Covers of Finitely Generated Acts over Monoids

Xiaoqin Zhang; Tingting Zhao

doi:10.20944/preprints202404.0268.v1

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Covers of Finitely Generated Acts over Monoids

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28 March 2024

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03 April 2024

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Abstract

In (Semigroup Forum 77: 325-338, 2008) Mahmoudi M. and Renshaw J. solved a study that covers of cyclic $S$-acts over monoids. This article is an attempt to initiate the covers of finitely generated $S$-acts. We give a necessary and sufficient condition for a monoid to have the properties that $n$-generated $S$-acts have strongly flat covers, Condition $(P)$ covers and projective covers. The main conclusions extend some known results. We show also that Condition $(P)$ covers of finitely generated $S$-acts are not unique, unlike the situation for strongly flat covers. Additionally, we demonstrate that 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.

Keywords:

cover

;

coproduct

;

finitely generated

;

$\mathcal{X}$-precover

Subject:

Computer Science and Mathematics - Algebra and Number Theory

MSC: 20M30; 20M50

1. Introduction and Preliminaries

Let S be a monoid. By a right S-act we mean a non-empty set A together with an action

A \times S \to A

given by

(a, s) \mapsto a s

such that for all

a \in A, s, t \in S, a 1 = a

and

a (s t) = (a s) t .

We refer the reader to [1,7,9] for all undefined terms concerning acts over monoids.

Mahmoudi and Renshaw introduced coessential epimorphisms and covers of cyclic S-acts in [9], gave a description of coessential epimorphisms in terms of congruence classes and a method of constructing covers from left unitary submonoids, and defined a strongly flat cover, Condition

(P)

cover, projective cover and provided a necessary and sufficient condition for a cyclic S-act to have a strongly flat cover, Condition

(P)

cover and projective cover. Now we restrict our attention to these notions.

Let S be a monoid. A right S-act B is called a cover of a right S-act A if there exists an epimorphism

f : B \to A

such that for any proper subact C of B the restriction

{f |}_{C}

is not an epimorphism. Moreover, we call an S-epimorphism

f : B \to A

coessential in [9] if for each S-act C and each S-map

g : C \to B

, if

f g

is an epimorphism, then g is an epimorphism. And we shall say that an act B together with an S-epimorphism

f : B \to A

is a X-cover of A if B satisfies property X and f is coessential. It is easy to see that

f : B \to A

is a cover of A if and only if it is a coessential epimorphism. Let

X

be a class of right S-acts. We assume that

X

is closed under isomorphisms, i.e., if

A \in X

and

B ≅ A

, then

B \in X

The concept of cover used in [5] is slightly different from that given above. (Enochs’ notion) For a right S-act A, an S-act

X \in X

is called an

X

-cover of A if there is a homomorphism

φ : X \to A

such that

(1) for any homomorphism

ψ

X^{'} \to A

with

X^{'} \in X

, there is a homomorphism f:

X^{'} \to X

with

ψ = φ f

. In other words, the following diagram Preprints 102502 i001

commutes;

(2) If an endomorphism

f : X \to X

is such that

φ = φ f

, then f must be an automorphism.

If (1) holds, we call

φ

X \to A

X

-precover.

Recall that an S-act

A_{S}

is called strongly flat if the tensor functor

A \otimes -

preserves pullbacks and equalizers. An S-act

A_{S}

is said to satisfy Condition

(P)

if for all

a, a^{'} \in A, s, s^{'} \in S

such that

a s = a^{'} s^{'}

, then there exist

a^{''} \in A, u, v \in S

with

a = a^{''} u, a^{'} = a^{''} v

and

u s = v s^{'}

. And recall that an S-act

A_{S}

satisfies Condition

(E)

a s = a s^{'}

for

a \in A, s, s^{'} \in S

, implies that there exist

a^{'} \in A, u \in S

such that

a = a^{'} u

and

u s = u s^{'}

. It is well known that an act A is strongly flat if and only if it satisfies Condition

(P)

and Condition

(E)

. We are principally concerned with finitely generated S-acts here and with the concepts of strongly flatness and Condition

(P)

and in these cases there are useful characterizations.

A monoid S is called right reversible if for all

s, t \in S

, there exist

p, q \in S

such that

p s = q t

. A monoid S is called left collapsible if for all

s, t \in S

, there exists

u \in S

such that

u s = u t

. A submonoid T of S is said to be left unitary if and only if whenever

t, t s \in T

then

s \in T

Throughout this paper, S denotes a monoid,

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉 = ⋃_{i = 1}^{n} a_{i} S

is said that A is a n-generated right S-act and

I = {1, 2, \dots, n}

. And

∐_{i = 1}^{n} a_{i} S

will denote a coproduct of cyclic subacts

a_{i} S

for all

i \in I

. We shall deal exclusively with right S-acts and simply refer to them as S-acts. We shall also consistently write our maps on the left so that

g f

means f followed by g.

The main results here are descriptions of finitely generated acts and coproducts in Section 2, and we provide a necessary and sufficient condition for a finitely generated S-act to have a cover in Section 3. Next in Section 4 we concentrate on strongly flat covers, Condition

(P)

covers and projective covers, and the main conclusions extend some known results in [9]. In Section 5 we study Enochs’ notion of cover in the category of acts over monoids and focus on

X

-precovers, where

X

is a class of S-acts closed under isomorphisms. Basic results on covers of acts over monoids can be found in [3,4,6,11].

