Quasi-exact Sequences of S-Acts

Abstract

In this paper, the notion of (short) quasi-exact sequence of S-acts is introduced. We study the behaviour of quasi-exact sequences in regard to some algebraic properties of S-acts including principal weak injectivity, principal weak flatness, regularity and torsion freeness. Moreover, some results concerning commutative diagrams of modules are generalized to acts over monoids.

1 Introduction and Preliminaries

The actions of a semigroup (monoid) S on sets, namely S-acts, have been used in many branches of mathematics like algebraic automata theory, mathematical linguistics, etc. There is a close relationship between automata and actions of semigroups. In fact, acts over free monoids are automata without outputs. For more information on applications of S-acts in automata theory, see [13]. Over the past three decades, an extensive theory of the properties of S-acts has been developed. A comprehensive survey of this area appeared in 2000 by Kilp et al. [12]. For acts over monoids in many cases, one tries to model a situation in modules over rings. The notion of exact sequence is a fundamental concept, and it has been widely used in mathematical theories such as groups, rings, modules, homological algebra and algebraic topology. The concept of exact sequence for S-acts was introduced by Chen [5] to study projectivity in the category of S-acts via exact functors. Moreover, Chen and Shum [6] studied the split Rees short exact sequences of S-acts and obtained some results different from the well-known results concerning split short exact sequences of modules. Recently, the notion of Rees short exact sequence of S-posets for a pomonois S has been considered in [10] and thereafter in a weaker form in [11]. In [7], Davvaz and Parnian–Garamaleky introduced the concept of quasi-exact sequences as a generalization of exact sequences for modules and Anvariyeh and Davvaz [2] extended further results about it (see also [1, 8]). Some results on exact sequences of semimodules can be found in [3, 4].

Here we introduce the concept of (short) quasi-exact sequence of S-acts using act congruences as a generalization of (short) exact sequences of S-acts. The behaviour of quasi-exact sequences relative to some algebraic properties of S-acts such as principal weak injectivity, principal weak flatness, regularity and torsion freeness is investigated. Furthermore, some generalization results concerning commutative diagrams in module theory to acts over monoids based on (quasi-) exact sequences are obtained.

Let S be a monoid. A non-empty set A is called a (right) S-act, usually denoted by $A_S$, if there exists a mapping $A\times S\rightarrow A$, $(a,s)\mapsto as$, satisfying the conditions $(as)t=a(st)$ and $a1=a$, for all $a\in A$ and $s,t \in S$. A non-empty subset $A'\subseteq A_S$ is called a subact of $A_S$ if $a's \in A'$ for all $s \in S$ and $a' \in A'$. Let $A_S$ and $B_S$ be two S-acts. A mapping $f:A_S\rightarrow B_S$ is called a homomorphism if $f(as)=f(a)s$ for all $a \in A_S, s \in S$. A homomorphism $f:A_S \rightarrow B_S$ is called a monomorphism (epimorphism) if it is injective (surjective). A non-empty subset I of S is called a right ideal of S if $IS\subseteq I$. We refer the reader to [9, 12] for more information on semigroups and acts over monoids.

Let $A_S$ be an S-act. An equivalence relation $\rho$ on A is called a congruence on $A_S$ if $a\rho a'$ implies that $(as)\rho (a's)$ for $a,a' \in A_S$, $s\in S$. The set of all congruences on $A_S$ is denoted by $Con (A_S)$. Let $\rho$ be a congruence on $A_S$. The factor set $A_S/\rho = \{[a]_\rho \mid a\in A\}$ becomes an S-act with the action $[a]_\rho s = [as]_\rho$ for every $a\in A, s\in S$ which is called the factor act of $A_S$ by $\rho$. Moreover, the canonical surjection $\pi _\rho : A_S \rightarrow A_S/\rho$ given by $\pi _\rho (a)=[a]_\rho$ is a homomorphism called a canonical epimorphism. Any subact $B_S \subseteq A_S$ defines the Rees congruence$\rho _B$ on A, by setting $a\rho _B a'$ if and only if $a=a'$ or $a,a' \in B$. We denote the resulting factor act by $A_S/B_S$ and call it the Rees factor act of $A_S$ by the subact $B_S$. Let $f:A_S \rightarrow B_S$ be an S-homomorphism. Then the kernel equivalence kerf of f defined by $a(kerf)a'$ if and only if $f(a)=f(a')$ for $a,a' \in A_S$ is a congruence called the kernel congruence of f. For any $a\in A_S$, by $\lambda _a$ we denote the homomorphism from $S_S$ into $A_S$ defined by $\lambda _a (s) = as$ for every $s\in S$. Similarly, if ${}_SA$ is an act and $a\in {}_SA$, then by $\rho _a$ we denote the homomorphism from ${}_SS$ into ${}_SA$ given by $\rho _a (s) = sa$ for every $s\in S$. The kernel congruences $ker\lambda _a$ and $ker\rho _a$ on $S_S$ and ${}_SS$ are called the annihilator congruences of a, respectively. The diagonal relation$\{(a,a)\mid a\in A\}$ and the universal relation$A\times A$ on a set A are denoted by ${{\Delta }_{A}}$ and ${{\nabla }_{A}}$, respectively. For any relations $\rho$ and $\sigma$, we use the notation $\rho \le \sigma$ if and only if $\rho \subseteq \sigma$. Throughout S stands for a monoid.

2 Quasi-exact Sequences and Some Algebraic Properties of S-Acts

In this section, first the notion of (short) quasi-exact sequences of S-acts is introduced. Then considering some basic algebraic properties of S-acts, we investigate the behaviour of such sequences with respect to these properties.

