EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6436 ISSN 1307-5543 – ejpam.com Published by New York Business Global On Po-injective and Po-surjective Wreath Product of Pomonoids Bana Al Subaiei1,∗, Ahlam Almulhim1, Aftab Hussain Shah2, Syed Ahtisham Ul Haq2 1 Department of Mathematics and Statistics, King Faisal University, Al-Ahsa, Saudi Arabia 2 Department of Mathematics, Central University of Kashmir, Ganderbal, India Abstract. Let R and S be pomonoids and RA be a left R-poset. The wreath product of the pomonoids R and S by RA is defined as the pomonoid T = R×F (A,S) while, the wreath product TC of the left R-poset RA with the left S-poset SB over the pomonoid T = R × F (A,S) is the left T -poset TC = RA× SB endowed with the monotone action given by (r, f)(a, b) = (ra, f(a)b), where (r, f) ∈ R×F (A,S) and (a, b) ∈ A×B. The po-injectivity and po-cancellative properties on the wreath product TC are studied and the relations between them are established. The relation between po-surjective property and other properties on the wreath product TC are also established. Finally the characterization of some properties of po-flatness such as po-torsion free, properties (P ), (E), (PE), and strongly flat have been examined on the wreath product TC and the relations among them have also been established. 2020 Mathematics Subject Classifications: 20-XX, 20M15, 06F05, 20M30 Key Words and Phrases: Wreath product, po-injective, po-surjective 1. Introduction In group theory the wreath product is a generalization of the semidirect product. The wreath product is a way to combine two groups, H and K, using the semidirect product. The key feature is that one group, say H acts on K in a specific way, and this action is a crucial part of the construction. The idea of wreath products has been extended to semigroups and posemigroups as well, allowing for a broader application of this construc- tion beyond just groups. The wreath product of semigroups is a generalization of the concept for groups, but with some modifications to accommodate the lack of inverses in semigroups. The construction involves not only the direct product of copies but also an action that reflects the interactions between the two semigroups. As with group wreath ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6436 Email addresses: banajawid@kfu.edu.sa (B. Al Subaiei), ahmulhem@kfu.edu.sa (A. Almulhim), aftab@cukashmir.ac.in (A. H. Shah), ahtishamulhaq1218@gmail.com (S. A. Ul Haq) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 2 of 14 products, the wreath product of semigroups allows for a structured approach to under- standing and constructing certain types of semigroups. It finds applications in areas like automata theory and formal languages, where semigroups play a significant role. Many researchers studied the concept of wreath product of semigroups (monoids) in numerous articles such as [1–5]. Many researchers are interested in generalizing results from the category of semigroups to that of posemigroups, as demonstrated by the work on [6–13]. Knauer and Mikhalev [14], initiated the study of wreath products of ordered semigroups by an ordered action. They considered three different types of ordered wreath products, specifically, order pre- serving, order reversing and preserving certain zigzag equivalence. Kilp, Knauer and Mikhalev [15] provided a characterization of torsion free wreath products of acts over the wreath product of monoids by describing injective, surjective and cancellative elements of the wreath product. In this paper, we extend the work on the ordered (monotone) wreath product of pomonoids over a poset by generalizing the work of Kilp, Knauer and Mikhalev on monoids in [15], and we will adopt their notations for the sake of simplicity for the reader. For the initial work on ordered wreath product of posets on pomonoids the reader is refered to [14, 16, 17]. A pomonoid S is a monoid endowed with a partial order usually denoted by ≤ such that it is compatible with the binary operation i.e., for any s, s′, t ∈ S, s ≤ s′ implies ts ≤ ts′ and st ≤ s′t. Therefore, for any pomonoid S and r, r′, p, p′ ∈ S, if r ≤ r′ and p ≤ p′, then rp ≤ r′p′. Let A and B be posets, a map f : A −→ B is called monotone if it preserves the order i.e., a ≤ a′ in A implies f(a) ≤ f(a′) in B. The set of all monotone mappings from A to B (resp. from A to A ) is usually denoted by F (A,B) (resp. F (A,A)) and it inherits a point-wise order as follows: f ≤ g ⇔ f(a) ≤ g(a), ∀ a ∈ A. For each b ∈ B, cb denotes the constant map on A with range {b} and such a map is clearly monotone. Let R be a pomoniod and A a poset. We say that A is a left R-poset if there exists a monotone map R × A −→ A such that 1a = a and (rp)a = r(pa) for all a ∈ A and r, p ∈ R. We denote a left R-poset by RA. The right R-poset is defined dually and it is denoted by AR. Let R and S be pomonoids. The wreath product of the pomonoids R and S by the left R-poset RA is the set T = R× F (A,S) endowed with the multiplication given by (r, f)(p, g) = (rp, fpg), where fpg(a) = f(pa)g(a) for all a ∈ A, r, p ∈ R, f, g ∈ F (A,S). The map fp means that fp(a) = f(pa). By [14] Proposition 2.1 the wreath product T is a semigroup, and it is a monoid if and only if S is a monoid. Again by [14] Proposition 2.4, T is a posemigroup with component- wise order. Knauer and Mikhalev [14] studied the ordered wreath product where they assumed F (A,S) to contain all montone (isotone), antimonotone (antitone), and zigzag B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 3 of 14 preserving mappings. In this paper, we only consider the monotone (isotone) case and thus under the notations introduced in [14], F (A,S) is precisely the set I(A,S). The wreath product TC of the left R-poset RA with the left S-poset SB over the pomonoid T = R×F (A,S) is the left T -poset TC = RA×SB endowed with the monotone action given by (r, f)(a, b) = (ra, f(a)b), for all (r, f) ∈ R× F (A,S) and (a, b) ∈ A×B. For simplicity, we will refer to the wreath product T = R× F (A,S) simply as T , and the wreath product T -poset TC = RA×SB as TC throughout the paper when the context is clear. In this paper, we first find the necessary conditions for an element of a pomonoid R to act po-injectively and to be po-cancellable respectively on the wreath product TC. We also establish necessary conditions on the wreath product T to act po-injectively and to be po- cancellable on the left T -poset TC. Finally we examine some of the well known properties of S−posets either on TC or on T . This particularly includes the properties such as po-surjectivity, po-torsion free, property (P ), property (E), property (PE), strongly flat, reversible, and solvable. 2. Po-injective action and left po-cancellability In this section we investigate the po-injective and po-cancellable properties on the wreath product of pomonoids and the relation between these properties. Let RA be a left R-poset and let r be any element of R. We say that r acts po- injectively on A if ra ≤ ra′ implies a ≤ a′, where a, a′ ∈ A. If this property holds for every r ∈ R, then it can be said that R acts po-injectively on A. The strong version of R being acting po-injectively on A is that of R being acting strongly po-injectively on A, which is defined as: R acts strongly po-injectively on A if for all r, r′ ∈ R, a, a′ ∈ A, if ra ≤ r′a′ and r ≤ r′ then a ≤ a′. An element r ∈ R is called left po-cancellable if for all t, t′ ∈ R, rt ≤ rt′ implies t ≤ t′. A pomoniod R is called left po-cancellative if every element in R is left po-cancellable. A pomonoid R is called strongly left po-cancellative if for all r, r′, t, t′ ∈ R, r ≤ r′ and rt ≤ r′t′ implies t ≤ t′. The right po-cancellable, right po-cancellative, and strongly right po-cancellative are defined dually. Let X be a left R-poset and Y a poset. Let T (X × Y ) be a left T -poset where T = R× F (A,S) is the wreath product defined above, and the action is defined through some monotone mapping α : F (A,S) × X × Y −→ Y such that (r, f)(x, y) = (rx, α(f, x, y)). This is equivalent to say that α satisfies the identities: (i) α(c1, x, y) = y. (ii) α(f, px, α(g, x, y)) = α(fpg, x, y), B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 4 of 14 for all x ∈ X, y ∈ Y, c1, f, g ∈ F (A,S) and p ∈ R. First we give necessary and sufficient condition for an element of T to act po-injectively on TX × Y Proposition 1. The element (r, f) ∈ T acts po-injectively on T (X × Y ) through α : F (A,S)×X × Y → Y if and only if the following conditions are satisfied. 