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 = u s

. In other words, U is a set of generating elements for

A_{S}

〈 U 〉 : = ⋃_{u \in U}^{} u S = A_{S}

, where

u S = {u s ∣ 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 [7].

Proposition 2.1.

Let S be a monoid and

A = ⋃_{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} = 〈 X_{i} 〉

. Then

A = 〈 ⋃_{i = 1}^{n} X_{i} 〉

. □

Proposition 2.2.

Let S be a monoid and

A = ⋃_{i = 1}^{n} A_{i}

an S-act,

A_{i}

are subacts of A for

i \in I

. And for

i \neq j \in I

, suppose that

A_{i} ⋂ 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.

Let

I = {1, 2}

, it is clear by Lemma 5.3 of [10]. If

I_{1} = {1, 2, \dots, n - 1}

the assertion is hold, that is, for

i \neq j \in I_{1}

A_{i} ⋂ A_{j}

is either finitely generated or empty, if A is finitely generated, then

A_{i}, i \in I

, is finitely generated. Assume that

I = {1, 2, \dots, n} = I_{1} ⋃ {n}

. If

A_{i} ⋂ A_{j} = \emptyset

i \neq j \in I

, obviously. Otherwise, take

j = n

, for

i \in I_{1}

A_{i} ⋂ A_{n}

is finitely generated, by Proposition 2.1,

(⋃_{i \in I_{1}}^{} A_{i}) ⋂ A_{n}

is finitely generated. Since

A = (⋃_{i \in I_{1}}^{} A_{i}) ⋃ A_{n}

is finitely generated,

⋃_{i \in I_{1}}^{} A_{i}

and

A_{n}

are finitely generated. By assumption,

A_{i}

i \in I_{1}

, is finitely generated. Therefore, for all

i \in I

A_{i}

is finitely generated. □

Corollary 2.1.

Let S be a monoid. S-act

A = ∐_{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 2.3.

Suppose that

f_{i} : B_{i} \to A_{i}, i \in I

are a family of S-morphisms, where

I = {1, 2, \dots, n}

. Let

B = ∐_{i = 1}^{n} B_{i}

A = ∐_{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}

are covers of

A_{i}

for all

i \in I

Proof. Sufficiency. Since

f_{i} : B_{i} \to A_{i}

i \in I

are 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

\bar{B}

of B such that

{f |}_{\bar{B}}

is epimorphic. Therefore, there is

j \in I

such that

B_{j}^{'} = B_{j} ⋂ \bar{B}

is a proper subact of

B_{j}

. Thus,

f_{j} |_{B_{j}^{'}}

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} (b_{j}^{'}) \neq a_{j}

for all

b_{j}^{'} \in B_{j}^{'}

. However, by the surjectivity of

{f |}_{\bar{B}}

, we obtain that

f (\bar{b}) = a_{j}

for some

\bar{b} \in \bar{B}

. Thus there exists

k \in I

such that

\bar{b} \in B_{k}

k \neq j

. Then

a_{j} = f (\bar{b}) \in A_{k}

which contradicts

a_{j} \in A_{j}

Necessity. Let

f : ∐_{i = 1}^{n} B_{i} \to ∐_{i = 1}^{n} A_{i}

be a coessential epimorphism. Then we obtain that

f_{i} : B_{i} \to A_{i}

are a family of epimorphisms for all

i \in I

. To show that

f_{i}

are 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}

B_{j}

. Thus

{f |}_{(∐_{i \neq j}^{} B_{i}) ∐_{}^{} C_{j}} : (∐_{i \neq j}^{} B_{i}) ∐_{}^{} C_{j} \to ∐_{i = 1}^{n} A_{i}

is an epimorphism. But

(∐_{i \neq j}^{} B_{i}) ∐_{}^{} C_{j}

is a proper subact of

∐_{i = 1}^{n} B_{i}

and by assumption

{f |}_{(∐_{i \neq j}^{} B_{i}) ∐_{}^{} C_{j}}

is not epimorphic, a contradiction. So

f_{i} : B_{i} \to A_{i}

are a family of coessential epimorphisms for all

i \in I

. □

Now we concentrate on covers of finitely generated of S-acts and we begin with some notations. For an element a of a right S-act A we denote by

R (a)

the set

{s \in S : a s = a}

, the submonoid of right identities of a, and it is clear that

R (a)

is a left unitary submonoid of S. We denote by

r (a, b)

the set

{(s, t) \in S \times S : a s = b t}

. In particular,

r (a) = r (a, a) = {(s, t) \in S \times S : a s = a t}

, the right annihilator congruence

ρ_{a}

of a. It is clear that

a S

is isomorphic to

S / ρ_{a}

under the S-isomorphism

a s \mapsto [s]

, where

[s]

denote the class of

s \in S

with respect to an equivalence relation

ρ

Lemma 2.1

([7], Proporsition III.13.14). Let

A_{S}

be a finitely generated S-act. If

A_{S}

satisfies Condition

(P)

, then

A_{S}

is a coproduct of cyclic subacts.

Let S be a monoid and

A = ∐_{i = 1}^{n} A_{i}

, where

A_{i}, i \in I

are right S-acts. Then A is strongly flat (resp. satisfies Condition

(P)

) if and only if

A_{i}

is strongly flat (resp. satisfies Condition

(P)

) for every

i \in I

. The following results turn out very helpful in the following section.

Proposition 2.4.

If the finitely generated S-act

A = 〈 a_{1}, \dots, a_{n} 〉

is strongly flat(resp. satisfis Conditon (P)), then

R (a_{i})

is a left collapsible(resp. right reversible) submonoid of S.