A sequence of S-acts and homomorphisms of the form

$$\begin{aligned} \cdots {\rightarrow }A_{-1} {\mathop {\rightarrow }\limits ^{f_{-1}}} A_0 {\mathop {\rightarrow }\limits ^{f_{0}}} A_1 {\mathop {\rightarrow }\limits ^{f_{1}}} A_2 {\mathop {\rightarrow }\limits ^{f_{2}}} A_3{\rightarrow } \cdots \end{aligned}$$

is said to be exact at $A_n$ provided that $(Imf_{n-1}\times Imf_{n-1}) \cup \Delta _{A_n} = kerf_n$. The above sequence is called exact if it is exact at $A_i$ for every integer i. An exact sequence $A_1 {\mathop {\rightarrow }\limits ^{f_{1}}} A_2 {\mathop {\rightarrow }\limits ^{f_{2}}} A_3$ at $A_2$ is called (Rees) short exact if $f_1$ is a monomorphism and $f_2$ is an epimorphism (see [5, 6]).

For any right ideal I of S, the sequence $I_S {\mathop {\hookrightarrow }\limits ^{i}} S_S {\mathop {\rightarrow }\limits ^{\pi _{\rho _I}}} S_S/I_S$ is clearly short exact.

Consider an exact sequence $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ at $A_S$ so that $(Imf\times Imf) \cup \Delta _{A_S} = kerg=(g\times g)^{-1}(\Delta _{C_S})$. It is natural to ask what does happen if we substitute a congruence $\rho$ on $C_S$ for the trivial congruence $\Delta _{C_S}$ in the above definition. This motivates us to introduce and study the new notion of quasi-exact sequence in the category of S-acts, as follows.

Definition 1

A sequence of S-acts and homomorphisms $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ is said to be $\rho$-exact at $A_S$ for a congruence $\rho$ on $C_S$ provided that $(Imf\times Imf) \cup \Delta _{A_S} = (g\times g)^{-1}(\rho )$ where $(g\times g)^{-1}(\rho )=\{(a,b)\in A_S\times A_S~|~(g(a),g(b))\in \rho \}$. If no confusion arises, we may use the term quasi-exact for $\rho$-exactness and denote the set $(g\times g)^{-1}(\rho )$ by $K(\rho )$. Clearly, $K(\rho )$ is a congruence on $A_S$ and $kerg \le K(\rho )$. In particular, $kerg=K(\Delta _{C_S})$. By a short quasi-exact sequence we mean a quasi-exact sequence $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ at $A_S$ for which f is a monomorphism and g is an epimorphism.

The following example illustrates that the two notions of exact and quasi-exact sequences of acts are different in general.

Example 1

• (i) For a non-singleton subact $B_S$ of $A_S$, the sequence $B_S {\mathop {\hookrightarrow }\limits ^{i}} A_S {\mathop {\rightarrow }\limits ^{id_A}} A_S$ is (short) $\rho _B$-exact but not (short) exact.

• (ii) Let $B_S,C_S$ be two subacts of $A_S$ with $C_S\subseteq B_S\subseteq A_S.$ Then the sequence $B_S {\mathop {\hookrightarrow }\limits ^{i}} A_S {\mathop {\rightarrow }\limits ^{\pi _{\rho _{C}}}} A_S/C_S$ is (short) $\rho _{B/C}$-exact. In particular, for a proper subact $C_S$ of $A_S$, the sequence $C_S {\mathop {\hookrightarrow }\limits ^{i}} A_S {\mathop {\rightarrow }\limits ^{\pi _{\rho _{C}}}} A_S/C_S$ is (short) exact but not (short) $\rho$-exact for any congruence $\rho \ne \Delta _{A/C}$ on $A_S/C_S$. Indeed, if $(a,b)\in K(\rho _{B/C})=\{ (a,b)\in A_S\times A_S~|~ (\pi _{\rho _{C}}(a),\pi _{\rho _{C}}(b)) \in \rho _{B/C}\}$, then $([a]_{\rho _{C}},[b]_{\rho _{C}}) \in \rho _{B/C}$ and so $[a]_{\rho _{C}}=[b]_{\rho _{C}}$ or $[a]_{\rho _{C}},[b]_{\rho _{C}} \in B/C$. In the first case, we have $a=b$ or $a,b \in C_S \subseteq B_S$ which implies that $(a,b) \in (Im i \times Im i) \cup \Delta _{A_S}$. In the second case, we get $a,b\in B_S$ and so $(a,b) \in Im i \times Im i$. Therefore, $K(\rho _{B/C}) \subseteq (Im i \times Im i) \cup \Delta _{A_S}$. The reverse inclusion is obvious.

Proposition 1

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a $\rho$-exact sequence at $A_S$ such that g is an epimorphism. Then the sequence is exact at $A_S$ if and only if $\rho = \Delta _{C_S}$.

Proof

Let $\rho = \Delta _{C_S}$. Since the sequence is $\rho$-exact at $A_S$, $(Imf\times Imf) \cup \Delta _{A_S} = K(\Delta _{C_S})= kerg$. Conversely, let the sequence be exact at $A_S$. Then, using hypothesis, $kerg=(Imf \times Imf)\cup \Delta _{A_S} = K(\rho )$. By surjectivity of g, let $(g(a),g(b))\in \rho$ for $a,b\in A_S$. Then $(a,b)\in K(\rho ) = kerg$ whence $(g(a),g(b))\in \Delta _{C_S}$. $\square$

Corollary 1

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a short $\rho$-exact sequence. Then the sequence is short exact if and only if $\rho = \Delta _{C_S}$.

An element $s\in S$ is called left(right) cancellable if $su=sv~(us=vs)$ for $u,v \in S$ implies that $u=v$. Let $s\in S$ and $a\in A_S$. The element a is called divisible bysin$A_S$ if there exists $b\in A_S$ such that $bs=a$. An act $A_S$ is said to be divisible if $Ac=A$ for any left cancellable element $c\in S$. It is clear from the definition that $A_S$ is divisible if and only if every element of $A_S$ is divisible by any left cancellable element of S.

