}}\\}$ of idempotents of $S$, and were subsequently developed and
applied to much broader classes. Specifically these directions are:

  * •

the use of Munn’s _fundamental inverse semigroup_ $T_{E}$, built from order-
isomorphisms of principal ideals of $E$ [93];

  * •

the Ehresmann-Schein-Nambooripad approach, which uses the _trace groupoid_ of
$S$ (which has identities $E$), together with the natural partial order on
$S$, to build an inductive groupoid from which $S$ can be recovered; a
comprehensive account of the genesis of this material is contained in Lawson’s
monograph [77];

  * •

McAlister’s theory of _proper_ (a.k.a.  _$E$ -unitary_) inverse semigroups
[86, 87].

Our current inspiration stems from this final direction, and we now describe
it in a little more detail, beginning with the key definition(s). The
following can be found, for example, on p55 of [17].

######  Definition 5.1.

Let $S$ be a semigroup, and $A\subseteq S$ an arbitrary subset. We say $S$ is
_$A$ -unitary_ if the following implications hold:

$ax\in A\ \Rightarrow\ x\in A\qquad\text{and}\qquad xa\in A\ \Rightarrow\ x\in
A\qquad\text{for all $a\in A$ and $x\in S$.}$ (5.2)

######  Remark 5.3.

Consider an _inverse_ semigroup $S$, with semilattice of idempotents $E=E(S)$.
Then $S$ is $E$-unitary when the implications in (5.2) hold with respect to
$A=E$. It is easy to check (when $A=E$) that either of the implications in
(5.2) follows from the other. Thus, $S$ is $E$-unitary if and only if

$\displaystyle ex\in E$ $\displaystyle\ \Rightarrow\ x\in E$ for all $e\in E$
and $x\in S$. It is again easy to check that this condition is equivalent to
$\displaystyle e=ex$ $\displaystyle\ \Rightarrow\ x\in E$ for all $e\in E$ and
$x\in S$. (5.4)

This all paves the way for the following:

######  Definition 5.5.

An inverse semigroup $S$ is _proper_ if it is $E$-unitary, where $E=E(S)$:
i.e., if (5.3) holds (cf. Remark 5.3).

######  Remark 5.6.

There are many other equivalent characterisations of proper ($E$-unitary)
inverse semigroups; see for example [64, Proposition 5.9.1]. One is for
$\mathscr{R}\cap\sigma$ (or equivalently $\mathscr{L}\cap\sigma$) to be
trivial, where $\mathscr{R}$ and $\mathscr{L}$ are two of Green’s relations,
and where $\sigma$ is the least group congruence. Thus, one can coordinatise
the elements of a proper inverse semigroup by an idempotent and a group
element. For broader classes, such as the left restriction semigroups we
consider below, being $E$-unitary is a weaker condition than being proper; the
definition of proper semigroups in these classes is given below.

In two landmark papers, McAlister showed that:

  * •

any inverse semigroup has a _proper cover_ [86]: i.e., a preimage under an
idempotent-separating morphism that is proper, and

  * •

any proper inverse semigroup $S$ is isomorphic to a so-called _P-semigroup_
[87]. The latter is built from an action of the group $G=S/\sigma$ (where here
$\sigma$ is the least group congruence on $S$) on a semilattice containing
$E$. Both the difficulty and the beauty of this approach is that the action of
$G$ on $E$ is not in general closed. It is also worth noting that while we
know from the outset that $S$ acts on $E$ (cf. Example 3.3(ii)), we also need
$S/\sigma$ to act on (a semilattice containing) $E$.

Subsequently, O’Carroll [98] showed that a proper inverse semigroup $S$ embeds
into a semidirect product $F\rtimes G$, where $F$ is a semilattice containing
$E$, and again $G=S/\sigma$. Immediately one sees a connection with (left)
action pairs, since a proper inverse semigroup $S$ is coordinatised by two
components, one from a semilattice and another from a group. Of course, $S$
might not be the internal product of these constituents.

There are many proofs of McAlister’s ‘P-theorem’, and of O’Carroll’s
subsequent embedding theorem. In addition to the original strategies, we
mention here the approaches of Munn [95], Billhardt [8] and Steinberg [111],
as well as the alternative strategy of Petrich and Reilly [99] using only
partial actions (which yields a somewhat differerent formulation, but which
nevertheless deserves mention here). A splendid account may be found in [78],
where [99] is referred to as ‘the maverick alternative’.

The approach of McAlister and O’Carroll for inverse semigroups prompted
analagous work for larger classes of semigroups; see for example [49, 113,
112, 115, 3, 83, 9, 75]. The extensions are, roughly speaking, in two
directions:

  * •

One is to keep the condition of $S$ being regular, but weaken the condition
that $E(S)$ forms a semilattice: e.g., it could be a left-regular band (these
satisfy the identities $x^{2}=x$ and $xyx=xy$).

  * •

The other direction is to drop the condition that $S$ is regular, and here the
main body of work retains commutativity of (certain) idempotents.

In particular, a theory analogous to that for inverse semigroups has been
developed for left restriction semigroups and the special (and in some sense
generating) classes of left ample and weakly left ample semigroups; see for
example [43, 46, 52, 51, 9]. The results of this chapter take this all much
further, and apply to semigroups and monoids arising from (weak) action pairs.

Here is the appropriate extension to left restriction semigroups of the notion
of a proper inverse semigroup [47, 57, 75].

######  Definition 5.7.

Let $S$ be a left restriction semigroup with semilattice of projections
$P=P(S)$. Define the congruence

$\displaystyle\sigma=\sigma_{S}$ $\displaystyle=\big{\\{}{(s,t)\in S\times
S}:{es=et\ (\exists e\in P)}\big{\\}}$ $\displaystyle=\big{\\{}{(s,t)\in
S\times S}:{es=ft\ (\exists e,f\in P)}\big{\\}}.$

We say $S$ is _proper_ if

$s=t\qquad\ \Leftrightarrow\ \qquad s^{+}=t^{+}\quad\text{and}\quad
s\mathrel{\sigma}t\qquad\qquad\text{for all $s,t\in S$.}$ (5.8)

######  Remark 5.9.

In fact, $\sigma=\nabla_{P}^{\sharp}$ is the least congruence on $S$ that
identifies all projections. As explained in [56], $\sigma$ is also a unary
semigroup congruence, meaning that ${(s,t)\in\sigma\ \Rightarrow\
(s^{+},t^{+})\in\sigma}$. The latter also follows quickly from (A2) and the
definition of $\sigma$, as $(P,S)$ is an action pair; cf. Proposition 4.37 and
Remark 4.39.

When $S$ is inverse (regarded as a left restriction semigroup with
$s^{+}=ss^{-1}$), $\sigma$ is the least group congruence on $S$. Definitions
5.5 and 5.7 are equivalent when $S$ is inverse; see for example [77] or [64,
Chapter 5]. It is worth noting that $s^{+}=t^{+}$ is equivalent to
$s\mathrel{\mathscr{R}}t$ in an inverse semigroup, where
$\mathrel{\mathscr{R}}$ is one of Green’s relations.

If $S$ is left ample (a left restriction semigroup that additionally satisfies
the quasi-identity $xz=yz\Rightarrow xz^{+}=yz^{+}$), then $\sigma$ is the
least right cancellative monoid congruence on $S$ [43]. If $S$ is weakly left
ample (a left restriction semigroup in which $P(S)=E(S)$), then $\sigma$ is
the least unipotent monoid congruence on $S$ [52].

######  Definition 5.10.

Let $S$ be a left restriction semigroup. A _cover_ of $S$ is a left
restriction semigroup $C$ together with a projection-separating surmorphism
$\psi:C\to S$. Such a $\psi$ is called a _covering morphism_ , and is required
to be a unary semigroup morphism, in the sense that $(xy)\psi=(x\psi)(y\psi)$
and $x^{+}\psi=(x\psi)^{+}$ for all $x,y\in C$. It follows quickly that $\psi$
restricts to an isomorphism $P(C)\to P(S)$. It is worth noting that when $C$
and $S$ are inverse semigroups (regarded as left restriction semigroups with
$x^{+}=xx^{-1}$), any semigroup morphism $C\to S$ automatically preserves the
-1 operation, and hence also the + operation.

Here are the formal statements of the results mentioned above:

######  Theorem 5.11 (cf. [86, 47, 9]).

Every left restriction semigroup (and in particular every inverse semigroup)
has a proper cover.

######  Theorem 5.12 (cf. [98, 75, 9]).

Let $S$ be a proper left restriction semigroup (in particular an inverse
semigroup) with semilattice of projections $P=P(S)$. Then $S$ embeds into a
semidirect product ${Q\rtimes(S/\sigma)}$, where $Q$ is a semilattice
containing $P$.

As we have already stated, our main goal in the current chapter is to devise a
suitable notion of a _proper_ (weak) action pair, and formulate and prove
corresponding extensions of Theorems 5.11 and 5.12 in this context; see
Theorems 5.39, 5.47, 5.60, 5.82 and 5.102. These more general results will
also lead to new proofs of Theorems 5.11 and 5.12.

####  5.2 From actions to action pairs

(Weak) action pairs are defined in terms of an action of a semigroup $S$ on
the monoid completion $U^{1}$ of a semigroup $U$, where $U$ and $S$ are
contained in a common over-monoid $M$. The purpose of the current section is
to show that the mere existence of an action (by morphisms) leads to an action
pair that mimicks the original action, in a technical sense made precise
below. This construction will be harnessed in the next section in order to
show that any semigroup arising from a (weak) action pair has a ‘proper
cover’.

For convenience of reference, we will gather the standing assumptions for this
section here, and also fix some notational conventions:

######  Assumption 5.13.

  1. (i)

We assume that $U$ and $S$ are semigroups, and that $S$ has a left action on
$U^{1}$ via semigroup morphisms, denoted $(s,u)\mapsto{}^{s}u$.