Proof.

Let A satisfy Conditon

(P)

. By Lemma 2.1,

a_{i} S

satisfies Conditon

(P)

for

i \in I

. Thus

R (a_{i})

is right reversible. And in a similar way we obtain that

R (a_{i})

is a left collapsible submonoid of S if A is strongly flat. □

Proposition 2.5.

Let T be a submonoid of a monoid S and A a right S-act,

a \in A

. If

r (a) = T \times T

, then

T = R (a)

Proof.

Since

r (a) = {(s, t) \in S \times S | a s = a t} = T \times T

, for any

t \in T

and

1 \in T

, then

(1, t) \in T \times T = r (a)

. Thus

a = a t

, that is,

t \in R (a)

. Therefore

T \subseteq R (a)

. It is clear that

R (a) \subseteq T

. □

Lemma 2.2.

Let S be a monoid and P a left collapsible(resp. right reversible) submonoid of S,

A = 〈 a_{1}, \dots, a_{n} 〉

is n-generated. If

r (a_{i}) = P \times P

, then

R (a_{i})

is left collapsible(resp. right reversible) and

∐_{i = 1}^{n} a_{i} S

is strongly flat(resp. satisfies Condition (P)).

Proof.

Since

r (a_{i}) = P \times P

P = R (a_{i})

by Proposition 2.5 and so

R (a_{i})

is left collapsible(resp. right reversible). Notice that in either case, P is right reversible and so defined by

b s = b t

if and only if there exists

p, q \in P

with

p s = q t

. Notice that

b = a_{i}

. In fact, if

a_{i} s = a_{i} t

then there exist

p, q \in P

with

p s = q t

and so

b s = b t

. Conversely, if

b s = b t

, then there are

p, q \in P

with

p s = q t

and

a_{i} = a_{i} p = a_{i} q

from

P \subseteq R (a_{i})

. So

a_{i} s = a_{i} p s = a_{i} q t = a_{i} t

and hence

b = a_{i}

. It is clear that

a_{i} S

satisfies Condition

(P)

. Then

∐_{i = 1}^{n} a_{i} S

satisfies Condition

(P)

. If in addition P is left collapsible, then

∐_{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 [9]. The following theorem concerns the covers of n-generated right S-acts.

Theorem 2.1.

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,

\bar{B} = {b_{1}, b_{2}, \dots, 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 = 〈 y_{1}, \dots, y_{k} 〉

k < n

. Then

f (z_{i}) = y_{i}

i = 1, \dots, k

. Therefore

B = ⋃_{i = 1}^{k} z_{i} S

, which contradicts that B is n-generated. Thus, A is n-generated.

Sufficiency. Suppose

A = 〈 a_{1}, \dots, a_{n} 〉

. Since f is an epimorphism, there exists

b_{i} \in B

such that

f (b_{i}) = a_{i}, i = 1, 2, \dots, n .

Take

C = ⋃_{i = 1}^{n} b_{i} S

, clearly

C \subseteq B

and

f (C) = 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 = 〈 x_{1}, \dots, x_{k} 〉, k < n

, and

f (B) = A

, we have A is generated by

f (x_{i}), i = 1, 2, \dots, k

, a contradiction. □

Let

A = ⋃_{i = 1}^{n} A_{i}

be n-generated and

B = ∐_{i = 1}^{n} B_{i}

. The following Proposition 2.6 shows that

f : B \to A

is a cover of A implies that

f_{i} : B_{i} \to A_{i}, i \in I

, are a family of covers of

A_{i}

Proposition 2.6.

Let

A = ⋃_{i = 1}^{n} A_{i}, B = ∐_{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

, are a family of covers of

A_{i}

Proof.

A_{i} ⋂ A_{j} = \emptyset

for any

i, j \in I

, it is clear by Proposition 2.3. Otherwise, suppose

a \in A_{i} ⋂ A_{j}, i \neq j, i, j \in I

. Since f is an epimorphism, for

a \in A_{i}

, there exists

b \in B_{i}

such that

f (b) = f_{i} (b) = a

. If

a \in A_{j}

, we have

f (b) = f_{j} (b) = a

for some

b \in B_{j}

. But

B_{i} ⋂ B_{j} = \emptyset

, a contradiction. Therefore

A_{i} ⋂ A_{j} = \emptyset

, we obtain that

A = ∐_{i = 1}^{n} A_{i}

. By Proposition 2.3,

f : B \to A

is a cover of A implies that

f_{i} : B_{i} \to A_{i}

are covers of

A_{i}

. □

Coversely, if B is n-generated and A is a disjoint union of n subacts, the argument always holds in Proposition 2.7.

Proposition 2.7.

Suppose that

B = ⋃_{i = 1}^{n} B_{i}

A = ∐_{i = 1}^{n} A_{i}

f_{i} : B_{i} \to A_{i}

i \in I

are 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}

are a family of covers of

A_{i}

for every

i \in I

Proof.