Theorem 1

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a $\rho$-exact sequence at $A_S$ such that g is an epimorphism. If $B_S$ and $C_S$ are divisible S-acts, then so is $A_S$.

Proof

Let $a\in A$ and c be a left cancellable element of S. Since g is an epimorphism and $C_S$ is divisible, there exists $a'\in A$ such that $g(a)=g(a')c=g(a'c)$. Hence $(a,a'c)\in kerg$ and so $(a,a'c)\in K(\rho )=(Imf\times Imf)\cup \Delta _{A_S}$. Thus $a=a'c$ or there exists $b\in B_S$ such that $a=f(b)$. In the second case, it follows from the divisibility of $B_S$ that there exists $b'\in B_S$ such that $b=b'c$. Therefore, $a=f(b)=f(b'c)=f(b')c$ and so $A_S$ is divisible. $\square$

Let A be an S-act. Then $A_S$ is called injective if for any monomorphism $\iota : B_S \rightarrow C_S$ and any homomorphism $f: B_S \rightarrow A_S$ there exists a homomorphism $\overline{f} : C_S \rightarrow A_S$ such that $f=\overline{f}\iota$. Also $A_S$ is said to be principally weakly injective if it is injective relative to all embeddings of principal right ideals into $S_S$.

Lemma 1

[12, Proposition 3.3.2] An S-act $A_S$ is principally weakly injective if and only if for any $s\in S$, $a\in A_S$ with $ker\lambda _s \le ker\lambda _a$, a is divisible by s in $A_S$, i.e., $a=zs$ for some $z\in A_S$.

Theorem 2

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a short $\rho$-exact sequence. If $B_S$ and $C_S$ are principally weakly injective S-acts, then so is $A_S$.

Proof

Assume that $a\in A_S$, $s\in S$ and $ker\lambda _s \le ker\lambda _a$. If $(u,v)\in ker\lambda _a$, then $au=av$ and so $g(a)u=g(au)=g(av)=g(a)v$ which implies that $(u,v)\in ker\lambda _{g(a)}$. Therefore, $ker\lambda _s \le ker\lambda _a \le ker\lambda _{g(a)}$. Since $C_S$ is principally weakly injective and g is an epimorphism, using Lemma 1, there exists $z\in A_S$ such that $g(a)=g(z)s=g(zs)$. Hence, $(a,zs) \in kerg$ and so $(a,zs)\in K(\rho )=(Imf\times Imf)\cup \Delta _{A_S}$. Thus $a=zs$ or there exists $b\in B_S$ such that $a=f(b)$. In the second case, let $(u,v)\in ker\lambda _a$. Then $au=av$ and so $f(bu)=f(b)u=f(b)v=f(bv)$. Since f is a monomorphism, $bu=bv$ and so $(u,v)\in ker\lambda _b$ which gives that $ker\lambda _s \le ker\lambda _a\le ker\lambda _b$. By the assumption and Lemma 1, there exists $y\in B_S$ such that $b=ys$. Therefore, $a=f(b)=f(y)s$, showing that $A_S$ is principally weakly injective by Lemma 1.$\square$

Recall (see [12, Definition 3.9.4]) that an act $A_S$ satisfies condition (E) if $as=as'$ for $a\in A_S, s,s'\in S$, implies that there exist $a'\in A_S, u\in S$ such that $a=a'u, us=us'$. This condition together with some other ones are useful in the study of flatness properties of S-acts. To get more information on these concepts, see [12].

Theorem 3

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a short $\rho$-exact sequence. If $B_S$ and $C_S$ satisfy condition (E), then so does $A_S$.

Proof

Suppose that $a\in A_S$, $s,s'\in S$ and $as=as'$. Then $g(a)s=g(as)=g(as')=g(a)s'$. Since $C_S$ satisfies condition (E) and g is an epimorphism, there exist $a'\in A_S, w\in S$ such that $g(a)=g(a')w=g(a'w)$ and $ws=ws'$ so that $(a,a'w)\in kerg$. Thus $(a,a'w)\in K(\rho )=(Imf\times Imf)\cup \Delta _{A_S}$ which implies that $a=a'w$ or there exists $b\in B_S$ such that $a=f(b)$. The first case together with $ws=ws'$ justifies the condition (E) for $A_S$. In the second case, since $as=as'$, $f(bs)=f(bs')$ and so by injectivity of f, $bs=bs'$. Hence, using hypothesis, there exist $b'\in B_S, z\in S$ such that $b=b'z$ and $zs=zs'$ which gives that $a=f(b)=f(b')z$ and $zs=zs'$. Therefore, $A_S$ satisfies condition (E). $\square$

An act $A_S$ is called principally weakly flat if the functor $A_S\otimes {}_S-$ preserves all embeddings of principal left ideals into S. An element $e\in S$ is called an idempotent if $e^2=e$. The set of all idempotents of S is denoted by E(S). An element $s\in S$ is called left (right) e-cancellable for an idempotent $e\in S$ if $s=se~(s=es)$ and $ker\lambda _s \le ker\lambda _e~(ker\rho _s\le ker\rho _e)$. A monoid S is called a right (left) PP monoid if every element $s\in S$ is left (right) e-cancellable for some idempotent $e\in S$. For more details on these notions, see [12].

Lemma 2

[12, Theorem 3.10.16] Let S be a left PP monoid. An act $A_S$ is principally weakly flat if and only if for every $a,a' \in A_S$ and $s\in S$, $as=a's$ implies that there exists $e\in E(S)$ such that $es=s$ and $ae=a'e$.

