Set-theoretic solutions to the Yang-Baxter equation and generalized semi-braces

This paper aims to introduce a construction technique of set-theoretic solutions of the Yang-Baxter equation, called strong semilattice of solutions. This technique, inspired by the strong semilattice of semigroups, allows one to obtain new solutions. In particular, this method turns out to be useful to provide non-bijective solutions of finite order. It is well-known braces, skew braces and semi-braces are closely linked with solutions. Hence, we introduce a generalization of the algebraic structure of semi-braces based on this new construction technique of solutions.

where a − denotes the inverse a in (B, •). If (B, +) is a left cancellative semigroup, then we call (B, +, •) a left cancellative left semi-brace. This was the original definition by Catino, Colazzo and Stefanelli [7]. It has been shown that left semi-braces, under some mild assumption, provide set-theoretic solutions of the Yang-Baxter equation.

Moreover, the associated solution is left non-degenerate if and only if the left semi-brace is left cancellative.
Out of algebraic interest Brzeziński introduced left trusses [4] and left semi-trusses [3]. A quadruple (B, +, •, λ) is called a left semi-truss if both (B, +) and (B, •) are semigroups and λ : B × B → B is a map such that a • (b + c) = (a • b) + λ(a, c). Clearly, the class of left semi-trusses contains all left semi-braces, rings, associative algebras and distributive lattices. This entails that it will prove difficult to present deep results on this class. However, one may examine large subclasses. In particular, Brzeziński [4] focused on left semi-trusses with (B, +) a left cancellative semigroup and (B, •) a group, and showed that such a left semi-truss is equivalent with a left cancellative semi-brace, thus providing set-theoretic solutions of the Yang-Baxter, albeit known ones. In [39] Miccoli introduced almost left semi-braces, a particular instance of left semi-trusses, and constructed set-theoretic solutions associated with this algebraic structure. In [18], Colazzo and Van Antwerpen continue this study focusing on the subclass of brace-like left semi-trusses, i.e., left semi-trusses in which the multiplicative semigroup is a group and which includes almost left semi-braces. Concerning solutions, they showed that the solution one can associate with an almost left semi-brace is already the associated solution of a left semi-brace. In particular, this shows that brace-like left semi-trusses will not yield a universal algebraic structure that produces set-theoretic solutions.
In this paper, we focus on a new algebraic structure that includes left semi-braces and is an instance of left semitrusses, which is on a different path with respect to brace-like left semi-trusses, called generalized left semi-brace. A triple (S, +, •) is called a generalized left semi-brace if (S, +) is a semigroup, (S, •) is a completely regular semigroup (or union of groups), and such that, for any a, b, c ∈ S. it holds where a − denotes the (group) inverse of a in (B, •). We prove that under some mild assumptions, generalized left semibraces provide solutions. In particular, elementary examples of generalized left semi-braces produce cubic solutions that cannot be obtained by skew lattices and left semi-brace. Also, we introduce a construction technique that provides generalized left semi-braces called the strong semilattice of generalized left semi-braces. This technique is inspired by the description of semigroups which are unions of groups due to Clifford [16].
Furthermore, we introduce a construction technique for solutions called the strong semilattice of solutions. This technique takes a family of disjoint sets {X α | α ∈ Y } indexed by a semilattice Y and solutions defined on these sets, then, under some assumptions of compatibility, it allows one to construct a solution on the union of the sets X α .
We prove that the solutions provide by the strong semilattice of left semi-braces are a particular instance of strong semilattice of solutions.
Finally, we prove that the strong semilattice of solutions is a useful tool to provide solutions of finite order. Indeed the strong semilattice of solutions of finite order is a solution of finite order. Moreover, a solution r is of finite order if there exist a non-negative integer i and a positive integer p such that r p+i = r i and the minimal integers that satisfy such relation are said to be index and period respectively. We show that it is possible to determine the index and the period of the semilattice of solutions {r α | α ∈ Y } as a function of the indexes and periods of r α . As a corollary of this result, we prove that solutions associated with strong semilattices of left semi-braces are not bijective, so they are clearly different from solutions obtained by left semi-braces.