Let

f : B \to A

be a cover of A. If there exist

i, j \in I

such that

B_{i} ⋂ B_{j} \neq \emptyset

, that is,

f (b_{i}) = f (b_{j}) = a

, where

b_{i} \in B_{i}, b_{j} \in B_{j}, a \in A

, then

{f |}_{(B_{i} ∖ {b_{i}}) ⋃ (⋃_{k \neq i}^{} B_{k})} : (B_{i} ∖ {b_{i}}) ⋃ (⋃_{k \neq i}^{} B_{k}) \to ∐_{i = 1}^{n} A_{i}

is an epimorphism, a contradiction. So f is one-to-one. Assume

b \in B_{i} ⋂ B_{j}

. Then

f (b) = f_{i} (b) \in A_{i}

, and

f (b) = f_{j} (b) \in A_{j}

which contradicts

A_{i} ⋂ A_{j} = \emptyset

. Hence

B_{i} ⋂ B_{j} = \emptyset

or the element in

B_{i} ⋂ B_{j}

has no image under the action of f. If the latter, then

{f |}_{B ∖ (B_{i} ⋂ B_{j})} : B ∖ (B_{i} ⋂ B_{j}) \to A

is an epimorphism, a contradiction. Therefore,

B = ∐_{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 = ∐_{i = 1}^{n} B_{i}

. By Proposition 2.3,

f : B \to A

is a cover of A if and only if

f_{i} : B_{i} \to A_{i}

are a family of covers of

A_{i}

. □

Remark 2.1.

From the above several propositions it is shown that if A and B are the union(coproduct) of nS-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}

are a family of covers of

A_{i}

for

i \in {1, 2, \dots, n}

Let S be a monoid and X an S-act. We say that X is Noetherian if every congruence on X is finitely generated, and we say that a monoid S is Noetherian if it is Noetherian as an S-act over itself.

Proposition 2.8.

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 [2], then B is finitely generated, and so is A by Theorem 2.1. All finitely generated S-acts over a Noetherian monoid are Noetherian, thus A is Noetherian. Coversely, it is obvious. □

3. Covers of Finitely Generated Acts

Let S be a monoid and

ρ, σ

right congruences on S. In Mahmoudi and Renshaw [9], it is proved that a cyclic act

S / ρ

has a cover

S / σ

if and only if

σ \subseteq ρ

and for all

u \in {[1]}_{ρ}

u S \cap {[1]}_{σ} \neq \emptyset

. The next there propositions are needed to get the result.

Proposition 3.1.

Let S be a monoid. Then

A^{'} = 〈 a_{1}^{'}, a_{2}^{'}, \dots, a_{n}^{'} 〉

is isomorphic to a finitely generated subact of

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

if and only if there exist

u_{i}, u_{p} \in S

, such that

r (a_{i}^{'}, a_{p}^{'}) = r (a_{i} u_{i}, a_{p} u_{p})

for any

i, p \in {1, 2, \dots, n}

Proof.

Let

f : A^{'} \to A

be an S-monomorphism and

I = {1, 2, \dots, n}

. For

a_{i}^{'} \in A^{'}

, there is

u_{j} \in S

such that

f (a_{i}^{'}) = 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 (a_{i}^{'}) = a_{i} u_{i}

for some

a_{i} \in {a_{1}, \dots, a_{n}}

and

u_{i} \in S

. And so

f (a_{i}^{'} s) = a_{i} u_{i} s

for any

s \in S

. Let

a_{i}^{'} s = a_{p}^{'} 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}^{'} s = a_{p}^{'} t

since f is monomorphic. Therefore

r (a_{i}^{'}, a_{p}^{'}) = r (a_{i} u_{i}, a_{p} u_{p})

Conversely, the mapping

f : A^{'} \to A

defined by

a_{i}^{'} s \mapsto a_{i} u_{i} s

is well-defined and f is an S-monomorphism. □

Lemma 3.1.

Let S be a monoid and

u_{i} \in S

. Consider the S-monomorphism

f : 〈 a_{1}^{'}, a_{2}^{'}, \dots, a_{n}^{'} 〉 \to 〈 a_{1}, a_{2}, \dots, a_{n} 〉

given by

f (a_{i}^{'} s) = a_{i} u_{i} s

s \in S

. Then f is onto if and only if

u_{i} S ⋂ R (a_{i}) \neq \emptyset

for every

i \in I

Proof.

Since f is an S-epimorphism. For any

a_{i} \in {a_{1}, \dots, a_{n}}

, there exist

a_{p}^{'} \in {a_{1}^{'}, \dots, a_{n}^{'}}

and

m \in S

such that

f (a_{p}^{'} m) = a_{i}

. And since

f (a_{p}^{'} m) = 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 ⋂ R (a_{i}) \neq \emptyset

Conversely, since

u_{i} S ⋂ R (a_{i}) \neq \emptyset

, there exists

s \in S

such that

a_{i} = a_{i} u_{i} s

for every

i \in I

. So

f (a_{i}^{'} s) = a_{i} u_{i} s = a_{i}

. It is easy to see that f is an epimorphism. □

Lemma 3.2.

Let S be a monoid,

B = 〈 b_{1}, b_{2}, \dots, b_{n} 〉

and

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

. If

f : B \to A

is a coessential epimorphism then there exists

u_{i} \in S

such that

g : B^{'} \to B

b_{i}^{'} s \mapsto b_{i} u_{i} s

, is isomorphic, where

B^{'} = 〈 b_{1}^{'}, b_{2}^{'}, \dots, b_{n}^{'} 〉

. And

f^{'} : B^{'} \to A

given by

f (b_{i}^{'} s) = a_{i} s

is a coessential S-epimorphism. In particular,

r (b_{i}^{'}, b_{p}^{'}) \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}, \dots, a_{n}}

, there exist

b_{i} \in {b_{1}, \dots, b_{n}}

and

u_{i} \in S

such that

f (b_{i} u_{i}) = a_{i}

. Suppose that

g : B^{'} \to B

given by

f (b_{i}^{'} s) = b_{i} u_{i} s

is a monomorphism whose composite with f is clearly onto. Since B is a cover of A,

B^{'} ≅ B

and so

f^{'} : B^{'} \to A

b_{i}^{'} s \mapsto a_{i} s

, is a coessential S-epimorphism. It then follows that

r (b_{i}^{'}, b_{p}^{'}) \subseteq r (a_{i}, a_{p})

. □

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

Theorem 3.1.