Theorem 4

Let S be a left PP monoid and E(S) be commutative. Let also $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a $\rho$-exact sequence at $A_S$ such that f is a monomorphism. If $B_S$ and $C_S$ are principally weakly flat S-acts, then so is $A_S$.

Proof

Suppose that $a,a'\in A_S$, $s\in S$ and $as=a's$. Then $g(a)s = g(a')s$. Since $C_S$ is principally weakly flat, it follows from Lemma 2 that there exists $e\in E(S)$ such that $g(ae)=g(a)e=g(a')e=g(a'e)$ and $es=s$ which implies that $(ae,a'e)\in kerg$ and so $(ae,a'e)\in K(\rho )=(Imf\times Imf) \cup \Delta _{A_S}$. Then $ae=a'e$ or there exist $b_1,b_2\in B_S$ such that $ae=f(b_1)$ and $a'e=f(b_2)$. In the second case, since $as=a's$ and $es=s$, $aes=a'es$ and then $f(b_1s)=f(b_2s)$ so that $b_1s=b_2s$ by injectivity of f. Using Lemma 2, since $B_S$ is principally weakly flat, there exists $h\in E(S)$ such that $hs=s$ and $b_1h=b_2h$ and so $f(b_1)h=f(b_2)h$ which gives that $aeh=a'eh$. Since E(S) is commutative, $eh\in E(S)$. Also we have $ehs=es=s$. Therefore, $A_S$ is principally weakly flat by Lemma 2. $\square$

Recall from [12] that $a\in A_S$ is said to be an act-regular element if there exists a homomorphism $f:aS\rightarrow S$ such that $af(a)=a$, and $A_S$ is called a regular act if every $a\in A_S$ is an act-regular element.

Lemma 3

[12, Proposition 3.19.2] Let $A_S$ be an S-act. An element $a\in A_S$ is act-regular if and only if there exists an idempotent $e\in S$ such that $ae=a$ and $as=at$ implies that $es=et$ for any $s,t\in S$.

Theorem 5

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a $\rho$-exact sequence at $A_S$ such that f is a monomorphism. If $B_S$ and $C_S$ are regular S-acts, then so is $A_S$.

Proof

Let $a\in A_S$. Since $C_S$ is regular, by Lemma 3, there exists $e\in E(S)$ such that $g(a)=g(a)e$ and $g(a)s=g(a)t$ implies that $es=et$ for any $s,t \in S$. Then $(a,ae) \in kerg$ and so $(a,ae)\in K(\rho )=(Imf\times Imf)\cup \Delta _{A_S}$ which gives that $a=ae$ or there exists $b\in B_S$ such that $a=f(b)$. In the first case, if $as=at$ for any $s,t\in S$, then $g(a)s=g(a)t$ and so $es=et$. In the second case, since $B_S$ is regular, it follows from Lemma 3 that there exists $h\in E(S)$ such that $b=bh$ and $bs=bt$ implies that $hs=ht$ for any $s,t\in S$. Hence, $a=f(b)=f(bh)=f(b)h=ah$. Suppose that $as=at$ for any $s,t \in S$. Then $f(bs)=f(b)s=f(b)t=f(bt)$ and so $bs=bt$ by injectivity of f whence $hs=ht$. Applying Lemma 3, $A_S$ is regular. $\square$

Recall from [12] that an S-act $A_S$ is called torsion free if for any $x,y \in A_S$ and any right cancellable element $c\in S$, if $xc=yc$, then $x=y$, i.e., $(x,y)\in \Delta _{A_S}$. In what follows, considering any congruence $\rho \ne \nabla _{A_S}$ on $A_S$ instead of $\Delta _{A_S}$ in the definition of torsion free act we give a new notion of $\rho$-torsion free act.

Definition 2

Let $A_S$ be an S-act and $\rho \in Con(A_S)\backslash \{\nabla _{A_S}\}$. We say that $A_S$ is $\rho$-torsion free if for any $x,y \in A_S$ and any right cancellable element $c\in S$, the equality $xc=yc$ implies that $(x,y)\in \rho$.

Clearly, for a non-singleton S-act $A_S$, $\Delta _{A_S}\ne \nabla _{A_S}$ and then $A_S$ is $\Delta _{A_S}$-torsion free if and only if it is torsion free. Moreover, torsion freeness implies $\rho$-torsion freeness for any $\rho \in Con(A_S)\backslash \{\nabla _{A_S}\}$. The following example shows that this implication is strict. For this, we give some preliminaries.

Let U be a subact of an S-act $A_S$. The amalgam$A_S\coprod ^{U} A_S$ is realized as the set $((A{\setminus }U) \times \{1,2\}) \cup U$, where the natural action on U is extended to the remaining elements by defining

$$\begin{aligned} (a,i)s=\left\{ \begin{array}{ll} (as,i)&{}\quad as\in A{\setminus } U\\ as &{}\quad as \in U\\ \end{array} \right. \end{aligned}$$

(1)

for every $a \in A{\setminus } U, s\in S$ and $i=1,2$ (see [12, Proposition 2.2.26]).

Example 2

Let $S=(\mathbb {N},\cdot )$ and consider the amalgam $A_S=\mathbb {N}\coprod ^{\mathbb {N}\backslash \{1\}}\mathbb {N}$. Then $(1,x)\not =(1,y)$ but $(1,x)2=2=(1,y)2$. Hence, $A_S$ is not torsion free. Now let

$$\begin{aligned} \rho =\Delta _{A_S}\cup \{((1,x),(1,y)),((1,y),(1,x))\}. \end{aligned}$$

It is easily seen that $\rho \in Con(A_S)\backslash \{\nabla _{A_S}\}$ and $A_S$ is $\rho$-torsion free.