For example, if $S$ has an action on $U$ itself, and if $U\not=U^{1}$, then
this can be extended to an action on $U^{1}$ by additionally defining
${}^{s}1=1$ for all $s\in S$. In this case, the action of $S$ on $U^{1}$ is by
monoid morphisms. More generally, however, we do not assume that the action in
(i) is by monoid morphisms.

  1. (ii)

If $S$ happens to be a monoid, then we assume in addition that the action of
$S$ on $U^{1}$ is monoidal: i.e., that the identity of $S$ acts identically.

Here, as usual, $U^{1}$ denotes $U$ if $U$ is a monoid or else
$U^{1}=U\cup\\{1\\}$, where $1$ is an adjoined identity. We use the same
convention for $S^{1}$, and we assume without loss of generality (and for
notational convenience) that the identities of $U^{1}$ and $S^{1}$ are the
same. But note that we do not assume that $U$ and $S$ are contained in a
common semigroup; in particular, we are not assuming at this point that
$(U,S)$ is a (weak) action pair. However, we will show that $(U,S)$ can indeed
be identified with an action pair $(\underline{U},\overline{S})$ in a suitable
monoid $M$.

The requirement for $S$ to act monoidally (if it is a monoid) in Assumption
5.13(ii) is necessary for the results of this section. In the next, we will
apply these results in the case that $(U,S)$ is a weak action pair, and we
noted in Remark 4.4 that whenever $S$ is a monoid, it acts monoidally on
$U^{1}$ in this case.

The action of $S$ on $U^{1}$ allows for the formation of the (external)
semidirect product $U\rtimes S$, as in Definition 3.4. As usual, and if
necessary, we can extend the action of $S$ on $U^{1}$ to a monoidal action of
$S^{1}$ on $U^{1}$ (it is already monoidal by assumption if $S=S^{1}$), and we
then have the semidirect product $U^{1}\rtimes S^{1}$, which contains
$U\rtimes S$ as a subsemigroup. Although $U^{1}\rtimes S^{1}$ might not be a
monoid (cf. Lemma 3.11), it follows from Proposition 3.32 (and monoidality of
the action of $S^{1}$ on $U^{1}$) that

$M=\big{\\{}{(u,s)\in U^{1}\rtimes S^{1}}:{u=us^{+}}\big{\\}}\leq U^{1}\rtimes
S^{1}$

is a monoid with identity $(1,1)$. Here, as usual, we write $s^{+}={}^{s}1$
for all $s\in S^{1}$, noting that $1^{+}=1$.

We begin by identifying two natural subsemigroups of $M$. To this end, we
define

$\underline{u}=(u,1)\qquad\text{and}\qquad\overline{s}=(s^{+},s)\qquad\text{for
$u\in U$ and $s\in S$,}$

and we set

$\underline{U}=\\{{\underline{u}}:{u\in
U}\\}\qquad\text{and}\qquad\overline{S}=\\{{\overline{s}}:{s\in S}\\}.$

It is clear that $\underline{U}\subseteq M$, while $\overline{S}\subseteq M$
follows from Lemma 3.1(i).

######  Lemma 5.14.

Given Assumption 5.13, and with the above notation, we have

  1.   2. (i)

$\underline{U}\leq M$ and $\underline{U}\cong U$,

  3. (ii)

$\overline{S}\leq M$ and $\overline{S}\cong S$.

###### Proof.

The first part is clear. The second follows quickly from the fact that for all
$s,t\in S$,

$\overline{s}\cdot\overline{t}=(s^{+},s)\cdot(t^{+},t)=(s^{+}\cdot{}^{s}(t^{+}),st)=({}^{s}(t^{+}),st)=((st)^{+},st)=\overline{st},$
(5.15)

where we used parts (iv) and (ii) of Lemma 3.1. ∎

For convenience in what follows, we also write $\underline{1}=(1,1)$, even if
$1\not\in U$. On a number of occasions during the next proof (and later in the
paper) we make use of the fact that
${\underline{u}\cdot\overline{s}=(us^{+},s)}$ for all $u\in U$ and $s\in S$;
in particular, $(u,s)=\underline{u}\cdot\overline{s}$ if $u=us^{+}$.

######  Proposition 5.16.

Given Assumption 5.13, and with the above notation,

  1. (i)

$(\underline{U},\overline{S})$ is an action pair in $M=\big{\\{}{(u,s)\in
U^{1}\rtimes S^{1}}:{u=us^{+}}\big{\\}}$,

  2. (ii)

$(\underline{U},\overline{S})$ is strong if and only if the action of $S$ on
$U^{1}$ is by monoid morphisms,

  3. (iii)

$\underline{U}\cdot\overline{S}=\big{\\{}{(u,s)\in U\rtimes
S}:{u=us^{+}}\big{\\}}$.

###### Proof.

(i). By Lemma 5.14, $\underline{U}$ and $\overline{S}$ are subsemigroups of
$M$. Since $\underline{U}\cong U$ and $\overline{S}\cong S$, we have an action
of $\overline{S}$ on
$\underline{U}^{1}=\underline{U}\cup\big{\\{}(1,1)\big{\\}}$, given by

${}^{\overline{s}}\underline{u}=\underline{{}^{s}u}=({}^{s}u,1)\qquad\text{for
$u\in U^{1}$ and $s\in S$.}$

Using Lemma 3.1(iv), one can easily check that
$\overline{s}\cdot\underline{u}=({}^{s}u,s)={}^{\overline{s}}\underline{u}\cdot\overline{s}$
for all $u\in U$ and $s\in S$. This shows that (A1) holds.

For (A2), we must show that

$\underline{u}\cdot\overline{s}=\underline{v}\cdot\overline{t}\ \Rightarrow\
\underline{u}\cdot\overline{s}^{+}=\underline{v}\cdot\overline{t}^{+}\qquad\text{for
all $u,v\in U^{1}$ and $s,t\in S$.}$

Noting that

$\overline{s}^{+}={}^{\overline{s}}\underline{1}=\underline{{}^{s}1}=\underline{s^{+}}=(s^{+},1),$
(5.17)

and similarly for $\overline{t}^{+}$, the required implication follows quickly
from the fact that

$\underline{u}\cdot\overline{s}=(us^{+},s),\qquad\underline{v}\cdot\overline{t}=(vt^{+},t),\qquad\underline{u}\cdot\overline{s}^{+}=(us^{+},1)\qquad\text{and}\qquad\underline{v}\cdot\overline{t}^{+}=(vt^{+},1).$

(ii). By Lemma 4.15(ii), $(\underline{U},\overline{S})$ is strong if and only
if $\overline{s}^{+}=\underline{1}$ for all $s\in S$. But by (5.17),

$\overline{s}^{+}=\underline{1}\ \Leftrightarrow\ (s^{+},1)=(1,1)\
\Leftrightarrow\ s^{+}=1\ \Leftrightarrow\ {}^{s}1=1.$

The assertion quickly follows.

(iii). First suppose $u\in U$ and $s\in S$ are such that $u=us^{+}$. Then

$(u,s)=(us^{+},s)=\underline{u}\cdot\overline{s}\in\underline{U}\cdot\overline{S}.$

Conversely, suppose ${\bf a}\in\underline{U}\cdot\overline{S}$, so that ${\bf
a}=\underline{v}\cdot\overline{s}$ for some $v\in U$ and $s\in S$. Then with
$u=vs^{+}\in U$, we have

${\bf a}=\underline{v}\cdot\overline{s}=(vs^{+},s)=(u,s)\qquad\text{and}\qquad
us^{+}=vs^{+}s^{+}=vs^{+}=u,$

using Lemma 3.1(i). ∎

####  5.3 Proper action pairs and a covering theorem

We are now almost ready to introduce the notion of a _proper_ action pair; see
Definitions 5.21 and 5.25. The main result of the current section is Theorem
5.39, which shows that any semigroup arising from a (weak) action pair is a
natural homomorphic image of a semigroup arising from a proper pair. The proof
of the theorem utilises the $(\underline{U},\overline{S})$ construction from
the previous section. Throughout the current section we comment on the special
case of left restriction semigroups, and we end by showing how the classical
Theorem 5.11 follows from our Theorem 5.39.

Consider a weak action pair $(U,S)$ in a monoid $M$. As ever, we denote the
action of $S$ on $U^{1}$ by $(s,u)\mapsto{}^{s}u$. We define

$S^{+}=\\{{s^{+}}:{s\in S}\\}=\\{{{}^{s}1}:{s\in S}\\},$

noting that this subset of $U^{1}$ need not be a subsemigroup. In any case, we
also define

$P=P(U,S)=\langle S^{+}\rangle,$

which is the subsemigroup of $U^{1}$ generated by $S^{+}$. Since $S^{+}$
consists of idempotents (cf. Lemma 3.1(i)), it follows that $P$ is an
idempotent-generated semigroup, and hence is a submonoid of $U^{1}$ if and
only if $1\in S^{+}$; indeed, the latter claim follows from [35, Lemma 2.1].

######  Lemma 5.18.

The subsemigroup $P$ of $U^{1}$ is closed under the action of $S$.

###### Proof.

By Lemma 3.1(ii), $S^{+}$ is closed under the action of $S$; so too therefore
is $\langle S^{+}\rangle=P$. ∎

######  Remark 5.19.

By Lemma 5.18 (and consulting Definition 4.2), it follows immediately that
$(P,S)$ is a weak action pair as well. If $(U,S)$ is a (strong) action pair,
then so too is $(P,S)$, by Lemma 4.18(ii); when $(U,S)$ is strong, Lemma
4.15(ii) gives $P=\\{1\\}$.