1) If x ≤ x′ and α(f, x, y) ≤ α(f, x′, y′) then y ≤ y′, where x, x′ ∈ X and y, y′ ∈ Y . 2) If rx ≤ rx′ and x ≰ x′ then for all y, y′ ∈ Y , α(f, x, y) ≰ α(f, x′, y′) where x, x′ ∈ X and r ∈ R. Proof. Let (r, f) ∈ T acts po-injectively on T (X × Y ) through α. 1) First suppose that x ≤ x′ in X and α(f, x, y) ≤ α(f, x′, y′) where y, y′ ∈ Y . Hence, for all (r, f) ∈ T (r, f)(x, y) = (rx, α(f, x, y)) ≤ (rx′, α(f, x′, y′)) = (r, f)(x′, y′). Since (r, f) acts po-injectively on T (X × Y ), it follows that (x, y) ≤ (x′, y′) and therefore, y ≤ y′ as required. 2) Suppose now that rx ≤ rx′ and x ≰ x′, where r ∈ R and x, x′ ∈ X. Also, suppose that α(f, x, y) ≤ α(f, x′, y′) for some y, y′ ∈ Y . Hence, (r, f)(x, y) = (rx, α(f, x, y)) ≤ (rx′, α(f, x′, y′)) = (r, f)(x′, y′). Again since (r, f) acts po-injectively on T (X × Y ), we must have (x, y) ≤ (x′, y′). Thus, x ≤ x′ and this contradicts the assumption x ≰ x′. Therefore, for all y, y′ ∈ Y we have α(f, x, y) ≰ α(f, x′, y′). For the other direction, assume the two conditions are satisfied. Suppose that (r, f)(x, y) ≤ (r, f)(x′, y′). This implies (rx, α(f, x, y)) ≤ (rx′, α(f, x′, y′)). Therefore, rx ≤ rx′ and α(f, x, y) ≤ α(f, x′, y′). The condition (2) implies x ≤ x′ and so from condition (1) we get y ≤ y′. Therefore, (x, y) ≤ (x′, y′). Hence (r, f) acts po-injectively on T , as required. In particular, by taking X = RA, Y = SB and defining α : F (A,S)×A× S −→ B by α(f, a, b) = f(a)b for all a ∈ A, b ∈ B, in Proposition 1 we get Theorem 1. However for the sake of clarity we have given its proof. Theorem 1. The element (r, f) ∈ T acts po-injectively on TC if and only if: 1) if a ≤ a′ and f(a)b ≤ f(a′)b′ where a, a′ ∈ A and b, b′ ∈ B, then b ≤ b′, and 2) if ra ≤ ra′ and a ≰ a′ where a, a′ ∈ A, then f(a)b ≰ f(a′)b′ for all b, b′ ∈ B. Proof. Let (r, f) ∈ T acts po-injectively on TC. 1) First suppose that a ≤ a′ in A and f(a)b ≤ f(a′)b where a, a′ ∈ A and b, b′ ∈ B. Then, (r, f)(a, b) = (ra, f(a)b) ≤ (ra′, f(a′)b′) = (r, f)(a′, b′). Since (r, f) acts po-injectively on TC =R A ×S B, we must have (a, b) ≤ (a′, b′) and so b ≤ b′, as required. 2) Now suppose that ra ≤ ra′ and a ≰ a′, where r ∈ R and a, a′ ∈ A. Also let f(a)b ≤ f(a′)b′ for some b, b′ ∈ B. Then, B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 5 of 14 (r, f)(a, b) = (ra, f(a)b) ≤ (ra′, f(a′)b′) = (r, f)(a′, b′). Since (r, f) acts po-injectively on TC we must have (a, b) ≤ (a′, b′). Thus, a ≤ a′ and this is a contradiction to the assumption in (2). Therefore, f(a)b ≰ f(a′)b′ for all b, b′ ∈ B. For the other direction, assume the given conditions (1) and (2) are satisfied. Let (r, f) ∈ T and suppose that (r, f)(a, b) ≤ (r, f)(a′, b′). So (ra, f(a)b) ≤ (ra′, f(a′)b′) which forces ra ≤ ra′ and f(a)b ≤ f(a′)b′. From condition (2) we have a ≤ a′ and this together with condition (1) gives b ≤ b′. Therefore (a, b) ≤ (a′, b′). Then, (r, f) acts po-injectively on TC, as required. In the unordered case, it has been shown in [15] that the analogue of condition (1) in Theorem 1 above is that f(a) acts injectively on B. However, in ordered case condition (1) does not imply that f(a) acts po-injectively on B and conversely it is not enough to have f(a) acting po-injectively or even strongly po-injectively on B to deduce condition (1). Therefore it will be interesting to find that under what conditions the result in ordered case is similar to that of unordered case. However we do have the following. Corollary 1. If (r, f) ∈ T acts po-injectively on TC then 1) f(a) acts po-injectively on B for any a ∈ A, and 2) if ra ≤ ra′ and a ≰ a′, then for any b, b′ ∈ B, f(a)b ≰ f(a′)b′. The proof of the following is similar to Theorem 1, however we add the proof for completeness. Proposition 2. The element (r, f) ∈ T is left po-cancellable if and only if the following two conditions hold. 1) p ≤ p′ and fpg ≤ fp′g ′ implies g ≤ g′, where p, p′ ∈ R and g, g′ ∈ F (A,S). 