We denote the set of all right cancellable elements of S by r(S). For an S-act $A_S$ and $X\subseteq A_S\times A_S$, the congruence on $A_S$ generated by X is denoted by $\rho (X)$ (see [12, Lemma 1.4.37]). We also set $X_A:=\{(x,y)\in A_S\times A_S\mid xc=yc \mathrm{~for ~some~} c\in r(S)\}$. If S is commutative, then one can easily show that $X_A$ is a congruence on $A_S$.

Remark 1

Let $A_S$ be an S-act. If $\lambda \in Con(A_S)\backslash \{\nabla _{A_S}\}$, then it follows from the definition that $A_S$ is $\lambda$-torsion free if and only if $X_A\subseteq \lambda$. Moreover, for any congruence $\rho$ on $A_S$ with $\rho \le \lambda$, if $A_S$ is $\rho$-torsion free, then it is $\lambda$-torsion free.

Proposition 2

Let $A_S$ be an S-act for which $\rho (X_A)\ne \nabla _{A_S}$. Then $A_S$ is $\rho (X_A)$-torsion free and $\rho (X_A)$ is the smallest congruence on $A_S$ satisfying this property.

Proof

It is straightforward.$\square$

Note that if $B_S {\mathop {\rightarrow }\limits ^{f}} A_S$ is a homomorphism and $\lambda \in Con(B_S)$, then $L(\lambda ):=(f\times f)(\lambda )\cup \Delta _{A_S}$ is a congruence on $A_S$ where $(f\times f)(\lambda )=\{(f(b_1),f(b_2))\mid (b_1,b_2)\in \lambda \}$.

Theorem 6

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be a $\rho$-exact sequence at $A_S$ where $\rho \in Con(C_S)\backslash \{\nabla _{C_S}\}$ such that f is a monomorphism. If $B_S$ is $\lambda$-torsion free for $\lambda \in Con(B_S)\backslash \{\nabla _{B_S}\}$ and $C_S$ is $\rho$-torsion free, then $A_S$ is $(L(\lambda )\cap K(\rho ))$-torsion free.

Proof

Since $\lambda \ne \nabla _{B_S}$ and f is injective, $L(\lambda )\ne \nabla _{A_S}$ and so $(L(\lambda )\cap K(\rho ))\in Con(A_S)\backslash \{\nabla _{A_S}\}$. It suffices to show that $X_A\subseteq L(\lambda )\cap K(\rho )$. If $(a_1,a_2)\in X_A$, then $a_1c=a_2c$ for some $c\in r(S)$ and so $g(a_1)c=g(a_2)c$. Since $C_S$ is $\rho$-torsion free, $(g(a_1),g(a_2))\in \rho$ which follows that $(a_1,a_2)\in K(\rho )=(Imf\times Imf)\cup \Delta _{A_S}$. Thus $a_1=a_2$ or there exist $b_1,b_2\in B_S$ such that $a_1=f(b_1)$ and $a_2=f(b_2)$. If $a_1=a_2$, then $(a_1,a_2)\in L(\lambda )\cap K(\rho )$. In the second case, since $a_1c=a_2c$, $f(b_1c)=f(b_2c)$ and so $b_1c=b_2c$ by injectivity of f. Since $B_S$ is $\lambda$-torsion free, $(b_1,b_2)\in \lambda$ and then $(f(b_1),f(b_2))\in (f\times f)(\lambda )$. This gives that $(a_1,a_2)\in L(\lambda )$. Therefore, $(a_1,a_2)\in L(\lambda )\cap K(\rho )$, as claimed. $\square$

Corollary 2

Let $B_S {\mathop {\rightarrow }\limits ^{f}} A_S {\mathop {\rightarrow }\limits ^{g}} C_S$ be an exact sequence at $A_S$ such that f is a monomorphism. If $B_S$ and $C_S$ are torsion free S-acts, then so is $A_S$.

3 Some Results Concerning Commutative Diagrams

Parallel to the theory of modules, in this section we generalize some known results dealing with commutative diagrams of modules based on exact sequences to that of (quasi-) exact sequences of S-acts.

Lemma 4

(Generalization of Short Five Lemma) Let the following diagram be a commutative diagram of S-acts and homomorphisms such that the first row is $\rho$-exact at $B_S$, the second row is $\rho '$-exact at $B'_S$, g is an epimorphism and $f'$ is a monomorphism:

$$\begin{aligned} \begin{array}{ccccc} A_S &{} {\mathop {\longrightarrow }\limits ^{f}} &{} B_S &{} {\mathop {\longrightarrow }\limits ^{g}} &{} C_S \\ \downarrow \alpha &{} &{} \downarrow \beta &{} &{} \downarrow \gamma \\ A'_S &{} {\mathop {\longrightarrow }\limits ^{f'}} &{} B'_S &{} {\mathop {\longrightarrow }\limits ^{g'}} &{} C'_S \end{array} \end{aligned}$$

• (i) If $\alpha$ and $\gamma$ are monomorphisms, then so is $\beta$.

• (ii) If $\alpha$ and $\gamma$ are epimorphisms, then so is $\beta$.

• (iii) If $\alpha$ and $\gamma$ are isomorphisms, then so is $\beta$.

Proof

• (i) Suppose that $\beta (b_1)=\beta (b_2)$ for $b_1,b_2\in B_S$. We have $\gamma g(b_1)=g'\beta (b_1)=g'\beta (b_2)=\gamma g(b_2)$ and so $g(b_1)=g(b_2)$ by injectivity of $\gamma$ which implies that $(b_1,b_2) \in kerg$. Hence, $(b_1,b_2)\in K(\rho )$ so that $b_1=b_2$ or there exist $a_1,a_2\in A_S$ such that $b_1 = f(a_1)$ and $b_2 = f(a_2)$. In the second case, we have $f'\alpha (a_1)=\beta f(a_1)=\beta f(a_2)=f'\alpha (a_2)$ which gives that $a_1=a_2$ by injectivity of $f'$ and $\alpha$ and so $b_1=b_2$.

