Restrictions on local embeddability into finite semigroups

In this paper the concept of local embeddability into finite structures (being LEF) for the class of semigroups is expanded with investigations of non-LEF structures, a closely related generalising property of local wrapping of finite structures (being LWF) and inverse semigroups. The established results include a description of a family of non-LEF semigroups unifying the bicyclic monoid and Baumslag--Solitar groups and establishing that inverse LWF semigroups with finite number of idempotents are LEF.


Introduction
The idea of approximating infinite structures by finite ones has been approached in many ways through different properties such as residual finiteness (an algebraic approach based on homomorphisms), pseudofiniteness (a model theory approach) and soficity (a more topological approach).
In particular, the notion of local embeddability into the class of finite (LEF, for short) structures, which in a sense incorporates all of the aforementioned paths, was initially introduced in [3] for groups, expanded to general structures in [1] and recently examined specifically in the class of semigroups in [5].
The main purpose of this work is to continue this examination.We produce more examples of non-LEF behaviour, investigate a closely related property which we name local wrapping of the class of finite (LWF, for short) semigroups, and study connections between general LEF semigroups and semigroups locally embeddable into the class of finite inverse semigroups (iLEF, for short).
Our paper is organised as follows.In Section 1 we provide the definition of LEF and LWF structures, demonstrate that the former implies the latter, recall connection between LEF groups and seigroups and also several examples.Section 2 covers numerous specific restrictions of being LEF.In Section 3 we prove general properties of LWF structures and establish that the key examples of non-LEF semigroups are also non-LWF.Finally, the results of Section 4 specify the theory of LEF and LWF objects to the class of inverse semigroups, and in particular demonstrate that inverse LWF semigroups with finite number of idempotents are also LEF.

Definitions and initial results
We begin with a definition of the main structures studied in this paper.Definition 1.A (semi)group S is called locally embeddable into the class of finite (semi)groups (an LEF (semi)group for short) if for every finite subset H of S there exists a finite (semi)group F H and an injective function f H : H → F H such that for all x, y ∈ H with xy ∈ H we have (xy)f H = (xf H )(yf H ).
The following definition encapsulates reason why a given non-LEF semigroup may fail the requirements of Definition 1.
Definition 2. We say that a finite set H with partially defined multiplication which satisfies associativity for any valid inputs is non-embeddable if there does not exist a finite (semi)group F H and an injective function f H : H → F H such that for all x, y ∈ H with xy ∈ H we have (xy)f H = (xf H )(yf H ).
We also recall the following result connecting LEF groups and semigroups.Proposition 1. [5, Proposition 1.2] A group is an LEF semigroup if and only if it is an LEF group.
It has been shown previously that the free groups and semigroups are LEF, while the bicyclic monoid B = a, b | ab = 1 is not LEF and the set {1, a, b, ba} with the partial multiplication inherited from B is non-emdeddable (see [5, Examples 1.5 and 1.7] for the semigroup and monoid results).
We introduce an additional notion closely related to the LEF property, inspired by [3].Definition 3. We say that a (semi)group S is locally wrapped by the class of finite (semi)groups (an LWF (semi)group for short) if for every finite subset H of S there exist a finite (semi)group D H and a function d H : D H → S such that H ⊆ D H d H and for all x ′ , y ′ ∈ D H with x ′ d H , y ′ d H ∈ H we have Proposition 2. Let S be an LEF (semi)group.Then it is LWF.
Proof.If S is finite, the statement is evident.Assume that S is infinite.
Consider a finite subset H of S. Let K = H ∪ H 2 , s ∈ S \ K and consider the (semi)group F K and the function f K which satisfy the requirements of Definition 1 for K.We claim that the (semi)group D H = F K and the function d H defined by satisfy the requirements of Definition 3. Indeed, D H is finite and we have In the group case, the converse is also true, which we will demonstrate in the section on inverse semigroups.

Being non-LEF
While the previous examples of non-LEF semigroups all contain idempotents, there exist non-LEF semigroups without them.
Example 1.Consider the semigroup A = a, b | aab = a .We claim that it has no idempotents and it is not LEF.