######  Remark 5.20.

When $S$ is a left restriction semigroup, $S^{+}=\langle S^{+}\rangle$ is
precisely the semilattice of projections $P(S)$. This explains our choice of
notation for $P=P(U,S)$. Recall from Proposition 4.37 (and Remark 4.39) that
$(P,S)$ is an action pair in $S$.

In order to give the definition of a proper action pair, we need an
appropriate congruence to play the role of $\sigma$ from Definition 5.7.

######  Definition 5.21.

Let $(U,S)$ be a weak action pair in a monoid $M$, and write $P=\langle
S^{+}\rangle$. Define the relation

$\displaystyle\kappa=\kappa(U,S)$ $\displaystyle=\big{\\{}{(s,t)\in S\times
S}:{ps=qt\ (\exists p,q\in P)}\big{\\}}$ $\displaystyle=\big{\\{}{(s,t)\in
S\times S}:{ps=qt\ (\exists p,q\in P^{1})}\big{\\}},$

and let $\sigma=\sigma(U,S)=\kappa^{\sharp}$ be the congruence on $S$
generated by $\kappa$.

######  Lemma 5.22.

The relation $\kappa$ is reflexive, symmetric and compatible. Consequently,
$\sigma$ is the transitive closure of $\kappa$.

###### Proof.

We just prove the first statement, as the second immediately follows.
Reflexivity, symmetry and right-compatibility of $\sigma$ are clear. For left-
compatibility, let $(s,t)\in\kappa$ and $x\in S$. So $ps=qt$ for some $p,q\in
P$. By (A1) we have ${}^{x}p\cdot xs=x\cdot ps=x\cdot qt={}^{x}q\cdot xt$.
Since ${}^{x}p,{}^{x}q\in P$, by Lemma 5.18, it follows that
$(xs,xt)\in\kappa$. ∎

The next result concerns a special case in which we have $\sigma=\kappa$. For
the statement, recall [22] that a semigroup $T$ is _right-reversible_ if any
two left ideals (equivalently, any two principal left ideals) of $T$ have non-
empty intersection: i.e., if for all $x,y\in T$, we have $ax=by$ for some
$a,b\in T$. Examples of right-reversible semigroups include semilattices (or
in fact arbitrary commutative semigroups), or more generally left-regular
bands; the latter satisfy the identity $xyx=xy$ (and $x=x^{2}$).

######  Lemma 5.23.

If $P$ is right-reversible, then $\sigma=\kappa$.

###### Proof.

It suffices by Lemma 5.22 to show that $\kappa$ is transitive (in this case).
So suppose $s\mathrel{\kappa}t\mathrel{\kappa}x$ for some $s,t,x\in S$. Then
$ps=qt$ and $p^{\prime}t=q^{\prime}x$ for some $p,p^{\prime},q,q^{\prime}\in
P$. By right-reversibility we have $aq=bp^{\prime}$ for some $a,b\in P$, and
then $ap\cdot s=aqt=bp^{\prime}t=bq^{\prime}\cdot x$, with $ap,bq^{\prime}\in
P$, so that $s\mathrel{\kappa}x$, as required. ∎

######  Remark 5.24.

As a special case, when $S$ is a left restriction semigroup, $P=\langle
S^{+}\rangle=P(S)$ is a semilattice, so $\sigma=\kappa$ coincides with the
relation $\sigma_{S}$ from Definition 5.7.

We can now give the definition of proper pairs and semigroups.

######  Definition 5.25.

We say a weak action pair $(U,S)$ is _proper_ if for all $u,v\in U^{1}$ and
$s,t\in S$,

$us=vt\qquad\ \Leftrightarrow\ \qquad us^{+}=vt^{+}\quad\text{and}\quad
s\mathrel{\sigma}t.$ (5.26)

If $(U,S)$ is proper, the semigroup $US$ is then said to be _$(U,S)$ -proper_.

The following simple fact will be used without explicit reference from now on:

######  Lemma 5.27.

Any proper weak action pair is an action pair.

###### Proof.

We need to show that (A2) holds: i.e., that $us=vt\ \Rightarrow\
us^{+}=vt^{+}$ for all $u,v\in U^{1}$ and $s,t\in S$. But this clearly follows
from (5.26). ∎

######  Remark 5.28.

Consider a _strong_ action pair $(U,S)$. Since $s^{+}=1$ for all $s\in S$, by
Lemma 4.15(ii), we have $P=S^{+}=\\{1\\}$, and it follows immediately that
$\kappa=\Delta_{S}$ (the equality relation on $S$), and hence that
$\sigma=\Delta_{S}$. Definition 5.25 then says that $(U,S)$ is proper
precisely when

$us=vt\qquad\ \Leftrightarrow\ \qquad u=v\quad\text{and}\quad
s=t\qquad\text{for all $u,v\in U^{1}$ and $s,t\in S$.}$

This then implies that the surmorphism $\pi:U\rtimes S\to US:(u,s)\mapsto us$
from Proposition 4.46 is injective, and hence an isomorphism.

Figure 1 gives a Venn diagram displaying the various classes of pairs studied
in the paper.

SAPPAPAPWAP

Figure 1: Venn diagram indicating inclusions among the classes of strong
action pairs (SAP), proper action pairs (PAP), action pairs (AP) and weak
action pairs (WAP).

Now that we have defined proper action pairs, we begin by showing that these
are indeed the ‘correct’ generalisation of proper left restriction semigroups.

######  Proposition 5.29.

If $S$ is a left restriction semigroup with semilattice of projections
$P=P(S)$, then the following are equivalent:

  1. (i)

$S$ is proper, as in Definition 5.7,

  2. (ii)

$(P,S)$ is a proper action pair, as in Definition 5.25.

###### Proof.

We have already noted in Remark 5.20 that $P=S^{+}=\langle S^{+}\rangle$ in
this case, and in Remark 5.24 that the definitions of $\sigma$ from
Definitions 5.7 and 5.21 coincide.

(i)$\ \Rightarrow\ $(ii). Suppose first that $S$ is proper, and let $u,v\in
P^{1}$ and $s,t\in S$. We must show that

$\displaystyle us=vt\qquad$ $\displaystyle\ \Leftrightarrow\ \qquad
us^{+}=vt^{+}\quad\text{and}\quad s\mathrel{\sigma}t.$ (5.30) Since $S$ is
proper, and since $us,vt\in S$, it follows from (5.8) that $\displaystyle
us=vt\qquad$ $\displaystyle\ \Leftrightarrow\
\qquad(us)^{+}=(vt)^{+}\quad\text{and}\quad us\mathrel{\sigma}vt.$ (5.31)

By Lemma 4.36, we have ${us^{+}=vt^{+}\ \Leftrightarrow\ (us)^{+}=(vt)^{+}}$.
It also follows quickly from the definition of $\sigma$ (in Definition 5.7)
that $s\mathrel{\sigma}t\ \Leftrightarrow\ us\mathrel{\sigma}vt$, keeping in
mind the fact that $P$ is a semilattice. Equivalence of (5.30) and (5.31) is
now clear.

(ii)$\ \Rightarrow\ $(i). Conversely, suppose $(P,S)$ is proper, and let
$s,t\in S$. We must show that

$s=t\qquad\ \Leftrightarrow\ \qquad s^{+}=t^{+}\quad\text{and}\quad
s\mathrel{\sigma}t.$

As the forwards implication is clear, suppose $s^{+}=t^{+}$ and
$s\mathrel{\sigma}t$. Then with $u=v=s^{+}=t^{+}\in P$, we have
$us^{+}=s^{+}=t^{+}=vt^{+}$. Since $(P,S)$ is proper, and since also
$s\mathrel{\sigma}t$, it follows from (5.26) that $us=vt$: i.e., $s=t$. ∎

The next result shows that when the action pair $(U,S)$ is proper, and when
$P=\langle S^{+}\rangle$ satisfies a certain kind of semigroup identity, the
congruence $\sigma$ can be described equationally, rather than by asserting
the existence of suitable elements of $P$.

######  Lemma 5.32.

Let $(U,S)$ be a proper action pair, and suppose $P$ satisfies a semigroup
identity of the form $f(x,y)x=g(x,y)y$, where $f(x,y),g(x,y)\in\\{x,y\\}^{*}$.
Then

$\sigma=\kappa=\big{\\{}{(s,t)\in S\times
S}:{f(s^{+},t^{+})s=g(s^{+},t^{+})t}\big{\\}}.$

###### Proof.

The identity implies that $P$ is right-reversible, so Lemma 5.23 gives
$\sigma=\kappa$. For the rest of the proof we write

$\sigma^{\prime}=\big{\\{}{(s,t)\in S\times
S}:{f(s^{+},t^{+})s=g(s^{+},t^{+})t}\big{\\}}.$

Since $f(s^{+},t^{+}),g(s^{+},t^{+})\in P^{1}$ for any $s,t\in S$, we clearly
have $\sigma^{\prime}\subseteq\sigma$.

Conversely, suppose $s\mathrel{\sigma}t$. Since $s^{+},t^{+}\in P$, we have
$f(s^{+},t^{+})\cdot s^{+}=g(s^{+},t^{+})\cdot t^{+}$. Combining this with
$s\mathrel{\sigma}t$, it follows from (5.26) that $f(s^{+},t^{+})\cdot
s=g(s^{+},t^{+})\cdot t$, whence $s\mathrel{\sigma}^{\prime}t$. ∎

######  Remark 5.33.

Since semilattices satisfy the identity $xy=yx$, it follows from Lemma 5.32
that

$\displaystyle\sigma=\kappa$ $\displaystyle=\big{\\{}{(s,t)\in S\times
S}:{t^{+}s=s^{+}t}\big{\\}}$ when $P$ is a semilattice. Similarly,
$\displaystyle\sigma=\kappa$ $\displaystyle=\big{\\{}{(s,t)\in S\times
S}:{s^{+}t^{+}s=s^{+}t}\big{\\}}$

if $P$ is a left-regular band.

