Covers of Finitely Generated Acts over Monoids

Abstract

This paper attempts to initiate the study of covers of finitely generated S-acts over monoids. We provide necessary and sufficient conditions for a monoid to ensure that n-generated S-acts have strongly flat covers, Condition $\left(P\right)$ covers, and projective covers. The main conclusions extend some known results. We also show that Condition $\left(P\right)$ covers of finitely generated S-acts are not unique, unlike strongly flat covers. Additionally, we demonstrate the property of Enochs’ $\mathcal{X}$-precover of S-act A, where $\mathcal{X}$ denotes a class of S-acts that are closed under isomorphisms.

1. Introduction and Preliminaries

Let S be a monoid. We call a nonempty set A a right S-act over S if we have a mapping $A\times S\to A$ given by $(a,s)\mapsto as$ such that for all $a\in A,s,t\in S,a1=a$ and $a\left(st\right)=\left(as\right)t.$ We refer the reader to [1,2,3] for all undefined terms concerning acts over monoids.

Mahmoudi and Renshaw, in [3], introduced coessential epimorphisms and covers of cyclic S-acts. They provided a characterization of coessential epimorphisms using congruence classes and left unitary submonoids. Furthermore, they defined various types of covers such as strongly flat covers, Condition $\left(P\right)$ covers, and projective covers. They also established necessary and sufficient conditions for cyclic S-acts to possess Condition $\left(P\right)$ covers, strongly flat covers, and projective covers.

In 2010, Khosravi et al. investigated the covers of Condition (P) and strongly flat covers for arbitrary S-acts over monoids in [4]. Much like a perfect monoid, an $SF$-perfect monoid was identified by condition (A) possessing a strongly flat cover for every cyclic right S-act, and a $\left(P\right)$-perfect monoid was defined by condition (A) and having having a Condition (P) cover for each cyclic right S-act. Additionally, the authors introduced a new discription for perfect monoids, defining them as monoids where each strongly flat right S-act has a projective cover.

Bailey and Renshaw introduced articles on covers of S-acts over monoids in 2013 and 2014, respectively. They considered free, torsion-free, divisible, and injective covers in [5]. As we all know, Enochs’s celebrated flat cover conjecture was ultimately proven in [6] in 2001. And in [1], they attempted to mimic some of Enochs’s work within the category of S-acts over monoids, primarily concentrating on strongly flat acts. Of course, the covers of [3] are slightly different from Enochs’ notion of covers.

Subsequently, several additional papers concerning covers have appeared (e.g., [7,8,9,10]). Recently, Irannezhad and Madanshekaf [11] researched covers of S-acts over monoids with Condition ($P^{\prime }$), and Qiao et al. introduced Condition ($P{F}^{″}$) covers in a similar way in [12]. Now, we restrict our attention to these notions.

Let S be a monoid. Recall [3] that a right S-act $B_S$ together with an S-epimorphism $f:B_S\to A_S$ is a cover of a right S-act $A_S$ if for any proper subact $C_S$ of $B_S$ the restriction ${f|}_C$ is not an S-epimorphism. Moreover, we call an S-epimorphism $f:B_S\to A_S$ coessential in [3] if $fg$ is an S-epimorphism for every S-act $C_S$ and every S-map $g:C_S\to B_S$, then g is an S-epimorphism. And we define an act $B_S$, along with an S-epimorphism $f:B_S\to A_S$, as an X-cover of $A_S$ if $B_S$ satisfies property X and f is coessential. One can readily observe that an S-morphism $f:B\to A$ being a cover of A is equivalent to f being a coessential epimorphism.

Let $\mathcal{X}$ be a class of right S-acts closed under isomorphisms, meaning if $X\in \mathcal{X}$ and $Y\cong X$, then $Y\in \mathcal{X}$. Now let us explore another concept of covers (Enochs’ notion). let A be a right S-act. Recall [1] that an S-act $X\in \mathcal{X}$ is called an $\mathcal{X}$-precover of A if there exists an S-homomorphism $\phi :X\to A$ such that, for every S-homomorphism $\psi$: $X^{\prime }\to A$, for $X^{\prime }\in \mathcal{X}$, there is a homomorphism f: $X^{\prime }\to X$ with $\psi =\phi f$. Alternatively, the following diagram

commutes. We define $\phi :X\to A$ as an $\mathcal{X}$-cover of A if the precover satisfies the condition that for any endomorphism $f:X\to X$ such that $\phi =\phi f$, f must be an automorphism.

Recall [2] that a right S-act $A_S$ is called strongly flat if it is both pullback flat and equalizer flat. It is a commonly acknowledged fact that an S-act A is strongly flat if, and only if, it satisfies both Condition (P) and Condition (E). Now, a right S-act $A_S$ is said to satisfy Condition $\left(P\right)$ if

$$(\forall a,a^{\prime }\in A)(\forall s,s^{\prime }\in S)(as=a^{\prime }{s}^{\prime }\Rightarrow (\exists a^{″}\in A)(\exists u,v\in S)(a=a^{″}u\wedge a^{\prime }=a^{″}v\wedge us=v{s}^{\prime })).$$

Recall that a right S-act $A_S$ satisfies Condition $\left(E\right)$ if

$$(\forall a\in A)(\forall s,s^{\prime }\in S)(as=a{s}^{\prime }\Rightarrow (\exists a^{\prime }\in A)(\exists u\in S)(a=a^{\prime }u\wedge us=u{s}^{\prime })).$$

In this paper, our primary focus lies on finitely generated S-acts, particularly on the notions of strong flatness and Condition (P), and we find valuable characterizations.

Let S be a monoid. By [3], it is said to be right reversible if there exist $x,y\in S$ such that $xs=yt$ for all $s,t\in S$. And it is said to be left collapsible if there is $x\in S$ such that $xs=xt$ for all $s,t\in S$. Let T be a submonoid of S. T is called a left unitary submonoid if and only if whenever $t,tr\in T$ then $r\in T$.

In [3], Mahmoudi M. and Renshaw J. solved a problem concerning the covers of cyclic S-acts over monoids. However, the problem of whether all S-acts have covers remains open. Recall [2] that every S-act has a unique decomposition into indecomposable subacts. If a finitely generated S-act satisfies Condition (P), then it is a coproduct of cyclic subacts. Based on the above facts, in this paper, we consider the covers of finitely generated S-acts and coproducts of S-acts and then study Condition $\left(P\right)$ covers, strongly flat covers, and projective covers. This research brings us closer to understanding the covers of any S-acts. Therefore, this study of the covers of finitely generated S-acts plays an important role.

The main results here are descriptions of finitely generated acts and coproducts of S-acts in Section 2, and we provide necessary and sufficient conditions for finitely generated S-acts to have covers in Section 3. Next, in Section 4, we concentrate on $\left(P\right)$-covers, strongly flat covers, and projective covers, and the main conclusions extend some known results in [3]. In Section 5, our exploration delves into Enochs’ notion of covers of acts over monoids, with a specific emphasis on $\mathcal{X}$-precovers.