Firstly note that every element of A can be expressed as b β0 ab β1 . . .ab βn a α , where β 0 , α, n ≥ 0 and β 1 , . . ., β n > 0 by reducing every aab to a.This expression is unique, as the rewriting system aab → a is noetherian (it reduces the length of the word, thus every rewriting chain is finite) and confluent (the rewriting rule cannot be applied to distinct intersecting subwords of a given word, which means that the result of the rewritings is the same regardless of their order).
Assume there is an idempotent e in A. It has the normal form as above, so we have equation b β0 ab β1 . . .ab βn a α b β0 ab β1 . . .ab βn a α = b β0 ab β1 . . .ab βn a α .
Continuing in the same manner, we must have α − β 0 − . . .− β n−1 + n > β n , otherwise there will be an irreducible suffix for e 2 different from the one for e.However, the value η = α − β 0 − . . .− β n−1 + n − β n is the difference between the number of a's and b's in e, which is preserved under the congruence generated by the relation a = aab.Since we have e 2 = e, it follows that 2η = η, i.e. η = 0.This is a contradiction, which allows us to conclude that the initial assumption is incorrect.Now we will prove that A is not LEF.Assume it is LEF.Consider the finite subset H = {a, b, ab, aba} of S and F H , f H satisfying the requirements of Definition 1 for H. Denote p = af H and q = bf H .We have Additionally, since F H is finite, there exist π, π ′ > 0 such that p π = p π+π ′ .By multiplying both sides by q π on the right we get pq = p π ′ .This means that pq commutes with p.However, (pq)p = (af H )(bf H )(af H ) = ((ab)f H )(af H ) = (aba)f H = af H = p = p(pq) by the injectivity of f H . Thus, we have another contradiction and A is not LEF.
The second part of the proof in the example above can be immediately generalised to the following statement.Proposition 3. Let S be a semigroup such that for some x, y ∈ S we have xxy = x and xyx = x, and let H = {x, y, xy, xyx}.Then H with multiplication inherited from S is non-embeddable, and, in particular, S is not LEF.
It has been proven that some of the natural semigroup transformations, such as adjoining a zero or taking direct products, preserve being LEF (see [5,Section 4]).The example below demonstrates that the same does not hold true for taking the power semigroup by utilizing the non-embeddable set described above.
is a free product of Z and Z 2 , which is residually finite, which means that it is LEF.
For the power semigroup P(S) consider the sets W a = {all the words in {a, b} obtained from a by the rewriting rule a → aab}, S a = {all of the elements of S presented by words in W a } and S b = {b}.We claim that S a S a S b = S a while S a S b S a = S a .
To see the former, note that S a S a S b ⊆ S a as for a 1 , a 2 obtained from a by using a → aab we have that a 1 a 2 b is also obtained from a by using this rule, as we can initially rewrite a into aab and then transform the first a into a 1 and the second a into a 2 .Additionally, any word obtained from a by using a → aab except for a has the form a 1 a 2 b since the first step is always a → aab, and a itself is presented by aabab ∈ S a S a S b .
To see the latter, consider the rewriting system abab → 1 and baba → 1.It is noetherian and locally confluent, which means that we can obtain a normal form for each word w in alphabet {a, b}.
Note that all the words in W A have form a α0 b β0 a α1 b β1 . . .a αn b βn where n ≥ 0, α 0 > β 0 and α i ≥ β i for 1 ≤ i ≤ n as a → aab preserves such form.Additionally, both abab → 1 and baba → 1 preserve such form.However, aba is irreducible and does not have such a form, which means that aba ∈ S a and allows us to conclude that S a S b S a = S a .This means that P(S) is not LEF by Proposition 3.
Naturally, there are more non-embeddable sets.
Proposition 5. Let S be an LEF semigroup, and x, y be elements of S such that yxxy = yx.Then either yxy = yx or (yxy)(yxx) = yx.
Proof.Consider the finite subset H = {x, y, yx, yxx, yxy, (yxy)(yxx)}.Let F H be a finite semigroup and f H : H → F H satisfying the requirements of Definition 1 for H. Denote p = xf H and q = yf H .