Basic tools on left semi-braces
Let us briefly present some basic background information regarding left semi-braces. Most of the content of this section appear in [31]. In particular, we provide different proof of [31,Corollary2.9] based on a result in semigroup theory due to Hickey [28] Throughout, 0 denotes the identity of the group (B, •). Moreover, we call (B, +) and (B, •) the additive semigroup and the multiplicative group of the left semi-brace (B, +, •), respectively. Furthermore, if the semigroup (B, +) has a pre-fix, pertaining to some property of the semigroup, we will also use this pre-fix with the left semi-brace. Hence, the left semi-braces introduced in [7], where one works under the restriction that the semigroup (B, +) is left cancellative, will be called left cancellative left semi-braces. Now, we recall that an element u of an arbitrary semigroup (S, +) is a middle unit of S if a + u + b = a + b, for all a, b ∈ S. Thus, u + u is idempotent but u itself need not be idempotent (see [17, p. 98]). This is not the case for the element 0 in the additive semigroup of a left semi-braces: the following proposition shows that 0 is an idempotent middle unit. Proposition 2. Let B be a left semi-brace. Then, the following hold: (1) 0 is a middle unit of (B, +); (2) 0 is an idempotent of (B, +); (4) B + 0 is a subgroup of (B, •);
(2) Since, by (1), 0 + 0 + 0 = 0 + 0, we have that Thus, since 0 is a middle unit, (4) By (2) it is clear that B + 0 is not empty. Moreover, by (1), if a, b ∈ B, it holds that To show the following theorem, let us recall that an arbitrary semigroup (S, +) is said to be a rectangular group if it is isomorphic to the direct product of a group and a rectangular band. For background and details on this topic we refer the reader to [17]. Proof. The thesis follows by [28,Corollary 3.5], which states that any completely simple semigroup with a middle unit is a rectangular group.
A special case in which the additive semigroup is completely simple is when 0 + B is a subgroup of (B, •) (see [31,Theorem 2.8]). For instance, this is the case when B is finite.
As a consequence of Theorem 3, we have that the additive semigroup (B, +) can be written as For the sake of completeness, let us introduce a further property of middle units of left semi-braces that holds without restrictions on the additive semigroup. At first, we recall that the additive semigroup of a left semi-brace B does not contain a zero element if B has at least two elements (see [31,Lemma 2.3]).
In the following, we prove that middle units of the additive structure of an arbitrary left semi-brace are idempotents. Proof. Since 0 is idempotent, we have that Since e is a middle unit, it follows that which is our assertion.
Then, M B is a subsemigroup of (B, +) that is a rectangular group. This is an interesting substructure of a left semibrace, which is beyond the purpose of this paper and shall be studied elsewhere.
Now, having as reference the [7, Example 2], we provide the following class of examples of completely simple left semi-braces that allow one to obtain solutions.
Example 5. Let (B, •) be a group with identity 0, f, g idempotent endomorphisms of (B, •) such that f g = gf . Let us consider the following operation for all a, b, c ∈ B. Thus, by [31, Proposition 2.14], the semigroup (B, +) is completely simple.
In addition, since ρ is an anti-homomorphism of the group (B, •), we obtain that the map r : r is explicitly given by Let us examine special cases of the previous class of examples. Firstly, observe that if f = id and g is not the constant map of value 0, then the semigroup (B, +) is not neither left nor right cancellative. Moreover, note that: for all a, b ∈ B, i.e., B coincides with the left cancellative left semi-brace provided in [7,Example 2]. Moreover, the solution associated to B is given by Case 2: if f = id, then for all a, b ∈ B. In this case, the semigroup (B, and so a = c. Moreover, it is easy to check that B is both right and left semi-brace. In addition, note that the solution associated to B is given by . i.e., every element is idempotent with respect to the sum. In this case, the semigroup (B, +) is a rectangular band where ker f = 0 + B and Im f = B + 0. Moreover, the solution r associated to B, given by is an idempotent solution, consistently with [31, Theorem 5.1] in the case in which the group G = {0}.
The following is a class of examples of completely simple left semi-braces such that, under suitable assumptions, give rise to solutions.
, the classical Zappa-Szép product of G and H (see [35]) that has identity (1, 1). Let ϕ be a map from G into H such that ϕ (1) = 1 and define the following operation on B is a sub-semigroup of (B, +) and it is also a left zero semigroup. Note also that (a, u) + (1, 1) = (a, uϕ (1) 1) = (a, u), i.e., (1, 1) is a right identity with respect to the sum. Furthermore, we have that One can check that ρ is an anti-homomorphism if and only if it holds holds, for all a, b ∈ G and u, v ∈ H. On the other hand, note that, if a, b ∈ G and u ∈ H, hence (1)