2) rp ≤ rp′ and p ≰ p′ implies fpg ≰ fp′g ′ for any g, g′ ∈ F (A,S). Proof. Let (r, f) ∈ T be left po-cancellable. 1) Suppose that p ≤ p′ in R and fpg ≤ fp′g ′ where g, g′ ∈ F (A,S). Therefore, (r, f)(p, g) = (rp, fpg) ≤ (rp′, fp′g ′) = (r, f)(p′, g′). Since (r, f) is left po-cancellable, we must have (p, g) ≤ (p′, g′). Thus g ≤ g′ as required. 2) Now assume that r, p, p′ ∈ R, rp ≤ rp′ and p ≰ p′. Also suppose that fpg ≤ fp′g ′ for some g, g′ ∈ F (A,S). Therefore, (r, f)(p, g) = (rp, fpg) ≤ (rp′, fp′g ′) = (r, f)(p′, g′). Since (r, f) is left po-cancellable, we have (p, g) ≤ (p′, g′). Thus, p ≤ p′ and we arrive at a contradiction. Therefore, fpg ≰ fp′g ′ for any g, g′ ∈ F (A,S) as required. For the other direction, assume the given conditions are satisfied. Suppose that (r, f)(p, g) ≤ (r, f)(p′, g′). So, (rp, fpg) ≤ (rp′, fp′g ′). Thus rp ≤ rp′ and fpg ≤ fp′g ′. From condition (2) p ≤ p′ and condition (1) implies that g ≤ g′. Therefore (p, g) ≤ (p′, g′). Hence (r, f) is left po-cancellable, as required. Recall from [16] that a free posemigroup F is a free semigroup F with order defined as: B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 6 of 14 a1a2 . . . an ≤ b1b2 . . . bt ⇔ n = t and ai ≤ bi where 1 ≤ i ≤ n. In Example 1.4 of [15] it has been shown that there exist left cancellable elements in the monoid T = R × F (A,S) for which the first components are not left cancellable in R. In the next example we show that the same example in [15] can be transformed in the ordered case by replacing the free semigroup by the free posemigroup, by defining suitable orders and carefully choosing the required monotone maps. Thus there exists a left po-cancellable element (r, f) ∈ T but r is not left po-cancellable in R. Example 1. Let A = {a, b} where a ≤ b and S =< u, v > ∪{1}, where u ≤ v, be the free pomonoid generated by u, v. Let P (A) = {ca, cb, 1} be the pomonoid which is also an A-poset under the action given by ca.x = a, cb.x = b for all x ∈ A. Consider T = P (A) × F (A,S). Let f ∈ F (A,S) such that f(a) = u and f(b) = v. We show that (ca, f) ∈ T is left po-cancellable while ca is not left po-cancellable, as clearly cb ≰ ca as b ≰ a while cacb ≤ caca. Now suppose that (ca, f)(h1, g1) ≤ (ca, f)(h2, g2) where (h1, g1), (h2, g2) ∈ T . Then, (cah1, fh1g1) ≤ (cah2, fh2g2). So for any a ∈ A, fh1g1(a) = f(h1(a))g1(a) ≤ f(h2(a))g2(a) = fh2g2(a). From the definition of free posemigroup and since the image of f is one letter word we get that h1 ≤ h2 and g1 ≤ g2. So (h1, g1) ≤ (h2, g2) and thus (ca, f) is left po-cancellable. Next we discuss the po-injectivity of the wreath product pomonoid T on the wreath product T -poset TC constructed above. Theorem 2. The pomonoid T acts po-injectively on TC if and only if the following con- ditions are satisfied. 1) a ≤ a′ and f(a)b ≤ f(a′)b′ implies b ≤ b′ where f ∈ F (A,S), a, a′ ∈ A, and b, b′ ∈ B. 2) R acts po-injectively on A. Proof. Let T acts po-injectively on TC. 1) Suppose that a ≤ a′ and f(a)b ≤ f(a′)b′, where a, a′ ∈ A and b, b′ ∈ B. By condition (1) of Theorem 1, we get that b ≤ b′. 2) Next suppose that ra ≤ ra′ where r ∈ R and a, a′ ∈ A. Assume that a ≰ a′. Using condition (2) in Theorem 1 we have ca(a)b ≰ ca′(a ′)b′ for all b, b′ ∈ B and this is a contradiction to T acts po-injectively on TC. Therefore, a ≤ a′ and so R acts po-injectively on A. Conversely assume that the two conditions are satisfied. Take any (r, f) ∈ T and (a, b), (a′, b′) ∈ T (A × B) and suppose that (r, f)(a, b) ≤ (r, f)(a′, b′). Therefore (ra, f(a)b) ≤ (ra′, f(a′)b′) so ra ≤ ra′ and f(a)b ≤ f(a′)b′. From condition (2) we have a ≤ a′. Since a ≤ a′ and f(a)b ≤ f(a′)b′, from condition (1) we get b ≤ b′. Hence (a, b) ≤ (a′, b′) and so T acts po-injectively on TC, as required. Corollary 2. The pomonoid T acts po-injectively on TC if and only if R acts po-injectively on A and S acts po-injectively on B. Proof. Suppose that T acts po-injectively on TC. By condition (2) of Theorem 2, R acts po-injectively on A. Take any s ∈ S and b, b′ ∈ B such that sb ≤ sb′. Clearly B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 7 of 14 (ra, sb) ≤ (ra, sb′) and so (r, cs)(a, b) ≤ (r, cs)(a, b ′). Since T acts