Perfection for semigroups

Abstract

We call a semigroup right perfect if every object in the category of unitary right acts over that semigroup has a projective cover. In this paper, we generalize results about right perfect monoids to the case of semigroups. In our main theorem, we will give nine conditions equivalent to right perfectness of a factorizable semigroup. We also prove that right perfectness is a Morita invariant for factorizable semigroups.

1. Preliminaries

Perfect and semiperfect rings with identity were introduced in [3] by Bass. Such rings have been studied in numerous articles. Perfect monoids were defined in [9] by Isbell, who proved that a monoid is left perfect if and only if it satisfies two conditions referred to as (A) and (D). Fountain [6] gave more conditions that are equivalent to left perfectness. In particular, he proved that a monoid is left perfect if and only if all of its weakly flat (in the sense of being pullback flat) unitary left acts are projective. Subsequently, a number of other papers related to perfect monoids or pomonoids have appeared (e.g., [2, 8, 10–12, 20]).

The purpose of this paper is to define (right) perfect semigroups and show that many descriptions of right perfect monoids can be transferred to the case of factorizable semigroups. We say that a semigroup is right perfect if the category of all its unitary right acts is perfect in the sense that each of its objects has a projective cover. In our main theorem, we will give nine different conditions that are equivalent to right perfectness of a factorizable semigroup. Some of these are familiar from the monoid case, but some of them are new. For example, a factorizable semigroup is right perfect if and only if all right sequence acts over it are projective.

It is well known that right perfectness and right semiperfectness are Morita invariants for rings with identity (Corollary 27.8 and Corollary 28.6 in [1]). We prove that right perfectness is also a Morita invariant for factorizable semigroups. Our results allow to conclude that the class of perfect semigroups contains all completely (0-)simple semigroups and all nilpotent semigroups.

Throughout this paper, S denotes a nonempty semigroup (if not stated otherwise), and S ¹ is the monoid obtained from S by adjoining an external identity 1 (no matter if S already has an identity element or not). All S-acts, subacts and ideals we consider are nonempty, unless stated otherwise explicitly.

An act A _S is called unitary if $A=AS$, that is, for every $a\in A$, there exist $a^{\prime}\in A$ and $s\in S$ such that $a=a^{\prime} s$. The category of unitary right S-acts with S-act homomorphisms as morphisms will be denoted by $\mathsf{UAct}_S$. Any S-act A _S can be made an S ¹-act by defining $a1=a$ for every $a\in A$.

Tensor products of acts over semigroups are defined in the same way as tensor products of acts over monoids (see [13, Construction 2.5.4]). The tensor product of a right act A _S and a left act ${_SB}$ is the quotient set $A\otimes_S B:=(A\times B)/\vartheta$, where ϑ is the smallest equivalence relation on A × B containing the set $\{((as,b),(a,sb))\mid a\in A,b\in B,s\in S\}$. The ϑ-class of a pair (a, b) is denoted by $a\otimes b$, so $A\otimes_S B = \{a\otimes b\mid a\in A,b\in B\}$. We will often write $A\otimes B$ instead of $A\otimes_S B$. Each right S-act A _S induces in a natural way a tensor functor $A\otimes _S -: {_S\mathsf{Act}}\to \mathsf{Set}$ (cf. [13, Construction 2.5.15]), where $ {_S\mathsf{Act}}$ is the category of all left S-acts and $\mathsf{Set}$ is the category of sets. We will use the properties of this functor to define different flatness properties of A _S.

An act A _S is called firm if the mapping $\mu_A:A\otimes S\to A,\ a\otimes s \mapsto as$ is bijective. The category of firm right S-acts is denoted by $\mathsf{FAct}_S$. The categories $\mathsf{UAct}_S$ and $\mathsf{FAct}_S$ are full subcategories of $\mathsf{Act}_S$, the category of all right S-acts. In all three categories, coproducts are disjoint unions.

A semigroup S is called firm if S _S is a firm right S-act. A semigroup S is called factorizable if $S^2=S$, i.e., any element in S is a product of two elements.

We will need the following result about firm acts.

Lemma 1.1. [16, Corollary 2.4]

If A _S is a unitary act over a factorizable semigroup, then $A\otimes _S S$ is a firm right S-act.

Let A _S be a right S-act. Finitely generated subacts of A _S are those of the form $FS^{1}:= \{fs \mid f\in F, s\in S^{1}\}$, where $F\subseteq A$ is a finite subset. An act A _S is called cyclic if there exists $a\in A$ such that $A= aS^1 = \{a\}\cup aS$. An S-act is locally cyclic if all of its finitely generated subacts are cyclic. It can be shown that A _S is locally cyclic if and only if

\begin{equation*} (\forall a,a^{\prime}\in A)(\exists a^{\prime\prime}\in A)(\exists s,s^{\prime}\in S^1)(a=a^{\prime\prime} s\wedge a^{\prime}=a^{\prime\prime} s^{\prime}). \end{equation*}

Essential epimorphisms are defined dually to essential monomorphisms.

Definition 1.2. An epimorphism $f: P\to A$ in a category $\mathcal{C}$ is called an essential epimorphism if, for every morphism $g:Q\to P$ in $\mathcal{C}$,

\begin{equation*} fg\ \mbox{is an epimorphism} \implies g\ \mbox{is an epimorphism.} \end{equation*}

It is easy to see that the composite of two essential epimorphisms is an essential epimorphism.

Definition 1.3. If $f:P\to A$ is an essential epimorphism in a category $\mathcal{C}$, then the object P is called a cover of the object A. If the object P is projective in $\mathcal{C}$, then it is called a projective cover of A.

Similar to Proposition 1.3 in [17], one can prove that epimorphisms in $\mathsf{UAct}_S$ are precisely the surjective morphisms.

Lemma 1.4. Let S be a semigroup. An epimorphism $f:P_S\to A_S$ in $\mathsf{UAct}_S$ is essential if and only if $f|_Q$ is not surjective for every unitary proper subact Q of P _S.

Proof. Necessity. Assume that $f:P_S\to A_S$ is an essential epimorphism in $\mathsf{UAct}_S$ and let Q be a unitary proper subact of P _S. Then the inclusion map $\iota: Q\to P$ is a morphism in $\mathsf{UAct}_S$ and $f|_Q = f\iota$. If $f\iota$ was surjective, then ι would be surjective due to essentiality of f.

Sufficiency. Let $f:P_S\to A_S$ and $g:U_S\to P_S$ be morphisms in $\mathsf{UAct}_S$ such that f and fg are surjective. It is easy to see that g(U) is a unitary subact of P _S and $f\vert _{g(U)}$ is surjective. By assumption, $g(U) = P$, so g is surjective and therefore f is essential.

Definition 1.5. We say that a semigroup S is:

• • right perfect if every object of $\mathsf{UAct}_S$ has a projective cover;

• • right semiperfect if every unitary cyclic right S-act has a projective cover in $\mathsf{UAct}_S$.

Dually, one can define left (semi)perfect semigroups. A semigroup is (semi)perfect if it is both right and left (semi)perfect.

Remark 1.6. We have defined right semiperfect semigroups as nonadditive analogues of right semiperfect rings (see [3]).

Remark 1.7. Note that if S is a monoid, then an act A _S is unitary if and only if $a1=a$ for every $a\in A$. Hence, S is right perfect in the sense of Definition 1.5 if and only if it is right perfect in the sense of [9] and [6]. Thus, the results in those articles are special cases of our more general approach.

Definition 1.8. (cf. [5])

A subsemigroup T of a semigroup S is called left unitary if, for every $s,t\in S$,

\begin{equation*} t,ts\in T \implies s\in T. \end{equation*}

In this paper, we will use the following conditions on a semigroup S.

Definition 1.9. A semigroup S satisfies Condition:

• • (A) if every right S-act satisfies the ascending chain condition (ACC) for cyclic subacts;

• • (D) if every left unitary subsemigroup of S contains a minimal right ideal generated by an idempotent;

• • ($M_L$) if S satisfies the descending chain condition (DCC) for principal left ideals of S.

2. Indecomposable and projective acts

An act is called indecomposable if it is not a disjoint union of two nonempty subacts.

Proposition 2.1. [4, Lemma 4]

Every right S-act is a disjoint union of indecomposable subacts.

In what follows, we will call the indecomposable subacts of A _S the indecomposable components of A _S. It is easy to see that each cyclic act is indecomposable. We will need the following result about projective acts.

Theorem 2.2. [4, Lemma 2]

Let S be a semigroup and let $\mathcal{C}$ be a full subcategory of $\mathsf{Act} _S$ which is closed under coproducts. Let $P_S = \bigsqcup_{i\in I} P_i$, where $P_i\in \mathcal{C}$ for every $i\in I$. Then P _S is projective in $\mathcal{C}$ if and only if P _i is projective in $\mathcal{C}$ for every $i\in I$.

Note that coproducts in $\mathsf{Act}_S$ are disjoint unions, and both $\mathsf{UAct}_S$ and $\mathsf{FAct}_S$ are closed under disjoint unions.

Lemma 2.3. Let e be an idempotent in a semigroup S. Then the right S-act eS is a projective object in $\mathsf{Act}_S$, $\mathsf{UAct}_S$ and $\mathsf{FAct}_S$.

Proof. By the dual of Lemma 3.1(1) in [18], the act eS is firm, thus also unitary. In all these categories, epimorphisms are precisely the surjective morphisms. Now, in any of these categories, we consider a morphism $f: eS\to B_S$ and an epimorphism $\pi: A_S \to B_S$. Denote $b:= f(e)$ and choose $a\in A$ such that $f(e) = \pi(a)$. Consider a homomorphism $g:eS\to A_S $ defined by

\begin{equation*} g(es):=aes,\quad s\in S. \end{equation*}

Then $\pi g =f$ because $ (\pi g)(es) = \pi(aes) = \pi(a)es = f(e)es = f(es). $

The last two results allow us to prove the following result.

Corollary 2.4. Every nilpotent semigroup is perfect. Every nilsemigroup is semiperfect.

Proof. If S is nilpotent, then $S^n = 0$ for some $n\in\mathbb N$. If A _S is unitary, then, for any $a\in A$, we can find $a^{\prime} \in A$ and $s_1,\ldots,s_n\in S$ such that

\begin{equation*} a=a^{\prime} s_n\ldots s_2s_1 = a^{\prime} 0 = a^{\prime} 00 = a0. \end{equation*}

So $aS^1= \{a\}$ is a one-element subact, which is isomorphic to the act $0S=\{0\}$. The last is projective in $\mathsf{UAct}_S$ by Lemma 2.3. Thus,

\begin{equation*} A_S = \bigsqcup_{a\in A} \{a\} \cong \bigsqcup_{a\in A} 0S \end{equation*}

is a projective act by Theorem 2.2, which is a projective cover of itself (with the identity mapping). Thus, all objects of $\mathsf{UAct}_S$ (and similarly ${_S\mathsf{UAct}}$) have projective covers and S is perfect.

