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ISRN Algebra
Volume 2011 (2011), Article ID 142403, 9 pages
http://dx.doi.org/10.5402/2011/142403
Research Article

Weakly Injective BCK-Modules

1Department of Mathematical Sciences, Saginaw Valley State University, 7400 Bay Road, University Center, MI 48710-0001, USA
2Department of Mathematics, University of Oregon, Eugene, OR 97403, USA

Received 7 June 2011; Accepted 21 July 2011

Academic Editors: V. K. Dobrev and Y.-H. Quano

Copyright © 2011 Olivier A. Heubo-Kwegna and Jean B. Nganou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We introduce the notion of weakly injective BCK-module and show that Baer's criterion holds for weakly injective BCK-modules but not for injective BCK-modules in general. We also provide examples and counterexamples of weakly injective BCK-modules.

1. Introduction

Inspired by Meredith's BCK-systems, Iséki and Imai introduced the notion of BCK-algebra in 1966. These pioneers developed major aspects of the theory in the late 1960s and the 1970s. They were soon joined by many other researchers to develop various aspects of the BCK-algebra theory. Since then, BCK-algebras have been a subject of intense research. The main approach of this development has been trying to build a theory that is parallel to the standard ring theory. In this order of ideas, Noetherian and Artinian BCK-algebras [1], BCK-modules [2], injective and projective BCK-modules [2], and fractions BCK-algebras [3] have recently been treated. So far, the only articles on BCK-modules have been [2, 4]. Considering the topics covered by these two articles, it is quite clear that very little is known about the theory of modules over BCK-algebras. For instance, the notion of injective modules over BCK-algebras was introduced in [2], but not a single example was treated. In classical ring theory, injective modules are studied using Baer's criterion and divisible modules. Unfortunately, as we will show, this criterion does not hold for injective BCK-modules, and there are no natural notion of divisible modules over BCK-algebras.

The main goal of this work is to shed some light on the notion of injective modules over BCK-algebras. We do this by introducing a new class of modules (weakly injective modules) that strictly contains the above class and obtain a Baer's criterion for this class. In order to achieve this goal, we found ourselves imposing a new axiom to BCK-modules.

Recall that the notion of left module over a bounded commutative BCK-algebra (𝑋,,0,1) was first introduced in 1994 by Abujabal et al. [4]. We consider the class of left BCK-modules that satisfy the following axiom in addition to the axioms of [4], 𝑥𝑦𝑚=𝑥𝑚+𝑦𝑚,(1.1) for all 𝑥,𝑦𝑋 and 𝑚𝑀 where 𝑥𝑦=(𝑥𝑦)(𝑦𝑥).

We will refer to BCK-modules of this class as BCK-modules of type 2. The consideration of this class is motivated not only by the fact that it makes BCK-modules more in line with modules over rings, but also the fact that the main results obtained by the previous authors remain valid for this class. Using this class of BCK-modules, we introduce weakly injective BCK-modules. We prove that weakly injective BCK-modules are Characterized by Baer's criterion, which we use to prove that over principal bounded implicative BCK's, every module is weakly injective [Corollary 3.9]. We use these characterizations to build examples of (weakly) injective modules over BCK-algebras and also find examples that prove that our Baer's criterion is the sharpest we can get.

2. Generalities on BCK-Modules

Recall that a BCK-algebra is an algebra (𝑋,,,0) satisfying for all 𝑥,𝑦,𝑧𝑋(i)(𝑥𝑧)(𝑥𝑦)𝑦𝑧, (ii)𝑥(𝑥𝑦)𝑦,(iii)0𝑥,(iv)𝑥𝑥,(v)𝑥𝑦 and 𝑦𝑥 implies 𝑥=𝑦,(vi)𝑥𝑦 if and only if 𝑥𝑦=0.