Throughout this paper, S denotes a monoid, $\mathcal{X}$ denotes a class of right S-acts closed under isomorphisms, $A=\langle a_1,a_2,\cdots ,a_n\rangle ={\displaystyle \bigcup _{i=1}^n}{a}_{i}S$ is said that A is a n-generated right S-act, and $I=\{1,2,\cdots ,n\}$. And ${\displaystyle \coprod _{i=1}^n}{a}_{i}S$ will denote a coproduct of cyclic subacts $a_{i}S$ for all $i\in I$. We focus solely on the right S-acts, often just called S-acts.

2. Finite Generation and Coproducts

This section aims to describe the finite generation and coproduct of S-acts. A subset U of an S-act A is a generating set for A if, for any $a\in A$, there exist $u\in U$, $s\in S$ such that $a=us$. In other words, U is a set of generating elements for $A_S$ if $\langle U\rangle :={\displaystyle \bigcup _{u\in U}}uS=A_S$, where $uS=\{us\mid s\in S\}$. An S-act A is said to be finitely generated (resp. cyclic) if it has a finite (resp. one-element) generating set. Further details may be found, for example, in [2].

Proposition 1.

Let S be a monoid and $A={\displaystyle \bigcup _{i=1}^n}{A}_i$ be S-act, $A_i$ is a subact of A. For every $i\in I$, $A_i$ is finitely generated, then A is finitely generated.

Proof. 

Let $A_i=\langle X_i\rangle$. Then, $A=\langle {\displaystyle \bigcup _{i=1}^n}{X}_i\rangle$. □

Lemma 1.

Let S be a monoid and $A={\displaystyle \bigcup _{i=1}^n}{A}_i$ an S-act, $A_i$ are subacts of A for $i\in I$. And for $i\ne j\in I$, suppose that $A_i\bigcap A_j$ is either finitely generated or empty. If A is finitely generated, then $A_i$ for all $i\in I$ is finitely generated.

Proof. 

It is clear by Lemma 5.3 of [13]. □

Corollary 1.

Let S be a monoid. S-act $A={\displaystyle \coprod _{i=1}^n}{A}_i$ is finitely generated if and only if S-act $A_i$ is finitely generated.

It is important for monoids to consider the covers of finitely generated acts. Now, we first aim to investigate the cover of coproducts of acts.

Proposition 2.

Suppose that $\{f_i:B_i\to A_i,i\in I\}$ is a family of S-morphisms, where $I=\{1,2,\cdots ,n\}$. Let $B={\displaystyle \coprod _{i=1}^n}{B}_i$, $A={\displaystyle \coprod _{i=1}^n}{A}_i$, and $f:B\to A$ satisfies ${f|}_{B_i}=f_i$. Then, $f:B\to A$ is a cover of A if and only if $f_i:B_i\to A_i$ is a cover of $A_i$ for every $i\in I$.

Proof. 

Sufficiency. Since $\{f_i:B_i\to A_i$, $i\in I\}$ is a family of epimorphisms, it is clear that $f:B\to A$ is epimorphic. Assume that f is not coessential, that is, there exists a proper subact $\overline{B}$ of B such that ${f|}_{\overline{B}}$ is epimorphic. Therefore, there is $j\in I$ such that $B_j^{\prime }=B_j\bigcap \overline{B}$ is a proper subact of $B_j$. Thus, $f_j{|}_{B_j^{\prime }}$ is not epimorphism since $f_j:B_j\to A_j$ is a coessential epimorphism, so there exists $a_j\in A_j$ such that $f_j\left(b_j^{\prime }\right)\ne a_j$ for all $b_j^{\prime }\in B_j^{\prime }$. However, by the surjectivity of ${f|}_{\overline{B}}$, we obtain that $f\left(\overline{b}\right)=a_j$ for some $\overline{b}\in \overline{B}$. Thus, there exists $k\in I$ such that $\overline{b}\in B_k$, $k\ne j$. Then $a_j=f\left(\overline{b}\right)\in A_k$ which contradicts $a_j\in A_j$.

Necessity. Let $f:{\displaystyle \coprod _{i=1}^n}{B}_i\to {\displaystyle \coprod _{i=1}^n}{A}_i$ be a coessential epimorphism. Then, we obtain that $\{f_i:B_i\to A_i,i\in I\}$ is a family of epimorphisms. To show that $f_i$ is coessential for all $i\in I$, assume that there exists a map $f_j:B_j\to A_j$ such that $f_j$ is not coessential, where $j\in I$. Then $f_j{|}_{C_j}:C_j\to A_j$ is an epimorphism for some proper subact $C_j$ of $B_j$. Thus, ${f|}_{({\displaystyle \coprod _{i\ne j}}{B}_i)\coprod C_j}:({\displaystyle \coprod _{i\ne j}}{B}_i)\coprod C_j\to {\displaystyle \coprod _{i=1}^n}{A}_i$ is an epimorphism. But $({\displaystyle \coprod _{i\ne j}}{B}_i)\coprod C_j$ is a proper subact of ${\displaystyle \coprod _{i=1}^n}{B}_i$ and by assumption ${f|}_{({\displaystyle \coprod _{i\ne j}}{B}_i)\coprod C_j}$ is not epimorphic, a contradiction. So, $\{f_i:B_i\to A_i,i\in I\}$ is a family of coessential epimorphisms. □

Now, we concentrate on covers of finitely generated S-acts, and we begin with some notations. For an element a of a right S-act A, we denote by $R\left(a\right)$ the set $\{s\in S:as=a\}$, the submonoid of right identities of a, and it is clear that $R\left(a\right)$ is a left unitary submonoid of S. We denote by $r(a,b)$ the set $\left\{\right(s,t)\in S\times S:as=bt\}$. In particular, $r\left(a\right)=r(a,a)=\left\{\right(s,t)\in S\times S:as=at\}$, the right annihilator congruence ${\rho }_a$ of a. It is clear that $aS$ is isomorphic to $S/{\rho }_a$ under the S-isomorphism $as\mapsto \left[s\right]$, where $\left[s\right]$ denote the class of $s\in S$ with respect to an equivalence relation $\rho$.

Lemma 2

([2], Proporsition III.13.14). Let $A_S$ be a finitely generated S-act. If $A_S$ satisfies Condition $\left(P\right)$, then $A_S$ is a coproduct of cyclic subacts.

Let S be a monoid and $A={\displaystyle \coprod _{i=1}^n}{A}_i$, where $A_i,i\in I$ are right S-acts. Then, A is strongly flat (resp. satisfies Condition $\left(P\right)$) if and only if $A_i$ is strongly flat (resp. satisfies Condition $\left(P\right)$) for every $i\in I$. The following results turn out to be very helpful in the following section.