Let S be a monoid and

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

a n-generated S-act. The map

f : 〈 b_{1}, b_{2}, \dots, b_{n} 〉 \to 〈 a_{1}, a_{2}, \dots, a_{n} 〉

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 (a_{i})

u_{i} S ⋂ R (b_{i}) \neq \emptyset

Proof.

Let

b_{i} s = b_{j} t

for

b_{i}, b_{j} \in {b_{1}, b_{2}, \dots, 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}, \dots, a_{n}}

, there exist

b_{j} \in {b_{1}, \dots, b_{n}}

and

u_{j} \in S

such that

f (b_{j} u_{j}) = a_{i} \in a_{i} S

, but

f (b_{j} u_{j}) = 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 (b_{i} u_{i}) = a_{i}

. Since

f : 〈 b_{1}, b_{2}, \dots, b_{n} 〉 \to 〈 a_{1}, a_{2}, \dots, a_{n} 〉

is a coessential epimorphism, by Lemma 3.2, there is a subact

〈 b_{1}^{'}, b_{2}^{'}, \dots, b_{n}^{'} 〉

such that

〈 b_{1}^{'}, b_{2}^{'}, \dots, b_{n}^{'} 〉 ≅ 〈 b_{1}, b_{2}, \dots, b_{n} 〉

and

{f |}_{〈 b_{1}^{'}, b_{2}^{'}, \dots, b_{n}^{'} 〉} : 〈 b_{1}^{'}, b_{2}^{'}, \dots, b_{n}^{'} 〉 \to 〈 a_{1}, a_{2}, \dots, a_{n} 〉

is a coessential epimorphism. So we have

b_{i} = b_{i}^{'} m

for some

m \in S

. And since

g : 〈 b_{1}^{'}, b_{2}^{'}, \dots, b_{n}^{'} 〉 \to 〈 b_{1}, b_{2}, \dots, b_{n} 〉

is isomorphic by Lemma 3.2, it follows that

b_{i} u_{i} m = b_{i}

, thus

u_{i} S ⋂ R (b_{i}) \neq \emptyset

as required.

Conversely, if the given conditions hold, then clearly f is well-defined. Let B be an S-subact of

〈 b_{1}, b_{2}, \dots, b_{n} 〉

and suppose that

{f |}_{B}

is onto. Then for

a_{i} \in {a_{1}, a_{2}, \dots, a_{n}}

, there exists

b_{p} u_{p} \in B

such that

f (b_{p} u_{p}) = a_{i}

and we obtain

p = i

, so

f (b_{i} u_{i}) = a_{i}

. However

f (b_{i} u_{i}) = a_{i} u_{i}

, and hence

u_{i} \in R (a_{i})

. By assumption, there exists

m \in S

such that

b_{i} = b_{i} u_{i} m

, in the sense that

b_{i} s = (b_{i} u_{i} m) s = (b_{i} u_{i}) (m s) \in B S \subseteq B

for an arbitrary

s \in S

. So

〈 b_{1}, b_{2}, \dots, b_{n} 〉 \subseteq B

, that is,

〈 b_{1}, b_{2}, \dots, b_{n} 〉 = B

, and f is a cover. □

n = 1

, we can easily obtain the Theorem 2.7 in [9].

Theorem 3.2.

Let S be a monoid and

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

a n-generated S-act. If the natural map

f : ⋃_{i = 1}^{n} S_{i} \to A

is a coessential epimorphism, where

S_{i} ≅ S

. Then

R (a_{i}), i \in I

, is a subgroup of S.

Proof.

Since

f : ⋃_{i = 1}^{n} S_{i} \to ⋃_{i = 1}^{n} a_{i} S

is a coessential epimorphism. Then for an arbitrary element

u_{i} \in R (a_{i})

u_{i} S ⋂ R (1) \neq \emptyset

. And

R (1) = {t \in S | 1 t = 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 (a_{i})

. Consequently,

R (a_{i})

is a subgroup of S. □

Proposition 3.2.

Let S be a monoid. Then the map

f : ⋃_{i = 1}^{n} S_{i} \to ⋃_{i = 1}^{n} a_{i} S

is a coessential epimorphism, where

S_{i} ≅ S

, if and only if

R (a_{i}), 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 : ⋃_{i = 1}^{n} S_{i} \to ⋃_{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 3.2,

R (a_{i})

is a subgroup of S.

Conversely, suppose that

R (a_{i})

is a subgroup of S. For any

u_{i} \in R (a_{i})

, there exists

u_{i}^{- 1} \in R (a_{i})

such that

u_{i} u_{i}^{- 1} = {1}

. Thus

u_{i} S ⋂ R (1) \neq \emptyset

. 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 3.1. □

Proposition 3.3.

Let S be a right simple semigroup with a 1 adjoined and

B = 〈 b_{1}, b_{2}, \dots, b_{n} 〉

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

. If

r (b_{i}, b_{p}) \subseteq r (a_{i}, a_{p})

for

i, p \in I

and

R (b_{i}) \neq {1}

. Then

f : B \to A

given by

b_{i} s \mapsto a_{i} s

is a coessential epimorphism.