Definitions and examples
Braces, skew braces, and semi-braces were introduced to study set-theoretic solutions of the Yang-Baxter equation.
The following definition generalizes these structure to the case in which the multiplicative structure is no more a group.
At first, we recall that a semigroup (S, •) is completely regular if for any element a of S there exists a (unique) Conditions (3) imply that a 0 := a • a − = a − • a is an idempotent element of (S, •).
for all a, b, c ∈ S. We call (S, +) and (S, •) the additive semigroup and the multiplicative semigroup of S, respectively.
A generalized right semi-brace is defined similarly, replacing condition (4) by for all a, b, c ∈ S.
A generalized two-sided semi-brace is a generalized left semi-brace that is also a generalized right semi-brace with respect to the same pair of operations.
Let us note that if S is a generalized left semi-brace and a ∈ S, then the map is an endomorphism of the semigroup (S, +) and λ a•b (x) = (a • b) 0 + λ a λ b (x), for all a, b, x ∈ S. Indeed, if a, b, x, y ∈ S, we have that and Of course, left semi-braces [7,31] are examples of generalized left semi-braces. Moreover, a generalized left semi-brace can be obtained from every completly regular semigroup.  (1) φ α,α is the identical automorphism of S α , for every α ∈ Y (2) φ β,γ φ α,β = φ α,γ , for all α, β, γ ∈ Y such that α ≥ β ≥ γ.
Then, S = {S α | α ∈ Y } endowed by the addition and the multiplication defined by for any a ∈ S α and b ∈ S β , is a generalized left semi-brace. Such a generalized left semi-brace is said to be the strong semilattice Y of generalized left semi-brace S α and is denoted by S = [Y ; S α , φ α,β ].

Solutions related to generalized left semi-braces
This section is devoted to provide a sufficient condition to obtain solutions through a generalized left semi-brace.
To this end, we recall that if S is a left cancellatice left semi-brace, then the map r : S × S → S × S given by for all a, b ∈ S, is a solution. Moreover, [31, Theorem 5.1] gives a sufficient condition to obtain that the map in (8) is still a solution for a left semi-brace, not necessarily left cancellative. In addition, in [10, Theorem 3], we state a necessary and sufficient condition to ensure that r is a solution.
Specifically, if (S, +, •) is a left semi-brace, then, the map r : holds, for all a, b, c ∈ S.
Let us remark that if S is a generalized left semi-brace with (S, +) a right zero semigroup, then the map r as in (8) is a solution if and only if it holds (a • b) 0 = (a 0 • b) 0 , for all a, b ∈ S. Observe that semigroups (S, •) satisfying such a condition, lie in the wide class of right cryptogroups, see [41]. In this way, we get new idempotent solutions that are of the form r (a, b) = a 0 , a • b , different from those obtained in [36,38,19,47].
Moreover, note that if (S, +, •) is the generalized left semi-brace of Example 9, we obtain that r (a, b) = easy to verify that r is a cubic solution, i.e., r 3 = r.
Our aim is to show that if S = [Y ; S α , φ α,β ] is a strong semilattice of generalized left semi-braces such that every S α satisfies condition (9), then the map in (8) is a solution. This result is a consequence of a more general construction technique on solutions we introduce in the following theorem.
Theorem 12. Let Y be a (lower) semilattice, {X α | α ∈ Y } a family of disjoint sets indexed by Y , and r α a solution on X α , for every α ∈ Y . For each pair α, β of elements of Y such that α ≥ β, let φ α,β : X α → X β be a map. If the following conditions are satisfied for all x ∈ X α and y ∈ X β , is a solution on X. We call the pair (X, r) the strong semilattice Y of solutions (X α , r α ).
The proof of Theorem 12 is technical, and for the sake of clarity, we present it in the next section. Now, as a consequence of this theorem, we obtain the following result.
Theorem 13. Let S = [Y ; S α , φ α,β ] be a strong semilattice of generalized left semi-braces. Then, if S α satisfies (9), for every α ∈ Y , then the map r S : S × S → S × S defined by for all a, b ∈ S, is a solution.