If $P$ satisfies an identity of the form
$f(x,y,z_{1},\ldots,z_{k})x=g(x,y,z_{1},\ldots,z_{k})y$, then one can
arbitrarily substitute $x$ and/or $y$ in place of the $z_{i}$ to obtain an
identity of the form ${f^{\prime}(x,y)x=g^{\prime}(x,y)y}$, and Lemma 5.32
then applies.

In Sections 5.4–5.7 we will be concerned with proper action pairs $(U,S)$,
when $U$ and $S$ are both submonoids of the over-monoid $M$. The
$\sigma$-class of the identity of $S$ will play an important role on many
occasions during these sections. The next lemma characterises the elements of
this class, and shows that they act on $U^{1}$ in a particularly simple way.
In the statement, we do not assume that $U$ is a submonoid of $M$, and we use
the notation $\hat{1}$ for consistency with later use.

######  Lemma 5.34.

Let $(U,S)$ be a proper action pair, with $S$ a submonoid of the over-monoid
$M$, and let $\hat{1}$ be the $\sigma$-class of $1\in S$.

  1. (i)

We have $\hat{1}=S\cap S^{+}=\\{{s\in S}:{s=s^{+}}\\}$.

  2. (ii)

For any $s\in\hat{1}$, and for any $u\in U^{1}$, we have ${}^{s}u=su=s^{+}u$.

###### Proof.

(i). We prove this part by showing that for all $s\in S$, the following are
equivalent:

  1.   2. (a)

$s\mathrel{\sigma}1$,

  3. (b)

$s\in S^{+}$,

  4. (c)

$s=s^{+}$.

(a)$\ \Rightarrow\ $(c). Taking $u=t=1$ and $v=s^{+}$, we have $us^{+}=vt^{+}$
and $s\mathrel{\sigma}t$. Since $(U,S)$ is proper, it follows from (5.26) that
$us=vt$: i.e., $s=s^{+}$.

(c)$\ \Rightarrow\ $(b). This is obvious.

(b)$\ \Rightarrow\ $(a). Suppose $s=t^{+}$ for some $t\in S$. Then since
$1\cdot s=t^{+}\cdot 1$, with $1,t^{+}\in S^{+}\subseteq P$, it follows that
$(s,1)\in\kappa\subseteq\sigma$; cf. Definition 5.21.

(ii). We have $s=s^{+}$ by part (i), so of course $su=s^{+}u$. Combining
$s=s^{+}$ with (A1) and Lemma 3.1(iv), we have $su={}^{s}u\cdot s={}^{s}u\cdot
s^{+}={}^{s}u$. ∎

The next lemma provides a handy criterion for an action pair to be proper. For
the statement, we say that a subsemigroup $V$ of a semigroup $T$ is _left-
dense_ in $T$ if the following condition holds:

$\displaystyle(\forall t\in T)\ (\exists a\in V)$ $\displaystyle\quad at\in
V.$ (5.35) Note, for example, that any right ideal is left-dense, but the
converse need not hold. (The term ‘left-dense’ is taken from [7, p. 98], where
the meaning is slightly different. There ‘$(\exists a\in V)$’ is replaced by
‘$(\exists a\in T)$’, and this results in a weaker condition. For example,
take $T$ to be an arbitrary group, $V$ a proper subgroup, and $t\in T\setminus
V$; any $a\in T$ such that $at\in V$ necessarily belongs to $T\setminus V$.)
Before stating the lemma, we claim that left-density is equivalent to the
ostensibly stronger condition: $\displaystyle(\forall s,t\in T^{1})\ (\exists
a\in V)$ $\displaystyle\quad as,at\in V.$ (5.36)

Obviously (5.36) implies (5.35). For the converse, suppose (5.35) holds, and
let $s,t\in T^{1}$. There exists $c\in V$ such that $ct\in V$ (this is obvious
if $t=1$, or follows from (5.35) otherwise). Since certainly $cs\in T$, there
also exists $b\in V$ such that $b\cdot cs\in V$. We then have $as,at\in V$ for
$a=bc\in V$. (The weaker version of left-density from [7] does not imply the
corresponding version of (5.36), as seen again with $T$ a group, $V$ a proper
subgroup, $s\in V$ and $t\in T\setminus V$.)

######  Lemma 5.37.

Suppose $(U,S)$ is an action pair.

  1. (i)

If $P$ is left-dense in $U$, then

$us=vt\qquad\Rightarrow\qquad us^{+}=vt^{+}\quad\text{and}\quad
s\mathrel{\sigma}t\qquad\qquad\text{for all $u,v\in U^{1}$ and $s,t\in S$.}$

  2. (ii)

If $\sigma=\Delta_{S}$, then

$us=vt\qquad\Leftarrow\qquad us^{+}=vt^{+}\quad\text{and}\quad
s\mathrel{\sigma}t\qquad\qquad\text{for all $u,v\in U^{1}$ and $s,t\in S$.}$

  3. (iii)

If $P$ is left-dense in $U$, and if $\sigma=\Delta_{S}$, then $(U,S)$ is
proper.

###### Proof.

It clearly suffices to prove (i) and (ii). For these, let $u,v\in U^{1}$ and
$s,t\in S$.

(i). Suppose $us=vt$. It follows from (A2) that $us^{+}=vt^{+}$. By (5.36)
there exists $p\in P$ such that $pu,pv\in P$, and it then follows from
$pus=pvt$ that $(s,t)\in\kappa\subseteq\sigma$.

(ii). Suppose $us^{+}=vt^{+}$ and $s\mathrel{\sigma}t$. Since
$\sigma=\Delta_{S}$, we have $s=t$. Combining this with ($+$1), it follows
that $us=us^{+}\cdot s=vt^{+}\cdot t=vt$. ∎

Consider again semigroups $S$ and $U$ for which $S$ has a left action on
$U^{1}$ by semigroup morphisms, and assume that $S$ acts monoidally if it
happens to be monoid; cf. Assumption 5.13. As in Section 5.2, we have the
semidirect products $U\rtimes S$ and $U^{1}\rtimes S^{1}$, and the
subsemigroups

$\underline{U}=\\{{\underline{u}}:{u\in
U}\\}\qquad\text{and}\qquad\overline{S}=\\{{\overline{s}}:{s\in S}\\}$

of $U^{1}\rtimes S^{1}$, where $\underline{u}=(u,1)$ and
$\overline{s}=(s^{+},s)$, for $u\in U$ and $s\in S$. We showed in Proposition
5.16 that $(\underline{U},\overline{S})$ is an action pair in the monoid

$M=\big{\\{}{(u,s)\in U^{1}\rtimes S^{1}}:{u=us^{+}}\big{\\}},$

and the next result takes this further. For the proof, we use the fact that
$\overline{s}^{+}=\underline{s^{+}}$ for $s\in S$; cf. (5.17). Together with
the fact that $S\to\overline{S}:s\mapsto\overline{s}$ is an isomorphism, it
follows that

$\overline{s}_{1}^{+}\cdots\overline{s}_{k}^{+}=\underline{s_{1}^{+}\cdots
s_{k}^{+}}\qquad\text{for all $s_{1},\ldots,s_{k}\in S$.}$

######  Proposition 5.38.

Given Assumption 5.13, and with the above notation, we have
$\sigma(\underline{U},\overline{S})=\Delta_{\overline{S}}$. Moreover, the
action pair $(\underline{U},\overline{S})$ is proper, so the semigroup

$\underline{U}\cdot\overline{S}=\big{\\{}{(u,s)\in U\rtimes
S}:{u=us^{+}}\big{\\}}$

is $(\underline{U},\overline{S})$-proper.

###### Proof.

Throughout the proof, we write $\kappa=\kappa(\underline{U},\overline{S})$ and
$\sigma=\sigma(\underline{U},\overline{S})$, and also

$P=P(\underline{U},\overline{S})=\langle\overline{S}^{+}\rangle\qquad\text{where}\qquad\overline{S}^{+}=\\{{\overline{s}^{+}}:{s\in
S}\\}=\\{{\underline{s^{+}}}:{s\in S}\\}.$

To show that $\sigma=\Delta_{\overline{S}}$ we must show that
$\kappa=\Delta_{\overline{S}}$, so suppose
$\overline{s}\mathrel{\kappa}\overline{t}$ for some $s,t\in S$. By definition,
this means that ${\bf p}\cdot\overline{s}={\bf q}\cdot\overline{t}$ for some
${\bf p},{\bf q}\in P$. Now, ${\bf
p}=\overline{s}_{1}^{+}\cdots\overline{s}_{k}^{+}=\underline{s_{1}^{+}\cdots
s_{k}^{+}}$ for some $s_{1},\ldots,s_{k}\in S$, so that ${\bf
p}=\underline{u}$, where $u=s_{1}^{+}\cdots s_{k}^{+}\in U^{1}$. Similarly,
${\bf q}=\underline{v}$ for some $v\in U^{1}$. But then

$(us^{+},s)={\bf p}\cdot\overline{s}={\bf q}\cdot\overline{t}=(vt^{+},t).$

It follows from this that $s=t$, whence $\overline{s}=\overline{t}$, and this
completes the proof that $\kappa=\Delta_{\overline{S}}$.