Proposition 3.

If the finitely generated S-act $A=\langle a_1,\cdots ,a_n\rangle$ is strongly flat (resp. satisfis Conditon (P)), then $R\left(a_i\right)$ is a left collapsible(resp. right reversible) submonoid of S.

Proof. 

Let A satisfy Condition $\left(P\right)$. By Lemma 2, $a_{i}S$ satisfies Condition $\left(P\right)$ for $i\in I$. Thus, $R\left(a_i\right)$ is right reversible. And, in a similar way, we obtain that $R\left(a_i\right)$ is a left collapsible submonoid of S if A is strongly flat. □

Proposition 4.

Let T be a submonoid of a monoid S and A a right S-act, $a\in A$. If $r\left(a\right)=T\times T$, then $T=R\left(a\right)$.

Proof. 

Since $r\left(a\right)=\left\{\right(s,t)\in S\times S|as=at\}=T\times T$, for any $t\in T$ and $1\in T$, then $(1,t)\in T\times T=r\left(a\right)$. Thus, $a=at$, that is, $t\in R\left(a\right)$. Therefore, $T\subseteq R\left(a\right)$. It is clear that $R\left(a\right)\subseteq T$. □

Lemma 3.

Let S be a monoid and P a left collapsible(resp. right reversible) submonoid of S, $A=\langle a_1,\cdots ,a_n\rangle$ is n-generated. If $r\left(a_i\right)=P\times P$, then $R\left(a_i\right)$ is left collapsible (resp. right reversible) and ${\displaystyle \coprod _{i=1}^n}{a}_{i}S$ is strongly flat(resp. satisfies Condition (P)).

Proof. 

Since $r\left(a_i\right)=P\times P$, $P=R\left(a_i\right)$ by Proposition 4, $R\left(a_i\right)$ is left collapsible (resp. right reversible). Notice that in either case, P is right reversible and so defined by $bs=bt$ if and only if there exist $p,q\in P$ with $ps=qt$. Notice that $b=a_i$. In fact, if $a_{i}s=a_{i}t$ then there exist $p,q\in P$ with $ps=qt$ and so $bs=bt$. Conversely, if $bs=bt$, then there are $p,q\in P$ with $ps=qt$ and $a_i=a_{i}p=a_{i}q$ from $P\subseteq R\left(a_i\right)$. So $a_{i}s=a_{i}ps=a_{i}qt=a_{i}t$ and hence $b=a_i$. It is clear that $a_{i}S$ satisfies Condition $\left(P\right)$. Then, ${\displaystyle \coprod _{i=1}^n}{a}_{i}S$ satisfies Condition $\left(P\right)$. If in addition P is left collapsible, then ${\displaystyle \coprod _{i=1}^n}{a}_{i}S$ is strongly flat. □

It is clear that any cover of a cyclic right S-act is cyclic by Lemma 2.3 of [3]. The following theorem concerns the covers of n-generated right S-acts.

Theorem 1.

Let $f:B_S\to A_S$ be a cover of $A_S$. Then, $B_S$ is n-generated if and only if $A_S$ is n-generated.

Proof. 

Necessity. Assume that B is n-generated, $\overline{B}=\{b_1,b_2,\cdots ,b_n\}$ is a generating set of B. Since $f:B\to A$ is an epimorphism, A is generated by n elements. Now, we are ready to prove n is the smallest. Suppose that $A=\langle y_1,\cdots ,y_k\rangle$, $k<n$. Then, $f\left(z_i\right)=y_i$, $i=1,\cdots ,k$. Therefore, $B={\displaystyle \bigcup _{i=1}^k}{z}_{i}S$, which contradicts that B is n-generated. Thus, A is n-generated.

Sufficiency. Suppose $A=\langle a_1,\cdots ,a_n\rangle$. Since f is an epimorphism, there exists $b_i\in B$ such that $f\left(b_i\right)=a_i,i=1,2,\cdots ,n.$ Take $C={\displaystyle \bigcup _{i=1}^n}{b}_{i}S$, clearly $C\subseteq B$ and $f\left(C\right)=A$. By the definition of cover, we have $C=B$. Therefore, B is generated by n elements. Now, we are ready to prove a result that n is the smallest. If $B=\langle x_1,\cdots ,x_k\rangle ,k<n$, and $f\left(B\right)=A$, we have that A is generated by $f\left(x_i\right),i=1,2,\cdots ,k$, a contradiction. □

Let $A={\displaystyle \bigcup _{i=1}^n}{A}_i$ be n-generated and $B={\displaystyle \coprod _{i=1}^n}{B}_i$. The following Proposition 5 shows that $f:B\to A$ is a cover of A implies that $\{f_i:B_i\to A_i,i\in I\}$ is a family of covers of $A_i$.

Proposition 5.

Let $A={\displaystyle \bigcup _{i=1}^n}{A}_i,B={\displaystyle \coprod _{i=1}^n}{B}_i$. Let $\{f_i:B_i\to A_i,i\in I\}$ be a family of S-morphisms, and $f:B\to A$ satisfies ${f|}_{B_i}=f_i$. Then, $f:B\to A$ is a cover of A implies $\{f_i:B_i\to A_i,i\in I\}$ is a family of covers of $A_i$.

Proof. 

If $A_i\bigcap A_j=\varnothing$ for any $i,j\in I$, it is clear by Proposition 2. Otherwise, suppose $a\in A_i\bigcap A_j,i\ne j,i,j\in I$. Since f is an epimorphism, for $a\in A_i$, there exists $b\in B_i$ such that $f\left(b\right)=f_i\left(b\right)=a$. If $a\in A_j$, we have $f\left(b\right)=f_j\left(b\right)=a$ for some $b\in B_j$. But $B_i\bigcap B_j=\varnothing$, a contradiction. Therefore, $A_i\bigcap A_j=\varnothing$, we obtain that $A={\displaystyle \coprod _{i=1}^n}{A}_i$. By Proposition 2, $f:B\to A$ is a cover of A implies that $f_i:B_i\to A_i$ is a cover of $A_i$. □

Conversely, if B is n-generated and A is a disjoint union of n subacts, the argument always holds in Proposition 6.

Proposition 6.

Suppose that $B={\displaystyle \bigcup _{i=1}^n}{B}_i$, $A={\displaystyle \coprod _{i=1}^n}{A}_i$, $\{f_i:B_i\to A_i,i\in I\}$ is a family of S-morphisms, and $f:B\to A$ satisfies ${f|}_{B_i}=f_i$. Then $f:B\to A$ is a cover of A if and only if $\{f_i:B_i\to A_i,i\in I\}$ is a family of covers of $A_i$.

Proof. 