Assume now that S is a nilsemigroup. If aS ¹ is a cyclic unitary act, then a = as for some $s\in S$. For s, there exists $n\in \mathbb{N}$ such that $s^n=0$. It follows that $a=as^n = a0$, and hence, $aS^1=\{a\} \cong 0S$, where 0 S is projective. Thus, S is right semiperfect and, similarly, it is also left semiperfect.

Nilpotent semigroups need not be factorizable in general. For example, any semigroup with more than one element and with zero multiplication is not factorizable.

Over a factorizable semigroup, projective acts can be described as follows. The description relies on Theorem 2 in [4], which generalizes a well-known characterization of projective acts over monoids.

Theorem 2.5. Let S be a factorizable semigroup. For $P_S\in \mathsf{Act}_S$, the following are equivalent:

1. (1) P _S is a projective object in $\mathsf{FAct}_S$;

2. (2) P _S is a projective object in $\mathsf{UAct}_S$;

3. (3) $P\cong \bigsqcup_{i\in I} e_iS$ for some idempotents $e_i\in S$, $i\in I$.

Proof. (1) $\Rightarrow$ (2). Assume that P _S is a projective object in $\mathsf{FAct}_S$. Clearly, P _S is an object in $\mathsf{UAct}_S$. We will prove that it is projective in $\mathsf{UAct}_S$. Consider a morphism $f:P_S\to B_S$ and an epimorphism $\pi: A_S\to B_S$ in $\mathsf{UAct}_S$. By Lemma 1.1, the acts in the diagram

are firm and $\pi\otimes 1_S$ is an epimorphism in $\mathsf{FAct}_S$ because the functor $-\otimes S$, as a left adjoint, preserves epimorphisms. Using projectivity of P _S, we can find a morphism $g:P_S\to A\otimes_S S$ such that the square commutes. Now $\mu_A g:P_S \to A_S$ is a morphism in $\mathsf{UAct}_S$. Take any $p\in P$. Then there exist $p_0\in P$, $a\in A$ and $s_0,s\in S$ such that $p=p_0s_0$ and $g(p) = a\otimes s$. Due to the commutativity of the square, we have

\begin{align*} \pi(a)\otimes s & = (\pi\otimes 1_S)(g(p)) = (f\otimes 1_S)(\mu_P^{-1}(p)) = (f\otimes 1_S)(p_0\otimes s_0) = f(p_0)\otimes s_0 \end{align*}

in $B\otimes_S S$. Applying the mapping µ _B to the equality $\pi(a)\otimes s = f(p_0)\otimes s_0$, we obtain $\pi(a)s = f(p_0)s_0$. Consequently,

\begin{equation*} (\pi\mu_A g)(p) = (\pi\mu_A)(a\otimes s) = \pi(as) = \pi(a)s = f(p_0)s_0 = f(p_0s_0) = f(p). \end{equation*}

We have shown that $\pi(\mu_A g) = f$, as needed.

(2) $\Rightarrow$ (3). This is a consequence of Theorem 2 in [4] because $\mathsf{UAct}_S$ satisfies the assumptions of that theorem.

(3) $\Rightarrow$ (1). By Lemma 2.3, the acts e _iS are projective in $\mathsf{FAct}_S$. A disjoint union of firm acts is firm; thus, P _S is an object of $\mathsf{FAct}_S$. It is projective in $\mathsf{FAct}_S$ due to Theorem 2.2.

A right S-act A _S is called (finite) limit flat (see [15, Definition 1.3]) if the functor $A\otimes _S -: {_S\mathsf{Act}}\to \mathsf{Set}$ preserves (finite) limits.

Proposition 2.6. The following are equivalent for an act A _S over a factorizable semigroup S:

1. (1) A _S is an indecomposable projective object in $\mathsf{UAct}_S$;

2. (2) A _S is unitary and limit flat;

3. (3) $A_S\cong eS$ for some idempotent $e\in S$;

4. (4) there exists $a\in A$ and an idempotent $e\in S$ such that $A_S=aS^1$, a = ae and

   \begin{equation*} (\forall s,t\in S^1)(as=at \implies es=et). \end{equation*}

Proof. (1) $\iff$ (3). This follows from Theorem 2.5 because cyclic acts are indecomposable.

(2) $\iff$ (3). This is Proposition 4.4 in [15].

(3) $\iff$ (4). This is due to Lemma 1.1 and Corollary 1.2 in [7].

The properties listed in item (4) may be assumed of any generator of a cyclic projective object in $\mathsf{UAct}_S$.

Corollary 2.7. An act $A_S= aS^1$ is a projective object in $\mathsf{UAct}_S$ if and only if there exists an idempotent $e\in S$ such that a = ae and as = at implies es = et for all $s,t\in S^1$.

Proof. Necessity. (This proof is similar to the proof of implication (ii) $\implies $ (iii) in [13, Corollary 3.17.9].) By Proposition 2.6, $aS^1\cong eS$ for some idempotent $e\in S$. Let $f:eS\to aS^1$ be an isomorphism. Then there exist $u\in S^1$ and $v\in S$ such that $f(e) = au$ and $f(ev)=a$. Now $f(e)=au=f(ev)u = f(evu)$ implies e = evu because f is injective. Also, $aue = f(e)e = f(e)=au$. Putting $z:=uev\in S$, we have $z^2 = (uev)(uev) = u(evu)ev = ueev = z$, so z is an idempotent.

Suppose that as = at, where $s,t\in S^1$. Then $f(evs) = f(ev)s = f(ev)t = f(evt)$, which implies evs = evt. Therefore, $zs=uevs = uevt = zt$.

Sufficiency. This follows from Proposition 2.6.

3. Sequence acts

Let us recall the construction of a sequence act over a semigroup (cf. [15, Construction 3.9]). A similar construction in the case of a monoid appeared already in the proof of Lemma 1 in [6].

Let $(s_i)_{i\in \mathbb{N}} \in S^{\mathbb{N}}$ be a sequence of elements of S. On the set

\begin{equation*} F= \mathbb{N}\times S^1 = \{(k,s)\mid k\in \mathbb{N}, s\in S^1\}, \end{equation*}

we define a right S-action by $ (k,s)z := (k,sz), $ $k\in \mathbb{N}$, $s\in S^1$ and $z\in S$. This gives us a right S-act F _S. On the set F, we define a binary relation ~ by

\begin{equation*} (k,s) \sim (k^{\prime},s^{\prime}) \iff (\exists n\geqslant k,k^{\prime})(s_n\ldots s_ks = s_n\ldots s_{k^{\prime}}s^{\prime}). \end{equation*}

Then ∼ is a congruence of F _S. Form the quotient act

\begin{equation*} M_S = F_S/\!\sim\, = \{[k,s]\mid k\in \mathbb{N}, s\in S^1 \}, \end{equation*}

where $[k,s]$ denotes the $\sim\,$-class of (k, s). We call such acts right sequence acts over S. Note that

\begin{equation*} s_{k+1}s_k\cdot 1 = s_{k+1}\cdot s_k \implies [k,1] = [k+1,s_k] = [k+1,1]s_k \in [k+1,1]S^1 \end{equation*}

for each $k\in \mathbb{N}$, so we see that M _S is unitary and it is a union of a chain of its cyclic subacts

\begin{equation*} [1,1]S^1 \subseteq [2,1]S^1 \subseteq [3,1]S^1 \subseteq \cdots . \end{equation*}

Lemma 3.1. Right sequence acts are locally cyclic (and hence indecomposable). A right sequence act M _S is cyclic if and only if $M_S= [k,1]S^1$ for some $k\in \mathbb{N}$.

Proof. Given any $[k,s],[\ell, t]\in M_S$, we have that $[k,s],[\ell, t] \in [\max\{k,\ell\},1]S^1$. Thus, right sequence acts are locally cyclic. It is well known that locally cyclic acts are indecomposable.

If M _S is cyclic, then $M = [m,u]S^1$ for some $u\in S^1$. By the above,

\begin{equation*} M_S = [m,u]S^1 = [m+1,s_mu]S^1 \subseteq [m+1,1]S^1, \end{equation*}

so $M_S = [m+1,1]S^1$. The converse is obvious.

Lemma 3.2. Let S be a semigroup. Each act of the form eS, $e^2=e\in S$, is isomorphic to a right sequence act.

Proof. Let M _S be the right sequence act constructed using the constant sequence $(e,e,\ldots)$. For every $k\in {\mathbb{N}}$,

\begin{equation*} e\cdot 1 = \underbrace{e \ldots e}_{k}\cdot e \implies (k,1)\sim (1,e) \implies [k,1] = [1,1]e \implies [k,1]S^1 = [1,1]S^1, \end{equation*}

so $M_S = [1,1]S^1$. Consider the S-act homomorphism

\begin{equation*} f: eS\to [1,1]S^1, \quad es\mapsto [1,es]. \end{equation*}

It is injective because

\begin{equation*} (1,es)\sim (1,et) \implies e\cdot es = e\cdot et \implies es=et \end{equation*}

for every $s,t\in S$, and it is surjective because

\begin{equation*} [1,s] = [2,es] = [2,e]es = [1,1]es = [1,es] = f(es) \end{equation*}

for every $s\in S^1$. Thus, f is an isomorphism.

Corollary 3.3. Let S be a factorizable semigroup. For S-act properties, we have the following implications:

\begin{equation*} \mbox{unitary limit flat} \implies \mbox{isomorphic to a right sequence act} \implies \mbox{unitary finite limit flat.} \end{equation*}

Proof. The first implication follows from Proposition 2.6 and Lemma 3.2. The second implication holds due to Proposition 4.3 in [15].

In our main theorem, we will show (among other things) that factorizable right perfect semigroups S are precisely those for which the converse of the first implication holds for every right S-act. Note also that according to this corollary, ‘right sequence act’ could be placed between ‘eS’ and ‘firm fin. lim. flat’ in the figure on page 90 of [15].

4. Condition (A)

A subset X of an act A _S is a set of generators if $A=XS^1$. A set X of generators is called independent ([6, p. 90]) if

\begin{equation*} \left(\forall x,x^{\prime} \in X\right)\left(x\in x^{\prime} S^1 \implies x=x^{\prime}\right). \end{equation*}

Lemma 4.1. Cf. [6, Lemma 2]

Let A _S be an act over a semigroup S satisfying (ACC) for cyclic subacts. Then A _S has an independent set of generators.