po-injectively on TC, it follows that (a, b) ≤ (a, b′). Thus b ≤ b′ and so S acts po-injectively on B as required. Conversely if R acts po-injectively on A and S acts po-injectively on B then conditions (1) and (2) of Theorem 2 are clearly satisfied and hence T acts po-injectively on TC. Proposition 3. If (r, f) ∈ T is left po-cancellable then for all a ∈ A s, s′ ∈ S and p, p′ ∈ R, rp ≤ rp′ and p ≰ p′ imply that f(pa)s ≰ f(p′a)s′ . Proof. Let (r, f) ∈ T be left po-cancellable. Assume that for all r, p, p′ ∈ R, rp ≤ rp′ and p ≰ p′ and there exist some s, s′ ∈ S such that f(pa)s ≤ f(p′a)s′ for all a ∈ A . Therefore fpcs ≤ fp′cs′ and so, (r, f)(p, cs) = (rp, fpcs) ≤ (rp′, fp′cs′) = (r, f)(p′, cs′). Since (r, f) is left po-cancellable it follows that (p, cs) ≤ (p′, cs′). Thus p ≤ p′ and this is a contradiction. Hence f(pa)s ≰ f(p′a)s′ for all s, s′ ∈ S and all a ∈ A. Proposition 4. The element (r, f) ∈ T is left po-cancellable if the following two conditions hold: 1) If a ≤ a′ and f(a)s ≤ f(a′)s′ then s ≤ s′. 2) If rp ≤ rp′ and p ≰ p′, then for all a ∈ A, s, s′ ∈ S, f(pa)s ≰ f(p′a)s′ where p, p′ ∈ R. Proof. Assume the two conditions are satisfied and let (r, f)(p, g1) ≤ (r, f)(p′, g2). Therefore, (rp, fpg1) ≤ (rp′, fp′g2) and so rp ≤ rp′ and fpg1 ≤ fp′g2. If p ≰ p′ then from condition (2) it follows that f(pa)s ≰ f(p′a)s′ for all a ∈ A and s, s′ ∈ S. Since g1(a), g2(a) ∈ S, the condition f(pa)s ≰ f(p′a)s′ contradictions the assumption fpg1 ≤ fp′g2 and so we must have p ≤ p′. Also, since pa ≤ p′a and fpg1(a) = f(pa)g1(a) ≤ f(p′a)g2(a) = fp′g2(a) for all a ∈ A so from condition (1) we get that g1 ≤ g2. Hence (r, f) is left po-cancellable, as required. Theorem 3. The wreath product T is left po-cancellative if and only if R is left po- cancellable and S is strongly left po-cancellative. Proof. Suppose that T is left po-cancellative. Therefore for all (r, f), (p, g) and (p′, g′) ∈ T the inequality (r, f)(p, g) ≤ (r, f)(p′, g′) implies that rp ≤ rp′ and fpg ≤ fp′g ′. By Proposition 3 we have p ≤ p′, proving that R is left po-cancellative. Let s, s′, t, t′ ∈ S such that s ≤ s′ and st ≤ s′t′. By taking fp = cs, fp′ = cs′ , g = ct and g′ = ct′ and using condition (1) of Propostion 2, we have t ≤ t′, proving that S is strongly left po-cancellative. Conversely assume that R is left po-cancellative and S is strongly left po-cancellative. Let (r, f), (p, g) and (p′, g′) ∈ T = R × F (A,S) be such that (r, f)(p, g) ≤ (r, f)(p′, g′). This implies rp ≤ rp′ and fpg ≤ fp′g ′. Left po-cancellative property of R forces p ≤ p′. For all a ∈ A, pa ≤ p′a and since f is monotone it follows that f(pa) ≤ f(p′a). From fpg ≤ fp′g ′ B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 8 of 14 we have f(pa)g(a) ≤ f(p′a)g′(a) for all a ∈ A. Since S is strongly left po-cancellative g(a) ≤ g′(a) for all a ∈ A and so g ≤ g′. Hence (p, g) ≤ (p′, g′) as required. Theorem 3 is only true for left po-cancellative. For the right po-cancellative we have the following result whose proof is straightforward and so we omitted it. Proposition 5. If the element (r, f) ∈ T is right po-cancellable then r is right po- cancellable in R. Proposition 6. If r ∈ R is not left po-cancellable then (r, c1) ∈ T is not left po-cancellable. Proof. Assume that r ∈ R is not left po-cancellable and (r, c1) is left po-cancellable. Then there exist p, p′ ∈ R such that rp ≤ rp′ and p ≰ p′. From Proposition 3, for all a ∈ A, s, s′ ∈ S, c1(pa)s ≰ c1(p ′a)s′. Therefore, s ≰ s′ for all s, s′ ∈ S which is impossible as s ≤ s. Therefore, (r, c1) in T is not left po-cancellable. 