Let $f:B\to A$ be a cover of A. If there exist $i,j\in I$ such that $B_i\bigcap B_j\ne \varnothing$, that is, $f\left(b_i\right)=f\left(b_j\right)=a$, where $b_i\in B_i,b_j\in B_j,a\in A$, then ${f|}_{(B_i\setminus \left\{b_i\right\})\bigcup ({\displaystyle \bigcup _{k\ne i}}{B}_k)}:(B_i\setminus \left\{b_i\right\})\bigcup ({\displaystyle \bigcup _{k\ne i}}{B}_k)\to {\displaystyle \coprod _{i=1}^n}{A}_i$ is an epimorphism, a contradiction. So, f is one-to-one. Assume $b\in B_i\bigcap B_j$. Then, $f\left(b\right)=f_i\left(b\right)\in A_i$, and $f\left(b\right)=f_j\left(b\right)\in A_j$ which contradicts $A_i\bigcap A_j=\varnothing$. Hence, $B_i\bigcap B_j=\varnothing$ or the element in $B_i\bigcap B_j$ has no image under the action of f. If the latter, then ${f|}_{B\setminus (B_i\bigcap B_j)}:B\setminus (B_i\bigcap B_j)\to A$ is an epimorphism, a contradiction. Therefore, $B={\displaystyle \coprod _{i=1}^n}{B}_i$. Conversely, let $f_i:B_i\to A_i$ be a family of covers of $A_i$, we obtain the same result $B={\displaystyle \coprod _{i=1}^n}{B}_i$. By Proposition 2, $f:B\to A$ is a cover of A if and only if $\{f_i:B_i\to A_i\}$ is a family of covers of $A_i$. □

Remark 1.

From the several propositions above, it is shown that if A and B are the union (coproduct) of n S-acts $A_i$ and $B_i$, respectively, then $f:B\to A$ is a cover of A if and only if $\{f_i:B_i\to A_i\}$ is a family of covers of $A_i$ for $i\in \{1,2,\cdots ,n\}$.

Let S be a monoid and X an S-act. Recall [2] that X is Noetherian if every congruence on X is finitely generated. Additionally, a monoid S is called Noetherian if it is Noetherian as an S-act over itself.

Proposition 7.

Let S be a Noetherian monoid and $f:B\to A$ be a cover. Then, A is Noetherian if and only if B is Noetherian.

Proof. 

Let $f:B\to A$ be a cover of A. If B is Noetherian. By Lemma 7.6 of [5], then B is finitely generated, and so is A by Theorem 1. All finitely generated S-acts over a Noetherian monoid are Noetherian, thus A is Noetherian. Conversely, it is obvious. □

3. Covers of Finitely Generated Acts

Let S be a monoid and $\rho ,\sigma$ right congruences on S. In Mahmoudi and Renshaw [3], it is proved that a cyclic act $S/\rho$ has a cover $S/\sigma$ if and only if $\sigma \subseteq \rho$ and for all $u\in {\left[1\right]}_{\rho }$, $uS\cap {\left[1\right]}_{\sigma }\ne \varnothing$. The next propositions are needed to obtain the result.

Proposition 8.

Let S be a monoid. Then, $A^{\prime }=\langle a_1^{\prime },a_2^{\prime },\cdots ,a_n^{\prime }\rangle$ is isomorphic to a finitely generated subact of $A=\langle a_1,a_2,\cdots ,a_n\rangle$ if and only if there exist $u_i,u_p\in S$, such that $r(a_i^{\prime },a_p^{\prime })=r(a_i{u}_i,a_p{u}_p)$ for any $i,p\in \{1,2,\cdots ,n\}$.

Proof. 

Let $f:A^{\prime }\to A$ be an S-monomorphism and $I=\{1,2,\cdots ,n\}$. For $a_i^{\prime }\in A^{\prime }$, there is $u_j\in S$ such that $f\left(a_i^{\prime }\right)=a_j{u}_j\in a_{j}S\subseteq A$, $i,j\in I$. Since f is a monomorphism, then $j=i$. Thus, we have $f\left(a_i^{\prime }\right)=a_i{u}_i$ for some $a_i\in \{a_1,\cdots ,a_n\}$ and $u_i\in S$. And so $f\left(a_i^{\prime }s\right)=a_i{u}_{i}s$ for any $s\in S$. Let $a_i^{\prime }s=a_p^{\prime }t$, $i,p\in I,s,t\in S$. Then, $a_i{u}_{i}s=a_p{u}_{p}t$. In addition, suppose that $a_i{u}_{i}s=a_p{u}_{p}t$, and we have $a_i^{\prime }s=a_p^{\prime }t$ since f is monomorphic. Therefore, $r(a_i^{\prime },a_p^{\prime })=r(a_i{u}_i,a_p{u}_p)$.

Conversely, the mapping $f:A^{\prime }\to A$ defined by $a_i^{\prime }s\mapsto a_i{u}_{i}s$ is well-defined, and f is an S-monomorphism. □

Lemma 4.

Let S be a monoid and $u_i\in S$. Consider the S-monomorphism $f:\langle a_1^{\prime },a_2^{\prime },\cdots ,a_n^{\prime }\rangle \to \langle a_1,a_2,\cdots ,a_n\rangle$ given by $f\left(a_i^{\prime }s\right)=a_i{u}_{i}s$, $s\in S$. Then, f is onto if and only if $u_{i}S\bigcap R\left(a_i\right)\ne \varnothing$ for every $i\in I$.

Proof. 

Since f is an S-epimorphism. For any $a_i\in \{a_1,\cdots ,a_n\}$, there exist $a_p^{\prime }\in \{a_1^{\prime },\cdots ,a_n^{\prime }\}$ and $m\in S$ such that $f\left(a_p^{\prime }m\right)=a_i$. And since $f\left(a_p^{\prime }m\right)=a_p{u}_{p}m$, it follows that $a_i=a_p{u}_{p}m\in a_{p}S$, namely $a_{i}S\subseteq a_{p}S$, a contradiction. Thus $p=i$ and so $a_i=a_i{u}_{i}m$, hence $u_{i}S\bigcap R\left(a_i\right)\ne \varnothing$.

Conversely, since $u_{i}S\bigcap R\left(a_i\right)\ne \varnothing$, there exists $s\in S$ such that $a_i=a_i{u}_{i}s$ for every $i\in I$. So $f\left(a_i^{\prime }s\right)=a_i{u}_{i}s=a_i$. It is easy to see that f is an epimorphism. □

Lemma 5.

Let S be a monoid, $B=\langle b_1,b_2,\cdots ,b_n\rangle$ and $A=\langle a_1,a_2,\cdots ,a_n\rangle$. If $f:B\to A$ is a coessential epimorphism, then there exists $u_i\in S$ such that $g:B^{\prime }\to B$, $b_i^{\prime }s\mapsto b_i{u}_{i}s$ is isomorphic, where $B^{\prime }=\langle b_1^{\prime },b_2^{\prime },\cdots ,b_n^{\prime }\rangle$. And $f^{\prime }:B^{\prime }\to A$ given by $f\left(b_i^{\prime }s\right)=a_{i}s$ is a coessential S-epimorphism. In particular, $r(b_i^{\prime },b_p^{\prime })\subseteq r(a_i,a_p)$.