Proof. Denote

\begin{equation*} X:= \left\{x\in A\mid (\forall x^{\prime} \in A)(xS^1\subseteq x^{\prime} S^1 \implies xS^1=x^{\prime} S^1)\right\}. \end{equation*}

We first show that X is a set of generators for A _S. It suffices to prove that $A\subseteq XS^1$. Take arbitrary $a\in A$ and consider the set

\begin{equation*} P_a = \left\{a^{\prime} S^1\mid a^{\prime} \in A, aS^1\subseteq a^{\prime} S^1\right\} \end{equation*}

as a poset with respect to inclusion. Since $aS^1\in P_a$, by assumption, this poset must have a maximal element $a_0S^1$. In particular, $aS^1 \subseteq a_0S^1$. We show that $a_0\in X$.

If $x^{\prime} \in A$ and $a_0S^1 \subseteq x^{\prime} S^1$, then also $aS^1 \subseteq x^{\prime} S^1$, so $x^{\prime} S^1\in P_a$. Due to maximality of $a_0S^1$, we have $a_0S^1 = x^{\prime} S^1$. Hence, $a_0\in X$ and $a\in a_0S^1 \subseteq XS^1$, as needed.

Define a relation ≈ on X by

\begin{equation*} x\approx x^{\prime} \iff xS^1\subseteq x^{\prime} S^1. \end{equation*}

It is clear by the choice of X that we have an equivalence relation.

From every ≈-class, we choose a representative and form a set X ^ʹ of those elements. We show that X ^ʹ is an independent set of generators for A _S. By the definition of ≈, for every $x \in X$, there exist $x^{\prime} \in X^{\prime}$ and $s \in S^1$ such that $x = x^{\prime} s$. Since X is a set of generators, X ^ʹ is also a set of generators.

To prove that X ^ʹ is independent, we suppose that $x\in x^{\prime} S^1$, where $x,x^{\prime} \in X^{\prime} \subseteq X$. Then $xS^1\subseteq x^{\prime} S^1$, which means that $x\approx x^{\prime}$. Now $x=x^{\prime}$ because each ≈-class contains precisely one element from X ^ʹ.

Definition 4.2. Cf. [2, Definition 4.1]

We say that a semigroup S is right $\mathcal{IC}$-perfect if every unitary act A _S has a cover $f:B_S\to A_S$, where B _S is unitary, and indecomposable components of B _S are cyclic. Such a cover will be called an $\mathcal{IC}$-cover.

Note that if $b=b^{\prime}s$, where $b,b^{\prime} \in B$ and $s\in S$, then b and b ^ʹ must be in the same indecomposable component. Therefore, indecomposable components of B _S must be unitary. Hence,

\begin{equation*} B_S = \bigsqcup_{i\in I} b_iS^1, \end{equation*}

where each $b_iS^1$ is a cyclic unitary act.

The following result is inspired by Result 1.2 in [9], by Lemma 1.3 in [10] and also by Lemma 2.2 and Theorem 5.2 in [2]. Among other things, it shows that Condition (A) can be given a description that does not refer to acts but only uses the elements of S.

Theorem 4.3. For a semigroup S, the following are equivalent:

1. (1) S satisfies Condition (A);

2. (2) every locally cyclic right S-act is cyclic;

3. (3) every right sequence act over S is cyclic;

4. (4) for every sequence $(s_i)_{i\in \mathbb{N}}\in S^{\mathbb{N}}$,

   \begin{equation*} (\exists k,m\in \mathbb{N})(\exists u\in S^1)(k\geqslant m+1 \ \mbox{and }\ s_k\ldots s_{m+1} = s_k\ldots s_{m+1}s_m u); \end{equation*}

5. (5) for every act A _S, there exists a set $\{A_i\mid i\in I\}$ of cyclic subacts of A _S such that

   (4.1)\begin{equation} A=\bigcup_{i\in I}A_i\quad \mbox{and }\quad (\forall j\in I)\left( A_j\not\subseteq \bigcup_{i\neq j}A_i\right); \end{equation}

6. (6) S is right $\mathcal{IC}$-perfect.

Proof. (1) $\implies$ (2). Suppose that A _S is a locally cyclic act which is not cyclic. Choose $a_1\in A$. Since A _S is not cyclic, there exists $b_1\in A\setminus a_1S^1$. Using that A _S is locally cyclic, we can find $a_2\in A$ and $u,v\in S^1$ such that $a_1=a_2u$ and $b_1=a_2v$. Note that $a_2\not\in a_1S^1$ because otherwise $b_1\in a_1S^1$. Thus, $a_1S^1 \subset a_2S^1$.

Again, $a_2S^1 \neq A$, and we can find an element $b_2\in A\setminus a_2S^1$. Continuing in this manner, we can construct a strictly increasing sequence

\begin{equation*} a_1S^1 \subset a_2S^1 \subset a_3S^1 \subset \cdots \end{equation*}

of cyclic subacts of A _S, which contradicts Condition (A).

(2) $\implies$ (3). Every right sequence act is locally cyclic because it is the union of an ascending chain of its cyclic subacts.

(3) $\implies$ (4). Assume that all right sequence acts over S are cyclic. Consider a sequence $(s_i)_{i\in \mathbb{N}}\in S^{\mathbb{N}}$ and the right sequence act

\begin{equation*} M_S = \left \{[k,s] \mid k\in\mathbb N,\ s\in S^1 \right \} \end{equation*}

determined by it. By assumption, M _S must be cyclic. By Lemma 3.1, there exists $m\in\mathbb N$ such that $M = [m,1]S^1$. Then $(m+1,1) \sim (m,u)$ for some $u\in S^1$, which means that there exists $k\geqslant m+1$ such that $ s_{k}\ldots s_{m+1} = s_k\ldots s_{m+1}s_mu. $

(4) $\implies$ (1). Suppose that A _S does not satisfy (ACC) for cyclic subacts. Then there exists a sequence $(a_i)_{i\in \mathbb{N}} \in A^{\mathbb{N}}$ such that

\begin{equation*} a_1S^1 \subset a_2S^1 \subset a_3S^1 \subset \cdots . \end{equation*}

Consequently, for every $i\in \mathbb{N}$, there exists $s_i\in S$ such that $a_i = a_{i+1}s_i$. For the sequence $(s_i)_{i\in \mathbb{N}}\in S^{\mathbb{N}}$, there exist $k,m\in \mathbb{N}$, $k\geqslant m+1$ and $u\in S^1$ such that $s_k\ldots s_{m+1} = s_k\ldots s_{m+1}s_m u$. Then

\begin{align*} a_{m+1} = a_{m+2}s_{m+1} &= a_{m+3}s_{m+2}s_{m+1} = a_{k+1}s_k\ldots s_{m+1} = a_{k+1}s_k\ldots s_{m+1}s_mu \\ &= a_k s_{k-1} \ldots s_{m+1}s_mu = a_{m+1}s_mu = a_mu \in a_mS^1. \end{align*}

It follows that $a_{m+1}S^1 = a_mS^1$, a contradiction.

(1) $\implies$ (5). Assume that S satisfies Condition (A). Consider an act A _S. By Lemma 4.1, A _S has an independent set of generators X. Then, clearly, $A= \bigcup_{x\in X} xS^1$. Take $y\in X$ and suppose that $yS^1 \subseteq \bigcup_{x\neq y} xS^1$. Then there exists $z\in X\setminus \{y\}$ such that $y\in zS^1$. Independence of X implies y = z, a contradiction. Hence,

\begin{equation*} yS^1 \not\subseteq \bigcup_{x\neq y} xS^1. \end{equation*}

(5) $\implies$ (6). Let A _S be a unitary act. By (5), there exist cyclic subacts $A_i = a_iS^1$, $i\in I$, such that Equation (4.1) holds. It is easy to see that each $a_iS^1$ is unitary.

Now $B_S:= \bigsqcup_{i\in I}a_iS^1$ is an act whose indecomposable components are cyclic and unitary. Consider the S-act homomorphism

\begin{equation*} f:B_S\to A_S,\qquad a_is\mapsto a_is. \end{equation*}

Suppose that f is not essential. Then there exists $j\in I$ and (possibly empty) proper subact C of $a_jS^1$ such that $f: C\sqcup \bigsqcup_{i\neq j}a_iS^1 \to A$ is surjective. In particular, there exists $b\in C\sqcup \bigsqcup_{i\neq j}a_iS^1$ such that $f(b)=a_j$. We have two possibilities.

1. 1) $b\in C$. Then $b=a_js$ for some $s\in S^1$. Hence, $ a_j= f(b) = f(a_js) = a_js = b\in C, $ which implies $a_jS^1 = C$, a contradiction.

2. 2) $b\in \bigsqcup_{i\neq j}a_iS^1$. Then $a_jS^1 = f(b)S^1 = f(bS^1) \subseteq \bigcup_{i\neq j} a_iS^1$, a contradiction.

(6) $\implies$ (3). Assume that S is right $\mathcal{IC}$-perfect and consider a right sequence act M _S determined by a sequence $(s_i)_{i\in \mathbb{N}}$. Since M _S is unitary by the comment preceding Lemma 3.1, it has a cover $f\colon B_S \to M_S$, where

\begin{equation*} B_S = \bigsqcup_{i\in I} b_iS^1. \end{equation*}

Fix an index $j\in I$. Let $f(b_j) = [k,s]\in M_S$. Then $f(b_jS^1) \subseteq [k,1]S^1$. Consider a number $\ell \gt k$. Suppose $[\ell,1]$ has an f-preimage in $b_iS^1$ for some i ≠ j. Then $[\ell,1]S^1\subseteq f(b_iS^1)$, which implies that f restricted to $B\setminus b_jS^1$ is surjective. But that contradicts essentiality of f. Thus, $[\ell,1]$ must have a preimage in $b_jS^1$. Let $[\ell,1] = f(b_ju)$ for some $u\in S^1$. Now

\begin{equation*} [\ell,1] = f(b_ju) = f(b_j)u = [k,s]u = [k,1]su \in [k,1]S^1, \end{equation*}

which implies $[\ell,1]S^1 = [k,1]S^1$. Since this equality holds for every $\ell \gt k$, we have shown that $M_S = [k,1]S^1$, so M _S is cyclic.

Remark 4.4. Condition (5) in Theorem 4.3 appears first in Lemma 2.2 of [2], but a similar condition was used in [10]. It is kind of interesting that this condition can be formulated in topological terms as follows.

If A _S is an act, then the set of its subacts (here including the empty subact) is a topology on the set A because any union or intersection of subacts is a subact. The set of all cyclic subacts is a basis for this topology. Now condition (5) says that this basis contains a subset which is a minimal cover (in the sense of topology) for A.