Let us demonstrate that (qpp) κ = qp κ+1 by induction on the power κ.The base.For κ = 1 the statement is immediate.The step.Assume the statement holds for k = κ 0 and consider κ = κ 0 + 1, κ 0 ≥ 1.We have In particular, it follows that As F H is finite, there exists κ, ρ > 0 such that By multiplying it by q κ on the right we get (qpp) ρ = qp.If ρ = 1, then another multiplication by q on the right gets us qp = qpq, which means yx = yxy by the injectivity of f H . Otherwise with the same multiplication we get (qpp) ρ−1 = qpq, meaning that qpq and qpp commute and (qpq)(qpp) = (qpp)(qpq) = qppq = qp, from which it follows that (yxy)(yxx) = yx.
The proposition above means that any set From the two previous propositions we can conclude the following.
Corollary 1.The semigroups T, A, B are not LEF.

Now that we know that
The following lemma is well-known, see for example [6, Proposition 1.1].
Lemma 1.Let F be a finite semigroup, s be an element of F and κ, ρ be such positive integers that s κ = s κ+ρ , κ is minimal possible and ρ is minimal possible for κ.Then for any κ ′ ≥ κ and Proposition 7. Let S be an LEF semigroup, x, y be elements of S such that Then there exists a power m such that m < n Let F H be a finite semigroup and f H : H → F H be a map satisfying the requirements of Definition 1 for H. Denote p = xf H and q = yf H .Note that by the multiplication properties and injectivity of f H we have (qp)(pq) n = (qp) n .It follows that for any α > 0 we have Let κ, ρ be such positive integers that (pq) κ = (pq) κ+ρ , κ is minimal possible and ρ is minimal possible for κ.Similarly, let λ, τ be such positive integers that (qp) λ = (qp) λ+τ , λ is minimal possible and τ is minimal possible for λ.
If we have ρ = 0(mod n), then set κ ′ to be n⌈ κ n ⌉.We have (qp) From this follows κ < n 2 + n and κ ′ < n 2 + n as well.Set γ to be the remainder of dividing κ + ρ by n.We have (qp)(pq) κ+n−γ = (qp)(pq) κ+n−γ+ρ = (qp) 1+ Set m ∈ N such that n ∤ m.We will prove that (ba)(ab) m (ba) = (ba) 2 (ab) m .Assume that we can get from the word w = (ba)(ab) m (ba) to the word w ′ = (ba) 2 (ab) m using a finite amount of rewritings Denote the chain of rewritings by w 1 = w, w 2 , . . ., w k = w ′ .Denote by γ the maximal power of ab which appears as a subword of any w i .
Let i 0 be the smallest among indices i such that w i ends with ab.It is possible only under rewriting ρ 2 , meaning that we have w i0 = g i0 (ba)(ab) n and w i0−1 = g i0 (ba) n where g i0 is some word in a, b.
Let i 1 be the smallest among indices i such that w i ends with (ba) n .Evidently i 1 ≤ i 0 − 1, so it is impossible to have w i1−1 = g i1 (ba)(ab) n and w i1 = g i1 (ba) n for some word g i1 in a, b.Additionally, we cannot obtain w i1 using ρ 2 as it cannot create subword (ba) n where there has not been one before.Thus, we must have w i1−1 = g i1 (ba)(ab) n (ba) j1 and w i1 = g i1 (ba) n+j1 for 1 ≤ j 1 < n.
Continuing in the same manner we can get words ending with (ba)(ab) αn (ba) j1 for any α > 0. Since m is not divisible by n, which means that there always will be a previous element and potential to increase the power in the middle.However, powers of ab in the chain are bound by γ, which gives us a contradiction.
From the two previous propositions and Corollary 1 we can infer the following.
Remark 1.It can be shown that the well-known Baumslag-Solitar groups fail the condition of Proposition 7 for x = a −1 and y = ab, meaning that they contain a non-embeddable subset which is a not a "group" one, i.e. the semigroup generated by it is not a group.