Proof. 

Since $f:B\to A$ is a coessential epimorphism, it follows that A and B are n-generated. For any $a_i\in \{a_1,\cdots ,a_n\}$, there exist $b_i\in \{b_1,\cdots ,b_n\}$ and $u_i\in S$ such that $f\left(b_i{u}_i\right)=a_i$. Suppose that $g:B^{\prime }\to B$ given by $f\left(b_i^{\prime }s\right)=b_i{u}_{i}s$ is a monomorphism whose composite with f is clearly onto. Since B is a cover of A, $B^{\prime }\cong B$ and so $f^{\prime }:B^{\prime }\to A$, $b_i^{\prime }s\mapsto a_{i}s$, is a coessential S-epimorphism. It then follows that $r(b_i^{\prime },b_p^{\prime })\subseteq r(a_i,a_p)$. □

We now present a fundamental theorem that yields the main result of this section.

Theorem 2.

Let S be a monoid and $A=\langle a_1,a_2,\cdots ,a_n\rangle$ a n-generated S-act. The map $f:\langle b_1,b_2,\cdots ,b_n\rangle \to \langle a_1,a_2,\cdots ,a_n\rangle$ given by $b_{i}s\mapsto a_{i}s$ is a coessential epimorphism if and only if $r(b_i,b_j)\subseteq r(a_i,a_j)$, $i,j\in I$, and for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\left(b_i\right)\ne \varnothing$.

Proof. 

Let $b_{i}s=b_{j}t$ for $b_i,b_j\in \{b_1,b_2,\cdots ,b_n\}$, $s,t\in S$ and $i,j\in I$. Then, $a_{i}s=a_{j}t$ because f is well-defined. Hence, $r(b_i,b_j)\subseteq r(a_i,a_j)$. For $a_i\in \{a_1,a_2,\cdots ,a_n\}$, there exist $b_j\in \{b_1,\cdots ,b_n\}$ and $u_j\in S$ such that $f\left(b_j{u}_j\right)=a_i\in a_{i}S$, but $f\left(b_j{u}_j\right)=a_j{u}_j\in a_{j}S$, we obtain $a_{i}S\subseteq a_{j}S$, a contradiction. Therefore, $j=i$ and so $f\left(b_i{u}_i\right)=a_i$. Since $f:\langle b_1,b_2,\cdots ,b_n\rangle \to \langle a_1,a_2,\cdots ,a_n\rangle$ is a coessential epimorphism, by Lemma 5, there is a subact $\langle b_1^{\prime },b_2^{\prime },\cdots ,b_n^{\prime }\rangle$ such that $\langle b_1^{\prime },b_2^{\prime },\cdots ,b_n^{\prime }\rangle \cong \langle b_1,b_2,\cdots ,b_n\rangle$ and ${f|}_{\langle b_1^{\prime },b_2^{\prime },\cdots ,b_n^{\prime }\rangle }:\langle b_1^{\prime },b_2^{\prime },\cdots ,b_n^{\prime }\rangle \to \langle a_1,a_2,\cdots ,a_n\rangle$ is a coessential epimorphism. So we have $b_i=b_i^{\prime }m$ for some $m\in S$. And since $g:\langle b_1^{\prime },b_2^{\prime },\cdots ,b_n^{\prime }\rangle \to \langle b_1,b_2,\cdots ,b_n\rangle$ is isomorphic by Lemma 5, it follows that $b_i{u}_{i}m=b_i$, thus $u_{i}S\bigcap R\left(b_i\right)\ne \varnothing$ as required.

Conversely, if the given conditions hold, then clearly f is well-defined. Let B be an S-subact of $\langle b_1,b_2,\cdots ,b_n\rangle$ and suppose that ${f|}_B$ is onto. Then, for $a_i\in \{a_1,a_2,\cdots ,a_n\}$, there exists $b_p{u}_p\in B$ such that $f\left(b_p{u}_p\right)=a_i$, and we obtain $p=i$, so $f\left(b_i{u}_i\right)=a_i$. However, $f\left(b_i{u}_i\right)=a_i{u}_i$, and hence $u_i\in R\left(a_i\right)$. By assumption, there exists $m\in S$ such that $b_i=b_i{u}_{i}m$, in the sense that $b_{i}s=\left(b_i{u}_{i}m\right)s=\left(b_i{u}_i\right)\left(ms\right)\in BS\subseteq B$ for an arbitrary $s\in S$. So $\langle b_1,b_2,\cdots ,b_n\rangle \subseteq B$, that is, $\langle b_1,b_2,\cdots ,b_n\rangle =B$, and f is a cover. □

If $n=1$, we can easily obtain the Theorem 2.7 in [3].

Theorem 3.

Let S be a monoid and $A=\langle a_1,a_2,\cdots ,a_n\rangle$ a n-generated S-act. If the natural map $f:{\displaystyle \bigcup _{i=1}^n}{S}_i\to A$ is a coessential epimorphism, where $S_i\cong S$. Then, $R\left(a_i\right),i\in I$, is a subgroup of S.

Proof. 

Since $f:{\displaystyle \bigcup _{i=1}^n}{S}_i\to {\displaystyle \bigcup _{i=1}^n}{a}_{i}S$ is a coessential epimorphism. Then, for an arbitrary element $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\left(1\right)\ne \varnothing$. And $R\left(1\right)=\{t\in S|1t=1\}=\{1\}$. We can verify that $u_i{s}_i=1$ for $s_i\in S$. Moreover, $a_i=a_i{u}_i$, it is easy to see that $a_i{s}_i=a_i{u}_i{s}_i=a_i$ and so $s_i\in R\left(a_i\right)$. Consequently, $R\left(a_i\right)$ is a subgroup of S. □

Proposition 9.

Let S be a monoid. Then, the map $f:{\displaystyle \bigcup _{i=1}^n}{S}_i\to {\displaystyle \bigcup _{i=1}^n}{a}_{i}S$ is a coessential epimorphism, where $S_i\cong S$, if and only if $R\left(a_i\right),i\in I$, is a subgroup of S and $r(1_i,1_p)\subseteq r(a_i,a_p)$ for $i,p\in I$.

Proof. 

Let $f:{\displaystyle \bigcup _{i=1}^n}{S}_i\to {\displaystyle \bigcup _{i=1}^n}{a}_{i}S$, $s\mapsto a_{i}s$ be a coessential epimorphism. Let $1_{i}s=1_{p}t$ for $s,t\in S$ and $i,p\in I$. We have $a_{i}s=a_{i}s$ since f is well-defined. By Theorem 3, $R\left(a_i\right)$ is a subgroup of S.