5. Condition (D)

In this section, we will prove that a factorizable semigroup satisfies Condition (D) if and only if it is right semiperfect. The following result can be proved precisely as Lemma 4.4 in [8].

Lemma 5.1. Let T be a left unitary subsemigroup of a semigroup S. Then

\begin{equation*} \left(\forall u,v\in T\right)\left(uT^1 \subseteq vT^1 \iff uS^1 \subseteq vS^1\right). \end{equation*}

Lemma 5.2. If a subsemigroup T of a semigroup S is left unitary, then it is a ρ-class of some right congruence ρ on S.

Proof. Assume that T is a left unitary subsemigroup of S. Let ρ be the right congruence on S generated by the set T × T. Take $t\in T$. Then $T\subseteq [t]_\rho$. We will prove that $[t]_\rho \subseteq T$. If $s\in [t]_\rho$, then $t\,\rho\, s$ and

\begin{equation*} \begin{array}{ccccccccccc} t & = & t_1s_1 & & t^{\prime} _2s_2 & = & t_3s_3\;\ldots\; t^{\prime}_{n-1}s_{n-1} & = & t_ns_n & & \\ & & t^{\prime}_1s_1 & = & t_2s_2 & & & & t^{\prime}_ns_n & = & s \\ \end{array} \end{equation*}

for some $t_i,t^{\prime}_i\in T$ and $s_i\in S^1$. If $s_1\neq 1$, then $s_1\in T$ because T is left unitary. Hence, $t^{\prime}_1s_1\in T$, both when $s_1\neq 1$ and when $s_1=1$. If $s_2\neq 1$, then $s_2\in T$, and so on. Thus, for every $i\in \{1,\ldots,n\}$, either $s_i=1$ or $s_i\in T$. It follows that $s=t^{\prime}_ns_n \in T$.

Lemma 5.3. If S is a semigroup and aS ¹ is a unitary cyclic S-act, then

\begin{equation*} T= \{s\in S \mid as=a\} \end{equation*}

is a left unitary subsemigroup of S.

Proof. Since aS ¹ is unitary, there exist $u\in S^1$ and $s\in S$ such that $a= (au)s = a(us)$. We see that $us\in T$, so T is nonempty, and it is clearly a subsemigroup of S. If $t,ts\in T$, then $as=(at)s = a(ts) = a$, so $s\in S$. Thus, T is left unitary.

Theorem 5.4. Cf. [9], Result 1.5 and Theorem 5.2 in [20]

For a factorizable semigroup S, the following assertions are equivalent:

1. (1) S is right semiperfect;

2. (2) S satisfies Condition (D).

Proof. (1) $\implies$ (2). (This is inspired by the proof of Proposition 4.6 in [8]. Differently from that proof, we cannot assume that R is the congruence class of the identity element, but with appropriate modifications, the proof will work.)

Let R be a left unitary subsemigroup of S. By Lemma 5.2, it is a ρ-class for some right congruence ρ on S, say $R=[t]$, $t\in R$. Then $[t]S^1$ is a cyclic subact of the right S-act $S/\rho$. It is a unitary S-act because $t,t^2\in R$ implies $[t]=[t]t$.

By assumption, $[t]S^1$ has a projective cover in $\mathsf{UAct}_S$. This must be a unitary cyclic projective act (see the proof of Lemma 4.1 in [8]). Due to Theorem 2.5, it must be of the form eS, where $e\in S$ is an idempotent. Also, there exists an essential epimorphism $f: eS\to [t]S^1$ in $\mathsf{UAct}_S$. Because of surjectivity of f, there exists $eu\in eS$ such that $[t] = f(eu)$. Since S is factorizable, euS is a unitary subact of eS. Now $eut\in euS$ and $[t]=[tt] = [t]t = f(eu)t = f(eut)$ imply that $f|_{euS}$ is surjective. Due to essentiality, we have the equality euS = eS. It follows that $e=eus^{\prime}$ for some $s^{\prime} \in S$. Putting $s:= s^{\prime} e\in S$, we have e = eus and s = se. Now

\begin{equation*} (su)(su) = (seu)(su) = s(eus)u = seu = su, \end{equation*}

so su is an idempotent. In addition,

\begin{equation*} [t] = f(eu) = f(eusu) = f(eu)su = [t]su = [tsu]. \end{equation*}

Since R is left unitary, $t,tsu\in R$ implies $su\in R$. Let us prove that suR ¹ is a minimal right ideal in R.

For this, we show that for every $r\in R$, $rR^1 \subseteq suR^1$ implies $rR^1 = suR^1$. Let $r\in R$. If $rR^1 \subseteq suR^1$, then $r=sur^{\prime}$ for some $r^{\prime} \in R^1$. Hence, $r= (su)^2r^{\prime} = (su)(sur^{\prime}) = sur$ and

\begin{equation*} f(eurt) = f(eu)rt = [t]rt = [trt] = [t] \end{equation*}

because $trt\in R$. Since eurS is a unitary subact of eS and $f|_{eurS}$ is surjective, $eurS = eS = eS^1$. Now

\begin{equation*} sS = seS = seurS = surS = rS \subseteq rS^1 = surS^1 \subseteq suS^1 \subseteq sS, \end{equation*}

which implies $rS^1 = suS^1$. Since $r,su\in R$, we conclude that $rR^1 = suR^1$ by Lemma 5.1.

(2) $\implies$ (1). Assume that S satisfies Condition (D) and consider a unitary cyclic S-act aS ¹. The fact that aS ¹ is unitary means that $a=at_0$ for some $t_0\in S$. Now

\begin{equation*} T:= \{t\in S\mid at=a\} \end{equation*}

is a left unitary subsemigroup of S by Lemma 5.3. Since S satisfies Condition (D), there exists an idempotent $e\in T$ (so ae = a) such that eT is a minimal right ideal in T. We will show that the unitary S-act eS is a projective cover for aS ¹ in $\mathsf{UAct}_S$.

For this, we consider the surjective S-act homomorphism

\begin{equation*} f: eS \to aS^1, \qquad es\mapsto aes=as. \end{equation*}

Suppose that B is a unitary subact of eS such that $f|_B$ is surjective. Then there exists $b\in B$ such that $f(b)=a$. Since $B\subseteq eS$, there exists $s\in S$ such that b = es. Observe that $a=f(es) = as$. Consequently, $s\in T$ and also $es\in T$. Now, using minimality of eT and Lemma 5.1,

\begin{align*} esT^1\subseteq eT & \implies esT^1=eT \implies esS^1=eS \implies eS = bS^1\subseteq B \subseteq eS \implies B=eS. \end{align*}

6. Condition $(M_L)$

Similar to Condition (A), Condition (M _L) has a description in terms of sequences of elements of S.

Proposition 6.1. A semigroup S satisfies Condition (M _L) if and only if

\begin{equation*} (\forall (s_i)_{i\in \mathbb{N}}\in S^{\mathbb{N}})(\exists n\in \mathbb{N}) (\forall m \gt n)(\exists u\in S^1) (us_m\ldots s_1 = s_n\ldots s_1). \end{equation*}

Proof. Necessity. Assume that S satisfies Condition (M _L). Consider a sequence $(s_i)_{i\in \mathbb{N}}\in S^{\mathbb{N}}$ and denote $t_i:= s_is_{i-1}\ldots s_1$ for each $i\in \mathbb{N}$. Since the descending chain

\begin{equation*} S^1t_1 \supseteq S^1t_2 \supseteq S^1t_3 \supseteq \cdots \end{equation*}

stabilises, there exists $n\in \mathbb{N}$ such that $S^1t_n = S^1t_m$ for every m > n. Hence, $t_n\in S^1 t_m$, which means that $us_m\ldots s_1 = s_n\ldots s_1$ for some $u\in S^1$.

Sufficiency. Suppose to the contrary that S does not satisfy Condition $(M_L)$. Then we have a strictly decreasing chain of principal left ideals

\begin{equation*} S^1t_1 \supset S^1t_2 \supset S^1t_3 \supset \cdots . \end{equation*}

For every $i\in \mathbb{N}$, there exists $s_{i+1}\in S^1$ such that $t_{i+1} = s_{i+1}t_i$. Since we assumed that $t_i\neq t_{i+1}$, we have $s_{i+1}\in S$. Also, put $s_1 := t_1$. By our assumption, there exists $n\in \mathbb{N}$ such that for every m > n, there exists $u\in S^1$ such that $us_m\ldots s_1 = s_n\ldots s_1$. If n = 1, then there exists $u\in S^1$ such that $us_2t_1 = t_1$, which implies $ut_2 = t_1$. Consequently, $S^1t_1 = S^1t_2$, a contradiction. Otherwise, let $m \gt n \gt 1$. Then $s_n\ldots s_1=u^{\prime} s_m\ldots s_1$ for some $u^{\prime} \in S^1$ and

\begin{equation*} t_n = s_nt_{n-1} = s_ns_{n-1}t_{n-2} = \cdots = s_ns_{n-1}\ldots s_1 = u^{\prime} s_ms_{m-1}\ldots s_1 = u^{\prime} t_m, \end{equation*}

whence $S^1t_n \subseteq S^1t_m$ and therefore $S^1t_n = S^1t_m$, a contradiction.

We will also need the following lemma.

Lemma 6.2. Cf. [6, Lemma 1]

If all right sequence acts over a semigroup S are projective, then S satisfies Condition (M _L).

Proof. We will employ Proposition 6.1. Consider a sequence $(s_i)_{i\in \mathbb{N}}$ and the right sequence act M _S determined by it. By assumption, M _S is projective. Hence, for the surjective homomorphism

\begin{equation*} \pi: \mathbb{N}\times S^1 = F_S\to M_S,\qquad (k,s)\mapsto [k,s], \end{equation*}

there exists a homomorphism $\mu: M_S\to F_S$ such that $\pi\mu = 1_M$.

Let $\mu([1,1]) = (k,s)\in \mathbb{N}\times S^1$. Then

\begin{equation*} [1,1] = (\pi\mu)([1,1]) = \pi(k,s) = [k,s]. \end{equation*}

Since $(1,1)\sim (k,s)$, there exists $n\geqslant k$ such that $ s_n\ldots s_1\cdot 1 = s_n\ldots s_k\cdot s. $ Let m > n and put $\mu([m+1,1]) = (r,c)$ for some $r\in \mathbb{N}$ and $c\in S^1$. Note that $(1,1)\sim (m+1,s_m\ldots s_1)$ because

\begin{equation*} s_{m+1}\ldots s_1\cdot 1 = s_{m+1}\cdot s_m\ldots s_1. \end{equation*}

Now

\begin{align*} (k,s) & = \mu([1,1]) = \mu ([m+1, s_m\ldots s_1]) = \mu([m+1,1])s_m\ldots s_1\\ &= (r,c)s_m\ldots s_1 = (r,cs_m\ldots s_1), \end{align*}

whence k = r and $s= cs_m\ldots s_1$. Putting $u:= s_n\ldots s_kc \in S$, we have

\begin{equation*} s_n\ldots s_1 = s_n\ldots s_ks = s_n\ldots s_kcs_m\ldots s_1 = us_m\ldots s_1, \end{equation*}

as needed.