Proof.

Since

R (b_{i}) \neq {1}

, we can verify that

R (b_{i}) \subseteq S ∖ {1} = T

, where T is a right simple semigroup. So

t T = T

for every

t \in T

. Then for

u_{i} \in R (a_{i})

u_{i} T ⋂ R (b_{i}) = T ⋂ R (b_{i}) = R (b_{i}) \neq \emptyset

. It is easy to see that f is a coessential epimorphism since

r (b_{i}, b_{p}) \subseteq r (a_{i}, a_{p})

. □

Remark 3.1.

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

⋃_{i = 1}^{n} S_{i}

as a cover, where

S_{i} ≅ S

. And so proper n-generated S-acts do not have unique covers. In fact, if the map

〈 b_{1}, b_{2}, \dots, b_{n} 〉 \to 〈 a_{1}, a_{2}, \dots, a_{n} 〉

is onto, then

〈 b_{1}, b_{2}, \dots, b_{n} 〉

is trivially a cover of

〈 a_{1}, a_{2}, \dots, a_{n} 〉

4. $N$ -Generated Flat Covers

In Mahmoudi and Renshaw [9] it is proved that an equivalent characterization of a cyclic act having flat properties covers, such as strongly flat, Condition

(P)

, projective. The next theorems are to investigate flatness covers of n-generated S-acts.

Theorem 4.1.

Let S be a monoid. Then the n-generated S-act

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

has a strongly flat cover B if and only if

B = ∐_{i = 1}^{n} b_{i} S

and

R (a_{i})

contains a left collapsible submonoid R such that for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

Proof.

Suppose that A has a strongly flat cover B. By Theorem 2.1,

B = 〈 b_{1}, b_{2}, \dots,

b_{n} 〉

. Then by Theorem 3.1 we can assume that

R = R (b_{i}) \subseteq R (a_{i})

and that for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

. Moreover, since B is strongly flat,

B = ∐_{i = 1}^{n} b_{i} S

by Lemma 2.1 and R is left collapsible by Proposition 2.4.

Conversely, suppose that

B = ∐_{i = 1}^{n} b_{i} S

and R is a left collapsible submonoid of

R (a_{i})

such that for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

. Define

r (b_{i}) = R \times R

, then

R \subseteq R (b_{i})

and further

B = ∐_{i = 1}^{n} b_{i} S

is strongly flat by Lemma 2.2. Define a map

f : 〈 b_{1}, b_{2}, \dots, b_{n} 〉 \to 〈 a_{1}, a_{2}, \dots, a_{n} 〉

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 is onto. Since

R \subseteq R (a_{i})

, it is easy to see that

R \times R \subseteq r (a_{i})

but

r (b_{i}) = R \times R

and hence

r (b_{i}) \subseteq r (a_{i})

. And

r (b_{i}, b_{j}) \subseteq r (a_{i}, a_{j})

for any

i \neq j, i, j \in I

, is not necessary to consider because

B = ∐_{i = 1}^{n} b_{i} S

. Further, for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

so f is coessential by Theorem 3.1. □

We can easily obtain the Theorem 3.2 of [9] if

n = 1

in Theorem 4.1.

Example 4.1.

Let S be a left cancellative monoid. Then n-generated S-act

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

has a strongly flat cover if and only if

R (a_{i})

i \in I

, is a subgroup of S and in this case

⋃_{i = 1}^{n} S_{i}

is a strongly flat cover of A, where

S_{i} ≅ S

. First notice that the only strongly flat cover of a n-generated S-act(assuming that it has one) is then

⋃_{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 [9]).

Theorem 4.2.

Let S be a monoid. Then the n-generated S-act

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

has a

(P)

-cover B if and only if

B = ∐_{i = 1}^{n} b_{i} S

and

R (a_{i})

contains a right reversible submonoid R such that for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

Proof.

Suppose that A has a

(P)

-cover,

B = 〈 b_{1}, b_{2}, \dots, b_{n} 〉

. Then by Theorem 3.1 we can assume that

R = R (b_{i}) \subseteq R (a_{i})

and that for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

. Moreover, since B satisfies Condition

(P)

B = ∐_{i = 1}^{n} b_{i} S

by Lemma 2.1 and R is right reversible by Proposition 2.4.

Conversely, suppose that

B = ∐_{i = 1}^{n} b_{i} S

and R is a right reversible submonoid of

R (a_{i})

such that for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

. Define

r (b_{i}) = R \times R

, then

R \subseteq R (b_{i})

and further

B = ∐_{i = 1}^{n} b_{i} S

satisfies Condition

(P)

by Lemma 2.2. Define a map

f : 〈 b_{1}, b_{2}, \dots, b_{n} 〉 \to 〈 a_{1}, a_{2}, \dots, a_{n} 〉

given by

b_{i} s \mapsto a_{i} s

and note that f is a well-defined S-epimorphism. Since

R \subseteq R (a_{i})

, it follows that

R \times R \subseteq r (a_{i})

but

r (b_{i}) = R \times R

and hence

r (b_{i}) \subseteq r (a_{i})

. And

r (b_{i}, b_{j}) \subseteq r (a_{i}, a_{j})

for any

i \neq j, i, j \in I

, is not necessary to consider because

B = ∐_{i = 1}^{n} b_{i} S

. Further, since for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

, then f is coessential by Theorem 3.1. □

Similarly, if

n = 1

, we can easily obtain the Theorem 4.2 of [9].