3. Po-surjective action and Po-flatness properties of posets This section will be devoted for studying the po-surjective and po-flatness properties of posets on the wreath product of pomonoids. The relationships among these properties have also been obtained in this section. Definition 1. (i) An element r ∈ R acts po-surjectively on the left R-poset RA if for every a ∈ A there exists a′ ∈ A such that ra′ ≤ a. (ii) R′ ⊆ R acts po-surjectively on A, when every r′ ∈ R′ acts po-surjectively on RA. (iii) For any fixed r ∈ R and a ∈ RA we define the set ar := {x ∈ A : rx ≤ a}. (iv) Let SB be a left S−poset, then we say that SB ≤ B if for all b ∈ B there exist s ∈ S and b′ ∈ B such that sb′ ≤ b. Proposition 7. The element (r, f) ∈ T acts po-surjectively on TC if and only if f(ar)B ≤ B for all a ∈ A. Proof. Suppose (r, f) ∈ T acts po-surjectively on TC. Therefore for every (a, b) ∈ TC it can be found that (a′, b′) ∈ TC such that (r, f)(a′, b′) ≤ (a, b). This implies ra′ ≤ a and f(a′)b′ ≤ b. Thus a′ ∈ ar and we have f(ar)B ≤ B. Conversely, suppose f(ar)B ≤ B. Specifically, the assumption implies that ar ̸= ϕ. Take (a, b) ∈ TC, choose a′ ∈ ar and b′ ∈ B where f(a′)b′ ≤ b. Hence, (r, f)(a′, b′) = (ra′, f(a′)b′) ≤ (a, b). Hence (r, f) ∈ T acts po-surjectively on TC, as required. Proposition 8. The element (r, f) ∈ T acts po-surjectively on TC if and only if s0 acts po-surjectively on SB for some s0 ∈ f(ar) for all a ∈ A. B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 9 of 14 Proof. Let a ∈ A and b ∈ B. Since (r, f) ∈ T acts po-surjectively on TC, for any (a, b) ∈ TC it can be found that (a′, b′) ∈ TC such that (r, f)(a′, b′) ≤ (a, b). Then, ra′ ≤ a and f(a′)b′ ≤ b implying a′ ∈ ar. Suppose f(a′) = s0. Thus for any b ∈ B there exists b′ ∈ B such that s0b ′ ≤ b for some s0 = f(a′) ∈ f(ar). Hence some s0 ∈ f(ar) acts po-surjectively on SB. Now suppose s0 ∈ f(ar) acts po-surjectively on SB. Therefore for any b ∈ B it can be found that b′ ∈ B such that s0b ′ ≤ b. Clearly, ar ̸= ∅ as there exists some a′ ∈ ar such that f(a′) = s0. Now for any (a, b) ∈ TC = RA × SB there exists some (a′, b′) ∈ Ra r × SB ⊆ RA × SB such that (r, f)(a′, b′) = (ra′, f(a′)b′) = (ra′, s0b ′) ≤ (a, b). Hence, (r, f) ∈ T = R× F (A,S) acts po-surjectively on TC. Theorem 4. The pomonoid T acts po-surjectively on TC if and only if R acts po- surjectively on RA and S acts po-surjectively on SB. Proof. Take (r, cs) ∈ T for any r ∈ R and s ∈ S. Since T acts po-surjectively on TC so for every (a, b) ∈T C it can be found that (a′, b′) ∈ TC such that (r, cs)(a ′, b′) ≤ (a, b) implying (ra′, sb′) ≤ (a, b) which further implies ra′ ≤ a and sb′ ≤ b for arbitrary r ∈ R, s ∈ S, a ∈ A and b ∈ B. Thus, R acts po-surjectively on RA and S acts po-surjectively on SB. Now suppose R acts po-surjectively on RA and S acts po-surjectively on SB. We need to show that T acts po-surjectively on TC. Assume on contrary that T doesn’t act po- surjectively on TC, there exists (r, f) ∈ T such that (r, f)TC ≰ TC. Therefore there exists (a, b) ∈ TC such that for every (a′, b′) ∈ TC we have (r, f)(a′, b′) ≰ (a, b) implying that (ra′, f(a′)b′) ≰ (a, b), the following cases arise: Case 1: If ra′ ≰ a and f(a′)b′ ≰ b, a contradiction to both R and S acting po-surjectively on RA and SB. Case 2: If ra′ ≰ a and f(a′)b′ ≤ b, a contradiction to R acting po-surjectively on RA. Case 3: If ra′ ≤ a and f(a′)b′ ≰ b, a contradiction to S acting po-surjectively on SB. Hence T acts po-surjectively on TC. Definition 2. A subset P ⊆ S is referred to as simultaneously right po-cancellable in S if sp ≤ s′p for any p ∈ P and s, s′ ∈ S implies s ≤ s′. Proposition 9. If the element (r, f) ∈ T is right po-cancellable, then r acts po-surjectively on RA, r is right po-cancellable in R, and for any a ∈ A the set f(A) is simultaneously right po-cancellable in S. Proof. Suppose (r, f) ∈ T is right po-cancellable. Assuming r is not po-surjectively on RA, by way of contadiction. Therefore, we can found that a ∈ A such that for every a′ ∈ A, ra′ ≰ a. Clearly, a ⋂ rA = ∅, for if x ∈ a ⋂ rA then x = a and x = ra′′ for some a′′ ∈ A implying ra′′ = a for some a′′ ∈ A implying ra′′ ≤ a for some a′′ ∈ A which is not true. Let g, g′ ∈ F (A,S) where g(a) ≰ g′(a) and g|rA ≤ g′|rA. Hence, we have (grf)(a ′) = g(ra′)f(a′) ≤ g′(ra′)f(a′) = (g′rf)(a ′) implying grf ≤ g′rf , where a′ ∈ A. Then, (1, g)(r, f) = (r, grf) ≤ (r, g′rf) = (1, g′)(r, f). Since by our assumption (r, f) ∈ T B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 10 of 14 is right po-cancellable so (1, g) ≤ (1, g′) which implies g ≤ g′ so g(x) ≤ g′(x) for all x ∈ A and g(a) ≤ g′(a) which is a contradiction. Thus, r acts po-surjectively on RA. The remainder can be proved using a similar argument to Proposition 2.4 in [15], with