In addition, if there exists an element 1 in 𝑋 such that 𝑥1 for all 𝑥𝑋, then 𝑋 is said to be bounded and we write 𝑁𝑥 for 1𝑥. Also, if 𝑥(𝑥𝑦)=𝑦(𝑦𝑥) for all 𝑥,𝑦𝑋, 𝑋 is said to be commutative. In addition, 𝑋 is called implicative if 𝑥(𝑦𝑥)=𝑥 for all 𝑥,𝑦𝑋. As proved in [5, Theorem 10], implicative BCK-algebras are commutative. A subset 𝐼 of a BCK-algebra 𝑋 is called an ideal of 𝑋 if it satisfies (i) 0𝐼 and (ii) for every 𝑥,𝑦𝑋 such that 𝑦𝐼 and 𝑥𝑦𝐼, then 𝑥𝐼.

As defined in [4, Definition 2.1], a left module over a bounded commutative BCK-algebra (𝑋,,0,1) is an Abelian group (𝑀,+) with a multiplication (𝑥,𝑚)𝑥𝑚 satisfying (i)(𝑥𝑦)𝑚=𝑥(𝑦𝑚) for all 𝑥,𝑦𝑋 and 𝑚𝑀,(ii)𝑥(𝑚+𝑛)=𝑥𝑚+𝑥𝑛 for all 𝑥𝑋 and 𝑚,𝑛𝑀,(iii)0𝑚=0 for all 𝑚𝑀,(iv)1𝑚=𝑚 for all 𝑚𝑀,

where 𝑥𝑦=𝑦(𝑦𝑥).

If in addition, 𝑀 satisfy the axiom (v) below, we call 𝑀 an 𝑋-module of type 2.(v)(𝑥𝑦)𝑚=𝑥𝑚+𝑦𝑚 for all 𝑥,𝑦𝑋 and 𝑚𝑀. Where 𝑥𝑦=(𝑥𝑦)(𝑦𝑥).

Our terminology type 2 is motivated by the fact that every 𝑋-module satisfying (𝑣) is as Abelian group, of exponent 2.

Recall [4, Lemma 2.4] that if 𝑋 is a bounded implicative BCK-algebra, then (𝑋,,) is a commutative ring. Therefore, 𝑋-module of type 2 are modules over the ring (𝑋,,).

If 𝑀 is a left 𝑋-module, a subset 𝑆 of 𝑀 is a submodule if (𝑆,+) is a subgroup of (𝑀,+) such that 𝑥𝑚𝑆 whenever 𝑥𝑋 and 𝑚𝑆.

Given two left 𝑋-modules 𝑀 and 𝑁, an 𝑋-module homomorphism from 𝑀 to 𝑁 is a map 𝑓𝑀𝑁 satisfying(i)𝑓(𝑚+𝑚)=𝑓(𝑚)+𝑓(𝑚) for all 𝑚,𝑚𝑀,(ii)𝑓(𝑥𝑚)=𝑥𝑓(𝑚) for all 𝑥𝑋 and 𝑚𝑀.

The set of all 𝑋-module homomorphisms from 𝑀 to 𝑁 is denoted by Hom𝑋(𝑀,𝑁) which has a natural structure of 𝑋-module via the multiplication (𝑥𝑓)(𝑚)=𝑥𝑓(𝑚).

We introduce the following definition.

Definition 2.1. A left 𝑋-module 𝑄 is weakly injective if for every left 𝑋-module 𝑀,𝑁 so that 𝑁 is of type 2, every injective homomorphism 𝑓𝑀𝑁 and every homomorphism 𝑔𝑀𝑄, there exists a homomorphism 𝜙𝑁𝑄 such that 𝜙𝑓=𝑔.

Note that injective 𝑋-modules as defined in [2] are clearly weakly injective. We have the following lemma whose some parts have been proved by other authors, but which we offer a proof here for the convenience of the reader.

