Regular semigroups weakly generated by one element

In this paper we study the regular semigroups weakly generated by a single element x, that is, with no proper regular subsemigroup containing x. We show there exists a regular semigroup F1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_{1}$$\end{document} weakly generated by x such that all other regular semigroups weakly generated by x are homomorphic images of F1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_{1}$$\end{document}. We define F1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_{1}$$\end{document} using a presentation where both sets of generators and relations are infinite. Nevertheless, the word problem for this presentation is decidable. We describe a canonical form for the congruence classes given by this presentation, and explain how to obtain it. We end the paper studying the structure of F1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_{1}$$\end{document}. In particular, we show that the ‘free regular semigroup FI2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\textrm{FI}}_2$$\end{document} weakly generated by two idempotents’ is isomorphic to a regular subsemigroup of F1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_{1}$$\end{document} weakly generated by {xx′,x′x}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{xx',x'x\}$$\end{document}.


Introduction
An element s of a semigroup S is called regular if sts = s for some t ∈ S. Note that s ′ = tst is a (von Neumann) inverse of s, that is, ss ′ s = s and s ′ ss ′ = s ′ .We denote by V (s) the set of all inverses of s in S. A regular semigroup is a semigroup with all elements regular.As usual, we denote by E(S) the set of idempotents of S.
A subset of a regular semigroup S may not generate a regular subsemigroup.Hence, the following notion of "weakly generated" seems natural: a regular subsemigroup T of S is weakly generated by a subset X if T has no proper regular subsemigroup containing X.Of course, this does not mean that T is effectively generated by X. Very often, the subsemigroup generated by X is a proper non-regular subsemigroup of T .In particular, we say that S is weakly generated by X if S has no proper regular subsemigroup containing X.We should also point out that the same set X may weakly generate several distinct regular subsemigroups of S.
The interest in studying the structure of regular semigroups weakly generated by a set X came first from the theory of e-varieties of regular semigroups (see [2,3] for the definition of e-variety).The existence of bifree objects on an e-variety V depends on the validity of the following property in V: for all S ∈ V and all matched subset A of S, there exists a unique regular subsemigroup of S weakly generated by A (see [7]).
In [6] we studied the structure of the regular semigroups weakly generated by idempotents.We proved the existence of a free regular semigroup FI(X) weakly generated by a set X of idempotents in the sense that FI(X) is weakly generated by X and all other regular semigroups weakly generated by X are homomorphic images of FI(X).We got FI(X) by introducing a presentation with both sets of generators and relations infinite.Despite that fact, the word problem for that presentation is decidable since a procedure to obtain a canonical representative for each congruence class was described.
If FI k denotes FI(X) for |X| = k, we proved that FI 2 contains copies of all FI k as subsemigroups.Thus, any regular semigroup generated by a finite set of idempotents must strongly divide FI 2 , that is, must be a homomorphic image of some regular subsemigroup of FI 2 .The case FI 1 is of no interest since any regular semigroup weakly generated by a single idempotent is just the trivial semigroup.
We intend to look now to the general non-idempotent case.In other words, we plan to see if there exists a regular semigroup F (X) weakly generated by a set X (not necessarily of idempotents) such that all other regular semigroups weakly generated by X are homomorphic images of F (X).The case where X is a singleton set is now distinct from the idempotent case: there are nontrivial regular semigroups weakly generated by a single element.So, in this paper, we focus only on the case where X is a singleton set and prove that F (X) exists for |X| = 1.This paper will follow the structure of [6].In fact, we will use the same terminology, but the concepts will be more complex here.For example, terms such as landscape and mountain will be used here again but to refer to words with a more intricate structure.The results we will present are also similar to the ones of [6], and some of the proofs use even the same arguments (or with just small adaptations), although they refer to more complex objects.For the sake of completeness, we decided to include those proofs again in this paper.
We will denote F (X) by F 1 for |X| = 1, and we will represent by x the unique element of X.In fact, we will usually not mention X but its unique element x instead.As stated above, the main goal of this paper is to prove the existence of F 1 .We begin by introducing F 1 using a presentation similarly to [6].It is an interesting exercise to compare the presentation for F 1 with the presentation for FI(X).We will see their resemblance, but also notice a new 'ingredient' appearing in F 1 , namely conjugation.We then prove that the presentation introduced here has decidable word problem by obtaining a canonical form for each congruence class.To attest the richness of the structure of F 1 , we will prove that F 1 has a regular subsemigroup weakly generated by the idempotents {xx ′ , x ′ x} isomorphic to FI 2 , where x ′ is the only inverse of x in F 1 .Since all regular semigroups weakly generated by a finite set of idempotents strongly divide FI 2 , they also strongly divide F 1 .