Given Proposition 5.16, it remains to show that $(\underline{U},\overline{S})$
is proper: i.e., that

$\underline{u}\cdot\overline{s}=\underline{v}\cdot\overline{t}\qquad\
\Leftrightarrow\
\qquad\underline{u}\cdot\overline{s}^{+}=\underline{v}\cdot\overline{t}^{+}\quad\text{and}\quad\overline{s}\mathrel{\sigma}\overline{t}\qquad\text{for
all $u,v\in U^{1}$ and $s,t\in S$.}$

The backwards implication follows from Lemma 5.37(ii), as
$\sigma(\underline{U},\overline{S})=\Delta_{\overline{S}}$. For the forwards
implication, suppose
$\underline{u}\cdot\overline{s}=\underline{v}\cdot\overline{t}$. Since
$(\underline{U},\overline{S})$ is an action pair, it follows from (A2) that
$\underline{u}\cdot\overline{s}^{+}=\underline{v}\cdot\overline{t}^{+}$. Also,
$(us^{+},s)=\underline{u}\cdot\overline{s}=\underline{v}\cdot\overline{t}=(vt^{+},t)$
gives $s=t$, so certainly $\overline{s}\mathrel{\sigma}\overline{t}$. ∎

We can now give the following covering theorem, which is the main result of
this section. In essence, it says that any semigroup $US$ arising from a weak
action pair $(U,S)$ is a homomorphic image of a semigroup
$U^{\prime}S^{\prime}$ arising from a _proper_ action pair
$(U^{\prime},S^{\prime})$, with $U^{\prime}\cong U$ and $S^{\prime}\cong S$.

######  Theorem 5.39.

Suppose $(U,S)$ is a weak action pair in a monoid $M$.

  1. (i)

There exists a proper action pair $(U^{\prime},S^{\prime})$ in a monoid
$M^{\prime}$, with $U^{\prime}\cong U$ and $S^{\prime}\cong S$, and a
surmorphism $\psi:U^{\prime}S^{\prime}\to US$.

  2. (ii)

If $S$ is a submonoid of $M$, then such a pair and surmorphism exist for
which:

     * •

$S^{\prime}$ is a submonoid of $M^{\prime}$,

     * •

$U^{\prime}$ is a subsemigroup of $U^{\prime}S^{\prime}$, and

     * •

the restriction $\psi\mathord{\restriction}_{U^{\prime}}$ is an isomorphism
$U^{\prime}\to U$.

  3. (iii)

If $s^{+}\in U$ for all $s\in S$ (including the case that $U$ is a submonoid
of $M$), then such a pair and surmorphism exist for which:

     * •

$S^{\prime}$ is a subsemigroup of $U^{\prime}S^{\prime}$, and

     * •

the restriction $\psi\mathord{\restriction}_{S^{\prime}}$ is an isomorphism
$S^{\prime}\to S$.

###### Proof.

(i). Since $(U,S)$ is a weak action pair, $S$ (and hence also $S^{1}$) acts on
$U^{1}$, and $S^{1}$ acts monoidally (cf. Remark 4.4). We then take

$M^{\prime}=\big{\\{}{(u,s)\in U^{1}\rtimes S^{1}}:{u=us^{+}}\big{\\}},\ \ \
U^{\prime}=\underline{U}=\\{{\underline{u}}:{u\in U}\\}\quad\text{and}\quad
S^{\prime}=\overline{S}=\\{{\overline{s}}:{s\in S}\\},$

as above. By Lemma 5.14, we have $U^{\prime}\cong U$ and $S^{\prime}\cong S$.
By Proposition 5.38, $(U^{\prime},S^{\prime})$ is a proper action pair in
$M^{\prime}$, and

$U^{\prime}S^{\prime}=\big{\\{}{(u,s)\in U\rtimes S}:{u=us^{+}}\big{\\}}.$

Define the map $\psi:U^{\prime}S^{\prime}\to US:(u,s)\mapsto us$. Since this
is the restriction to $U^{\prime}S^{\prime}$ of the morphism $\pi:U\rtimes
S\to US$ from Proposition 4.46, it follows that $\psi$ is a morphism. To
demonstrate surjectivity, let $x\in US$, so that $x=us$ for some $u\in U$ and
$s\in S$. As in Remark 4.4, we have $s=s^{+}s$, and so
$(\underline{u}\cdot\overline{s})\psi=(us^{+},s)\psi=us^{+}\cdot s=us=x$, with
$\underline{u}\cdot\overline{s}\in U^{\prime}S^{\prime}$.

(ii). Suppose $S$ is a submonoid of $M$. Clearly then
$S^{\prime}=\overline{S}$ is a submonoid of $M^{\prime}$, so that
${U^{\prime}=U^{\prime}\cdot\overline{1}\leq U^{\prime}S^{\prime}}$. Since
$\psi\mathord{\restriction}_{U^{\prime}}$ maps each
$\underline{u}=(u,1)\mapsto u\cdot 1=u$, it is clearly an isomorphism
$U^{\prime}\to U$.

(iii). The stated assumption ensures that $U^{\prime}S^{\prime}$ contains each
$\overline{s}=(s^{+},s)$, with $s\in S$: i.e., that $S^{\prime}\leq
U^{\prime}S^{\prime}$. Since $\psi$ maps each $\overline{s}=(s^{+},s)\mapsto
s^{+}s=s$, the final assertion follows. ∎

######  Remark 5.40.

We have already noted that a semigroup $US$ arising from a (weak) action pair
$(U,S)$ need not contain $U$ itself, or even an isomorphic copy of $U$; see
Example 4.32(iii). Moreover, $U^{\prime}$ is not contained in
$U^{\prime}S^{\prime}$ (in the above construction) if $S$ is not a submonoid
of $M$. Thus, $S$ being a submonoid is a necessary assumption in part (ii) of
Theorem 5.39, at least for the particular choice of $U^{\prime}=\underline{U}$
and $S^{\prime}=\overline{S}$ used in the proof.

In part (iii), note that the stronger assumption of $U$ being a submonoid of
$M$ implies that $U^{\prime}$ is a submonoid of $M^{\prime}$.

As an application of Theorem 5.39, we show how to deduce Theorem 5.11. We
begin with the case of left restriction _monoids_ :

######  Corollary 5.41 (cf. [9, Theorem 6.4]).

Every left restriction monoid has a proper cover.

###### Proof.

Consider a left restriction monoid $M$, and the action pair $(P,M)$ in $M$,
where $P$ is the semilattice of projections (cf. Proposition 4.37). Let
$(P^{\prime},M^{\prime})=(\underline{P},\overline{M})$ be the proper action
pair constructed in the proof of Theorem 5.39, and let

$C=P^{\prime}M^{\prime}=\big{\\{}{(u,s)\in P\rtimes M}:{u=us^{+}}\big{\\}}.$

So $\psi:C\to PM=M:(u,s)\mapsto us$ is a (monoid) surmorphism, and
$\psi\mathord{\restriction}_{P^{\prime}}$ is an isomorphism ${P^{\prime}\to
P}$.

It is routine to check that $C$ is left restriction under the operation
$(u,s)^{+}=(u,1)$. Thus, $P(C)=\big{\\{}{(u,1)}:{u\in
P}\big{\\}}=\\{{\underline{u}}:{u\in P}\\}=P^{\prime}$. Consequently, $\psi$
is projection-separating, since $\psi\mathord{\restriction}_{P^{\prime}}$ is
an isomorphism.

It remains to check that $\psi$ respects the unary operations of $S$ and $C$.
But for any $(u,s)\in C$ we use Lemma 4.36 to calculate

$(u,s)^{+}\psi=(u,1)\psi=u=us^{+}=(us)^{+}=((u,s)\psi)^{+},$

as required. ∎

We now show how the general case follows.

###### Proof of Theorem 5.11.

Let $S$ be a left restriction semigroup. If $S$ is a monoid, then Corollary
5.41 applies, so we assume otherwise. Let $M=S^{1}=S\sqcup\\{1\\}$, so that
$M$ is also left restriction (where we additionally define $1^{+}=1$). Let
$C=P^{\prime}M^{\prime}$ be the proper cover of $M$ from the proof of
Corollary 5.41, with covering morphism $\psi:C\to M:(u,s)\mapsto us$. Since
$M\setminus\\{1\\}=S\leq M$, is follows that

${\bf c}\psi=1\ \Leftrightarrow\ {\bf c}={\bf 1}\qquad\text{for all ${\bf
c}\in C$,}$ (5.42)

where we write ${\bf 1}=(1,1)$ for the identity of $C$.

Now we let $D=\\{{{\bf c}\in C}:{{\bf c}^{+}\not={\bf 1}}\\}$. By Lemma
4.42(iii), $D$ is a subsemigroup of $C$. Moreover, $D$ is closed under +, as

${\bf c}\in D\ \Rightarrow\ ({\bf c}^{+})^{+}={\bf c}^{+}\not={\bf 1}\
\Rightarrow\ {\bf c}^{+}\in D,$

by (L6). It follows that $D$ is a proper left restriction semigroup. By
(5.42), the restriction $\Psi=\psi\mathord{\restriction}_{D}$ maps $D$ into
$S=M\setminus\\{1\\}$, as

${\bf c}\psi=1\ \Rightarrow\ 1=1^{+}=({\bf c}\psi)^{+}={\bf c}^{+}\psi\
\Rightarrow\ {\bf c}^{+}={\bf 1}\ \Rightarrow\ {\bf c}\not\in D.$

In fact, $\Psi:D\to S$ is surjective, since for any $s\in S$ we have $s={\bf
c}\psi$ for some ${\bf c}\in C$, and we must in fact have ${\bf c}\in D$;
indeed, if ${\bf c}^{+}={\bf 1}$, then we would have $1={\bf 1}\psi={\bf
c}^{+}\psi=({\bf c}\psi)^{+}=s^{+}$, contradicting the fact that
$S=M\setminus\\{1\\}$ is closed under +. Thus, $\Psi$ is a surmorphism $D\to
S$, and it is projection-separating since $\psi$ is. ∎

######  Remark 5.43.

Consider an _inverse_ monoid $M$, regarded as a left restriction monoid under
$s^{+}=ss^{-1}$. So $P=P(M)=E(M)$ is the semilattice of (all) idempotents of
$M$; as usual, we denote this semilattice by $E$. Let $C$ and $\psi$ be as in
the proof of Corollary 5.41, so that

$C=\big{\\{}{(e,s)\in E\rtimes M}:{e=es^{+}}\big{\\}}=\big{\\{}{(e,s)\in
E\rtimes M}:{e\leq s^{+}}\big{\\}}$

is a proper left restriction monoid under $(e,s)^{+}=(e,1)$, and $\psi:C\to
M:(e,s)\mapsto es$ is a covering morphism. (In the above, $\leq$ is the
_natural partial order_ on $M$, defined by $s\leq t\ \Leftrightarrow\ s\in
Et$. Note that for $e,f\in E$ we have $e\leq f\ \Leftrightarrow\ e=ef$.) It is
routine to show the following:

  * •

$C$ is inverse, with inversion given by $(e,s)^{-1}=({}^{s^{-1}}e,s^{-1})$,

  * •

$E(C)=\big{\\{}{(e,s)}:{e,s\in E,\ e=es}\big{\\}}=\big{\\{}{(e,s)}:{e,s\in E,\
e\leq s}\big{\\}}$.