Proposition 4.1.

Let S be a group. Then every n-generated S-act

A (A \neq S)

has at least two

(P)

-covers but only has a unique strongly flat cover.

Proof.

Let

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

. Since S is a group, then all S-acts satisfy condition

(P)

and so in particular A is a

(P)

-cover of itself. But

R (a_{i}), i \in I

, is a subgroup of S, thus

⋃_{i = 1}^{n} S_{i}

is a cover of A by Proposition 3.2, where

S_{i} ≅ S

. While the latter part follows immediately from the Example 4.1. □

Lemma 4.1

([8], Corollary 3.8). A right S-act

P_{S}

is projective if and only if

P = ∐_{i \in I}^{} P_{i}

where

P_{i} ≅ e_{i} S

for idempotents

e_{i} \in S

i \in I

Theorem 4.3.

Let S be a monoid and

A = 〈 a_{1}, a_{2}, \dots, a_{n} 〉

a n-generated S-act. Then the following are equivalent:

(1)

A has a projective cover;

(2)

There exits a submonoid R of

R (a_{i})

which has a left zero element and for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

(3)

The submonoid

R (a_{i})

of S has a minimal right ideal generated by an idempotent.

Proof. (1)⇒(2) Let

B = 〈 b_{1}, b_{2}, \dots, b_{n} 〉

be a projective cover of A. Then by Theorem 3.1, take

R = R (b_{i})

, for all

u_{i} \in R (a_{i})

u_{i} S ⋂ R \neq \emptyset

. Moreover by Lemma 4.1 there exists

e_{i} \in E (S)

with

B = ∐_{i = 1}^{n} b_{i} S ≅ ∐_{i = 1}^{n} e_{i} S

and thus obviously

e_{i} \in R (e_{i})

and

R (e_{i}) = R (b_{i})

. If

u_{i} \in R (e_{i})

, then

e_{i} u_{i} = e_{i}

and so

e_{i}

is a left zero element of

R (b_{i})

(2)⇒(3) Assume that

z \in R

is a left zero element of R and consider the right ideal

z R (a_{i})

R (a_{i})

. If this is not minimal, then there exists

u_{i} \in R (a_{i})

with

z u R (a_{i}) \neq z R (a_{i})

. By assumption there exists

s \in S

with

u_{i} s \in R \subseteq R (a_{i})

, and since

R (a_{i})

is a left unitary submonoid, then

s \in R (a_{i})

. Since z is a left zero element of R,

z = z (u_{i} s) \in z u_{i} R (a_{i})

and so

z R (a_{i}) = z u_{i} R (a_{i})

, which is a contradiction. Therefore the submonoid

z R (a_{i})

z R (a_{i})

is a minimal right ideal generated by an idempotent.

(3)⇒(1) Suppose that

e_{i} \in E (S)

with

e_{i} R (a_{i})

is a minimal right ideal of

R (a_{i})

. Clearly

∐_{i = 1}^{n} e_{i} S

is projective. Define the map

f : ∐_{i = 1}^{n} e_{i} S \to ⋃_{i = 1}^{n} a_{i} S

f (e_{i} s) = 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 (a_{i})

, then

e_{i} u_{i} R (a_{i}) = e_{i} R (a_{i})

by the minimality of

e_{i} R (a_{i})

, and so

e_{i} = e_{i} u_{i} t

for some

t \in R (a_{i})

. But then

u_{i} t \in R (e_{i})

and so

u_{i} S ⋂ R (e_{i}) \neq \emptyset

. Hence the result follows by Theorem 3.1. □

n = 1

in Theorem 4.3, we can get the Theorem 5.2 in [9].

5. $X$ -Precovers

In this section we study Enochs’ notion of cover [1] in the category of acts over monoids and focus on

X

-precovers and

X

-covers, where

X

will be a class of S-acts closed under isomorphisms. For example, we denote the class of all projective S-acts by

P

. Clearly the concept of cover by Enochs’ is slightly different from that the coessential cover.

Proposition 5.1.

Let S be a monoid and let

X

satisfy the property that

X \in X \Leftrightarrow X \otimes M \in X

for

_{S} M_{S} \in S

-act-S. If A has an

X

-precover, then

A \otimes M

has an

X

-precover.

Proof.

Let

g : C \to A

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 X

and let

h \otimes 1 : X \otimes M \to A \otimes M

. Now, by the hypothesis,

X \in X

, so, since C is an

X

-precover of A, there exists

f \in

Hom

_{S} (X, C)

such that

g f = 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) = g f \otimes 1 = h \otimes 1

. Thus,

A \otimes M

has an

X

-precover. □

Proposition 5.2.

Let A be an S-act and

X

a class of S-acts. Any retract of any act satisfying

X

satisfies

X

. Then the

X

-precover of A, if it exists, is the

X

-precover of the retract of A.

Proof.

Please see the following diagram Preprints 102502 i002

Let

f : C \to A

be the

X

-precover and B the retract of A. Then

α : A \to B

and

β : B \to A

satisfy

α β = 1_{B}

. For S-map

ϕ : M \to B

M \in X

, by the

X

-precover property, there exists an S-map

ψ : M \to C

such that

f ψ = β ϕ

. We claim that

α f ψ = α β ϕ = 1_{B} ϕ = ϕ .

Hence,

α f : C \to B

is the

X

-precover of B. □

Proposition 5.3.