7. Pullback flatness

Definition 7.1. A right S-act A _S is called pullback flat if the functor $A\otimes _S -: {_S\mathsf{Act}}\to \mathsf{Set}$ preserves pullbacks.

Since pullbacks are finite limits, every finite limit flat act is pullback flat.

We will need the following conditions (introduced first for acts over monoids in [21]) and the following theorem.

• (P) If $as = a^{\prime} s^{\prime}$ for some $a,a^{\prime} \in A$ and $s,s^{\prime} \in S$, then there exist $a^{\prime\prime} \in A$ and $u,v\in S$ such that

  \begin{equation*} a=a^{\prime\prime} u,\quad a^{\prime}=a^{\prime\prime} v,\quad us = vs^{\prime}. \end{equation*}

• (E) If $as = as^{\prime}$ for some $a\in A$ and $s,s^{\prime}\in S$, then there exist $a^{\prime} \in A$ and $u\in S$ such that

  \begin{equation*} a=a^{\prime} u,\quad us=us^{\prime}. \end{equation*}

Theorem 7.2. [15, Theorem 3.5]

Let A _S be a unitary act over a semigroup S. Then A _S is pullback flat if and only if it satisfies both Condition (P) and Condition (E).

Proposition 7.3. A cyclic S-act aS ¹ is unitary and pullback flat if and only if

\begin{equation*} (\forall s,t\in S^1)(as= at \implies (\exists u\in S)(a = au \wedge us = ut)). \end{equation*}

Proof. Necessity. Let $A=aS^1$ be unitary and pullback flat. Take $s,t\in S^1$ such that as = at. Since aS ¹ is unitary, a = av for some $v\in S$. Then $a(vs)=a(vt)$, where $vs,vt\in S$. By Theorem 7.2, aS ¹ satisfies Condition (E), so there exists $w\in S$ such that a = aw and $w(vs) = w(vt)$. Putting $u:= wv$, we have a = au and us = ut.

Sufficiency. By assumption, there exists $u\in S$ such that a = au, thus aS ¹ is unitary. That aS ¹ satisfies Condition (E) follows immediately from the assumption. To prove Condition (P), suppose that $(as_1)s = (as_2)s^{\prime}$, where $s_1,s_2\in S^1$ and $s,s^{\prime} \in S$. By assumption, there exists $u\in S$ such that a = au and $us_1s=us_2s^{\prime}$. Now $as_1= a(us_1)$ and $as_2 = a(us_2)$, where $us_1,us_2\in S$.

Lemma 7.4. Cf. [6, Lemma 3]

Let A _S be a unitary pullback flat act which satisfies (ACC) for cyclic subacts. If A _S is indecomposable, then it is cyclic.

Proof. Take any $x\in A$ and consider the set

\begin{equation*} P_x := \{bS^1 \mid b\in A,\ xS^1 \subseteq bS^1\} \end{equation*}

as a poset with respect to inclusion. Since $xS^1 \in P_x$, we have $P_x\neq\emptyset$, so by assumption, there exists a maximal element $aS^1\in P_x$. In particular, $xS^1\subseteq aS^1$. Consider the set $Y := A\setminus aS^1$. Assume for a contradiction that $Y\neq\emptyset$. Now, A _S is a union of its subacts aS ¹ and YS ¹. Since A _S is indecomposable,

\begin{equation*} aS^1 \cap YS^1 \neq \emptyset , \end{equation*}

which means that there exist $y\in Y$ and $s, \, t \in S^1$ such that as = yt.

On the one hand, if t = 1, then $y=as\in aS^1$, a contradiction. On the other hand, suppose s = 1. The equality a = yt implies that $xS^1 \subseteq aS^1 \subseteq yS^1$. It follows that $yS^1 \in P_x$ and $aS^1 = yS^1$ due to maximality. So again, we must have $y\in aS^1$, which is impossible. Thus, we must have $s,t\in S$.

Since A _S is a unitary pullback flat act, by Theorem 7.2, it satisfies Condition (P). Hence, there exist $a^{\prime} \in A$ and $u,v \in S$ such that $a = a^{\prime} u$, $y = a^{\prime} v$ and us = vt. We conclude that $aS^1 \subseteq a^{\prime} S^1$. It follows that $a^{\prime} S^1 \in P_x$, so $aS^1 = a^{\prime} S^1$ due to maximality. Now, for some $z \in S^1$, $a^{\prime} = az$ and therefore $ y = a^{\prime} v = azv, $ whence $y\in aS^1$, a contradiction. Thus, $Y=\emptyset$ and $A=aS^1$ is cyclic, as required.

Corollary 7.5. If a semigroup S satisfies Condition (A), then every unitary pullback flat S-act is a disjoint union of cyclic unitary pullback flat subacts.

Proof. Assume that S satisfies Condition (A) and let A _S be unitary and pullback flat. By Proposition 2.1,

\begin{equation*} A= \bigsqcup_{i\in I} A_i, \end{equation*}

where A _i is an indecomposable subact of A _S for every $i\in I$. It is clear by Theorem 7.2 that the A _i are pullback flat. Since S satisfies Condition (A), each A _i satisfies (ACC) for cyclic subacts. We see that the A _i satisfy all assumptions of Lemma 7.4, so they must be cyclic.

Lemma 7.6. Cf. [6, Lemma 5 and Lemma 4]

If a factorizable semigroup S satisfies either Condition (D) or Condition (M _L), then every unitary cyclic pullback flat right S-act is projective.

Proof. Let $A_S=aS^1$ be a unitary cyclic pullback flat S-act. By Lemma 5.3, the set $T= \{s\in S \mid as=a\}$ is a left unitary subsemigroup of S. If S satisfies Condition (D), then T contains a minimal right ideal eT, where $e\in T$ is an idempotent. If S satisfies Condition $(M_L)$, then the set

\begin{equation*} \mathcal{I} = \{S^1t \mid t\in T\} \end{equation*}

of principal left ideals of S contains a minimal element $S^1e$ for some element $e\in T$. In both cases, it suffices to prove that condition (4) of Proposition 2.6 is satisfied.

Assume that as = at for some $s,t\in S^1$. Since aS ¹ is pullback flat, by Proposition 7.3, there exists $u\in S$ such that a = au and us = ut. The equality au = ae implies that a = av and vu = ve for some $v\in S$. Note that $u,v\in T$.

First, let S satisfy Condition (D). By Lemma 8.12 in [5], Te is a minimal left ideal of T. It follows that

\begin{equation*} Tvu \subseteq Tu \cap Te \subseteq Te, \end{equation*}

whence $Te=Tvu\subseteq Tu$ due to minimality. Therefore, e = wu for some $w\in T$. We conclude that $es = wus = wut = et$, as needed.

Second, assume S satisfies Condition $(M_L)$. Since T is a subsemigroup, $vu\in T$ and $S^1vu \in \mathcal I$. The equality vu = ve implies $S^1vu\subseteq S^1e$. We have $S^1vu = S^1e$ due to minimality and e = wvu for some $w\in S^1$. This implies $es = wvus = wvut = et$. In particular, $a1 = ae$ implies that $e1 = ee$ by the previous argument, so e is an idempotent.

8. Condition (K)

Definition 8.1. Cf. [11, Definition 1.6]

We say that a subsemigroup T of S is left collapsible if T ¹ is a left collapsible submonoid of S ¹, that is,

\begin{equation*} (\forall t,t^{\prime} \in T^1)(\exists u\in T^1)(ut=ut^{\prime}). \end{equation*}

Condition (K) for monoids was introduced by Kilp in [11] and used later in [12]. We use it for semigroups in the following form.

Definition 8.2. A semigroup S satisfies Condition (K) if every left collapsible subsemigroup of S has a left zero.

Proposition 8.3. Cf. [11, Theorem 2.3]

A factorizable semigroup S satisfies Condition (K) if and only if every unitary cyclic pullback flat right S-act is projective.

Proof. Necessity. Suppose S satisfies Condition (K). Let aS ¹ be unitary pullback flat and let $T= \{s\in S \mid as=a\}$. If $t,t^{\prime} \in T^1$, then $at=at^{\prime}$. By Proposition 7.3, there exists $u\in S$ such that a = au and $ut=ut^{\prime}$. In particular, $u\in T$. We have shown that T ¹ is a left collapsible submonoid of S ¹.

By assumption, T has a left zero e. We will check condition (4) of Proposition 2.6. We know that a = ae. Suppose that as = at, $s,t\in S^1$. Then there exists $u\in S$ such that a = au and us = ut. Then $u\in T$, eu = e and $es=eus=eut = et$.

Sufficiency. Assume that every unitary cyclic pullback flat right S-act is projective. Let $T\subseteq S$ be a left collapsible subsemigroup of S. Then by definition $P:=T^1$ is a left collapsible submonoid of S ¹. By Lemma 2.1 in [11], the relation ρ, defined by

\begin{equation*} s\,\rho\, t \iff (\exists p,q\in P)(ps=qt), \end{equation*}

is a right congruence on the monoid S ¹. In particular, $1\,\rho\,p$ for all $p\in P$. The cyclic S ¹-act $S^1/\rho = [1]_\rho\, S^1$ is pullback flat.

It turns out that $S^1/\rho = [1]_\rho\, S^1$ is also a cyclic unitary pullback flat S-act. Since T is nonempty, we can choose $t_0\in T\subseteq S$. If $[1]_\rho\, s = [1]_\rho\, t$, $s,t\in S^1$, then Condition (E) implies that there exists $u\in S^1$ such that $[1]_\rho = [1]_\rho\, u$ and us = ut. If it happens that u = 1, then still $[1]_\rho = [1]_\rho\, t_0$ and $t_0s=t_0 t$. By Proposition 7.3, $S^1/\rho$ is a unitary pullback flat S-act.