Lemma 2.2. Let 𝑋 be a bounded implicative BCK-algebra with unit 1. Then, for all 𝑥,𝑦,𝑧𝑋, (i)𝑥𝑦=𝑥𝑁𝑦,(ii)𝑥(𝑥𝑦)=𝑥𝑦, (iii)𝑥(𝑦𝑧)=(𝑥𝑦)(𝑥𝑧), (iv)(𝑥𝑦)(𝑦𝑥)=𝑥𝑦,(v)(𝑥𝑦)𝑧=(𝑥𝑧)(𝑥𝑧).

Proof. (i) From [5, Proposition 6], we have 𝑥𝑁𝑦𝑥𝑦. In addition, 𝑥𝑦𝑥 and 𝑥𝑦1𝑦=𝑁𝑦, so 𝑥𝑦𝑥𝑁𝑦. Thus, 𝑥𝑁𝑦=𝑥𝑦. Therefore, 𝑥𝑦=𝑥𝑁𝑁𝑦=𝑥𝑁𝑦.
For (ii), let 𝑥,𝑦𝑋. By (i), we do have 𝑥(𝑥𝑦)=𝑥(𝑥𝑁𝑦)=𝑁𝑦𝑥=𝑥𝑦.(2.1) For (iii), let 𝑥,𝑦,𝑧𝑋. We first prove that 𝑥(𝑦𝑧)=(𝑥𝑦)𝑧.(2.2) In fact, =𝑥(𝑦𝑧)=(𝑦𝑧)𝑁𝑥(𝑦𝑁𝑥)𝑧=(𝑥𝑦)𝑧.(2.3)
Now, we use (ii) and (2.2) to show (iii)=(𝑥𝑦)(𝑥𝑧)=(𝑥𝑁𝑦)(𝑥𝑧)(𝑥(𝑥𝑧))𝑁𝑦=(𝑥𝑧)𝑁𝑦by(ii)=(𝑥𝑁𝑦)𝑧=(𝑥𝑦)𝑧=𝑥(𝑦𝑧)by(2.2).(2.4)(iv) We have [][](𝑥𝑦)(𝑦𝑥)=(𝑥𝑦)(𝑦𝑥)(𝑦𝑥)(𝑥𝑦)=((𝑥(𝑦𝑥))𝑦)((𝑦(𝑥𝑦))𝑥)by5,Equation(3)=(𝑥𝑦)(𝑦𝑥)since𝑋isimplicative=𝑥𝑦.(2.5)(v) Recall that being a bounded implicative BCK-algebra, (𝑋,,) is a distributive lattice [5, Theorem 12]. We have 𝑥𝑦𝑧=((𝑥𝑦)(𝑦𝑥))𝑧=((𝑥𝑦)𝑧)((𝑦𝑥)𝑧)=((𝑥𝑧)(𝑦𝑧))((𝑦𝑧)(𝑥𝑧))by(ii)=(𝑥𝑧)(𝑦𝑧).(2.6)

Proposition 2.3. Every bounded implicative BCK-algebra has a natural structure of 𝑋-module of type 2. Furthermore, under this structure, every ideal of 𝑋 is a submodule of 𝑋.

Proof. Consider the operational system (𝑋,). Then, by [4, Proposition 2.5], (𝑋,) is an Abelian group and together with the multiplication (𝑥,𝑦)𝑥𝑦=𝑥𝑦 satisfying (i), (ii), (iii), and (iv) of the definition of 𝑋-module above. It remains to verify that (𝑋,) satisfies (v). But this is straightforward from the definition of the multiplication and Lemma 2.2 (v).
As for the proof that an ideal of 𝑋 is a submodule, the argument is identical to the one of [2, Theorem 2.1].

Example 2.4. Consider the bounded implicative BCK, 𝑋=𝒫() with the standard operations. Consider 𝑀1=Maps(,) and 𝑀2=Maps(,2), then under the multiplication (𝐴,𝑓)1𝐴𝑓 (where 1𝐴 is the characteristic function of 𝐴), 𝑀1 is an 𝑋-module that is not of type 2 while 𝑀2 is an 𝑋-module of type 2.