(To show that $C$ is inverse, one shows that it is regular, and $E(C)$ is
commutative. To verify the above claim about $(e,s)^{-1}$, one needs to note
that

$(e,s)\in C\ \Rightarrow\ e=es^{+}\ \Rightarrow\
{}^{s}({}^{s^{-1}}e)={}^{ss^{-1}}e={}^{s^{+}}e=s^{+}e(s^{+})^{-1}=s^{+}es^{+}=es^{+}s^{+}=e.)$

We note, however, that $C$ is not necessarily a proper inverse monoid, even
though it is a proper left restriction monoid. In fact, one can easily show
that $C$ is a proper inverse monoid if and only if $S$ is a proper inverse
monoid.

The key point here is that $P(C)=\big{\\{}{(e,1)}:{e\in E}\big{\\}}$ is
generally not equal to $E(C)$, so that $(E(C),C)$ might not be a proper action
pair, even though $(P(C),C)$ always is; cf. Proposition 4.37. (In fact,
$E(C)=P(C)\ \Leftrightarrow\ E=\\{1\\}$, as $(e,e)\in E(C)\setminus P(C)$ for
any $e\in E\setminus\\{1\\}$.) Similarly, $\psi$ is generally not idempotent-
separating, even though it is projection-separating, as we have $(e,s)\psi=e$
for any idempotent $(e,s)$ of $C$. (So again, $\psi$ is idempotent-separating
if and only if $E=\\{1\\}$. As we have noted on a number of occasions,
$E=\\{1\\}$ is equivalent to $M$ being a group.)

####  5.4 Embedding theorems for proper monoids I

In this section, and the subsequent three sections, we turn to the task of
finding an appropriate generalisation of Theorem 5.12 in the context of proper
action pairs. That is, we consider a semigroup $US$ arising from a proper pair
$(U,S)$, and seek to embed $US$ in a semidirect product
$\mathcal{U}\rtimes(S/\sigma)$, where $\mathcal{U}$ is some semigroup
containing $U$, and where $\sigma$ is the congruence on $S$ given in
Definition 5.21. For technical reasons, we will need to assume that $U$ and
$S$ are both submonoids of the over-monoid $M$. However, this does not end up
hampering us, in the sense that we can still derive (the full semigroup
version of) Theorem 5.12 as a corollary.

The main result of the current section is Theorem 5.47. This states that a
$(U,S)$-proper monoid $M=US$ embeds in a suitable semidirect product
$\mathcal{U}\rtimes(S/\sigma)$, under the assumption that the submonoid
$P=\langle S^{+}\rangle$ of $U$ (see Definition 5.21) is central in $U$. This
centrality condition is quite natural. For example, when $M$ is a proper left
restriction monoid, arising from the proper action pair $(P,M)$, we have
already observed that $P=U$ is the semilattice of projections of $M$, and is
therefore commutative, and hence central in itself. After proving Theorem
5.47, we deduce the monoid version of Theorem 5.12 in Corollary 5.58, and then
show that the general case follows.

The centrality assumption just discussed is in fact not necessary to obtain an
embedding of the desired kind. Indeed, in the next section we prove Theorem
5.60, which is essentially the same as Theorem 5.47 but with the centrality
assumption removed. We do, however, have strong motivation for stating and
proving the two separate embedding results. First, the construction of the
monoid $\mathcal{U}$ is somewhat more abstract in the general case. By
contrast, the monoid $\mathcal{U}$ used in the current section is more
transparent, and the corresponding semidirect product
$\mathcal{U}\rtimes(S/\sigma)$ is in fact an (unrestricted) wreath product
$I\operatorname{wr}{}(S/\sigma)$, in the sense of [105, 90], for a suitable
monoid $I$, as we explain in a little more detail below. Another advantage of
the current construction is that when $M$ is a proper left restriction monoid,
$\mathcal{U}$ is easily seen to be a semilattice, and as noted above this
allows us to deduce Theorem 5.12 as a corollary.

In this section, and the next three, an important role will be played by
certain special factorisations of elements of a semigroup $US$ arising from a
(weak) action pair $(U,S)$:

######  Definition 5.44.

Let $(U,S)$ be a weak action pair. A _natural factorisation_ for an element
$a\in US$ is a pair $(u,s)\in U\times S$ such that $a=us$ and $u=us^{+}$.

The next result lists some special properties of these factorisations. Note
that we can think of a natural factorisation of $a\in US$ as an element of the
semigroup

$\big{\\{}{(u,s)\in U\rtimes S}:{u=us^{+}}\big{\\}}$

that maps to $a$ under the surmorphism $\pi:U\rtimes S\to US$ from Proposition
4.46; cf. Remark 4.47. The first two parts of the next result can be proved
with reference to this viewpoint, but we give simple direct proofs for
convenience.

######  Lemma 5.45.

Let $(U,S)$ be a weak action pair.

  1. (i)

Every element of $US$ has a natural factorisation.

  2. (ii)

If $(u,s)$ and $(v,t)$ are natural factorisations of elements $a$ and $b$ of
$US$, respectively, then $(u\cdot{}^{s}v,st)$ is a natural factorisation of
$ab$.

  3. (iii)

If $(U,S)$ is an action pair, and if $(u,s)$ and $(v,t)$ are natural
factorisations for the same element of $US$, then $u=v$.

  4. (iv)

If $(U,S)$ is a proper action pair, and if $(u,s)$ and $(v,t)$ are natural
factorisations for the same element of $US$, then $s\mathrel{\sigma}t$.

###### Proof.

(i). Let $a\in US$. We certainly have $a=vs$ for some $v\in U$ and $s\in S$,
and we take $u=vs^{+}$. Then since $s=s^{+}s$ (cf. Remark 4.4) we have
$us=vs^{+}s=vs=a$, while $us^{+}=vs^{+}s^{+}=vs^{+}=u$, by Lemma 3.1(i).

(ii). We obtain $ab=u\cdot{}^{s}v\cdot st$ right from (A1). Combining Lemma
3.1(ii) with $v=vt^{+}$, we have

$u\cdot{}^{s}v\cdot(st)^{+}=u\cdot{}^{s}v\cdot{}^{s}(t^{+})=u\cdot{}^{s}(vt^{+})=u\cdot{}^{s}v.$

(iii). Suppose $us=vt$, where $(u,s)$ and $(v,t)$ are natural. By (A2) we have
$us^{+}=vt^{+}$, and since $u=us^{+}$ and $v=vt^{+}$ (by naturality), it
follows that $u=v$.

(iv). This follows immediately from (5.26). ∎

######  Remark 5.46.

On many occasions, Lemma 5.45 will allow us to define a morphism $\psi$ whose
domain $US$ arises from a proper action pair $(U,S)$. Given $a\in US$, parts
(i), (iii) and (iv) allow us to unambiguously define $a\psi$, by explaining
how to construct it from $u$ and the $\sigma$-class of $s$ for _any_ natural
factorisation $(u,s)$ of $a$. Part (ii) will be useful in showing that such a
map $\psi$ is a morphism.

We are now almost ready to state our first embedding theorem. The theorem
concerns a proper action pair $(U,S)$ in a monoid $M$, where $U$ and $S$ are
both submonoids. Without loss of generality, we may assume that $M=US$, so
that $M$ is $(U,S)$-proper. We continue to use the notation $S^{+}$,
$P=\langle S^{+}\rangle$ and $\sigma=\kappa^{\sharp}$ from Section 5.3. So $P$
is an idempotent-generated submonoid of $U$, and $\sigma$ is a congruence on
$S$. For $s\in S$ we write $\hat{s}$ for the $\sigma$-class of $s$.

As we have already noted, the following theorem concerns the case in which
$P=\langle S^{+}\rangle$ is central in $U$, meaning that $pu=up$ for all $p\in
P$ and $u\in U$. This is of course equivalent to the generating set $S^{+}$
being central in $U$.

######  Theorem 5.47.

Let $M=US$ be a $(U,S)$-proper monoid, and suppose $P=\langle S^{+}\rangle$ is
central in $U$. Then there exists a semidirect product
$\mathcal{M}=\mathcal{U}\rtimes(S/\sigma)$ and an embedding
$\psi:M\to\mathcal{M}$ such that:

  1. (i)

$\mathcal{U}$ contains a subsemigroup $U^{\prime}$ isomorphic to $U$,

  2. (ii)

$\psi\mathord{\restriction}_{U}:U\to U^{\prime}\rtimes\\{\hat{1}\\}$ is an
isomorphism,

  3. (iii)

$\mathcal{U}$ is commutative if $U$ is commutative.