By assumption, $[1]_\rho\, S^1$ is projective. Using Corollary 2.7, we know that there exists $e\in E(S)$ such that $[1]_\rho = [1]_\rho e$ and $[1]_\rho s= [1]_\rho t$ implies es = et for all $s,t\in S^1$. In other words, $1\,\rho\, e$ and $s\,\rho\, t$ implies es = et for all $s,t\in S^1$. From $1\,\rho\, e$, we obtain $p,q\in P$ such that p = qe. Let $r\in P$ be arbitrary. Then $1\,\rho\, r$, which implies e = er, and therefore $pr=qer=qe=p$. Thus, p is a left zero for P. If p = 1, then $1=qe=qee=e$, which contradicts the fact that $e\in S$. Thus, $p\in T$ and p is a left zero for T.

Corollary 8.4. If S is factorizable, then Condition (M _L) implies Condition (K).

Proof. This follows from Proposition 8.3 and Lemma 7.6.

9. The main theorem

Our main theorem is the following.

Theorem 9.1. For a factorizable semigroup S, the following are equivalent:

1. (1) S is right perfect;

2. (2) S is right $\mathcal{IC}$-perfect and right semiperfect;

3. (3) S satisfies both Condition (A) and Condition (D);

4. (4) S satisfies both Condition (A) and Condition (M _L);

5. (5) S satisfies both Condition (A) and Condition (K);

6. (6) every right sequence act over S is projective in $\mathsf{UAct}_S$;

7. (7) every finite limit flat right S-act is projective in $\mathsf{UAct}_S$;

8. (8) every unitary pullback flat right S-act is projective in $\mathsf{UAct}_S$;

9. (9) every finite limit flat right S-act is limit flat;

10. (10) every right sequence act over S is limit flat.

Proof. (1) $\Rightarrow$ (2). If a semigroup S is right perfect, then it is clearly right semiperfect. If $f: \bigsqcup_{i\in I} e_iS \to A_S$ is a projective cover for a unitary act A _S, then its indecomposable components e _iS, $i\in I$, are cyclic and unitary. Hence, S is also right $\mathcal{IC}$-perfect.

(2) $\Rightarrow$ (1) Let A _S be a unitary S-act. Since S is $\mathcal{IC}$-perfect, there exists a cover $g:B_S\to A_S$, where $B_S = \bigsqcup_{i\in I}b_iS^1$ is a disjoint union of unitary cyclic subacts. Semiperfectness of S gives a unitary projective cover $f_i :P_i\to b_iS^1$ for each $i\in I$.

Then $P_S:= \bigsqcup_{i\in I} P_i$ is a unitary projective act by Theorem 2.5 and

\begin{equation*} f: P_S\to B_S, \qquad x \mapsto f_i(x)\quad \mbox{if } x\in P_i \end{equation*}

is a surjective homomorphism of right S-acts. We will show that f is essential.

Suppose that Q is a proper subact of P _S. Then there exists $i\in I$ such that $Q_i:= Q\cap P_i$ is a proper (possibly empty) subact of P _i. Hence, $f_i|_{Q_i}: Q_i\to b_iS^1$ is not surjective because f _i is essential. But then also $f|_Q$ is not surjective because the preimages of elements of $b_iS^1$ can only come from P _i. Thus, f is an essential epimorphism. Hence, $gf: P_S\to A_S$ is an essential epimorphism and P _S is a projective cover for A _S.

(2) $\Leftrightarrow$ (3). This follows from Theorem 5.4 and Theorem 4.3.

(6) $\Rightarrow$ (3). Assume that all right sequence acts are projective. Since they are indecomposable (see Lemma 3.1), unitary and projective, they must be cyclic. Hence, S satisfies Condition (A) by Theorem 4.3.

We will prove that S satisfies Condition (D). The proof of this implication is inspired by the proof of Theorem 6.2 in [8].

Let T be a left unitary subsemigroup of S. Take any $t\in T$. We will prove that the principal right ideal tT ¹ of T contains an idempotent. By the assumption, the right sequence act M _S, determined by the constant sequence $(t)\in S^{\mathbb N}$, is projective. Since M _S is indecomposable, it must be cyclic. From Lemma 3.1, we conclude that there exists $i\in {\mathbb{N}}$ such that $M =[i,1]S^1$. By Corollary 2.7, there exists $e^2=e\in S$ such that $[i,1]=[i,e]$ and

(9.1)\begin{equation} (\forall x,y\in S^1)([i,x] = [i,y] \implies ex=ey). \end{equation}

From $[i,1]S^1 \subseteq [i+1,1]S^1 \subseteq M_S = [i,1]S^1$, we conclude that $[i,1]S^1 = [i+1,1]S^1$. Hence, there exists $z\in S^1$ such that $[i+1,1] = [i,z]$. We have the diagram

with 1-generated free S-acts $(i,1)S^1$ and $(i+1,1)S^1$ and with right S-act homomorphisms defined by

\begin{align*} \alpha(i,s) &:= (i+1,ts), \\ \beta(i+1,s) & := [i+1,s] = [i,zs], \\ \tau([i,s]) & := (i,es), \\ \psi & := \alpha\tau\beta: (i+1,1)S^1 \to (i+1,1)S^1. \end{align*}

Note that τ is well defined due to the implication (9.1).

Denoting $h:= tez$, since $[i+1,t] = [i,1]$, we have that

\begin{align*} (\beta\alpha\tau)([i,z]) & = (\beta\alpha)(i,ez) = \beta(i+1,tez) = [i+1,tez] = [i+1,t]ez = [i,1]ez \\ & = [i,e]z = [i,1]z = [i,z] \end{align*}

and

\begin{align*} \psi^2(i+1,1) & = (\psi\alpha\tau\beta)(i+1,1) = (\psi\alpha\tau)([i+1,1]) = (\alpha\tau\beta \alpha\tau)([i,z]) \\ & = (\alpha\tau)([i,z]) = (\alpha\tau\beta)(i+1,1) = \psi(i+1,1) = (\alpha\tau)([i,z]) \\ &= \alpha(i,ez) = (i+1,tez) = (i+1,h). \end{align*}

Hence,

\begin{align*} (i+1,h) & = \psi^2(i+1,1) = \psi(i+1,h) = \psi(i+1,1)h = (i+1,h)h = (i+1,h^2), \end{align*}

which yields $h^2=h$. The equalities $[i+1,1]=[i,z]$ and $[i,1]=[i,e]$ imply that $t^n=t^{n+1}z$ and $t^m=t^me$ for some $m,n\in \mathbb{N}$. Since T is a left unitary subsemigroup, $z,e\in T$ and we have an idempotent $h\in tT^1$.

Now consider a chain

\begin{equation*} e_1T^1 \supseteq e_2T^1 \supseteq e_3T^1 \supseteq \cdots \end{equation*}

of principal right ideals of T generated by idempotents $e_i\in T$. By Lemma 5.1, we have a chain

\begin{equation*} e_1S^1 \supseteq e_2S^1 \supseteq e_3S^1 \supseteq \cdots . \end{equation*}

Similar to the proof of Theorem 6.2 in [8], we can find idempotents $g_i\in S$ such that

\begin{equation*} S^1g_1 \supseteq S^1g_2 \supseteq S^1g_3 \supseteq \cdots , \end{equation*}

$e_iS^1 = g_iS^1$ and $g_{i+1}g_i = g_{i+1} = g_ig_{i+1}$ for every $i\in \mathbb{N}$.

By Lemma 6.2, S satisfies Condition (M _L). Thus, there exists $n\in \mathbb{N}$ such that $S^1g_i = S^1g_{i+1}$ for every $i\geqslant n$. Now $g_i=g_ig_{i+1} = g_{i+1}$; therefore, $g_i=g_{i+1}$ and $e_iS^1=e_{i+1}S^1$ for every $i\geqslant n$. By Lemma 5.1, it follows that $e_iT^1 = e_{i+1}T^1$ for every $i\geqslant n$. We have shown that T satisfies the DCC for principal right ideals generated by idempotents.

Suppose that T does not have a minimal principal right ideal. Take any $b_1\in T$. Then $b_1T^1$ contains an idempotent e ₁ and $b_1T^1 \supseteq e_1T^1$. Since $e_1T^1$ is not a minimal ideal, there exists $b_2\in T$ such that $e_1T^1 \supset b_2T^1$. In this way, we get an infinite descending chain

\begin{equation*} b_1T^1 \supseteq e_1T^1 \supset b_2T^1 \supseteq e_2T^1 \supset b_3T^1 \supseteq e_3T^1 \supset \cdots \end{equation*}

of principal right ideals of T, where $e_i\in E(T)$. Thus, we also have a chain

\begin{equation*} e_1T^1 \supset e_2T^1 \supset e_3T^1 \supset \cdots , \end{equation*}

a contradiction. It follows that T has a minimal principal right ideal bT ¹. As it contains an idempotent e, we have $bT^1=eT^1$ due to minimality.

(4) $\Rightarrow$ (5). By Corollary 8.4.

(5) $\Rightarrow$ (6). Assume that S satisfies both Condition (A) and Condition (K). By Theorem 4.3, every right sequence act M _S is cyclic. Due to Corollary 3.3, every M _S is unitary and finite limit flat, hence also pullback flat. By Proposition 8.3, M _S is projective.

(6) $\Rightarrow$ (4). By Lemma 6.2, S satisfies Condition (M _L). We show that S satisfies Condition (A) using Theorem 4.3. Take any sequence $(s_i)_{i\in \mathbb{N}} \in S^{\mathbb{N}}$ and consider the right sequence act M _S determined by it. By assumption, it is projective. It is also indecomposable; hence, by Proposition 2.6, there exists an S-isomorphism $f:M_S\to eS$, where $e\in E(S)$ is some idempotent. Then there exists $[m,s]\in M$ such that $f([m,s]) = e$. Also, there exists $s^{\prime} \in S$ such that $f([m+1,1]) = es^{\prime} $. Now

\begin{equation*} f([m+1,1]) = f([m,s])s^{\prime} = f([m,ss^{\prime}]) \implies [m+1,1] = [m,ss^{\prime}]. \end{equation*}

Thus, there exists $k\geqslant m+1$ such that

\begin{equation*} s_k\ldots s_{m+1} = s_k\ldots s_{m+1}s_mss^{\prime}. \end{equation*}

(3) $\lor$ (4) $\Rightarrow$ (8). Let A _S be a unitary pullback flat act. Using Condition (A), by Corollary 7.5, we conclude that $ A_S = \bigsqcup_{i\in I}A_i, $ where A _i is a cyclic unitary pullback flat act for every $i\in I$. We have by Lemma 7.6 that each A _i is projective when S satisfies either Condition (D) or Condition $(M_L)$. Thus, A _S is projective due to Theorem 2.2.

(8) $\Rightarrow$ (7). Since pullbacks are finite limits, each finite limit act is pullback flat.