This paper is organized as follows.In the next section, we recall some basic concepts needed for this paper.We include also a brief description of the construction of FI 2 since we will need it later.We change also the terminology used in [6] so that it does not overlap with the terminology used here.In Section 3 we introduce a presentation and define F 1 .We present also a solution for the word problem for this presentation.We prove that F 1 is a regular semigroup weakly generated by x in Section 4. To achieve this result we need to characterize first the Green's relations on F 1 .Since we characterize the Green's relations in this section, we end it by describing also the idempotents, the inverses and the natural partial order.However, as it will become evident, this description will be just theoretical and not very useful for practical purposes.In Section 5, we show that all regular semigroups weakly generated by x are homomorphic images of F 1 .Finally, in the last section, we prove that F 1 has a regular subsemigroup weakly generated by {xx ′ , x ′ x} isomorphic to FI 2 ; thus concluding that all regular semigroups generated by a finite set of idempotents strongly divide F 1 .

Preliminaries
Given a semigroup S, S 1 denotes the monoid obtained by adding an identity element to S if necessary.The quasi-orders ≤ L , ≤ R and ≤ J on S are defined as follows: Thus, the Green's equivalence relations L , R and J are just the relations where ≥ L , ≥ R and ≥ J are respectively the dual relations of ≤ L , ≤ R and ≤ J .Let H = L ∩ R and D = L ∨ R be the other two Green's relations.For a regular semigroup S, we can use S instead of S 1 in the previous definitions.We will denote by K a the K -class of the element a ∈ S, for K ∈ {H , L , R, D, J }.
A crucial concept for this paper is also the notion of sandwich set S(e, f ) of two idempotents e and f of S. This concept has a few equivalent definitions.We list some of them next: S(e, f ) = V (ef ) ∩ E(f Se) = f V (ef )e = {g ∈ E(S) | f g = g = ge and egf = ef } .
The sandwich set S(e, f ) is nonempty precisely when ef is a regular element of S. Thus, S(e, f ) is always nonempty if S is regular.In fact, S(e, f ) is always a rectangular band and a subsemigroup of S, whenever it is nonempty.
There is another interesting property of these sets: if e, e 1 , f, f 1 ∈ E(S) are such that e L e 1 and f R f 1 , then S(e, f ) = S(e 1 , f 1 ).Hence, for regular semigroups, one can extend the definition of sandwich set to all elements of S as follows: for a, b ∈ S, let S(a, b) = S(a ′ a, bb ′ ) for some (any) a ′ ∈ V (a) and b ′ ∈ V (b).
If S is a regular semigroup, there is also another important relation on S, the natural partial order ≤ : s ≤ t ⇔ s = et = tf for some e, f ∈ E(S) .
A useful fact about the natural partial order is that we can choose the idempotents e and f such that e R s L f .
Let X be a nonempty set.As usual, we denote by X + the free semigroup on X.The elements of X are called letters in this context, while the elements of X + are called words.The content of a word u is the set of letters from X that occur in u, and the length of u is the number of letters (counting repetitions) that occur in u.For u ∈ X + , let σ(u) and τ (u) denote the first and the last letter of u, respectively.
Next, we recall the construction of FI 2 introduced in [6] for any set X.As explained before, we need to modify the notation and terminology used in [6] so that it does not overlap with the one used in this paper.Hence h, H and ̺ will replace the symbols g, G and ρ used in [6].Given a triple h = (a, b, c), we denote a, b and c by h r , h c and h l , respectively.
Let X = {e, f } be a two element set, and consider where 1 is a new symbol not in X.We identify each x ∈ X with the triple (1, x, 1) ∈ H 1 .Thus x r = x l = 1.Recursively for i > 1, assume that H i−1 is a set of triples and let Note that H 2 = {h 2,1 , h 2,2 } for h 2,1 = (e, 1, f ) and h 2,2 = (f, 1, e), and Consider the following two relations on the free semigroup H + : Let FI 1 2 be the semigroup given by the presentation H, ̺ e ∪ ̺ s , that is, FI 1  2 = H + /̺ where ̺ is the smallest congruence on H + containing ̺ e ∪ ̺ s .Note that ̺ e just tells us that FI 1  2 is an idempotent generated monoid with identity element 1̺.Using both ̺ e and ̺ s , one can show easily that (h r h c )̺ and (h c h l )̺ are idempotents of FI 1  2 .Then ̺ s just turns h̺ into an element of the sandwich set S((h r h c )̺, (h c h l )̺).The semigroup FI 2 is defined as FI 1 2 \{1̺} ({1̺} is the group of units of FI 1 2 ).We can also easily see that 1̺ is constituted by all words of H + with content {1}.