Strong semilattices of set-theoretical solutions
This section aims to provide a proof of Theorem 12 and to give some examples of strong semilattices of solutions.
Furthermore, we focus on analyzing strong semilattices of solutions with finite order.
Strong semilattices of solutions (X α , r α ) allows one to produce examples of solutions with finite order if solutions r α are.
2. Let X := X α ∪ X β be a semilattice of sets such that α > β, c a fixed element of X β , and φ α,β (x) = c, for every x ∈ X α . Let r α be the twist map on X α and r β the idempotent solution on X β defined by r β (x, y) := (x, c), for all x, y ∈ X β . Then, if x, y ∈ X α , we have that hence the assumptions of Theorem 12 are satisfied. Moreover, the strong semilattice of solutions (X, r) is such that r 3 = r. Indeed, if x ∈ X α and y ∈ X β , since r (x, y) = r β (φ α,β (x) , y) = r β (c, y) = (c, c), it follows that r 2 (x, y) = r β (c, c) = (c, c) = r (x, y) .
Therefore, we obtain that r 3 = r.
3. Let X := X α ∪ X β be a semilattice of sets such that α > β, c a fixed element of X β , and φ α,β (x) = c, for every x ∈ X α . Let f be an idempotent map from X β into itself, f = id X β , and r α the map from X α × X α into itself defined by r α (x, y) := (f (x) , x), for all x, y ∈ X α . Thus, r α is a solution such that r 3 α = r 2 α . Let r β be the idempotent solution defined by r β (x, y) = (x, c), for all x, y ∈ X β . Then, if x, y ∈ X α , we obtain that thus the hypotheses of Theorem 12 are satisfied. Moreover, the strong semilattice of solutions (X, r) is such that r 3 = r 2 . Indeed, if x ∈ X α and y ∈ X β , since r (x, y) = r β (φ α,β (x) , y) = r β (c, y) = (c, c), it follows that r 2 (x, y) = r β (c, c) = (c, c) = r (x, y) and clearly r 3 (x, y) = r 2 (x, y). Therefore, r 3 = r 2 .
To investigate strong semilattices of solutions with finite order, we need the notions of the index and the period of a solution r that are i (r) := min j | j ∈ N 0 , ∃ l ∈ N r j = r l , j = l p (r) := min k | k ∈ N, r i (r)+k = r i (r) , respectively. These definitions of the index and the order are slightly different from the classical ones (cf. [30, p. 10]), but they are functional to distinguish bijective solutions from non-bijective ones. For more details, we refer the reader to [10].
Therefore, r n+1 = r and clearly the index of r is 1 by the assumption on i. Now, assume that i = 0 and that i = i (r γ ), for a certain γ ∈ Y . If r n = r h , for a positive integer h, in particular we have that r n γ = r h γ . Since n = p (r γ )q + i for a certain natural number q, it follows that r p (rγ )+i γ = r i γ = r p (rγ )q+i γ = r n γ = r h γ , hence i ≤ h and so i (r) = i. Now, we proceed to determine the period of r. If r m = r i (r) for a natural number m, then r m α = r i (r) α , for every α ∈ Y . Consequently, p (r α ) divides m − i (r), for every α ∈ Y , thus lcm {p (r α ) | α ∈ Y } divides m − i (r), i.e., n − i (r) divides m − i (r). Therefore, n − i (r) ≤ m − i (r) and hence p (r) = n − i (r). Conversely, suppose that the solution r is with finite order, set i := i (r), and p := p (r). If α ∈ Y , since φ α,α = id Xα×Xα we obtain that r p+i α = r i α . Therefore, r α is a solution with finite order. Clearly, we have that i (r α ) is less than i and p (r α ) divides p.
Let us note that if (X, r) is a strong semilattice of non-bijective solutions r α such that i (r α ) = i and p (r α ) = n, for every α ∈ Y , then r is still a solution of index i and period n, also in the case of an infinite semilattice Y . Indeed, one can prove this statement by similar computations used for the non-bijective case in the proof of Theorem 15.