(7) $\Rightarrow$ (6). Every right sequence act M _S satisfies Condition (E) by Lemma 3.10 in [15]. It is easy to see that M _S satisfies Condition (LC): if $a,a^{\prime} \in M$, then there exist $a^{\prime\prime}\in M$ and $u,v\in S$ such that $a=a^{\prime\prime} u$ and $a^{\prime} =a^{\prime\prime} v$. By Theorem 4.2 in [15], right sequence acts are finite limit flat. We also know that they are unitary. By assumption, they are projective.

(7) $\Leftrightarrow$ (9). A finite limit flat act A _S must satisfy Condition (LC) by Theorem 4.2 in [15]. This implies that A _S is indecomposable. By Proposition 2.6, A _S is an indecomposable projective in $\mathsf{UAct}_S$ if and only if it is a unitary limit flat act.

(6) $\Leftrightarrow$ (10). Let M _S be a right sequence act. Then M _S is limit flat if and only if $M_S\cong eS_S$ for some idempotent $e\in S$ by Proposition 4.4 in [15]. In other words, a right sequence act over a factorizable semigroup is limit flat if and only if it is projective.

Remark 9.2. 1. Condition (4) in Theorem 9.1 is important because together with Theorem 4.3 and Proposition 6.1, it shows that perfectness of a factorizable semigroup S can be verified by checking two conditions that are formulated in terms of sequences of elements of S. These conditions are internal to S and do not refer to any categories.

2. If S is factorizable and right perfect, then any right sequence act over S is projective. In particular, S must contain at least one idempotent.

It is easy to check that Rees matrix semigroups (with zero) over a group satisfy both Condition (A) and Condition $(M_L)$ (and their duals). Thus, we have the following result.

Corollary 9.3. Cf. [2, Corollary 3.13]

Every completely $(0\text{-})$simple semigroup is perfect.

10. Morita invariance

Recall that semigroups S and T are called Morita equivalent if the categories $\mathsf{FAct}_S$ and $\mathsf{FAct}_T$ are equivalent (see [18]). A Morita invariant is a property that is shared by all Morita equivalent semigroups. In this section, we will prove that right semiperfectness, right $\mathcal{IC}$-perfectness and right perfectness are Morita invariants on the class of factorizable semigroups. For this, we need to examine essential epimorphisms in $\mathsf{UAct}_S$ and $\mathsf{FAct}_S$.

Proposition 10.1. Let S be a semigroup and $A_S\in \mathsf{UAct}_S$. Then µ _A is an essential epimorphism in $\mathsf{UAct}_S$.

Proof. Since A _S is unitary, µ _A is surjective, and it is easy to see that $A\otimes _S S$ is unitary. Let $U\subseteq A\otimes S$ be a unitary subact and consider the diagram

\begin{equation*} U_S \overset{\iota}\longrightarrow A\otimes _SS \overset{\mu_A}\longrightarrow A_S \end{equation*}

in $\mathsf{UAct}_S$, where ι is the embedding. Suppose $\mu_A\iota$ is surjective. By Lemma 1.4, it suffices to show that ι is surjective, that is, $U=A\otimes S$. Take $a\otimes s\in A\otimes S$. By surjectivity of $\mu_A\iota$, there exist $a^{\prime} \in A$ and $s^{\prime} \in S$ such that $a^{\prime} \otimes s^{\prime} \in U$ and $ a = (\mu _A\iota)(a^{\prime} \otimes s^{\prime}) = a^{\prime} s^{\prime}. $ Now

\begin{equation*} a\otimes s = a^{\prime} s^{\prime} \otimes s = a^{\prime} \otimes s^{\prime} s = (a^{\prime} \otimes s^{\prime})s \in US \subseteq U. \end{equation*}

Thus, ι is surjective, as required.

Proposition 10.2. Let S be a factorizable semigroup. For a morphism $f:P_S\to A_S$ in $\mathsf{FAct}_S$, the following are equivalent:

1. (1) f is an essential epimorphism in $\mathsf{UAct}_S$;

2. (2) f is an essential epimorphism in $\mathsf{FAct}_S$.

Proof. (1) $\Rightarrow$ (2). This is clear.

(2) $\Rightarrow$ (1). Let $U\subseteq P_S$ be a unitary subact and let $\iota :U\to P$ be the embedding. Suppose $f\iota$ is surjective. By Lemma 1.4, it suffices to show that ι is surjective, that is, U = P. According to Lemma 1.1, $U\otimes _SS$ is a firm right S-act. Consider the diagram

\begin{equation*} U\otimes _SS \overset{\mu _U}\longrightarrow U_S \overset{\iota}\longrightarrow P_S \overset{f}\longrightarrow A_S. \end{equation*}

Since µ _U is surjective, $f\iota\mu _U$ is a surjective morphism in $\mathsf{FAct}_S$. Also, $\iota\mu _U$ is a morphism in $\mathsf{FAct}_S$. Then $\iota\mu _U$ is surjective due to essentiality of f. It follows that ι is surjective.

Proposition 10.3. The following are equivalent for a factorizable semigroup S:

1. (1) every cyclic act in $\mathsf{UAct}_S$ has a projective cover;

2. (2) every cyclic act in $\mathsf{FAct}_S$ has a projective cover.

Proof. (1) $\Rightarrow$ (2). Take a cyclic act $aS^1 \in\mathsf{FAct}_S$. Since aS ¹ is unitary, it has a cyclic projective cover $f:eS \to aS^1$ in $\mathsf{UAct}_S$ (note that a cover of a cyclic act must be cyclic). Recall that eS is firm (see Lemma 2.3). Consider a diagram

\begin{equation*} B_S \overset{g}\longrightarrow eS \overset{f}\longrightarrow aS^1 \end{equation*}

in $\mathsf{FAct}_S$. If fg is surjective, then g must be surjective because this diagram is also in $\mathsf{UAct}_S$ and f is an essential epimorphism in $\mathsf{UAct}_S$. Thus, f is an essential epimorphism in $\mathsf{FAct}_S$.

(2) $\Rightarrow$ (1). Take a cyclic act $aS^1\in \mathsf{UAct}_S$. By Lemma 1.1, $aS^1\otimes_S S$ is firm. Since aS ¹ is unitary, there exist $s\in S$ and $u\in S^1$ such that $a=(au)s$. Then

\begin{equation*} aS^1 \otimes_S S = (au\otimes s)S^1. \end{equation*}

The inclusion $\supseteq$ is clear. On the other hand, if $v\in S^1$ and $t\in S$, then

\begin{equation*} av\otimes t = (au)sv\otimes t = au\otimes svt = (au\otimes s)vt\in (au\otimes s)S^1. \end{equation*}

Therefore, $aS^1 \otimes_S S$ is a cyclic right S-act.

By assumption, there exists a projective cover $f: eS\to aS^1\otimes S$ in $\mathsf{FAct}_S$. By Proposition 10.2, f is also an essential epimorphism in $\mathsf{UAct}_S$. Proposition 10.1 implies that $\mu _{aS^1}: aS^1\otimes S \to aS^1$ is an essential epimorphism in $\mathsf{UAct}_S$. Therefore,

\begin{equation*} eS \xrightarrow[]{\mu _{aS^1}f} aS^1 \end{equation*}

is a projective cover in $\mathsf{UAct}_S$.

Having a projective cover is a purely categorical property; thus, it is preserved by equivalence functors. Also, indecomposability is a categorical property: an act is indecomposable if it is not a coproduct of two noninitial objects (the initial object is just the empty act). However, for cyclicity, there is no obvious categorical description. Still, cyclic acts are preserved under tensor multiplication functors coming from certain Morita contexts.

Definition 10.4. [22]

A Morita context is a six-tuple $(S,T,{_SP_T}, {_TQ_S}, \theta,\phi)$, where S and T are semigroups, ${_SP_T}\in {_S\mathsf{Act}_T}$ and ${_TQ_S}\in {_T\mathsf{Act}_S}$ are biacts and

\begin{equation*} \theta: {_S(P\otimes Q)_S} \to {_SS_S}, \qquad \phi: {_T(Q\otimes P)_T} \to {_TT_T} \end{equation*}

are biact homomorphisms such that for every $p,p^{\prime} \in P$ and $q,q^{\prime} \in Q$, we have

\begin{equation*} \theta(p\otimes q)p^{\prime} =p\phi(q\otimes p^{\prime})\quad\mbox{and}\quad q\theta(p\otimes q^{\prime})=\phi(q\otimes p)q^{\prime}. \end{equation*}

This context is called unitary if $_SP_T$ and $_TQ_S$ are unitary biacts, meaning that $SP=P=PT$ and $TQ=Q=QS$.

By Theorem 5.9 in [14], two firm semigroups S and T are Morita equivalent if and only if they are connected by a unitary Morita context with surjective mappings θ and ϕ.

Proposition 10.5. Let S and T be firm semigroups connected by a unitary Morita context $(S,T,{_SP_T},{_TQ_S},\theta,\phi)$ with surjective mappings. Then the functors

\begin{equation*} -\otimes {_SP_T}: \mathsf{FAct}_S\to \mathsf{FAct}_T\quad\mbox{and}\quad -\otimes {_TQ_S}: \mathsf{FAct}_T\to \mathsf{FAct}_S \end{equation*}

take cyclic acts to cyclic acts.

Proof. As in Proposition 3.16 of [18], one can prove that $-\otimes {_SP_T}: \mathsf{FAct}_S\to \mathsf{FAct}_T$ and $-\otimes {_TQ_S}: \mathsf{FAct}_T\to \mathsf{FAct}_S$ are equivalence functors inverse to each other (see also Theorem 5.9 in [14]). We will prove that $-\otimes {_SP_T}$ takes cyclic acts to cyclic acts. The same is true for $-\otimes {_TQ_S}$, so cyclic acts will correspond to each other under these functors.

Consider a cyclic act $aS^1\in \mathsf{FAct}_S$. We will prove that $aS^1 \otimes_S P_T$ is cyclic. Since aS ¹ is unitary, there exists $s\in S$ such that as = a. Using surjectivity of θ, we can find $p_s\in P$ and $q_s\in Q$ such that $s=\theta(p_s\otimes q_s)$. We will prove that

\begin{equation*} aS^1\otimes_S P_T = (a\otimes p_s)T^1. \end{equation*}

The inclusion $(a\otimes p_s)T^1\subseteq aS^1\otimes_S P_T$ is clear. To prove the converse, we note that

\begin{align*} au\otimes p & = asu\otimes p = a\otimes sup = a\otimes \theta(p_s\otimes q_s)up = a\otimes p_s\phi(q_s\otimes up) \\ & = (a\otimes p_s)\phi(q_s\otimes up) \in (a\otimes p_s)T^1 \end{align*}

for every $u\in S^1$ and $p\in P$.