Example 2.5. For every 𝑋-modules 𝑀 and 𝑁 so that 𝑁 is of type 2, the 𝑋-module Hom𝑋(𝑀,𝑁) is also of type 2. In particular, for every 𝑋-module 𝑀, Hom𝑋(𝑀,𝑋) is of type 2.

Remark 2.6. (𝑋,) is an Abelian group of exponent 2; therefore, finite bounded implicative BCK-algebras have order a power of 2. This is not surprising as such BCKs are Boolean algebras [5, Theorem 12].

3. Injective BCK-Modules and Baer's Criterion

𝑋 will denote a bounded implicative BCK-algebra with unit 1. In addition, the term 𝑋-module will refer to left 𝑋-module. We start by the following lemma which is crucial for the proof of Baer's criterion.

Lemma 3.1. Let N be an 𝑋-module of type 2 and 𝑀 a submodule of 𝑁. For every 𝑛𝑁, define 𝐼𝑛={𝑥𝑋𝑥𝑛𝑀}.(3.1) Then, 𝐼𝑛 is an ideal of 𝑋 for all 𝑛𝑁.

Proof. Let 𝑛𝑁, then(i)0𝑛=0𝑀 as 𝑀 is a submodule; therefore, 0𝐼𝑛.(ii)Let 𝑥,𝑦𝑋 such that 𝑥𝑦𝐼𝑛 and 𝑦𝐼𝑛. As 𝑋 is implicative, 𝑦𝑥=𝑦𝑁𝑥=𝑁𝑥𝑦 by Lemma 2.2 (i). Hence, (𝑦𝑥)𝑛=(𝑁𝑥𝑦)𝑛=𝑁𝑥(𝑦𝑛) which is in 𝑀 as 𝑦𝑛𝑀 and 𝑀 is a submodule, thus 𝑦𝑥𝐼𝑛. Therefore, 𝑥𝑦,𝑦𝑥,𝑦 are all in 𝐼𝑛. Thus (𝑥𝑦)𝑛,(𝑦𝑥)𝑛 and 𝑦𝑛 are all in 𝑀, hence as 𝑀 is a submodule, then (𝑥𝑦)𝑛+(𝑦𝑥)𝑛+𝑦𝑛𝑀. But from the axiom (v) of 𝑋-module, it follows that 𝑛=(𝑥𝑦)𝑛+(𝑦𝑥)𝑛+𝑦𝑛=(𝑥𝑦)(𝑦𝑥)𝑦𝑥𝑦𝑦𝑛byLemma2.2(iv)=𝑥𝑛since𝑦𝑦=0.(3.2)Thus, 𝑥𝑛𝑀, consequently 𝑥𝐼𝑛 as desired.
Whence 𝐼𝑛 is an ideal of 𝑋 as stated.

Remark 3.2. Given an 𝑋-module 𝑀 and 𝑚𝑀, then the set 𝑋𝑚={𝑥𝑚𝑥𝑋} is a submodule of 𝑀, the submodule generated by 𝑚.

Theorem 3.3 (Baer's Criterion). Let 𝑄 be an 𝑋- module.
Then, 𝑄 is weakly injective if and only if for every ideal 𝐼 of 𝑋, every 𝑋-module homomorphism from 𝐼𝑄 extends to a homomorphism from 𝑋𝑄.