respect to the order relation. Theorem 5. If the pomonoid T is right po-cancellative, then R acts po-surjectively on RA and, R and S are right po-cancellative. The proof follows by similar argument as Theorem 2.6 in [15] with respect to the order relation and by using Proposition 9 above. In [11, 13, 18, 19] many of po-flatness properties such as po-torsion free, property (E), and property (P) in the category of R-posets were considered. Definition 3. Let RA be a left R-poset. We say that: (i) RA is po-torsion free over R if every left po-cancellable element of R acts po-injectively on RA. (ii) RA satisfies property (P ) if ra ≤ r′a′ where r, r′ ∈ R and a, a′ ∈ A, then a = ua′′, a′ = u′a′′ with ru ≤ r′u′ for some a′′ ∈ A, u, u′ ∈ R. (iii) RA satisfies property (E) if ra ≤ r′a where r, r′ ∈ R and a ∈ A, then a = ua′′ with ru ≤ r′u for some a′′ ∈ A, u ∈ R. (iv) RA satisfies property (PE) if ra ≤ r′a′ where r, r′ ∈ R and a, a′ ∈ A, then a ≤ ua′′ and u′a′′ ≤ a′ with ru ≤ r′u′ for some a′′ ∈ A, u, u′ ∈ R. (v) RA is strongly flat if it satisfies properties (P ) and (E). Theorem 6. If the left T -poset TC is po-torsion free then RA and SB are po-torsion free. Proof. Assume that TC is po-torsion free. Suppose that r ∈ R is a left po-cancellable. We show that it acts po-injectively on A. Let ra ≤ ra′ where a, a′ ∈ A. We first show that (r, c1) is left po-cancellable by using Proposition 4. If a1 ≤ a2 and c1(a1)s ≤ c1(a2)s ′, then s ≤ s′, and thus the first condition holds. For the second condition, assume that rp ≤ rp′ and p ≰ p′. However, this is impossible as r is left po-cancellable. Thus (r, c1) is left po-cancellable. Since RA ×S B is po-torsion free, (r, c1) acts po-injectively on RA ×S B. Therefore for any b ∈ b, (r, c1)(a, b) = (ra, b) ≤ (ra′, b) = (r, c1)(a ′, b). So (a, b) ≤ (a′, b) and then a ≤ a′. Thus r acts po-injectively on A and hence RA is po-torsion free. Now to prove that SB is po-torsion free suppose that s ∈ S is left po-cancellable and sb ≤ sb′ for some b, b′ ∈ B. We again use Proposition 4 to show that (1, cs) is left po- cancellable. If a ≤ a′ and cs(a)s ≤ cs(a ′)s′, then ss ≤ ss′. As s is left po-cancellable, s ≤ s′. B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 11 of 14 So the first condition holds. Suppose that p ≤ p′ and p ≰ p′, but this is impossible, so the second condition holds. Thus (1, cs) is left po-cancellable, so (1, cs) acts po-injectively on RA×S B. Therefore for any b ∈ B, (1, cs)(a, b) = (a, sb) ≤ (a, sb′) = (1, cs)(a, b ′). So (a, b) ≤ (a, b′), and thus b ≤ b′. Thus s acts po-injectively on B and hence SB is po-torsion free as required. Recall that in unordered case, when TC = RA× SB is torsion free, then A and B are both torsion free. Therefore the above theorem generalises this result to the ordered case. Proposition 10. For the left T -poset TC the following are true. (i) If TC satisfies the property (P ) then so does RA. (ii) If TC satisfies the property (E) then so does RA. (iii) If TC is strongly flat then so is RA. (iv) If TC satisfies the property (PE) then so does RA. Proof. 1) Suppose that T -poset TC satisfies property (P ) and let ra ≤ r′a′, where r, r′ ∈ R and a, a′ ∈ A. Now for any b ∈ B, (r, c1)(a, b) = (ra, b) ≤ (r′a′, b) = (r′, c1)(a ′, b). Since TC satisfies property (P) there exists some (a′′, b′′) ∈ TC such that (a, b) = (r1, f1)(a ′′, b′′) = (r1a ′′, f1(a ′′)b′′) and (a′, b) = (r2, f2)(a ′′, b′′) = (r2a ′′, f2(a ′′)b′′) with (r, c1)(r1, f1) ≤ (r′, c1)(r2, f2). Therefore a = r1a ′′, a′ = r2a ′′ and (rr1, (c1)r1f1) ≤ (r′r2, (c1)r2f2). It follows that rr1 ≤ r′r2. Hence RA satisfies property (P). 2) It can be proved by an argument similar to case (1). 3) It follows from cases (1) and (2). 