Proposition 10.6. Let S be a factorizable semigroup. Then the functors

\begin{equation*} - \otimes _SS_{S\otimes S} :\mathsf{FAct}_S \to \mathsf{FAct}_{S\otimes S} \quad\mbox{and}\quad -\otimes _{S\otimes S}S_S : \mathsf{FAct}_{S\otimes S} \to \mathsf{FAct}_S \end{equation*}

are inverse equivalence functors which take cyclic acts to cyclic acts.

Proof. By Proposition 4.9 in [16], these functors are inverse equivalence functors. We also recall that $S\otimes S$ is considered as a semigroup with the multiplication $(s\otimes t)(u\otimes v) = st\otimes uv$, and the action of $S_{S\otimes S}$ is $s(u\otimes v)=suv$.

Let $aS^1\in \mathsf{FAct}_S$ be cyclic. We want to show that $aS^1\otimes _S S_{S\otimes S}$ is a cyclic $S\otimes S$-act. Suppose $a=(au)s$ for some $u\in S^1$ and $s\in S$. We will prove that

\begin{equation*} aS^1\otimes _S S_{S\otimes S} = (au\otimes s)(S\otimes S)^1. \end{equation*}

On the one hand,

\begin{equation*} (au\otimes s)(S\otimes S)^1 \subseteq (aS^1\otimes _SS)(S\otimes S)^1 = aS^1\otimes _SS. \end{equation*}

Conversely, suppose that $av\otimes t \in aS^1\otimes _S S_{S\otimes S}$. Since S is factorizable, $t=s_1s_2$ for some $s_1,s_2\in S$. Hence,

\begin{align*} av\otimes t &= ausv\otimes t = au\otimes svt = au\otimes svs_1s_2 = au\otimes (s(vs_1\otimes s_2)) \\ & = (au\otimes s)(vs_1 \otimes s_2) \in (au\otimes s)(S\otimes S)^1. \end{align*}

Let now $a(S\otimes S)^1 \in \mathsf{FAct}_{S\otimes S}$ be cyclic. We want to show that $a(S\otimes S)^1 \otimes _{S\otimes S}S_S$ is a cyclic S-act. Since $a(S\otimes S)^1$ is unitary, $a=a(s_1\otimes s_2)$ for some $s_1,s_2\in S$. We will prove that

\begin{equation*} a(S\otimes S)^1 \otimes _{S\otimes S}S_S = (a \otimes s_1s_2)S^1. \end{equation*}

The inclusion $\supseteq$ is clear. Conversely, let $a(u_1\otimes u_2)\otimes s \in a(S\otimes S)^1 \otimes_{S\otimes S} S_S$, where $u_1,u_2,s\in S$. Then

\begin{align*} a(u_1\otimes u_2)\otimes s &= a(s_1\otimes s_2)(u_1\otimes u_2) \otimes s = a(s_1s_2\otimes u_1u_2) \otimes s = a \otimes (s_1s_2\otimes u_1u_2)s \\ & = a\otimes s_1s_2u_1u_2s = (a\otimes s_1s_2)u_1u_2s \in (a \otimes s_1s_2)S^1. \end{align*}

Our first theorem about invariants is the following.

Theorem 10.7. Right semiperfectness is a Morita invariant for factorizable semigroups.

Proof. Suppose S and T are Morita equivalent factorizable semigroups. Equivalently, by Proposition 4.9 in [16], $S\otimes S$ and $T\otimes T$ are Morita equivalent firm semigroups. According to Theorem 5.9 in [14], there exists a unitary Morita context with bijective mappings connecting $S\otimes S$ and $T\otimes T$. Consequently,

\begin{align*} S\ \mbox{is right semiperfect} & \iff \mbox{all cyclic objects in}\ UAct_{S}\ \mbox{have a projective cover}\\ (\text{by Proposition }10.3) & \iff \mbox{all cyclic objects in}\ FAct_{S}\ \mbox{have a projective cover}\\ (\text{by Proposition }10.6) & \iff \mbox{all cyclic objects in}\ FAct_{S\otimes S}\ \mbox{have a projective cover}\\ (\text{by Proposition }10.5) & \iff \mbox{all cyclic objects in}\ FAct_{T\otimes T}\ \mbox{have a projective cover}\\ (\text{by Proposition }10.6) & \iff \mbox{all cyclic objects in}\ FAct_{T}\ \mbox{have a projective cover}\\ (\text{by Proposition }10.3) & \iff \mbox{all cyclic objects in}\ UAct_{T}\ \mbox{have a projective cover}\\ & \iff T\ \mbox{is right semiperfect.} \end{align*}

Proposition 10.8. The following are equivalent for a factorizable semigroup S:

1. (1) every act in $\mathsf{UAct}_S$ has an $\mathcal{IC}$-cover;

2. (2) every act in $\mathsf{FAct}_S$ has an $\mathcal{IC}$-cover.

Proof. (1) $\Rightarrow$ (2) Take $A_S\in\mathsf{FAct}_S$ and let

\begin{equation*} B_S:=\bigsqcup_{i\in I} b_iS^1 \overset{f}\longrightarrow A_S \end{equation*}

be an $\mathcal{IC}$-cover in $\mathsf{UAct}_S$. By Lemma 1.1, $B\otimes _SS$ is firm. The indecomposable components of B are unitary. Hence, for every $i\in I$, there exist $u_i\in S^1$ and $s_i\in S$ such that $b_i=(b_iu_i)s_i$. Then

\begin{equation*} B\otimes_S S = \left ( \bigsqcup _{i\in I} b_iS^1 \right ) \otimes_S S = \bigsqcup _{i\in I} (b_iS^1\otimes_S S) = \bigsqcup _{i\in I} (b_iu_i\otimes s_i)S^1. \end{equation*}

The acts $(b_iu_i\otimes s_i)S^1$ are indecomposable because they are cyclic.

Consider the diagram

\begin{equation*} B\otimes S \overset{\mu _B}\longrightarrow B_S \overset{f}\longrightarrow A_S. \end{equation*}

By Proposition 10.1, µ _B is essential in $\mathsf{UAct}_S$. Thus, the composition $f\mu _B$ is an essential epimorphism in $\mathsf{UAct}_S$. Since $f\mu _B$ is also a morphism in $\mathsf{FAct}_S$, it is essential in $\mathsf{FAct}_S$ by Proposition 10.2.

(2) $\Rightarrow$ (1) Take $A_S\in\mathsf{UAct}_S$. Then $A\otimes _SS$ is firm by Lemma 1.1. By assumption, there exists an $\mathcal{IC}$-cover

\begin{equation*} C_S:=\bigsqcup _{i\in I}c_iS^1 \overset{f}\longrightarrow A\otimes _SS \end{equation*}

in $\mathsf{FAct}_S$. Then f is an essential epimorphism in $\mathsf{UAct}_S$ by Proposition 10.2. By Proposition 10.1, µ _A is essential in $\mathsf{UAct}_S$. Therefore, $\mu _Af:C_S \to A_S$ is an $\mathcal{IC}$-cover in $\mathsf{UAct}_S$.

Theorem 10.9. Right $\mathcal{IC}$-perfectness is a Morita invariant for factorizable semigroups.

Proof. Let S and T be factorizable semigroups and $F:\mathsf{FAct}_S \to \mathsf{FAct}_T$ an equivalence functor that takes cyclic acts to cyclic acts. Suppose that $A_S\in \mathsf{FAct}_S$ has an $\mathcal{IC}$-cover

\begin{equation*} f: \bigsqcup_{i\in I} b_iS^1 \to A_S. \end{equation*}

Since F preserves coproducts and essential epimorphisms,

\begin{equation*} F(f): \bigsqcup_{i\in I} F(b_iS^1) \cong F\left(\bigsqcup_{i\in I} b_iS^1\right) \to F(A_S) \end{equation*}

is an $\mathcal{IC}$-cover of $F(A_S)$, where $F(b_iS^1)\in \mathsf{FAct}_S$ are cyclic acts (and hence indecomposable). So $F(A_S)$ also has an $\mathcal{IC}$-cover. If F has an inverse equivalence functor G, which also takes cyclic acts to cyclic acts, then we can prove that A _S has an $\mathcal{IC}$-cover if and only if $F(A_S)$ has an $\mathcal{IC}$-cover.

Let now S and T be as in the proof of Theorem 10.7. Then

\begin{align*} S\ \mbox{is right}\ \mathcal{IC}\mbox{-perfect } & \iff \mbox{all objects in}\ \mathsf{UAct}_S\ \mbox{have an}\ \mathcal{IC}\mbox{-cover} \\ (\text{by Proposition }10.8)& \iff \mbox{all objects in}\ \mathsf{FAct}_S\ \mbox{have an}\ \mathcal{IC}\mbox{-cover} \\ (\text{by Proposition }10.6)& \iff \mbox{all objects in}\ \mathsf{FAct}_{S\otimes S}\ \mbox{have an}\ \mathcal{IC}\mbox{-cover}\\ (\text{by Proposition }10.5)& \iff \mbox{all objects in}\ \mathsf{FAct}_{T\otimes T}\ \mbox{have an}\ \mathcal{IC}\mbox{-cover} \\ (\text{by Proposition }10.6)& \iff \mbox{all objects in}\ \mathsf{FAct}_T\ \mbox{have an}\ \mathcal{IC}\mbox{-cover} \\ (\text{by Proposition }10.8)& \iff \mbox{all objects in}\ \mathsf{UAct}_T\ \mbox{have an}\ \mathcal{IC}\mbox{-cover} \\ & \iff T\ \mbox{is right}\ \mathcal{IC}\mbox{-perfect. } \\ \end{align*}

Corollary 10.10. Right perfectness and perfectness are Morita invariants for factorizable semigroups.

Proof. Let S and T be Morita equivalent factorizable semigroups. Then $\mathsf{FAct}_S$ and $\mathsf{FAct}_T$ are equivalent categories. By Theorem 10.7, Theorem 10.9 and Theorem 9.1, it follows that right perfectness is a Morita invariant.

From Remark 4.12 in [16] we know that also ${_S\mathsf{FAct}}$ and ${_T\mathsf{FAct}}$ are equivalent categories. If now S is left perfect, then the duals of the proofs of Theorem 10.7 and Theorem 10.9 yield that T is also left perfect. It follows that perfectness is a Morita invariant.

We saw that completely simple semigroups are perfect. This fact can also be concluded from Morita invariance of perfectness in case of factorizable semigroups.

Corollary 10.11. Completely simple semigroups are perfect.

Proof. Clearly, groups are perfect semigroups. Factorizable semigroups Morita equivalent to a given group are precisely Rees matrix semigroups over that group [19].