Conversely, suppose that $R\left(a_i\right)$ is a subgroup of S. For any $u_i\in R\left(a_i\right)$, there exists $u_i^{-1}\in R\left(a_i\right)$ such that $u_i{u}_i^{-1}=\left\{1\right\}$. Thus, $u_{i}S\bigcap R\left(1\right)\ne \varnothing$. In addition, since $r(1_i,1_p)\subseteq r(a_i,a_p)$ for $i,p\in I$, we can verify that f is a coessential epimorphism by Theorem 2. □

Proposition 10.

Let S be a right simple semigroup with a 1 adjoined and $B=\langle b_1,b_2,\cdots ,b_n\rangle$, $A=\langle a_1,a_2,\cdots ,a_n\rangle$. If $r(b_i,b_p)\subseteq r(a_i,a_p)$ for $i,p\in I$ and $R\left(b_i\right)\ne \left\{1\right\}$. Then $f:B\to A$ given by $b_{i}s\mapsto a_{i}s$ is a coessential epimorphism.

Proof. 

Since $R\left(b_i\right)\ne \left\{1\right\}$, we can verify that $R\left(b_i\right)\subseteq S\setminus \left\{1\right\}=T$, where T is a right simple semigroup. So $tT=T$ for every $t\in T$. Then for $u_i\in R\left(a_i\right)$, $u_{i}T\bigcap R\left(b_i\right)=T\bigcap R\left(b_i\right)=R\left(b_i\right)\ne \varnothing$. It is easy to see that f is a coessential epimorphism since $r(b_i,b_p)\subseteq r(a_i,a_p)$. □

Remark 2.

It follows from the above that covers of finitely generated S-acts need not be unique. If S is a group then every n-generated S-act has ${\displaystyle \bigcup _{i=1}^n}{S}_i$ as a cover, where $S_i\cong S$. And so proper n-generated S-acts do not have unique covers. In fact, if the map $\langle b_1,b_2,\cdots ,b_n\rangle \to \langle a_1,a_2,\cdots ,a_n\rangle$ is onto, then $\langle b_1,b_2,\cdots ,b_n\rangle$ is trivially a cover of $\langle a_1,a_2,\cdots ,a_n\rangle$.

4. $\mathit{N}$-Generated Flat Covers

In Mahmoudi and Renshaw [3], it is proved that an equivalent characterization of a cyclic act having flat properties covers, such as strongly flat, Condition $\left(P\right)$, and projective. The next theorems investigate the flatness covers of n-generated S-acts.

Theorem 4.

Let S be a monoid. Then, the n-generated S-act $A=\langle a_1,a_2,\cdots ,a_n\rangle$ has a strongly flat cover B if and only if $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ and $R\left(a_i\right)$ contains a left collapsible submonoid R such that for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$.

Proof. 

Suppose that A has a strongly flat cover B. By Theorem 1, $B=\langle b_1,b_2,\cdots ,$$b_n\rangle$. Then, by Theorem 2, we can assume that $R=R\left(b_i\right)\subseteq R\left(a_i\right)$ and that for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$. Moreover, since B is strongly flat, $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ by Lemma 2, and R is left collapsible by Proposition 3.

Conversely, suppose that $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ and R is a left collapsible submonoid of $R\left(a_i\right)$ such that for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$. Define $r\left(b_i\right)=R\times R$, then $R\subseteq R\left(b_i\right)$ and further $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ is strongly flat by Lemma 3. Define a map $f:\langle b_1,b_2,\cdots ,b_n\rangle \to \langle a_1,a_2,\cdots ,a_n\rangle$ given by $b_{i}s\mapsto a_{i}s$ and note that f is a well-defined S-epimorphism. To see this, first notice that if f is well-defined, then it is clearly an S-map which it is on. Since $R\subseteq R\left(a_i\right)$, it is easy to see that $R\times R\subseteq r\left(a_i\right)$, but $r\left(b_i\right)=R\times R$ and hence $r\left(b_i\right)\subseteq r\left(a_i\right)$. And $r(b_i,b_j)\subseteq r(a_i,a_j)$ for any $i\ne j,i,j\in I$, is not necessary to consider because $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$. Further, for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$ so f is coessential by Theorem 2. □

We can easily obtain the Theorem 3.2 of [3] if $n=1$ in Theorem 4.

Example 1.

Let S be a left cancellative monoid. Then, n-generated S-act $A=\langle a_1,a_2,\cdots ,a_n\rangle$ has a strongly flat cover if and only if $R\left(a_i\right)$, $i\in I$, is a subgroup of S, and in this case, ${\displaystyle \bigcup _{i=1}^n}{S}_i$ is a strongly flat cover of A, where $S_i\cong S$. First, notice that the only strongly flat cover of a n-generated S-act (assuming that it has one) is then ${\displaystyle \bigcup _{i=1}^n}S$. Notice also that not all finitely generated S-acts need have a strongly flat cover (see, for example, Remark 3.6 in [3]).

Theorem 5.

Let S be a monoid. Then, the n-generated S-act $A=\langle a_1,a_2,\cdots ,a_n\rangle$ has a $\left(P\right)$-cover B if and only if $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ and $R\left(a_i\right)$ contains a right reversible submonoid R such that for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$.

Proof. 

Suppose that A has a $\left(P\right)$-cover, $B=\langle b_1,b_2,\cdots ,b_n\rangle$. Then, by Theorem 2, we can assume that $R=R\left(b_i\right)\subseteq R\left(a_i\right)$ and that for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$. Moreover, since B satisfies Condition $\left(P\right)$, $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ by Lemma 2, and R is right reversible by Proposition 3.

Conversely, suppose that $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ and R is a right reversible submonoid of $R\left(a_i\right)$ such that for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$. Define $r\left(b_i\right)=R\times R$, then $R\subseteq R\left(b_i\right)$ and further $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$ satisfies Condition $\left(P\right)$ by Lemma 3. Define a map $f:\langle b_1,b_2,\cdots ,b_n\rangle \to \langle a_1,a_2,\cdots ,a_n\rangle$ given by $b_{i}s\mapsto a_{i}s$ and note that f is a well-defined S-epimorphism. Since $R\subseteq R\left(a_i\right)$, it follows that $R\times R\subseteq r\left(a_i\right)$, but $r\left(b_i\right)=R\times R$ and hence $r\left(b_i\right)\subseteq r\left(a_i\right)$. And $r(b_i,b_j)\subseteq r(a_i,a_j)$ for any $i\ne j,i,j\in I$, is not necessary to consider because $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S$. Further, since for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$, then f is coessential by Theorem 2. □

Similarly, if $n=1$, we can easily obtain Theorem 4.2 of [3].

Proposition 11.

Let S be a group. Then, every n-generated S-act $A(A\ne S)$ has at least two $\left(P\right)$-covers but only has a unique strongly flat cover.

Proof. 