A natural avenue of further study is to understand the nature of nonembeddable sets: while it is clear that expansions of non-embeddalbe sets are non-embeddable and existence of some non-embeddable sets inside a semigroup implies the existence of certain others, the classification of such sets or even deciding whether there is a finite number of "independent" non-embeddable sets remain open problems.

LEF and LWF
While all LEF semigroups are LWF, we can demonstrate that only some of the non-LEF semigroups discussed above are non-LWF.To do this, we will require several auxiliary definitions and results.Definition 4. Consider an LWF semigroup S and its finite subset H.We say that the pair of semigroup D H and function d H satisfying the requirements of Definition 3 is tight if there exists no such function r H : We can establish a stricter structural property for tight pairs.Definition 5. Let S be an infinite LWF semigroup, H = {h 1 , . . ., h t } be its finite subset, D H , d H be a tight pair and {h ′ 1 , . . ., h ′ t } be a set of elements of Define the function r H : D H → S as follows: Our goal is to demonstrate that D H , r H satisfy the properties of Definition 3. Firstly, since meaning that the product of the accurate products representing them (which is accurate itself) equals to x ′ y ′ , allowing us to conclude that (x ′ y ′ )r H = (x ′ y ′ )d H . Thus we have ( Thus, the pair D H , d H is not tight as H(r The resulting contradiction means that our assumption was incorrect and U is empty.Let x ′ , y ′ be arbitrary elements of by the property of d H and by the same property any power of x ′ y ′ maps to 1. As D H is finite, there exists a power n such that (x ′ y ′ ) n is idempotent.Consider elements a ′ = (x ′ y ′ ) 2n−1 x ′ and b ′ = y ′ (x ′ y ′ ) n .By the multiplicative property we have We want to demonstrate that a ′ b ′ acts as an identity for the subsemigroup a ′ , b ′ of D H , while b ′ a ′ is not an identity for this subsemigroup.The latter follows from the former and the fact that a ′ b ′ = b ′ a ′ as they have different images (1 and ba, respectively) under d H .
To prove the former, consider a ′ .By Proposition 10 applied to set of pre- Our goal is to prove that for all x, y ∈ H such that xy ∈ H we have Q x Q y = Q xy , where Q x , Q y and Q xy are the sets of pre-images of x, y, xy under d H and the product on the left-hand side is taken in the sense of the power semigroup P(D H ). Once we have established that, it will follow that F H = P(D H ) and f H : H → F H defined by h → Q h satisfy the requirements of Definition 1 as F H is finite, Q h are distinct for different h and the multiplication property is exactly described above.However, by Proposition 3 it is impossible to find such F H and f H . First, let us prove that

Consider the multiplication table of the elements of
From the multiplication property we have that Q a Q b ⊆ Q ab .To see the reverse inclusion, consider z ′ ∈ Q ab .By Proposition 10 there exists an accurate product of elements in {a where both u ′ and v ′ are accurate products, so in particular From the multiplication property we have that Q a Q ab ⊆ Q a .To see the reverse inclusion, consider z ′ ∈ Q a .By Proposition 10 there exists an accurate product of elements in the set Finally, we need to prove that Note that it follows from Proposition 10 that Q aba = Q ab Q a ∪ Q aba Q ab similarly to the above.We will show Q ab Q a = Q aba Q ab from which the initial equality will follow.
Assume there exist . Denote this property by ( * ).By Proposition 10 we know that w ′ = u ′ v ′ where both u ′ and v ′ are in Hd −1 H , which leaves us only the possibility Let us demonstrate that D H , d H is not tight by constructing another function r H as follows: Our goal is to demonstrate that D H , r H satisfies the properties of Definition 3.
Since ( * ) is only applicable to elements of Q aba and a ′ b ′ a ′ does not satisfy ( * ) we have D H r H ⊇ H. Additionally, for all x ′ , y ′ ∈ D H with x ′ r H , y ′ r H ∈ H we have x ′ r H = x ′ d H , y ′ r H = y ′ d H , and since x ′ and y ′ do not satisfy ( * ), x ′ y ′ does not satisfy ( * ) as well according to the observation above.Thus (x ′ y ′ )r H = (x ′ y ′ )d H and we have This means that our assumption is incorrect and no element of Q aba satisfies ( * ), i.e.