• (ii) Let $b'\in B'_S$. Then $g'(b')\in C'_S$ and there exists $c\in C_S$ such that $\gamma (c)=g'(b')$ by surjectivity of $\gamma$. Also surjectivity of g gives that there exists $b\in B_S$ such that $c=g(b)$. Hence, $g'\beta (b)=\gamma g(b)=\gamma (c)= g'(b')$ so that $(\beta (b),b') \in kerg'$ and then $(\beta (b),b') \in K(\rho ')$ which implies that $\beta (b)=b'$ or there exists $a'\in A'_S$ such that $b'= f'(a')$. In the second case, there exists $a\in A_S$ such that $a'=\alpha (a)$ by surjectivity of $\alpha$ and so $\beta f(a)=f'\alpha (a)=f'(a')=b'$.

• (iii) It follows from (i) and (ii). $\square$

Theorem 7

Let the following diagram of S-acts and homomorphisms

$$\begin{aligned} \begin{array}{ccccc} A_S &{} {\mathop {\longrightarrow }\limits ^{f}} &{} B_S &{} {\mathop {\longrightarrow }\limits ^{g}} &{} C_S \\ \downarrow \alpha &{} &{} \downarrow \beta &{} &{} \downarrow \gamma \\ A'_S &{} {\mathop {\longrightarrow }\limits ^{f'}} &{} B'_S &{} {\mathop {\longrightarrow }\limits ^{g'}} &{} C'_S \end{array} \end{aligned}$$

be commutative and have exact rows at $B_S$ and $B'_S$, respectively.

• (i) If $\beta$ is a monomorphism and $\alpha$ and g are epimorphisms, then $\gamma$ is a monomorphism.

• (ii) If $\beta$ is an epimorphism and $\gamma$ and $f'$ are monomorphisms, then $\alpha$ is an epimorphism.

Proof

• (i) Consider any $c_1,c_2\in C_S$ with $\gamma (c_1)=\gamma (c_2)$. Since g is an epimorphism, $c_1=g(b_1)$ and $c_2=g(b_2)$ for some $b_1,b_2\in B_S$ which implies that $\gamma g(b_1)=\gamma g(b_2)$ and so $g'\beta (b_1)=g'\beta (b_2)$. Hence, $(\beta (b_1),\beta (b_2))\in kerg'=(Imf'\times Imf')\cup \Delta _{B'_S}$ and then $\beta (b_1)=\beta (b_2)$ or $\beta (b_1)=f'(a_1')$ and $\beta (b_2)= f'(a_2')$ for some $a_1',a_2'\in A_S'$. In the first case, since $\beta$ is a monomorphism, $b_1=b_2$ and so $c_1=g(b_1)=g(b_2)=c_2$. In the second case, since $\alpha$ is an epimorphism, there exist $a_1,a_2 \in A_S$ such that $\beta (b_1)=f'\alpha (a_1)$ and $\beta (b_2)=f'\alpha (a_2)$. Thus $\beta (b_1)=\beta f(a_1)$ and $\beta (b_2)=\beta f(a_2)$ and so $b_1=f(a_1)$ and $b_2=f(a_2)$ by injectivity of $\beta$. Then $(b_1,b_2)\in (Imf\times Imf)\le kerg$ and so $(b_1,b_2)\in kerg$. Therefore, $c_1=g(b_1)=g(b_2)=c_2$.

• (ii) If $|A'_S|=1$, then there is nothing to prove. Let $a_1'\in A'_S$. Take any $a_2'\in A'_S$ with $a_1'\not =a_2'$. Since $\beta$ is an epimorphism and $f'(a_1'),f'(a_2')\in B_S'$, $f'(a_1')=\beta (b_1)$ and $f'(a_2')=\beta (b_2)$ for some $b_1,b_2\in B_S$. Thus $g'f'(a_1')=g'\beta (b_1)$ and $g'f'(a_2')=g'\beta (b_2)$. Since $Imf'\times Imf' \le kerg'$, $g'\beta (b_1)=g'f'(a_1')=g'f'(a_2')=g'\beta (b_2)$ and so $\gamma g(b_1)=\gamma g(b_2)$. Since $\gamma$ is a monomorphism, $g(b_1)=g(b_2)$ and so $(b_1,b_2)\in kerg = (Imf\times Imf)\cup \Delta _{B_S}$ which implies that $b_1=b_2$ or there exists $a_1\in A_S$ such that $b_1=f(a_1)$. The first case is a contradiction, because $f'(a_1')=\beta (b_1)=\beta (b_2)=f'(a_2')$ and so $a_1'=a_2'$ by injectivity of $f'$. In the second case, $\beta (b_1)=\beta f(a_1)$ and so $f'(a_1')=\beta (b_1)=f'\alpha (a_1)$. Since $f'$ is a monomorphism, $a_1'=\alpha (a_1)$. Therefore, $\alpha$ is an epimorphism. $\square$

Theorem 8

Let the following diagram of S-acts and homomorphisms be commutative and have short exact rows:

$$\begin{aligned} \begin{array}{ccccc} A_S' &{} {\mathop {\longrightarrow }\limits ^{\alpha '}} &{} A_S &{} {\mathop {\longrightarrow }\limits ^{\alpha }} &{} A_S'' \\ \downarrow f' &{} &{} \downarrow f &{} &{} \downarrow f'' \\ B_S' &{} {\mathop {\longrightarrow }\limits ^{\beta '}} &{} B_S &{} {\mathop {\longrightarrow }\limits ^{\beta }} &{} B_S''\\ \downarrow g' &{} &{} \downarrow g &{} &{} \downarrow g'' \\ C_S' &{} {\mathop {\longrightarrow }\limits ^{\gamma '}} &{} C_S &{} {\mathop {\longrightarrow }\limits ^{\gamma }} &{} C_S'' \end{array} \end{aligned}$$

If the middle column is short exact, then the last column is short exact if and only if the first column is short exact.