We build towards the proof of the theorem with a series of preliminary
results. For the remainder of this section, we assume that $M=US$ is a
$(U,S)$-proper monoid (with $U$ and $S$ submonoids), and that $P=\langle
S^{+}\rangle$ is central in $U$.

We begin with a result that gives a convenient equational formulation of
$\sigma$.

######  Lemma 5.48.

The monoid $P$ is a semilattice, and consequently $\sigma=\big{\\{}{(s,t)\in
S\times S}:{t^{+}s=s^{+}t}\big{\\}}$.

###### Proof.

Commutativity of $P$ follows from centrality. Since every element of $S^{+}$
is an idempotent (cf. Lemma 3.1(i)), so too therefore is every element of $P$.
The assertion regarding $\sigma$ follows from Lemma 5.32; cf. Remark 5.33. ∎

In what follows, it will be convenient to write

$\mathcal{S}=S/\sigma=\\{{\hat{s}}:{s\in S}\\}.$

So we must construct a monoid $\mathcal{U}$, a semidirect product
$\mathcal{M}=\mathcal{U}\rtimes\mathcal{S}$, and a suitable embedding
${\psi:M\to\mathcal{M}}$. The monoid $\mathcal{U}$ will in fact be of the form
$I^{\mathcal{S}}$ for some monoid $I$, and then $\mathcal{U}\rtimes S$ will be
an _unrestricted wreath product_
$I\operatorname{wr}\mathcal{S}=I^{\mathcal{S}}\rtimes\mathcal{S}$ in the sense
of [105, 90]. Since we will be studying a different kind of (transformational)
wreath product in Chapter 9, and since we wish to emphasise the monoid
$\mathcal{U}$ itself in our construction (rather than $I$), we will not speak
explicitly of (any kind of) wreath products in the current chapter. However,
in order to prepare the reader for our construction we briefly recall the
definition of the unrestricted wreath product $V\operatorname{wr}T$, for
semigroups $V$ and $T$. First, $V^{T}$ denotes as usual the semigroup of all
functions $T\to V$ under componentwise product, which we denote by $\star$.
Given $f,g\in V^{T}$, the product $f\star g\in V^{T}$ is defined by $x(f\star
g)=xf\cdot xg$ for $x\in T$ (with the product $xf\cdot xg$ taken in $V$).
Next, $T$ has a left action on $V^{T}$ by semigroup morphisms, given by
$(t,f)\mapsto{}^{t}\\!f$, where the latter is defined by $x({}^{t}\\!f)=(xt)f$
for $x\in T$. This allows for the formation of the semidirect product
$V^{T}\rtimes T$, as in Definition 3.4, and this is taken as the definition of
$V\operatorname{wr}T$. As noted above, we will no longer refer explicitly to
this kind of wreath product, even though the semidirect product
$\mathcal{M}=\mathcal{U}\rtimes S$ will be of this form.

The construction of $\mathcal{U}$ begins with the _power semigroup_ of $U$:

${\mathcal{P}}(U)=\\{{V}:{V\subseteq U}\\}\qquad\text{with set product}\qquad
V\cdot W=\\{{vw}:{v\in V,\ w\in W}\\}\qquad\text{for $V,W\subseteq U$.}$
(5.49)

We then take $I$ to be the principal ideal of ${\mathcal{P}}(U)$ generated by
$P$. Since ${\mathcal{P}}(U)$ is a monoid with identity $\\{1\\}$, the ideal
$I={\mathcal{P}}(U)\cdot P\cdot{\mathcal{P}}(U)$ consists of all subsets of
$U$ of the form $V\cdot P\cdot W$, with $V,W\subseteq U$. In fact, since $P$
is a central submonoid of $U$, we have

$\displaystyle I$ $\displaystyle={\mathcal{P}}(U)\cdot
P=P\cdot{\mathcal{P}}(U)=\\{{V\cdot P=P\cdot V}:{V\subseteq U}\\},$ and
moreover $\displaystyle I$ $\displaystyle=P\cdot{\mathcal{P}}(U)\cdot
P=\\{{V\in{\mathcal{P}}(U)}:{V=V\cdot P=P\cdot V}\\}.$

Thus, $I$ is a (local) monoid with identity $P$ (and zero $\varnothing$). We
note, however, that $I$ is generally not a _submonoid_ of ${\mathcal{P}}(U)$,
as the respective identities of these monoids are $P$ and $\\{1\\}$.

We now let

$\mathcal{U}=I^{\mathcal{S}}$

be the set of all functions $\mathcal{S}\to I$. We denote the componentwise
product in $\mathcal{U}$ by $\star$. So for $f,g\in\mathcal{U}$, we have
$f\star g:\mathcal{S}\to I$, defined by

$\hat{x}(f\star g)=\hat{x}f\cdot\hat{x}g\qquad\text{for each $x\in S$.}$

Note here that $\hat{x}f$ and $\hat{x}g$ are both elements of $I$, and in
particular of ${\mathcal{P}}(U)$, so $\hat{x}f\cdot\hat{x}g$ denotes the set
product in $I$. It is easy to see that $\star$ is associative, so that
$\mathcal{U}$ is a semigroup. In fact, $\mathcal{U}$ is a monoid, whose
identity ${\bf 1}$ is the constant map with image $\\{P\\}$: i.e.,
$\hat{x}{\bf 1}=P$ for all $x\in S$.

The first part of the next result verifies the third item in Theorem 5.47. The
second part concerns the special case in which $U=P$; since $P$ is central in
$U$, certainly $U=P$ is commutative in this case.

######  Lemma 5.50.

  1. (i)

If $U$ is commutative, then so too is $\mathcal{U}$.

  2. (ii)

If $U=P$ (is commutative), then $U$ and $\mathcal{U}$ are both semilattices.

###### Proof.

(i). If $U$ is commutative, then obviously ${\mathcal{P}}(U)$ is as well. But
then so too is its subsemigroup $I$, and hence also
$\mathcal{U}=I^{\mathcal{S}}$.

(ii). We already observed in Lemma 5.48 that $P$ is a semilattice. Since
$\mathcal{U}=I^{\mathcal{S}}$, it is enough to show that $I$ is a semilattice.
An arbitrary element of $I$ is a set of the form $P\cdot V\cdot P$ for some
$V\subseteq U=P$. These sets are precisely the ideals of $P$, and since the
set product of two ideals of a semilattice is their intersection, we see that
$I$ is the $\cap$-semilattice of ideals of $P$. ∎

To define the semidirect product $\mathcal{M}=\mathcal{U}\rtimes\mathcal{S}$,
we need an action of $\mathcal{S}$ on $\mathcal{U}$, and this is also
componentwise. For $s\in S$ and $f\in\mathcal{U}$, we define
${}^{\hat{s}}\\!f\in\mathcal{U}$ by

$\hat{x}({}^{\hat{s}}\\!f)=(\hat{x}\hat{s})f\qquad\text{for all $x\in S$.}$

Clearly ${}^{\hat{s}}\\!f$ is well defined: i.e., the above definition does
not depend on the choice of representative $s$ of the $\sigma$-class
$\hat{s}$. It is also a routine matter to check that we have a monoidal action
by monoid morphisms: i.e., that

${}^{\hat{s}}({}^{\hat{t}}\\!f)={}^{\hat{s}\hat{t}}\\!f,\qquad{}^{\hat{s}}(f\star
g)={}^{\hat{s}}\\!f\star{}^{\hat{s}}\\!g,\qquad{}^{\hat{1}}\\!f=f\qquad\text{and}\qquad{}^{\hat{s}}{\bf
1}={\bf 1}\qquad\text{for all $s,t\in S$ and $f,g\in\mathcal{U}$.}$

We may therefore form the semidirect product

$\mathcal{M}=\mathcal{U}\rtimes\mathcal{S}=\big{\\{}{(f,\hat{s})}:{f\in\mathcal{U},\
s\in S}\big{\\}}\qquad\text{with
operation}\qquad(f,\hat{s})\cdot(g,\hat{t})=(f\star{}^{\hat{s}}\\!g,\hat{s}\hat{t}).$

By Lemma 3.11(iii), $\mathcal{M}$ is a monoid with identity $({\bf
1},\hat{1})$. (As noted above, our monoid $\mathcal{M}$ is an unrestricted
wreath product $I\wr\mathcal{S}=I^{\mathcal{S}}\rtimes\mathcal{S}$.)

Now that we have defined the monoid $\mathcal{M}=\mathcal{U}\rtimes S$, we
wish to construct an embedding ${\psi:M\to\mathcal{M}}$. As in Remark 5.46,
$\psi$ will be defined in terms of natural factorisations. Specifically, given
a natural factorisation $(u,s)$ for an element $a\in M$, we will define
$a\psi=(f_{u},\hat{s})$, for a certain $f_{u}\in\mathcal{U}$. We now turn to
the definition of these elements $f_{u}$, and establish the important
properties of the mapping $u\mapsto f_{u}$.

To this end, let $u\in U$. For $s\in S$, we define the subset

$V_{\hat{s},u}=\\{{{}^{t}u}:{t\in\hat{s}}\\}\subseteq U.$ (5.51)

We then define the function $f_{u}\in\mathcal{U}$ by

$\hat{s}f_{u}=V_{\hat{s},u}\cdot P\qquad\text{for all $s\in S$.}$

Finally, we define the map

$\phi:U\to\mathcal{U}:u\mapsto f_{u}.$ (5.52)

######  Lemma 5.53.

The map $\phi:U\to\mathcal{U}$ is a monomorphism.