This together with the initial observation concludes the proof.
It remains an open question whether all LWF semigroups are LEF.

Inverse semigroups
The class of inverse semigroups can be seen as an intermediate between general semigroups and groups.
Definition 6.A semigroup S is called inverse if for every element x ∈ S there exists a unique element y ∈ S such that xyx = x and yxy = y.This element y is denoted by x −1 .
One of the classical examples of the inverse semigroups are symmetric inverse monoids (semigroups) which are the sets of partial bijections on a given set Σ with semigroup operation being the composition.
The structural theorem for the inverse semigroups similar to the Cayley Theorem in the group case is known as Wagner-Preston theorem.
Theorem 1. [4, Theorem 5.1.7]Let S be an inverse semigroup.Then there exists a symmetric inverse semigroup I X and a monomorphism φ from S into I X .
Similarly to the group and semigroup cases, we can define local embeddability into finite for the class of inverse semigroups.Definition 7.An inverse semigroup S is called locally embeddable into the class of inverse finite semigroups (an iLEF semigroup for short) if for every finite subset H of S there exists a finite inverse semigroup H and an injective function Proposition 13.An inverse semigroup is an LEF semigroup if and only if it is an iLEF semigroup.
Proof.Let S be an iLEF semigroup.It is immediate that it is an LEF semigroup as well.
Let S be an inverse semigroup which is LEF.Let H be a finite subset of S. We can expand H to the set K Consider the finite semigroup F K 3 and function f K 3 satisfying Definition 1 for K 3 .Without the loss of generality we can consider F K 3 to be a full transformation semigroup T Σ of a finite set Σ as we can embed F K 3 in T Σ and carry f K 3 over to the bigger semigroup.We will shorten f K 3 to f for brevity in the rest of the proof.Now set F (i) to be the inverse semigroup consisting of all partial bijections on Σ (the symmetric inverse monoid on Σ) and f (i) to be a map from K 3 to F (i) such that for x ∈ K 3 the image xf (i) is a partial bijection between Im x −1 f and Im xf induced by restricting xf on Im x −1 f .We will prove that (i) , f (i)  satisfy the conditions of Definition 7 for the set K (and, in turn, for H ⊆ K).
Firstly, we need to demonstrate that f (i) is well-defined, meaning that restricting xf on Im x −1 f provides a bijection between Im x −1 f and Im xf , i.e. f (i) sends elements of K to the inverse semigroup F (i) .
Take arbitrary a, b meaning that a(xf ) and b(xf ) are separated by the function (x −1 f ), i.e. distinct.Thus, the restriction of xf to Im x −1 f is an injective map.
To prove that this restriction is surjective, note that xx −1 x = x, correspondingly (xf )(x −1 f )(xf ) = xf .This, in particular, means that the image of the restriction of xf to Im(xf )(x −1 f coincides with the image of xf , and, since Im(xf )(x −1 f ) ⊂ Im x −1 f , we get that the image of the restriction of xf to Im x −1 f indeed coincides with Im xf as well.
Note that for e ∈ K such that e = e 2 (and correspondingly e = e −1 ) we have ef (i) to be the identity function on Im ef .This means for x ∈ K we have (xf (i) ) −1 = x −1 f (i) as x −1 f (i) is by construction the bijection between Im xf and Im x −1 f which is inverse to xf (i) due to (xf )(x −1 f ) = (xx −1 )f and (x −1 f )(xf ) = (x −1 x)f being identity functions on Im x −1 f and Im xf respectively.
Secondly, we need to prove that f (i) is injective on K. Assume that for x, y ∈ H we have xf (i) = yf (i) .This means that both hf (i) and yf (i) provide the same partial bijection of Σ (in particular, Im xf = Im yf and Im x −1 f = Im y −1 f ), and, by the observation above, x −1 f (i) and y −1 f (i) provide the same inverse partial bijection too.Consider the element (xf )(y and as x, y, xy −1 , xy −1 x ∈ K 3 , we get xy −1 x = x by Definition 1. Similarly we can get y −1 xy −1 = y −1 , which allows us to conclude that x −1 = y −1 and x = y. Finally, we need to prove that ∀x, y ∈ K with xy ∈ K it holds that (xy)f (i) = (xf (i) )(yf (i) ).