Proof

Suppose the last column is short exact. We show that so is the first column. Injectivity of $f'$ follows clearly from injectivity of f and $\alpha '$ and commutativity. We claim that $(Imf'\times Imf')\cup \Delta _{B_S'}=kerg'$. Let $(b_1',b_2')\in Imf'\times Imf'$. Then there exist $a_1',a_2'\in A_S'$ such that $b_1'=f'(a_1')$ and $b_2'=f'(a_2')$ which implies that $g\beta 'f'(a_1')=\gamma ' g'(b_1')$ and $g\beta 'f'(a_2')=\gamma ' g'(b_2')$. So we get $gf\alpha '(a_1')=\gamma ' g'(b_1')$ and $gf\alpha '(a_2')=\gamma ' g'(b_2')$ (as $\beta ' f' = f\alpha '$). Since $Imf\times Imf \le kerg$, $gf\alpha '(a_1')=gf\alpha '(a_2')$ and so $\gamma ' g'(b_1')=\gamma ' g'(b_2')$ which implies that $g'(b_1')=g'(b_2')$ by injectivity of $\gamma '$ which means $(b_1',b_2')\in kerg'$. Hence, $(Imf'\times Imf')\cup \Delta _{B_S'} \subseteq kerg'$. For the reverse inclusion, let $(b_1',b_2')\in kerg'$. Then $g'(b_1')=g'(b_2')$ and so $\gamma ' g'(b_1')=\gamma ' g'(b_2')$ whence $g\beta '(b_1')=g\beta '(b_2')$ (as $\gamma ' g'=g\beta '$) so that $(\beta '(b_1'),\beta '(b_2'))\in kerg=(Imf\times Imf)\cup \Delta _{B_S}$. Hence, $\beta '(b_1')=\beta '(b_2')$ or $\beta '(b_1')=f(a_1)$ and $\beta '(b_2')=f(a_2)$ for some $a_1,a_2\in A_S$.

In the first case, since $\beta '$ is a monomorphism, $b_1'=b_2'$. In the second case, we have $\beta \beta '(b_1')=\beta f(a_1)$ and $\beta \beta '(b_2')=\beta f(a_2)$. Since $Im\beta '\times Im\beta ' \le ker\beta$, $\beta \beta '(b_1')=\beta \beta '(b_2')$ and so $\beta f(a_1)=\beta f(a_2)$ which gives that $f''\alpha (a_1)=f''\alpha (a_2)$ (as $\beta f= f''\alpha$) so that $\alpha (a_1)=\alpha (a_2)$ by injectivity of $f''$. Thus $(a_1,a_2)\in ker\alpha = (Im\alpha '\times Im\alpha ')\cup \Delta _{A_S}$ and so $a_1=a_2$ or there exist $a_1',a_2'\in A_S'$ such that $a_1=\alpha '(a_1')$ and $a_2=\alpha '(a_2')$.

If $a_1=a_2$, then $\beta '(b_1')= f(a_1)=f(a_2)=\beta '(b_2')$ and so $b_1'=b_2'$ by injectivity of $\beta '$. In the second case, $f(a_1)=f\alpha '(a_1')$ and $f(a_2)=f\alpha '(a_2')$ and so $\beta '(b_1')=f(a_1)=\beta 'f'(a_1')$ and $\beta '(b_2')=f(a_2)=\beta 'f'(a_2')$ (as $f\alpha ' = \beta ' f'$) which follows that $b_1'=f'(a_1')$ and $b_2'=f'(a_2')$ by injectivity of $\beta '$. Therefore, $kerg'\subseteq (Imf'\times Imf')\cup \Delta _{B_S'}$.

Finally, we show that $g'$ is an epimorphism. If $|C'_S|=1$, then there is nothing to prove. Let $c_1'\in C'_S$. Take any $c_2'\in C'_S$ with $c_1'\not =c_2'$. Since g is an epimorphism, there exist $b_1,b_2\in B_S$ such that $\gamma '(c_1')=g(b_1)$ and $\gamma '(c_2')=g(b_2)$ and then $\gamma \gamma '(c_1')=\gamma g(b_1)$ and $\gamma \gamma '(c_2')=\gamma g(b_2)$. Since $Im\gamma '\times Im\gamma '\le ker\gamma$, $\gamma g(b_1)=\gamma \gamma '(c_1')=\gamma \gamma '(c_2')= \gamma g(b_2)$ and so $g''\beta (b_1)=g''\beta (b_2)$ (as $\gamma g= g''\beta$) which implies that $(\beta (b_1),\beta (b_2))\in kerg''= (Imf''\times Imf'')\cup \Delta _{B_S''}$. Hence, $\beta (b_1)=\beta (b_2)$ or there exist $a_1'',a_2''\in A_S''$ such that $\beta (b_1)=f''(a_1'')$ and $\beta (b_2)=f''(a_2'')$.

Case 1 Let $\beta (b_1)=\beta (b_2)$. Then $(b_1,b_2)\in ker\beta =(Im\beta '\times Im\beta ')\cup \Delta _{B_S}$ and so $b_1=b_2$ or there exist $b_1',b_2'\in B_S'$ such that $b_1=\beta '(b_1')$ and $b_2=\beta '(b_2')$.