###### Proof.

We begin by showing that $\phi$ is injective. To do so, suppose $f_{u}=f_{v}$
for some $u,v\in U$. In particular, $\hat{1}f_{u}=\hat{1}f_{v}$, so that

$u={}^{1}u\cdot 1\in V_{\hat{1},u}\cdot
P=\hat{1}f_{u}=\hat{1}f_{v}=V_{\hat{1},v}\cdot P.$

It follows that $u={}^{t}v\cdot p$ for some $t\in\hat{1}$ and $p\in P$.
Combining this with Lemma 5.34(ii), and centrality of $P$, it follows that
$u=t^{+}v\cdot p=t^{+}p\cdot v\in Pv$, as $t^{+}p\in P$. By symmetry, $v\in
Pu$. Since $P$ is a semilattice (cf. Lemma 5.48), it follows from Lemma 2.5
that $u=v$, as required.

To show that $\phi$ is a morphism, suppose $u,v\in U$. We must show that
$f_{uv}=f_{u}\star f_{v}$: i.e., that

$\displaystyle\hat{s}f_{uv}$ $\displaystyle=\hat{s}f_{u}\cdot\hat{s}f_{v}$ for
all $s\in S$. By definition, and using centrality of $P$, this reduces to
showing that $\displaystyle V_{\hat{s},uv}\cdot P$
$\displaystyle=V_{\hat{s},u}\cdot V_{\hat{s},v}\cdot P$ for all $s\in S$.
(5.54)

For the fowards inclusion, let $x\in V_{\hat{s},uv}\cdot P$. Then
$x={}^{t}(uv)\cdot p$ for some $t\in\hat{s}$ and $p\in P$, and so

$x={}^{t}u\cdot{}^{t}v\cdot p\in V_{\hat{s},u}\cdot V_{\hat{s},v}\cdot P.$

Conversely, let $x\in V_{\hat{s},u}\cdot V_{\hat{s},v}\cdot P$, so that
$x={}^{t}u\cdot{}^{z}v\cdot p$ for some $t,z\in\hat{s}$ and $p\in P$. Since
$t\mathrel{\sigma}z$ (as $t,z\in\hat{s}$), it follows from Lemma 5.48 that
$z^{+}\cdot t=t^{+}\cdot z$. Lemma 4.25 then gives

$z^{+}\cdot{}^{t}v=t^{+}\cdot{}^{z}v.$ (5.55)

But then

$\displaystyle x={}^{t}u\cdot{}^{z}v\cdot p$ $\displaystyle={}^{t}u\cdot
t^{+}\cdot{}^{z}v\cdot p$ by Lemma 3.1(iv) $\displaystyle={}^{t}u\cdot
z^{+}\cdot{}^{t}v\cdot p$ by (5.55) $\displaystyle={}^{t}u\cdot{}^{t}v\cdot
z^{+}p$ by centrality of $P$ $\displaystyle={}^{t}(uv)\cdot z^{+}p\in
V_{\hat{s},uv}\cdot P.$

This completes the proof of (5.54), and hence of the lemma. ∎

It follows that $\phi$ is an isomorphism from $U$ to its image, which is the
subsemigroup

$U^{\prime}=\\{{f_{u}}:{u\in U}\\}\leq\mathcal{U}.$

Note that while $U^{\prime}$ is a monoid (as $U$ is), it is generally not a
submonoid of $\mathcal{U}$, as the identities of $U^{\prime}$ and
$\mathcal{U}$ are $f_{1}$ and ${\bf 1}$, respectively.

We can now tie together the loose ends. The next proof uses the natural
factorisations from Definition 5.44.

###### Proof of Theorem 5.47.

We define the map

$\psi:M=US\to\mathcal{M}=\mathcal{U}\rtimes\mathcal{S}$

by

$a\psi=(f_{u},\hat{s})\qquad\text{for any natural factorisation $(u,s)$ of
$a\in M=US$.}$

This is well defined by Lemma 5.45; cf. Remark 5.46.

We begin by showing that $\psi$ is injective. To do so, suppose $a\psi=b\psi$
for some $a,b\in M$, and fix natural factorisations $(u,s)$ and $(v,t)$ for
$a$ and $b$, respectively. Then $(f_{u},\hat{s})=a\psi=b\psi=(f_{v},\hat{t})$,
so that $f_{u}=f_{v}$ and $\hat{s}=\hat{t}$. The former implies $u=v$, by
Lemma 5.53. Combining this with naturality, we obtain $us^{+}=u=v=vt^{+}$.
Since also $s\mathrel{\sigma}t$ (as $\hat{s}=\hat{t}$), and since $(U,S)$ is
proper, it follows from (5.26) that $us=vt$: i.e., $a=b$.

Clearly ${u\psi=(f_{u},\hat{1})}$ for all $u\in U$, so
$\psi\mathord{\restriction}_{U}$ maps $U$ bijectively onto
$U^{\prime}\rtimes\\{\hat{1}\\}$, and we have already observed that
$U^{\prime}\cong U$. We also observed in Lemma 5.50 that $\mathcal{U}$ is
commutative if $U$ is.

It remains to show that $\psi$ is a morphism. So fix some $a,b\in M$, with
natural factorisations $(u,s)$ and $(v,t)$. We must show that
$(ab)\psi=a\psi\cdot b\psi$. We first observe that

$a\psi\cdot
b\psi=(f_{u},\hat{s})\cdot(f_{v},\hat{t})=(f_{u}\star{}^{\hat{s}}\\!f_{v},\hat{s}\hat{t}).$

By Lemma 5.45(ii), $ab$ has natural factorisation $(u\cdot{}^{s}v,st)$, and so

$(ab)\psi=(f_{u\cdot{}^{s}v},\hat{s}\hat{t}).$

Thus, the proof will be complete if we can show that
$f_{u}\star{}^{\hat{s}}\\!f_{v}=f_{u\cdot{}^{s}v}$. To do so, let $x\in S$ be
arbitrary. Then

$\hat{x}f_{u\cdot{}^{s}v}=V_{\hat{x},u\cdot{}^{s}v}\cdot
P\qquad\text{and}\qquad\hat{x}(f_{u}\star{}^{\hat{s}}\\!f_{v})=\hat{x}f_{u}\cdot(\hat{x}\hat{s})f_{v}=V_{\hat{x},u}\cdot
V_{\hat{x}\hat{s},v}\cdot P,$

again using centrality of $P$. So we must show that

$V_{\hat{x},u\cdot{}^{s}v}\cdot P=V_{\hat{x},u}\cdot V_{\hat{x}\hat{s},v}\cdot
P.$ (5.56)

For the forwards inclusion, let $c\in V_{\hat{x},u\cdot{}^{s}v}\cdot P$, so
that $c={}^{y}(u\cdot{}^{s}v)\cdot p$ for some $y\in\hat{x}$ and $p\in P$.
Then $c={}^{y}u\cdot{}^{ys}v\cdot p$, with ${}^{y}u\in V_{\hat{x},u}$ and
${}^{ys}v\in V_{\hat{x}\hat{s},v}$ (as $y\mathrel{\sigma}x\ \Rightarrow\
ys\mathrel{\sigma}xs$).

Conversely, let $d\in V_{\hat{x},u}\cdot V_{\hat{x}\hat{s},v}\cdot P$, so that
$d={}^{y}u\cdot{}^{z}v\cdot p$ for some $y\in\hat{x}$, $z\in\hat{x}\hat{s}$
and $p\in P$. Since $z\mathrel{\sigma}xs\mathrel{\sigma}ys$, Lemma 5.48 gives
$(ys)^{+}\cdot z=z^{+}\cdot ys$. It then follows from Lemma 4.25 that

$(ys)^{+}\cdot{}^{z}v=z^{+}\cdot{}^{ys}v.$ (5.57)

But then

$\displaystyle d={}^{y}u\cdot{}^{z}v\cdot p$
$\displaystyle={}^{y}(us^{+})\cdot{}^{z}v\cdot p$ as $u=us^{+}$, by naturality
$\displaystyle={}^{y}u\cdot{}^{y}(s^{+})\cdot{}^{z}v\cdot p$
$\displaystyle={}^{y}u\cdot(ys)^{+}\cdot{}^{z}v\cdot p$ by Lemma 3.1(ii)
$\displaystyle={}^{y}u\cdot z^{+}\cdot{}^{ys}v\cdot p$ by (5.57)
$\displaystyle={}^{y}u\cdot{}^{ys}v\cdot z^{+}p$ by centrality of $P$
$\displaystyle={}^{y}(u\cdot{}^{s}v)\cdot z^{+}p.$

This completes the proof of (5.56), and hence of the lemma, since
${}^{y}(u\cdot{}^{s}v)\in V_{\hat{x},u\cdot{}^{s}v}$ and ${z^{+}p\in P}$. ∎

This time we may deduce Theorem 5.12 as a corollary. Again we begin with the
monoid case:

######  Corollary 5.58.

Let $S$ be a proper left restriction monoid with semilattice of projections
${P=P(S)}$. Then $S$ can be embedded in a semidirect product
$Q\rtimes(S/\sigma)$, where $Q$ is a semilattice containing $P$.

###### Proof.

We aim to apply Theorem 5.47 to the pair $(U,S)$ where $U=P$. We first observe
that $(P,S)$ is indeed a proper action pair, by Proposition 5.29. Since $P$ is
a semilattice, it is certainly central in $U=P$. Theorem 5.47 then does indeed
apply, and it tells us that $S=PS$ embeds in a semidirect product
$\mathcal{U}\rtimes\mathcal{S}$, where $\mathcal{U}$ is a monoid containing
(an isomorphic copy) of $U=P$, and where $\mathcal{S}=S/\sigma$. By Lemma
5.50(ii), $\mathcal{U}$ is a semilattice. ∎