Proof. ) This direction is obvious as 𝑋 is an 𝑋-module of type 2, and every ideal of 𝑋 is an 𝑋-module [Proposition 2.3].
) Assume that for every ideal 𝐼 of 𝑋, every 𝑋-module homomorphism from 𝐼𝑄 extends to a homomorphism from 𝑋𝑄. Consider 0𝑀𝑓𝑁 and 𝑀𝑔𝑄, where 𝑁 is type 2. Let X-Mod be the set of X-modules. Consider =(||𝐶,𝜙)𝐶𝑋-Mod,𝑀𝐶𝑁;𝜙𝐶𝑄;𝜙𝑀=𝑔.(3.3) First, note that since (𝑀,𝑔). Define on 𝑋 the relation by (𝐶1,𝜙1)(𝐶1,𝜙1) if 𝐶1𝐶2 and 𝜙2|𝐶1=𝜙1. Then, is easily verified to be an order on . The usual argument also show that every chain in (,) has an upper bound, and therefore, by the Zorn's lemma, (,) has a maximal element (𝐷,𝜑).
We show that 𝐷=𝑁, and therefore, 𝜑 would be the required extension of 𝑔.
By definition, we have 𝐷𝑁. Conversely, let 𝑛𝑁, then by Lemma 3.1, as 𝑁 is type 2, the set 𝐼𝑛={𝑥𝑋𝑥𝑛𝐷} is an ideal of 𝑋. Define, 𝛼𝐼𝑛𝑄 by 𝛼(𝑥)=𝜑(𝑥𝑛). Then 𝛼(𝑥𝑦)=𝜑((𝑥𝑦)𝑛)=𝜑(𝑥𝑛+𝑦𝑛)=𝜑(𝑥𝑛)+𝜑(𝑦𝑛)=𝛼(𝑥)+𝛼(𝑦). In addition 𝛼(𝑥𝑦)=𝛼(𝑥𝑦)=𝜑((𝑥𝑦)𝑛)=𝜑(𝑥(𝑦𝑛))=𝑥𝜑(𝑦𝑛)=𝑥𝛼(𝑦). Hence, 𝛼 is an 𝑋-module homomorphism, and by hypothesis 𝛼 extends to 𝛽𝑋𝑄.
Define 𝜑𝐷+𝑋𝑛𝑄 by 𝜑(𝑑+𝑥𝑛)=𝜑(𝑑)+𝛽(𝑥). We need to verify that 𝜑 is a well-defined homomorphism (which clearly extends 𝜑).
For the well-definition, suppose 𝑑1+𝑥1𝑛=𝑑2+𝑥2𝑛, then 𝑑1𝑑2=𝑥1𝑛+𝑥2𝑛=(𝑥1𝑥2)𝑛. So, (𝑥1𝑥2)𝑛𝐷, hence (𝑥1𝑥2)𝐼𝑛. Now using the fact that 𝜑 is a homomorphism, we obtain 𝜑𝑑1𝑑𝜑2𝑥=𝜑1𝑥2𝑛𝑥=𝛼1𝑥2𝑥=𝛼1𝑥+𝛼2𝑥=𝛽1𝑥+𝛽2.(3.4) Hence, 𝜑(𝑑1)+𝛽(𝑥1)=𝜑(𝑑1)+𝛽(𝑥1), because 𝛽(𝑥1)+𝛽(𝑥1)=0. Therefore, 𝜑(𝑑1+𝑥1𝑛)=𝜑(𝑑2+𝑥2𝑛) and 𝜑 is well defined. Next, we check that 𝜑 is a homomorphism.
Let 𝑑,𝑑𝐷 and 𝑥,𝑥𝑋, then 𝜑𝑑(𝑑+𝑥𝑛)++𝑥𝑛=𝜑𝑑+𝑑+𝑥𝑛+𝑥𝑛=𝜑𝑑+𝑑+𝑥𝑥𝑛=𝜑𝑑+𝑑+𝛽𝑥𝑥𝑑=𝜑(𝑑)+𝜑𝑥+𝛽(𝑥)+𝛽=𝜑(𝑑+𝑥𝑛)+𝜑𝑑+𝑥𝑛,𝜑(𝑥(𝑑+𝑥𝑛))=𝜑𝑥𝑑+𝑥(𝑥𝑛)=𝜑𝑥𝑑+𝑥𝑥𝑛𝑥=𝜑𝑑𝑥+𝛽𝑥=𝑥𝜑(𝑑)+𝑥𝛽(𝑥)=𝑥𝜑(𝑑+𝑥𝑛).(3.5) Thus, 𝜑 is a homomorphism as needed. Whence (𝐷,𝜑)(𝐷+𝑋𝑛,𝜑) and by the maximality of (𝐷,𝜑), we obtain 𝐷=𝐷+𝑋𝑛, so 𝑛𝐷 which shows that 𝐷=𝑁 as required.

