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.

Proof. By the dual of Lemma 3.1(1) in [Reference Lawson18], 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). $

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.

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 [Reference Chen and Shum4] 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.

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

(2) $\iff$ (3). This is Proposition 4.4 in [Reference Laan, Reimaa, Tart and Teor15].

(3) $\iff$ (4). This is due to Lemma 1.1 and Corollary 1.2 in [Reference Fountain7].

Proof. Necessity. (This proof is similar to the proof of implication (ii) $\implies $ (iii) in [Reference Kilp, Knauer and Mikhalev13, 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.

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.

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.

Proof. The first implication follows from Proposition 2.6 and Lemma 3.2. The second implication holds due to Proposition 4.3 in [Reference Laan, Reimaa, Tart and Teor15].

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.

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.

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 $s=t^{\prime}_ns_n \in T$.

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.

Proof. (1) $\implies$ (2). (This is inspired by the proof of Proposition 4.6 in [Reference Gould and Shaheen8]. 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 [Reference Gould and Shaheen8]). 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*}

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.

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.

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 $us_1,us_2\in S$.

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.

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.

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 [Reference Clifford and Preston5], 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.

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 [Reference Kilp11], 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.

Proof. This follows from Proposition 8.3 and Lemma 7.6.

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 [Reference Gould and Shaheen8].

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 [Reference Gould and Shaheen8], 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 [Reference Laan, Reimaa, Tart and Teor15]. 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 [Reference Laan, Reimaa, Tart and Teor15], 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 [Reference Laan, Reimaa, Tart and Teor15]. 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 [Reference Laan, Reimaa, Tart and Teor15]. In other words, a right sequence act over a factorizable semigroup is limit flat if and only if it is projective.

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.

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.

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 $\mathsf{UAct}_S$.

Proof. As in Proposition 3.16 of [Reference Lawson18], 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 [Reference Laan, Márki and Reimaa14]). 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 $p\in P$.

Proof. By Proposition 4.9 in [Reference Laan and Reimaa16], 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*}

Proof. Suppose S and T are Morita equivalent factorizable semigroups. Equivalently, by Proposition 4.9 in [Reference Laan and Reimaa16], $S\otimes S$ and $T\otimes T$ are Morita equivalent firm semigroups. According to Theorem 5.9 in [Reference Laan, Márki and Reimaa14], 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*}

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 $\mathsf{UAct}_S$.

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*}

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 [Reference Laan and Reimaa16] 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.

Proof. Clearly, groups are perfect semigroups. Factorizable semigroups Morita equivalent to a given group are precisely Rees matrix semigroups over that group [Reference Lepik19].