Let

X

be a class of acts and

D \in X

. Consider the following commutative diagram of S-acts: Preprints 102502 i003

g : C \to A

is an

X

-precover of S-act A, then

h : D \to A

is also a

X

-precover of A.

Proof.

Let

ψ : M \to A

be an S-homomorphism for

M \in X

. By the

X

-precover property, there exists an S-map

ϕ : M \to C

such that

g ϕ = ψ

. Hence,

h f ϕ = ψ

from

h f = g

. So

h : D \to A

is a

X

-precover of A. □

Theorem 5.1.

Let

X

be a class of S-acts and

f : A \to B

be a monomorphism. Consider the following commutative diagram of S-acts: Preprints 102502 i004

h : P \to B

is an

X

-precover of B, then

g : P \to A

is an

X

-precover of A.

Proof.

For every S-map

h^{'} : P^{'} \to A

, where

P^{'} \in X

. Since

h : P \to B

is an

X

-precover of B, there exists an S-map

φ : P^{'} \to P

with

f g φ = h φ = f h^{'}

. Then

g φ = 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 5.2.

Theorem 5.2.

Let S be a monoid and

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: Preprints 102502 i004

g : P \to A

is a

P

-precover of A, then

h : P \to B

is a

P

-precover of B.

Proof.

Let

g : P \to A

P

-precover of A and

f : A \to B

an S-epimorphism. To show that

h : P \to B

is a

P

-precover of B, we assume that any S-map

φ : P^{'} \to B

, where

P^{'} \in P

. By assumption, there exists

ψ : P^{'} \to A

such that

f ψ = φ

. Since

g : P \to A

is a

P

-precover of A, we obtain that

g ϕ = ψ

for some

ϕ : P^{'} \to P

. So

φ = f ψ = f g ϕ = h ϕ

. Therefore,

h : P \to B

is a

P

-precover of B. □

Remark 5.1.

It is clear that if

g : P \to A

is an

X

-cover of A, then

h : P \to B

is an

X

-cover of B in Theorem 5.2. A question that could be brought up is whether Theorem 5.2 is valid for free S-acts. It is easy to answer this question positively.

Theorem 5.3.

Let S be a monoid and

X

a class of S-acts. If

A \in X

is an

X

-precover of B and

C \in X

is an

X

-precover of A. Then C is an

X

-precover of B.

Proof.

Suppose that

f : A \to B

is an

X

-precover of B and

g : C \to A

is an

X

-precover of A. For any S-map

h : P \to B

, for

P \in X

, there is an S-map

φ : P \to A

with

f φ = h

since

f : A \to B

is an

X

-precover of B. Similarly, because

g : C \to A

is an

X

-precover of A, we have

g ψ = φ

for some S-map

ψ : P \to C

. Thus, we can verify that

h = f φ = (f g) ψ

. So

f g : C \to B

is an

X

-precover of B, that is, C is an

X

-precover of B. □

Proposition 5.4.

Let S be a monoid and B a right S-act. Let

X

be a class of S-acts closed under factor acts. If

f : A \to B

is a

X

-precover for B, then

f^{'} : A / ρ \to B

is a

X

-precover for B, where

ρ \subseteq k e r (f)

is a congruence on A.

Proof.

Let

ρ

be a congruence on A contained in

k e r (f)

and

A \in X

. First note that

A / ρ \in X

. By Homomorphism Theorem for acts, there exists a homomorphism

f^{'} : A / ρ \to B

, defined by

f^{'} (a ρ) = f (a)

, with

f^{'} π = f

, where

π : A \to A / ρ

. For each morphism

h : C \to B

with

C \in X

, there exists a morphism

g : C \to A

such that

f g = h

. Thus

f^{'} (π g) = h

, that is,

f^{'} : A / ρ \to B

is a

X

-precover for B. □

Proposition 5.5.

Let

X

be a class of S-acts which is closed under subacts. And suppose that all subacts of B have

X

-covers. If

A \in X

and

f : A \to B

is an

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

X

-cover of C.

Proof.

By assumption, we note that

f^{- 1} (C) \in X

. Thus

k : f^{- 1} (C) \to C

is an

X

-precover of C. Let

h : A_{0} \to C

be an

X

-cover of C. Then we have the commutative diagram Preprints 102502 i005

such that

k φ = h

and

h ψ = k

, and so

h ψ φ = h

. Since h is an

X

-cover of C, it follows that

ψ φ

is an automorphism of

A_{0}

. Thus

φ

is monomorphism. Take

A^{*} = φ (A_{0}) ≅ A_{0}

is an S-subact of A, then

g = h ψ : A^{*} \to C

is an

X

-cover of C. □

6. Conclusions

Based on the covers of cyclic S-acts over monoids(see [9]), we have introduced the projective covers (resp. strongly flat covers,

(P)

-covers) of the finitely generated S-acts over monoids in this work. Then we pose the following questions for consideration:

1.What is the equivalent conditions of the Condition

(P^{'})

-cover and

(P F^{''})

-cover of the finitely generated S-acts?

2.What is the flat covers of finitely presented S-acts?

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

References

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Covers of Finitely Generated Acts over Monoids

Abstract

Keywords:

Subject:

1. Introduction and Preliminaries

2. Finite Generation and Coproducts

3. Covers of Finitely Generated Acts

4. N -Generated Flat Covers

5. X -Precovers

6. Conclusions

Institutional Review Board Statement

Informed Consent Statement

Conflicts of Interest

References

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4. $N$ -Generated Flat Covers

5. $X$ -Precovers