By construction, we have xf (i) to be the bijection between Im x −1 f and Im xf induced by xf , yf (i) to be the bijection between Im y −1 f and Im yf induced by yf , and (xy)f (i) to be the bijection between Im((xy) −1 )f and Im(xy)f induced by (xy)f .Note that Im((xy This image is a subset of Im x −1 f , meaning that xf (i) acts on it exactly as xf .Additionally, as (y (i) acts on the former exactly as yf .This allows us to conclude that for a ∈ Im((xy) −1 )f we have i.e. (xy)f (i) is the restriction of (xf (i) )(yf (i) ) onto Im((xy) −1 )f .Since the function (xf (i) )(yf (i) ) is a bijection and the size of its image is less than or equal to We can also apply the notion of being LWF to inverse semigroups in a similar manner.
Definition 8.An inverse semigroup S is called locally wrapped by the class of inverse finite semigroups (an iLWF semigroup for short) if for every finite subset H of S there exists a finite inverse semigroup D H and a function d H : D H → S, such that H ⊂ D H d H and for all x ′ , y In order to establish a partial correspondence between iLEF and iLWF semigroups we require the following results.
Remark 2. As the proofs of Proposition 9 and Proposition 10 do not change the wrapping semigroup D H , we can apply them to iLWF case as well.
For the next set of lemmas we will consider symmetrised subsets K of inverse semigroups, i.e. such that K = K −1 and K ⊇ KK −1 , K ⊇ K −1 K.Note that for any given finite H, the set K = H ∪ H −1 ∪ HH −1 ∪ H −1 H is symmetrised.We will also limit ourselves to infinite semigroups as finite case is trivial (all structures are LEF/LWF).Lemma 2. Let S be an infinite iLWF semigroup, K be its finite symmetrised subset, and D K , d K be a tight pair for K. Then for all w Proof.We will denote D K and d K as D and d for brevity.
Consider an arbitrary h ∈ K which is not an idempotent.There exists x ′ ∈ D such that x ′ d = h.Since K is symmetrised, h −1 is also in K, which means that there exists y ′ ∈ D such that yd = h −1 .Since D is finite, there exists a positive n such that (x ′ y ′ ) n is a idempotent.Additionally, x ′ y ′ maps to hh −1 under d by the multiplication property, which means (uv) m for any positive m maps to hh −1 as well.Thus, by the same property, u ′ = (x ′ y ′ ) n x ′ maps to h and v ′ = y ′ (x ′ y ′ ) 2n−1 maps to h −1 .We claim that u ′ , v ′ are inverse to each other.To see this, note that Thus, we can find inverse pre-images for h and h −1 under d.
Consider an arbitrary h ∈ K which is an idempotent.There exists z ′ ∈ D such that z ′ d = h.Since D is finite, there exists a positive n such that (z ′ ) n is a idempotent and by multiplicative properties (z ′ ) n maps to h under d.Thus, we can find an idempotent pre-image for idempotent h ∈ K under d.Now let us apply Proposition 10 for the set of pre-images (such pre-images exist by the arguments above).Now consider w ′ ∈ Kd −1 .By the proposition there exists an accurate product h ′ i1 . . .h ′ i k equal to w ′ .To finish the proof we will use an induction on k in order to demonstrate that (w ′ d) −1 = w ′−1 d.
The base.For k = 1 the statement follows immediately from our choice of K ′ .