4) Suppose that TC satisfies property (PE) and let ra ≤ r′a′, where r, r′ ∈ R and a, a′ ∈ A. Then for any b ∈ B, (r, c1)(a, b) = (ra, b) ≤ (r′a′, b) = (r′, c1)(a ′, b). Since TC satisfies property (PE) there exists (a′′, b′′) ∈ TC such that (a, b) ≤ (r1, f1)(a ′′, b′′) = (r1a ′′, f1(a ′′)b′′) ⇒ a ≤ r1a ′′ and (r2a ′′, f2(a ′′)b′′) = (r2, f2)(a ′′, b′′) ≤ (a′, b) ⇒ r2a ′′ ≤ a′ B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 12 of 14 with (r, c1)(r1, f1) ≤ (r′, c1)(r2, f2). Thus, (rr1, (c1)r1f1) ≤ (r′r2, (c1)r2f2) and so rr1 ≤ r′r2. Hence RA satisfies property (PE) as required. The concepts of reversible and weakly reversible were considered in the literature, see for example [18, 20]. Next we study these concepts for the case of wreath product and obtain some crucial results. A pomonoid R is said to be left reversible if for every r, r′ ∈ R, rR ∩ r′R ̸= ∅. If Z is a subset of a poset Y , the down-set (Z] of Y is (Z] = {y ∈ Y |y ≤ z for some z ∈ Z}. A pomonoid R is called weakly left reversible if for every r, r′ ∈ R, rR ∩ (r′R] ̸= ∅. Theorem 7. The wreath product T is left reversible if and only if R and S are left reversible. Proof. Let r, r′ ∈ R. Suppose that T is left reversible. Then, (r, c1)T ∩ (r′, c1)T ̸= ∅. So there exist (r1, f1) and (r2, f2) ∈ T such that (r, c1)(r1, f1) = (r′, c1)(r2, f2) and so (rr1, (c1)r1f1) = (r′r2, (c1)r2f2). Therefore rr1 = r′r2 and thus rR ∩ r′R ̸= ∅. Hence R is left reversible. Let s, s′ ∈ S. As (r, cs)T ∩ (r, cs′)T ̸= ∅, there exist (r1, f1), (r2, f2) ∈ T such that (r, cs)(r1, f1) = (r, cs′)(r2, f2). Thus (rr1, (cs)r1f1) = (rr2, (cs′)r2f2) and so (cs)r1f1 = (cs′)r2f2. Take any a ∈ A, then cs(r1a)f1(a) = cs′(r2a)f2(a). This implies sf1(a) = s′f2(a). As f1(a), f2(a) ∈ S, sS ∩ s′S ̸= ∅. Hence S is left reversible. For the other direction, let (r, f), (p, g) ∈ T be arbitrary. Since R is left reversible, there exist r′, r′′ ∈ R such that rr′ = pr′′. Also, as S is left reversible and f(r′a), g(r′′a) ∈ S, there exist s′, s′′ ∈ S such that f(r′a)s′ = g(r′′a)s′′, so f(r′a)cs′(a) = g(r′′a)cs′′(a) for all a ∈ A, then fr′cs′(a) = gr′′cs′′(a) for all a ∈ A, which implies fr′cs′ = gr′′cs′′ . Now, (r, f)(r′, cs′) = (rr′, fr′cs′) = (pr′′, gr′′cs′′) = (p, g)(r′′, cs′′), so (r, f)T ∩ (p, g)T ̸= ∅. Hence, T is left reversible. Proposition 11. If T is weakly left reversible, then R and S are weakly left reversible. Proof. Let r, r′ ∈ R. As T is weakly left reversible, (r, c1)T ∩ ((r′, c1)T ] ̸= ∅. So, we can found (p, g) ∈ T where (p, g) = (r, c1)(r1, f1) and (p, g) ≤ (r′, c1)(r2, f2) for some (r1, f1), (r2, f2) ∈ T . Now (p, g) = (rr1, (c1)r1f1) and (p, g) ≤ (r′r2, (c1)r2f2). Therefore p = rr1 and p ≤ r′r2. Thus p ∈ rR ∩ (r′R], and so R is weakly left reversible. Next take any s, s′ ∈ S. Again as T is weakly left reversible, (1, cs)T ∩ ((1, cs′)T ] ̸= ∅. So there exists (q, h) ∈ T such that (q, h) = (1, cs)(p1, h1) and (q, h) ≤ (1, cs′)(p2, h2) for some (p1, h1), (p2, h2) ∈ T . Now (q, h) = (p1, (cs)p1h1) and (q, h) ≤ (p2, (cs′)p2h2). So h = (cs)p1h1 and h ≤ (cs′)p2h2. Take any a ∈ A, then h(a) = cs(p1a)h1(a) and h(a) ≤ cs′(p2a)h2(a). So h(a) = sh1(a) and h(a) ≤ s′h2(a). Thus h(a) ∈ sS ∩ (s′S], and so S is weakly left reversible, as required. B. Al Subaiei et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6436 13 of 14 Recall that a posemigroup S is termed left solvable if for any u, v ∈ S there exist s ∈ S such that su = v. However, if for any u, v ∈ S there exist a unique s ∈ S such that su = v then S is called left uniquely solvable. The right solvable and uniquely solvable is defined dually. As known that left (resp. right) uniquely solvable semigroup is called left (resp. right) group. Theorem 8. The posemigroup T is a left solvable if and only if R and S are both left solvable. Proof. Suppose that T is a left solvable. For all p, t ∈ R, we know that (p, g), (t, k) ∈ T . Then there exist (r, f) ∈ T such that (r, f)(p, g) = (t, k). Hence, (rp, fpg) = (t, k). Then, rp = t and so R is left solvable. Now, for any s, s′ ∈ S we know that (p, cs), (t, cs′) ∈ T . Hence, there exist (r, f) ∈ T such that (r, f)(p, cs) = (t, cs′). Hence, (rp, fpcs) = (t, cs′). So fpcs(a) = f(pa)cs(a) = f(pa)s = cs′(a) = s′. Therefore, S is left solvable. Now, suppose that R and S are both left solvable. Suppose that (p, g), (t, k) ∈ T . Since R is left solvable there exist r ∈ R such that rp = t. Since, g(a), k(a) ∈ S for any a ∈ A and since S is left solvable then there exist s ∈ S such that sg(a) = k(a). Hence, cs(a)g(a) = k(a). Then (cs)pg(a) = cs(pa)g(a) = cs(a)g(a) = k(a). 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