A nontrivial i-mountain (called mountain in [6]) is a word In [6] we proved that each ̺-class of FI(X) contains a unique nontrivial i-mountain and solved the word problem for FI(X) by given a process to construct it.In this paper we will use the term mountain to refer to more complex words.
Let i-M (denoted by M(X) in [6] for |X| = 2) denote the set of all nontrivial i-mountains of H + .In [6] we defined an operation ⊙ in i-M that turns i-M into a semigroup isomorphic to FI 2 .The concept of i-river (called river in [6]) and the process of uplifting of i-rivers (called uplifting of rivers in [6]) were central.In this paper we will use the terms river and uplifting of rivers again.Their definitions will be very similar to the ones introduced in [6].They differ only on the 'context' where they are defined.It will be immediate to see what i-river and uplifting of i-rivers mean from the notions of river and uplifting of rivers introduced here, but the readers may consult [6] for further details.
The semigroup FI 2 is regular and weakly generated by the two idempotents e and f (identifying each letter with the corresponding ̺-class).It has the following universal property: all regular semigroups weakly generated by two idempotents are homomorphic images of FI 2 (under a homomorphism that sends e̺ and f ̺ into those two idempotents).In Section 6 we show that FI 2 is embedded into the semigroup F 1 that we will construct in the next section.

The presentation G, R
This section is devoted to the construction of F 1 and, after some initial terms are introduced, we will divide it into subsections for a better organization.As we will see, the construction of F 1 resembles that of FI 2 , and it will be easier to define and work with the monoid F 1 1 obtained from F 1 by adding a new identity.In contrast to the FI 2 case, we will work now with 5-tuples since more information is needed to be included in those tuples.Conjugation will have now an important role too and the two extra entries will contain information about when and where to apply conjugation.
We begin by defining a presentation G, R for F 1 1 .From now on, and if nothing is said in contrary, X will be the singleton set {x}.Let A be the set {1, x, x ′ }.We endow A with a natural involution ′ by setting (x ′ ) ′ = x and 1 ′ = 1.We call anchors to the elements of A. As should be expected, x ′ will represent the inverse of x in F 1  1 , while 1 will correspond to its identity element.For that reason, we will make some abuse of terminology from now on by calling a ′ the inverse (anchor) of a, for any a ∈ A.
The generator set G will contain the set A. All other elements from G will be special 5-tuples.We will denote the set of all those 5-tuples by G 5 .Thus G = A ∪ G 5 .The set G 5 will be defined recursively as the union of sets G i for i ≥ 1.In fact, each G i will be the union of two other sets G i,e and G i,d .We will describe these sets in more detail below.
We will use the notation g = (g l , g la , g c , g ra , g r ) to refer to the entries of a 5-tuple g of G 5 .The entries g la and g ra will be always anchors.For that reason, they will be called the left anchor and the right anchor of g, respectively.The entries g l , g c and g r will be called the left, middle and right entries of g.The entries g l and g r will be 5-tuples of G i−1 if g ∈ G i for i > 1; and the entry g c will be a 5-tuple of As we can see, the left, middle and right entries of a 5-tuple g ∈ G 5 are usually 5-tuples again.If we need to refer to an entry of g l for example, we will use the abbreviation g ls to refer to the entry (g l ) s of g l , for s ∈ {l, la, c, ra, r}.But, some entries of g l may be again 5-tuples.In general, we convene that r} and s n ∈ {l, la, c, ra, r}, whenever the right side of this equality makes sense.We consider also g 1 = g.
To help following the remaining of this section, we will divide it into subsections.In the first two subsections we define the sets G i,e and G i,d , respectively.It is important that the reader pays some attention to the subtleties involved in these definitions as they may be important to understand some of the arguments presented during the paper.In the following subsection we introduce the set of relations R.This set will give us an interpretation for the 5-tuples of G 5 .In the forth subsection some special words are described.Finally, the operation of uplifting of rivers is introduced in the fifth subsection and used in the last subsection to present a solution to the word problem for this presentation.
Analyzing the structure of the 5-tuples of G i,e in more detail for i > 1, we see that g ∈ G i,e if and only if (i) g l = g r ∈ G i−1,e , (ii) g c = g l 2 = g lr , and (iii) either {g la , g ra } = {1, x} if i even or {g la , g ra } = {1, x ′ } if i odd.Hence, the set G i,e has twice the number of elements of G i−1,e , and so G i,e has 2 i−1 elements.Further, The 5-tuples of G e will correspond to the 5-tuples of G with the same left and right entry.