The step.Assume the statement holds for k = 1, . . ., k 0 − 1, k 0 ≥ 2 and consider k = k 0 .As the product h ′ i1 . . .h ′ i k 0 is accurate and k 0 ≥ 2, there exists an index j, 1 ≤ j < k 0 such that h ′ i1 . . .h ′ ij is accurate and h ′ ij+1 . . .h ′ i k 0 is accurate.Denote the value of the first product by u ′ and the value of the second product by v ′ .It holds that u and by the induction hypothesis we have (u Definition 9. Let S be an infinite iLWF semigroup, H be its finite subset and D H , d H be a tight pair for H.We say that h H in the natural partial order on D H . Lemma 3. Let S be an infinite iLWF semigroup, K be its finite symmetrised subset, D K , d K be a tight pair for K and h ′ ∈ D K be an h-minimal element for some h ∈ K. Then h ′−1 is h −1 -minimal.
Let us define a map γ from D K to I. Consider an arbitrary element x ′ of D K .Let ∆ and Λ denote its domain and range as a partial bijection on Σ.We will construct the element x ′ γ of I by specifying its range and image ∆ ′ and Λ ′ and the connection between them.
Finally, let x ′ γ be the map between ∆ ′ and Λ ′ which maps elements exactly as x ′ , except for changing all b ∈ (Σ 2 \ Σ 1 ) from domain/range into bη.Define (x ′ γ)d * K := x ′ d K .Note that d * K is well-defined as γ is a 1-to-1 (we can recover x ′ from x ′ γ by replacing all of bη with b and keeping the relations).
Let K ′ be the set {h ′ 1 , . . ., h ′ t } such that each h ′ i is h i -minimal, if h i is idempotent then h ′ i is also idempotent and if h i = h −1 j then h ′ i = h ′−1 j .We claim that γ is an injective homomorphism from T = K ′ into I.
Consider an h-minimal element h ′ from K ′ and its domain Υ 1 and range Υ 2 as a partial bijection on Σ.Note that if Υ 1 ∩ (Σ 2 \ Σ 1 ) = ∅ then it is impossible for the equality (h ) −1 and h ′ h ′−1 is an hh −1 -minimal element by Lemma 5. Thus, γ acts on K ′ exactly by renaming all of the elements of the set Σ 2 \ Σ 1 in the domains and ranges to the respective elements of Ω under η, which evidently preserves the semigroup structure and has the injective property.
Note that f ′ maps to f ′ under γ as f ′ f ′−1 e ′ = e ′ and (f ′ f ′−1 e ′ )d K = e ′ d K .This means that f ′ does not belong to T γ as none of the generators in the set K ′ γ contain elements of Σ 2 \ Σ 1 in their domains or ranges.However, f ′ ∈ T as by Proposition 10 applied to the set K ′ of pre-images of K as D K , d K is tight.This is a contradiction, meaning that there are no idempotents e ′ , f ′ ∈ ed −1 K with e ′ < f ′ in the natural partial order inside D K .This immediately implies that there are no idempotents e ′ , f ′ ∈ ed −1 K with e ′ = f ′ as for such a pair we have e ′ f ′ ∈ ed −1 K , e ′ f ′ is an idempotent and at least one of e ′ f ′ < e ′ and e ′ f ′ < f ′ holds.Now we can demonstrate the final result of this section.
Theorem 2. Let S be a countable iLWF semigroup with a finite number of idempotents.Then S is iLEF.
Proof.The statement evidently holds if S is finite.
Assume that S is infinite and let H = {h 1 , . . ., h t } be its finite subset.Consider the finite set K = H∪H −1 ∪E(S), where E(S) is the set of idempotents of S. The set K is symmetrised.Consider its tight pair D K , d K .By Wagner-Preston theorem D K embeds into the symmetric inverse monoid on Σ.
Define F H to be the power semigroup of D K .We will demonstrate that F H and the function f H : H → F H defined by h i → h i d −1 K M where M = E(S)q −1 K satisfy the properties of Definition 1.
Evidently, F H is finite since D K is finite.

Proposition 4 .