The theory of modules over BCK-algebras displays some real pathologies as the remark below explains. Before the remark, a couple of definitions.

Definition 3.4. As defined in [6], an element 𝑥 of a BCK-algebra 𝑋 is called a zero-divisor if there exists a nonzero element 𝑦 in 𝑋 such that 𝑥𝑦=0. If 𝑋 has nontrivial zero-divisors, then 𝑋 is called cancellative. These correspond to domains in ring theory.

Remark 3.5. The natural approach for understanding injective modules over rings consists of establishing the relationship with divisible modules. Unfortunately, what should be the natural equivalent of divisible modules over BCK-algebras turns out to be useless. In fact, it is straightforward to see that the only cancellative implicative BCK-algebra is {0,1} so that every module over such is always divisible.

Remark 3.6. Recall [7, Theorem 3] that if 𝑋 is a BCK-algebra (not necessarily implicative) and 𝑎𝑋, the ideal of 𝑋 generated by 𝑎 is denoted by 𝑎is given by {𝑥𝑋𝑛>0;𝑥𝑎𝑛=0}. In the case when 𝑋 is implicative, this simplifies to 𝑎={𝑥𝑋𝑥𝑎=0}={𝑥𝑋𝑥𝑎}.

Definition 3.7. A BCK-algebra 𝑋 is principal if every ideal of 𝑋 is generated by one element.

Example 3.8. (1) 𝑋={0,1,2,,𝜔} as defined in [5, Example 1] is a principal BCK-algebra. In fact, it is easy to see that the only ideals of 𝑋 are 0,𝑋 and {0,1,2,}. Note that 𝑋 is not implicative.
(2) The BCK-algebra 𝐵423 from [8, Appendix] is bounded implicative and principal.
We now deduce from the above Baer's criterion that all modules over principal bounded implicative BCK-algebras are weakly injective.

Corollary 3.9. Let 𝑀 be an 𝑋-module and suppose that 𝑋 is bounded implicative and principal. Then, 𝑀 is weakly injective. In particular, every bounded implicative and principal BCK-algebra is weakly injective as a module over itself.

Proof. Let 𝑀 be an 𝑋-module, with 𝑋 bounded implicative and principal, 𝐼 an ideal of 𝑋, and 𝑓𝐼=𝑎𝑀 an 𝑋-homomorphism. Define 𝑋𝑀 by (𝑥)=𝑓(𝑎𝑥). It is clear that is an 𝑋-homomorphism; in fact, using Lemma 2.2 (v), we obtain 𝑥1𝑥2𝑥=𝑓𝑎1𝑥2=𝑓𝑎𝑥1𝑎𝑥2=𝑓𝑎𝑥1+𝑓𝑎𝑥2𝑥=1𝑥+2,(𝛾𝑥)=𝑓(𝑎(𝛾𝑥))=𝑓(𝛾(𝑎𝑥))=𝛾𝑓(𝑎𝑥)=𝛾(𝑥).(3.6) We claim that extends 𝑓. In fact, let 𝑥𝐼=𝑎. Then, 𝑥=𝑎𝑥. So, (𝑥)=𝑓(𝑎𝑥)=𝑓(𝑥). Hence, 𝑀 is weakly injective by Baer's criterion.

Example 3.10. Consider 𝐵423 as above which is a bounded implicative and principal BCK-algebra. By Corollary 3.9, 𝐵423 is weakly injective as a module over itself.