If $b_1=b_2$, then $\gamma '(c_1')=g(b_1)=g(b_2)=\gamma '(c_2')$ and so $c_1'=c_2'$ by injectivity of $\gamma '$ which is a contradiction. Otherwise, we get $\gamma '(c_1')=g(b_1)=g\beta '(b_1')=\gamma 'g'(b_1')$ and so $c_1'=g'(b_1')$ by injectivity of $\gamma '$.

Case 2 Assume that $\beta (b_1)=f''(a_1'')$ and $\beta (b_2)=f''(a_2'')$. Since $\alpha$ is an epimorphism, $a_1''=\alpha (a_1)$ and $a_2''=\alpha (a_2)$ for some $a_1,a_2\in A_S$ and so $\beta (b_1)=f''\alpha (a_1)$ and $\beta (b_2)=f''\alpha (a_2)$ which implies that $\beta (b_1)=\beta f(a_1)$ and $\beta (b_2)=\beta f(a_2)$ (as $f''\alpha =\beta f$). Hence, $(b_1,f(a_1)),(b_2,f(a_2))\in ker\beta = (Im\beta '\times Im\beta ')\cup \Delta _{B_S}$.

Subcase 2a Let $b_1=f(a_1)$ and $b_2=f(a_2)$. We have $g(b_1)=gf(a_1)$ and $g(b_2)=gf(a_2)$. Since $Imf\times Imf\le kerg$, $g(b_1)=gf(a_1) = gf(a_2)=g(b_2)$ and so $\gamma '(c_1')=\gamma '(c_2')$ whence $c_1'=c_2'$ by injectivity of $\gamma '$ which is a contradiction.

Subcase 2b Let $b_1=f(a_1)$ and there exist $b_1',b_2'\in B_S'$ such that $b_2=\beta '(b_1')$ and $f(a_2)=\beta '(b_2')$. Then $g(b_1)=gf(a_1)$ and $gf(a_2)=g\beta '(b_2')$ and so $g(b_1)=g\beta '(b_2')$ which implies that $\gamma '(c_1')=\gamma 'g'(b_2')$ (as $g\beta '=\gamma 'g'$) so that $c_1'=g'(b_2')$ by injectivity of $\gamma '$.

Subcase 2c Let $b_2=f(a_2)$ and there exist $b_1',b_2'\in B_S'$ such that $b_1=\beta '(b_1')$ and $f(a_1)=\beta '(b_2')$. So $g(b_1)=g\beta '(b_1')$ and then $\gamma '(c_1')=\gamma 'g'(b_1')$ which gives that $c_1'=g'(b_1')$ by injectivity of $\gamma '$.

Subcase 2d Let there exist $b_1',b_2',b_3',b_4'\in B_S'$ such that $b_1=\beta '(b_1')$, $f(a_1)=\beta '(b_2')$, $b_2=\beta '(b_3')$ and $f(a_2)=\beta '(b_4')$. In this case, we proceed exactly as in Subcase 2c.

Similarly, one can show that if the first column is exact then so is the last column. $\square$

Theorem 9

Let the following diagram be a commutative diagram of S-acts and homomorphisms such that the first row is $\rho$-exact at $B_S$, the second row is exact at $B'_S$, $f'$ is a monomorphism and g is an epimorphism:

$$\begin{aligned} \begin{array}{ccccc} A_S &{} {\mathop {\longrightarrow }\limits ^{f}} &{} B_S &{} {\mathop {\searrow }\limits ^{g}} &{} \\ \downarrow \alpha &{} &{} \downarrow \beta &{} &{} C_S \\ A'_S &{} {\mathop {\longrightarrow }\limits ^{f'}} &{} B'_S &{} {\mathop {\nearrow }\limits ^{g'}} &{} \end{array} \end{aligned}$$

Then $\alpha$ is an epimorphism if and only if so is $\beta$.

Proof

Let $\alpha$ be an epimorphism and $b'\in B_S'$. Since g is an epimorphism, $g'(b')=g(b)$ for some $b\in B_S$ and so $g'(b')=g'\beta (b)$ which implies that $(b',\beta (b))\in ker g'=(Imf'\times Imf')\cup \Delta _{B_S'}$. Hence, $b'=\beta (b)$ or there exists $a'\in A_S'$ such that $b'=f'(a')$. In the second case, it follows from surjectivity of $\alpha$ that there exists $a\in A_S$ such that $b'=f'\alpha (a)=\beta f(a)$. Therefore, $\beta$ is an epimorphism. Conversely, let $\beta$ be an epimorphism. If $|A'_S|=1$, then we are done. Let $a_1'\in A'_S$. Take any $a_2'\in A'_S$ with $a_1'\not =a_2'$. Then $f'(a_1'),f'(a_2')\in B_S'$. Since $\beta$ is an epimorphism, $f'(a_1')=\beta (b_1)$ and $f'(a_2')=\beta (b_2)$ for some $b_1,b_2\in B_S$ and so $g'f'(a_1')=g'\beta (b_1)=g(b_1)$ and $g'f'(a_2')=g'\beta (b_2)=g(b_2)$. Since $Imf'\times Imf'\le ker g'$, $g(b_1)=g'f'(a_1')=g'f'(a_2')=g(b_2)$ and so $(b_1,b_2)\in ker g$ whence $(b_1,b_2)\in K(\rho )=(Imf\times Imf)\cup \Delta _{B_S}$. Thus $b_1=b_2$ or there exists $a_1\in A_S$ such that $b_1=f(a_1)$. In the first case, $f'(a_1')=\beta (b_1)=\beta (b_2)=f'(a_2')$ and so $a_1'=a_2'$ by injectivity of $f'$ which is a contradiction. Otherwise, $f'(a_1')=\beta (b_1)=\beta f(a_1)=f'\alpha (a_1)$ which implies that $a_1'=\alpha (a_1)$ by injectivity of $f'$. Consequently, $\alpha$ is an epimorphism. $\square$