Let $A=\langle a_1,a_2,\cdots ,a_n\rangle$. Since S is a group, then all S-acts satisfy condition $\left(P\right)$ and so in particular A is a $\left(P\right)$-cover of itself. But $R\left(a_i\right),i\in I$ is a subgroup of S; thus, ${\displaystyle \bigcup _{i=1}^n}{S}_i$ is a cover of A by Proposition 9, where $S_i\cong S$, while the latter part follows immediately from the Example 1. □

According to Proposition 11, we have an obvious example, if $S=N$ under multiplication, then the only subgroup is $\left\{1\right\}$. Then, $2N$ and N are $\left(P\right)$-covers of $2N$. However, $2N$ has a strongly flat cover N if $R\left(2\right)=\left\{1\right\}$.

Lemma 6

([14], Corollary 3.8). A right S-act $P_S$ is projective if and only if $P={\displaystyle \coprod _{i\in I}}{P}_i$, where $P_i\cong e_{i}S$ for idempotents $e_i\in S$, $i\in I$.

Theorem 6.

Let S be a monoid and $A=\langle a_1,a_2,\cdots ,a_n\rangle$ a n-generated S-act. Then, the following are equivalent:

(1) 

A has a projective cover;

(2) 

There exists a submonoid R of $R\left(a_i\right)$ which has a left zero element and for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$.

(3) 

The submonoid $R\left(a_i\right)$ of S has a minimal right ideal generated by an idempotent.

Proof. 

(1)⇒(2) Let $B=\langle b_1,b_2,\cdots ,b_n\rangle$ be a projective cover of A. Then by Theorem 2, take $R=R\left(b_i\right)$, for all $u_i\in R\left(a_i\right)$, $u_{i}S\bigcap R\ne \varnothing$. Moreover by Lemma 6 there exists $e_i\in E\left(S\right)$ with $B={\displaystyle \coprod _{i=1}^n}{b}_{i}S\cong {\displaystyle \coprod _{i=1}^n}{e}_{i}S$ and thus obviously $e_i\in R\left(e_i\right)$ and $R\left(e_i\right)=R\left(b_i\right)$. If $u_i\in R\left(e_i\right)$, then $e_i{u}_i=e_i$ and so $e_i$ is a left zero element of $R\left(b_i\right)$.

(2)⇒(3) Assume that $z\in R$ is a left zero element of R and consider the right ideal $zR\left(a_i\right)$ of $R\left(a_i\right)$. If this is not minimal, then there exists $u_i\in R\left(a_i\right)$ with $zuR\left(a_i\right)\ne zR\left(a_i\right)$. By assumption, there exists $s\in S$ with $u_{i}s\in R\subseteq R\left(a_i\right)$, and since $R\left(a_i\right)$ is a left unitary submonoid, then $s\in R\left(a_i\right)$. Since z is a left zero element of R, $z=z\left(u_{i}s\right)\in z{u}_{i}R\left(a_i\right)$, and so $zR\left(a_i\right)=z{u}_{i}R\left(a_i\right)$, which is a contradiction. Therefore, the submonoid $zR\left(a_i\right)$ of $zR\left(a_i\right)$ is a minimal right ideal generated by an idempotent.

(3)⇒(1) Suppose that $e_i\in E\left(S\right)$ with $e_{i}R\left(a_i\right)$ is a minimal right ideal of $R\left(a_i\right)$. Clearly ${\displaystyle \coprod _{i=1}^n}{e}_{i}S$ is projective. Define the map $f:{\displaystyle \coprod _{i=1}^n}{e}_{i}S\to {\displaystyle \bigcup _{i=1}^n}{a}_{i}S$ by $f\left(e_{i}s\right)=a_{i}s$. Then f is well-defined since if $e_{i}s=e_{j}t$, $i,j\in I$ and $s,t\in S$, then $i=j$ and $a_{i}s=a_i{e}_{i}s=a_i{e}_{i}t=a_{i}t$. In addition, it is also coessential since if $u_i\in R\left(a_i\right)$, then $e_i{u}_{i}R\left(a_i\right)=e_{i}R\left(a_i\right)$ by the minimality of $e_{i}R\left(a_i\right)$, and so $e_i=e_i{u}_{i}t$ for some $t\in R\left(a_i\right)$. But then $u_{i}t\in R\left(e_i\right)$ and so $u_{i}S\bigcap R\left(e_i\right)\ne \varnothing$. Hence, the result follows by Theorem 2. □

If $n=1$ in Theorem 6, we can obtain the Theorem 5.2 in [3].

5. $\mathcal{X}$-Precovers

In this section, we study Enochs’ notion of cover [1] in the category of acts over monoids and focus on $\mathcal{X}$-precovers and $\mathcal{X}$-covers, where $\mathcal{X}$ is a class of S-acts closed under isomorphisms. For example, we denote the class of all projective S-acts by $\mathcal{P}$. Clearly the concept of cover by Enochs’ notion is slightly different from that of the coessential cover.

Proposition 12.

Let S be a monoid and let $\mathcal{X}$ satisfy the property that $X\in \mathcal{X}\iff X\otimes M\in \mathcal{X}$ for ${}_{S}M_S\in S$-act-S. If A has an $\mathcal{X}$-precover, then $A\otimes M$ has an $\mathcal{X}$-precover.

Proof. 

Let $g:C\to A$ be $\mathcal{X}$-precover of A. Define $g\otimes 1:C\otimes M\to A\otimes M$ to be the obvious induced map. Let $X\otimes M\in \mathcal{X}$ and let $h\otimes 1:X\otimes M\to A\otimes M$. Now, by the hypothesis, $X\in \mathcal{X}$, so, since C is an $\mathcal{X}$-precover of A, there exists $f\in$ ${\mathrm{Hom}}_S(X,C)$ such that $gf=h$. So define $f\otimes 1:X\otimes M\to C\otimes M$ to be the induced map, and clearly $(g\otimes 1)(f\otimes 1)=gf\otimes 1=h\otimes 1$. Thus, $A\otimes M$ has an $\mathcal{X}$-precover. □

Proposition 13.

Let A be an S-act and $\mathcal{X}$ a class of S-acts. Any retract of any act satisfying $\mathcal{X}$ satisfies $\mathcal{X}$. Then, the $\mathcal{X}$-precover of A, if it exists, is the $\mathcal{X}$-precover of the retract of A.

Proof. 

Please see the following diagram

Let $f:C\to A$ be the $\mathcal{X}$-precover and B the retract of A. Then $\alpha :A\to B$ and $\beta :B\to A$ satisfy $\alpha \beta =1_B$. For S-map $\varphi :M\to B$, $M\in \mathcal{X}$, by the $\mathcal{X}$-precover property, there exists an S-map $\psi :M\to C$ such that $f\psi =\beta \varphi$. We claim that $\alpha f\psi =\alpha \beta \varphi =1_B\varphi =\varphi .$ Hence, $\alpha f:C\to B$ is the $\mathcal{X}$-precover of B. □

Proposition 14.