4. Examples

This section is devoted to constructing examples.

Example 4.1 (A non weakly injective BCK-Module). Consider 𝑆 any infinite set and the BCK-algebra 𝕏=𝒫(𝑆) with the natural operations. Consider ||𝑋||𝐼=𝑋𝑆<.(4.1) Then, 𝐼 is clearly an ideal of 𝕏 and, therefore, an 𝕏-module by Proposition 2.3.
Claim 4.2. 𝐼 is not weakly injective.
To see this, it is enough to produce an 𝕏-module homomorphism 𝜑𝐼𝐼 that does not extend to 𝕏.
For this, consider any finite complement subset 𝐴 of 𝑆 and 𝜑𝐼𝐼 defined by 𝜑(𝑋)=𝑋𝐴. Since distributes over and is associative, it follows that 𝜑 is an 𝕏-module homomorphism.

We assert that there is no homomorphism 𝜑𝕏𝐼 such that 𝜑|𝐼=𝜑. In fact, by contradiction, suppose there is such an extension. Then, for every 𝑋𝑆, we have 𝑋=(𝑋𝐴)(𝑋𝐴𝐶); therefore, since 𝜑 is a homomorphism, then 𝜑(𝑋)=𝜑(𝑋𝐴)𝜑𝑋𝐴𝐶=𝑋𝜑(𝐴)𝑋𝜑𝐴𝐶=𝑋𝜑(𝐴)𝜑𝐴𝐶=𝑋𝜑(𝐴)𝜑𝐴𝐶𝐴=𝜑𝐶.=(4.2) Let 𝐵=𝜑(𝐴), then since 𝐴𝐴𝐶= and 𝜑 is a homomorphism, then 𝐴𝐶𝐵=, so 𝐵𝐴. Note that 𝐵𝐴, because if 𝐵=𝐴, then 𝜑(𝐴)=𝐴𝐴=𝐴 and 𝐴𝐼. Therefore, there exists an element 𝑎 of 𝐴 that is not in 𝐵. We have 𝜑({𝑎})=𝜑({𝑎}), that is {𝑎}𝐴={𝑎}𝐵 which is a contradiction.

Whence, 𝐼 is an 𝕏-module that is not weakly injective (much less injective).

Example 4.3 (A weakly injective BCK-module that is not injective). 𝐵211 denotes the unique BCK-algebra with two elements: 0,1.
First, observe that every Abelian group (𝑀,+) has a natural structure of 𝐵211-module via 0𝑚=0 and 1𝑚=𝑚 for all 𝑚𝑀. We consider and in this view as 𝐵211-modules. Consider the inclusion and 𝑓𝐵211 defined by 𝑓(2𝑘)=0 and 𝑓(2𝑘+1)=1. Then 𝑓 is a 𝐵211-module homomorphism. If 𝐵211 was injective, then, there would exist a homomorphism 𝑓𝐵211 such that 𝑓(𝑚)=𝑓(𝑚) for all 𝑚. But such an extension would satisfy 1=𝑓(1)=𝑓(1)=𝑓(1/2)+𝑓(1/2)=0, which is a contradiction. Therefore, as a module over itself, 𝐵211 is not injective.
On the other hand, 𝐵211 is clearly weakly injective, as it has only two ideals making Baer's criterion obvious. Or more directly, use Corollary 3.9, since 𝐵211 is principal.

Remark 4.4. The existence of weakly injective modules that are not injective (see Example 4.3), shows that Baer's criterion does not characterize injective BCK-modules. That is, being able to extend homomorphisms from ideals to the whole BCK-algebra is weaker than being injective. Note also that the first example shows that not every type 2 module is weakly injective. Finally, the proof of Baer's criterion clearly works in the subcategory of 𝑋-modules of type 2 so that injective objects in this category are characterized by the criterion.

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