The monoid S = M on a, b | abab = 1, baba = 1 is an LEF semigroup, while the power semigroup P(S) is not an LEF semigroup.Proof.Note that S is a group with presentation Gp a, b | abab = 1 or with an equivalent presentation Gp a, c | c 2 = 1 (with a sequence of Tietze transformations Gp a, b | abab y, yx, yxx, yxy, (yxy)(yxx)} which fails the stated property is non-embeddable.Proposition 6.For the semigroup T = Sg a, b | baab = ba , the semigroup A = Sg a, b | aab = a and the bicyclic monoid B = M on a, b | ab = 1 there exist elements x, y such that yxxy = yx and neither yxy = yx nor (yxy)(yxx) = yx.Proof.For the bicyclic monoid consider x = a, y = b.We have baab = B ba, but bab = B b = B ba and babbaa = b 2 a 2 = B ba.For A and T it is enough to note that the natural homomorhisms φ : T → B with aφ = a, bφ = b and ψ : A → B with aψ = a, bψ = b separate ba, bab and babbaa, while in both of them baab = ba.

Proposition 11 .
The bicyclic monoid B = a, b | ab = 1 is not an LWF semigroup.Proof.Assume that B is LWF.Consider the set H = {1, a, b, ba} and consider a finite semigroup D H and a map d H : D H → B satisfying the requirements of Definition 3 such that D H , d H is a tight pair.
there exists an accurate product of these pre-images equal to a ′ .Since b ′ a ′ b ′ = a ′ b ′ , a ′ b ′ acts as an identity on the right for each of these pre-images, meaning that the same is true for their product a ′ , i.e.a ′ a ′ b ′ = a ′ .Similarly, a ′ b ′ b ′ = b ′ .As we already know that a ′ b ′ a ′ = a ′ and b ′ a ′ b ′ = b ′ , a ′ b ′ is indeed an identity for a ′ , b ′ .Thus, by [2, Lemma 1.31] the semigroup a ′ , b ′ is isomorphic to B, meaning that the semigroup D H cannot be finite.This contradicts our initial assumption.Proposition 12.The semigroup A = a, b | aab = a is not LWF.Proof.Assume that A is LWF.Consider its subset H = {a, b, ab, aba} and a finite semigroup D H and a map d H : D H → A satisfying the requirements of Definition 3 for H such that D H , d H is a tight pair.

2
Fix an arbitrary pre-image a ′ of a and b ′ of b under d H .By the multiplication property, a ′ a ′ b ′ is a pre-image of a, a ′ b ′ is a pre-image of ab and a ′ b ′ a ′ is a preimage of aba under d H as well.

.
Proposition 9. Let S be a LWF semigroup and H be a finite subset of S. Then there exist a semigroup D H and a map d H : D H → S satisfying the requirements of Definition 3 such that (D H , d H ) is a tight pair.Proof.Consider a finite semigroup D H and a map d ′ H : D H → S satisfying the requirements of Definition 3. If the pair D H , d ′ H is not tight, there exists r H such that the pair D H , r H also satisfies the requirements of the definition and H(r H ) −1 H(d ′ H ) −1 .Note that H(d H ) −1 is finite as a subset of a finite structure D H .If D H , r H is not tight, we can continue this process inductively with finding another function r ′ H such that D H , r ′ H also satisfies the requirements of the definition and H(r ′ H ) −1 H(r H ) −1 , and so on.Due to the fact that the size of the pre-image of H cannot be smaller than |H|, the process will stop eventually, resulting in a tight pair D H , d H .
One of the following holds: Choose an arbitrary s ∈ S \ H and denote by U the subset of H(d H ) −1 consisting of elements which cannot be presented with an accurate product.Assume that U is non-empty.
Proposition 10.Let S be an infinite LWF semigroup, H = {h 1 , . . ., h t } be its finite subset, D H , d H be a tight pair and {h ′ 1 , . . ., h ′ t } be a set of elements of D H such that h i = h ′ i d H . Then for every element w ′ ∈ H(d H ) −1 there exist a product h ′ i1 . . .h ′ i k equal to w ′ , k ≥ 1, i 1 , . . ., i k ∈ {1, . . ., t} which is accurate.Proof.
where both u ′ and v ′ are accurate products, so in particularu ′ d H , v ′ d H ∈ H and by the multiplication property (u ′ d H )(v ′ d H ) = z ′ d H = a.Following the multiplication table, this means that u ′ d H = a and v ′ d H = ab, and z