By construction, both g l 2 and g lr are equal to g c for each g ∈ G i,e with i ≥ 2. Of course, they have different meanings: g c represents the middle entry of g, while g l 2 and g lr represent the left and right entries of g l , respectively.In the special words that we will need to consider later, the 5-tuples of G will have to be 'anchored' to each other.For that, we need to attribute a side to g c inside g l in order to identify 'the anchor of g c in g l with respect to g'.Note that only one of the elements g l 2 a and g lra is equal to (g la ) ′ .For each g, we define Therefore, the anchors g laa and (g la ) ′ are equal.We reinforce that the purpose of the notation l a is to identify which anchor g l 2 a or g lra is equal to (g la ) ′ , and that l a differs according to g.Hence, the use of l a only makes sense when we refer to a specific 5-tuple g ∈ G i with i ≥ 2. Note also that g la and g c have the same 'value'.However, when we write g la , we want to emphasize not only its 'value', but also a specific entry of the 5-tuple g l , either its left or its right entry.
In a similar manner we will use also the notation r a .For each g ∈ G i,e with i ≥ 2, rl if g rla = (g ra ) ′ .We alert again that the use of r a only makes sense when associated with a specific 5-tuple g ∈ G i,e with i ≥ 2.

3.2.
The sets G i,d .We now introduce the 5-tuples of G with distinct left and right entries.They will be described recursively in the sets G i,d for i ≥ 1.In contrast with the previous case, the left and right entries of a 5-tuple g from G i,d may belong not only to G i−1,d , but also to G i−1,e .Thus, we set G i = G i,e ∪ G i,d .As we will see next, the definition of the sets G i,d is more intricate than the definition of the sets G i,e .
In fact, the definition of the sets G i,d is only interesting for i > 2, since we set G 1,d and G 2,d as empty sets.Basically, what it says is that for i = 1 and i = 2 there are no 5-tuples in G with distinct left and right entries.Now, assume that i > 2 and that G j,d is defined for j < i.Consequently, the sets G j are also defined.Let G i,d be the set of all 5-tuples ) ′ or g c , g la = g lr , (g lra ) ′ ; (iii) g c , g ra = g rl , (g rla ) ′ or g c , g ra = g r 2 , (g r 2 a ) ′ .From (i) we know that the left entry of g is distinct from its right entry.Note that (ii) tells us, in particular, that g c is the left or the right entry of g l .In fact, it tells us something stronger: either g c is the left entry of g l and the left anchor of g is the inverse of the left anchor of g l , or g c is the right entry of g l and the left anchor of g is the inverse of the right anchor of g l .Condition (iii) is just the dual of (ii) with respect to g r .
Next, we list the elements of G 3,d as an example to help understand these sets.It has already 8 elements: ) .
As the reader may have notice, we have attributed a 'name' only to the elements of the first column.The reason is because only these elements will be important for Section 6.The elements from the second column will not be used there.It is also a combinatorial exercise to compute the cardinality of G i,d and G i .Since we do not need that information for this paper, we leave that computation for the reader.We just point out the obvious fact that the size of the sets G i,d , G i,e and G i increases exponentially with i.
Like in the previous subsection, for each g ∈ G i,d , we need to attribute a side to g c inside g l .In this case, there are two distinct situations accordingly to g l ∈ G i−1,e or g l ∈ G i−1,d .If g l ∈ G i−1,e , then g l 2 and g lr are both equal to g c .In this case, due to condition (ii) above, we can define l a in the same manner as we did in the previous subsection.If g l ∈ G i−1,d , then the definition of l a is more natural since only one of left and right entries of g l is equal to g c : let l a = l 2 if g l 2 = g c or let l a = lr if g lr = g c .The notation r a is now introduced similarly (see the previous subsection).We alert once more that these two notations only make sense when referring to a specific 5-tuple g.
By definition of both G i,d and G i,e , observe that g laa = (g la ) ′ and g raa = (g ra ) ′ in all cases where the notation makes sense (that is, except for g = g xx ′ ).This observation will be used frequently in this paper without further notice.