Let $\mathcal{X}$ be a class of acts and $D\in \mathcal{X}$. Consider the following commutative diagram of S-acts:

If $g:C\to A$ is an $\mathcal{X}$-precover of S-act A, then $h:D\to A$ is also a $\mathcal{X}$-precover of A.

Proof. 

Let $\psi :M\to A$ be an S-homomorphism for $M\in \mathcal{X}$. By the $\mathcal{X}$-precover property, there exists an S-map $\varphi :M\to C$ such that $g\varphi =\psi$. Hence, $hf\varphi =\psi$ from $hf=g$. So $h:D\to A$ is a $\mathcal{X}$-precover of A. □

Theorem 7.

Let $\mathcal{X}$ be a class of S-acts and $f:A\to B$ be a monomorphism. Consider the following commutative diagram of S-acts:

If $h:P\to B$ is an $\mathcal{X}$-precover of B, then $g:P\to A$ is an $\mathcal{X}$-precover of A.

Proof. 

For every S-map $h^{\prime }:P^{\prime }\to A$, where $P^{\prime }\in \mathcal{X}$. Since $h:P\to B$ is an $\mathcal{X}$-precover of B, there exists an S-map $\phi :P^{\prime }\to P$ with $fg\phi =h\phi =f{h}^{\prime }$. Then, $g\phi =h$ since f is a monomorphism, and we are done. □

Dually, if $f:A\to B$ is an S-epimorphism, we obtain the following Theorem 8.

Theorem 8.

Let S be a monoid and $\mathcal{P}$ the class of all projective S-acts. Let $f:A\to B$ be an S-epimorphism. Consider the following commutative diagram of S-acts:

If $g:P\to A$ is a $\mathcal{P}$-precover of A, then $h:P\to B$ is a $\mathcal{P}$-precover of B.

Proof. 

Let $g:P\to A$ be $\mathcal{P}$-precover of A and $f:A\to B$ an S-epimorphism. To show that $h:P\to B$ is a $\mathcal{P}$-precover of B, we assume that any S-map $\phi :P^{\prime }\to B$, where $P^{\prime }\in \mathcal{P}$. By assumption, there exists $\psi :P^{\prime }\to A$ such that $f\psi =\phi$. Since $g:P\to A$ is a $\mathcal{P}$-precover of A, we obtain that $g\varphi =\psi$ for some $\varphi :P^{\prime }\to P$. So $\phi =f\psi =fg\varphi =h\varphi$. Therefore, $h:P\to B$ is a $\mathcal{P}$-precover of B. □

Remark 3.

It is clear that if $g:P\to A$ is an $\mathcal{X}$-cover of A, then $h:P\to B$ is an $\mathcal{X}$-cover of B in Theorem 8. A question that could be brought up is whether Theorem 8 is valid for free S-acts. It is easy to answer this question positively.

Theorem 9.

Let S be a monoid and $\mathcal{X}$ a class of S-acts. If $A\in \mathcal{X}$ is an $\mathcal{X}$-precover of B and $C\in \mathcal{X}$ is an $\mathcal{X}$-precover of A. Then, C is an $\mathcal{X}$-precover of B.

Proof. 

Suppose that $f:A\to B$ is an $\mathcal{X}$-precover of B and $g:C\to A$ is an $\mathcal{X}$-precover of A. For any S-map $h:P\to B$, for $P\in \mathcal{X}$, there is an S-map $\phi :P\to A$ with $f\phi =h$ since $f:A\to B$ is an $\mathcal{X}$-precover of B. Similarly, because $g:C\to A$ is an $\mathcal{X}$-precover of A, we have $g\psi =\phi$ for some S-map $\psi :P\to C$. Thus, we can verify that $h=f\phi =\left(fg\right)\psi$. So $fg:C\to B$ is an $\mathcal{X}$-precover of B, that is, C is an $\mathcal{X}$-precover of B. □

Proposition 15.

Let S be a monoid and B a right S-act. Let $\mathcal{X}$ be a class of S-acts closed under factor acts. If $f:A\to B$ is a $\mathcal{X}$-precover for B, then $f^{\prime }:A/\rho \to B$ is a $\mathcal{X}$-precover for B, where $\rho \subseteq ker\left(f\right)$ is a congruence on A.

Proof. 

Let $\rho$ be a congruence on A contained in $ker\left(f\right)$ and $A\in \mathcal{X}$. First note that $A/\rho \in \mathcal{X}$. By the Homomorphism Theorem for acts, there exists a homomorphism $f^{\prime }:A/\rho \to B$, defined by $f^{\prime }\left(a\rho \right)=f\left(a\right)$, with $f^{\prime }\pi =f$, where $\pi :A\to A/\rho$. For each morphism $h:C\to B$ with $C\in \mathcal{X}$, there exists a morphism $g:C\to A$ such that $fg=h$. Thus $f^{\prime }\left(\pi g\right)=h$, that is, $f^{\prime }:A/\rho \to B$ is a $\mathcal{X}$-precover for B. □

Proposition 16.

Let $\mathcal{X}$ be a class of S-acts which is closed under subacts. And suppose that all subacts of B have $\mathcal{X}$-covers. If $A\in \mathcal{X}$ and $f:A\to B$ is an $\mathcal{X}$-precover and C is a proper subact of B. Then, there exists an S-subact $A^{*}$ of A and a homomorphism $g:A^{*}\to C$ such that it is an $\mathcal{X}$-cover of C.

Proof. 

By assumption, we note that $f^{-1}\left(C\right)\in \mathcal{X}$. Thus, $k:f^{-1}\left(C\right)\to C$ is an $\mathcal{X}$-precover of C. Let $h:A_0\to C$ be an $\mathcal{X}$-cover of C. Then, we have the commutative diagram

such that $k\phi =h$ and $h\psi =k$, and so $h\psi \phi =h$. Since h is an $\mathcal{X}$-cover of C, it follows that $\psi \phi$ is an automorphism of $A_0$. Thus, $\phi$ is monomorphism. Take $A^{*}=\phi \left(A_0\right)\cong A_0$ as an S-subact of A, then $g=h\psi :A^{*}\to C$ is an $\mathcal{X}$-cover of C. □

6. Conclusions

Covers of cyclic S-acts are proven in [3]. However, the study did not consider monoids over which covers exist for all right S-acts. In order to further the study, in this article, we investigated covers of finitely generated S-acts over monoids and obtained an equivalent characterization of essential epimorphisms of n-generated S-acts in Theorem 3.4. We then investigated whether n-generated S-acts have Condition $\left(P\right)$ covers, strongly flat covers, or projective covers in this work. Then, we pose the following questions for consideration:

• What are the equivalent conditions of the Condition $\left(P^{\prime }\right)$-cover and $\left(P{F}^{″}\right)$-cover of the finitely generated S-acts?
• What are the flat covers of finitely presented S-acts?