A quick inspection to the definition of g la and g ra allow us to conclude also that g la , g ra ∈ {1, x} if υ(g) even, while g la , g ra ∈ {1, x ′ } if υ(g) odd.In fact, if g ∈ G e , we can say more: {g la , g ra } = {1, x} if υ(g) even, while , we may have g la = g ra as can be observed in some elements of G 3,d .Note that the set G i,e is also contained in Thus G 5 is the set of all 5-tuples of G, and it is divided into the set G e of all 5-tuples of G with the same left and right entries, and the set G d of all 5-tuples of G with distinct left and right entries.The height υ(g) of an element g ∈ G ′ is defined as follows: 3.3.The relation R. Let us look now to the relation R. We will denote by F 1  1 the semigroup given by the presentation G, R , that is, F 1 1 = G + /ρ where ρ is the smallest congruence containing R. We will write u ≈ v to indicate that the words u and v of G + are ρ-equivalent.We will use also the notation [u] to refer to the ρ-class uρ.We want: [1] to be the identity element of F 1 1 ; and (iii) [g] to be an idempotent for all g ∈ G 5 .Thus, we include in R the relation We immediately see now that we could have omitted g xx ′ from G.However, its inclusion gives a better feeling about the construction of G 5 : this list derives form the 'idempotent' xx ′ , and not from x or x ′ .Furthermore, we will see that the congruence classes of the 5-tuples of G 5 constitute a transversal set for the D-classes of F 1 .We should also point out that we could have replaced xx ′ with x ′ x in the theory developed in this paper (we just needed to adapt the definition of G 5 for the x ′ x case).But the definition of R is not yet finished.We still need to include the conditions that relate g with its entries.For that, we need to explain carefully some notation we will use in order to avoid future ambiguities.
Given three letters g 1 , g ∈ G ′ and a ∈ A, the triplet g 1 ag is left anchored if (g 1 , a) = (g l , g la ) or (g 1 , a) = (g r , g ra ) .The notion of right anchored is not quite the dual: the triplet gag Observe that in the definition of left anchored, the anchor a is one of the anchors of g, while in the definition of right anchored, the anchor a is the inverse of one of the anchors of g.In both cases, we say that g 1 is anchored to g.Finally, a triplet For s ∈ {l, r}, we may denote the anchored triplets g s g sa g and g(g sa ) ′ g s by g s • g and g • g s , respectively.Note that this notation may become ambiguous if we don't indicate explicitly the value of s.For example, if g ∈ G e and h = g l , then the expression h • g is ambiguous.Note that h is also equal to g r and, therefore, h • g may refer to either g l g la g or g r g ra g, with g la = g ra .Hence, we must be careful when using this 'dot' notation.
There is another instance where this 'dot' notation may be useful but ambiguous too if we don't clarify it, namely g c • g s and g s • g c for s ∈ {l, r}.Note that g c may be equal to both g sl and g sr .We define Clearly, these triplets are anchored by the way they are defined.
Next, we introduce the (non-anchored) triplets g L and g R .Let g L = (g la ) ′ g l g la and g R = (g ra ) ′ g r g ra .Note that In fact, we can say also that g L = xg l x ′ if υ(g) even, and g L = x ′ g l x if υ(g) odd.Similar conclusions are also true for g R .Note now that The next observation is also useful.Since g r • g • g l = g r g ra g(g la ) ′ g l and Before we continue, this is a good place to make a pause and explain the main difference between the theory developed here and the one developed in [6].This explanation may help also the readers familiar with [6] following this paper.We have already mentioned the close resemblance between the structure of these two papers.The main difference is on what we will consider to be landscapes and mountains.In [6], they were words from H + where any two consecutive letters had a special connection (see Section 2).Here, they will be words constituted by alternating letters from G ′ and A such that any triplet subword with two letters from G ′ and one from A is anchored.The notation introduced above will help us, in some cases, to give a less cumbersome aspect to those words.
Let ρ s be the following relation on G + : We set R = ρ e ∪ ρ s .Observe that [1] is composed by all words from G + of content {1}.
The next result gives a list of properties about the product in F 1  1 and terminates with an interpretation for ρ s .We already know that [g] is an idempotent of F 1  1 due to the definition of ρ e .We will prove that [g c g R ] and [g L g c ] are also idempotents of F 1  1 .Then, the definition of ρ s is just to assure that by definition of ρ e and ρ s .The proof of (i) is now complete by duality.Then (ii) follow obviously from (i).Further, [g L g] and [g R g] also belong to L [g] .Then, since (g la ) ′ g l • g = g L g and (g ra ) ′ g r • g = g R g, we must have both [g l • g] and [g r • g] in L [g] too.Thus (iii) holds by duality.
Note now that Finally, to prove (vi), we already know that 3.4.Special words.To facilitate the organization of this paper, if nothing is said in contrary, we will use g and h (with possible indices) to refer to elements of G ′ , a and b (with possible indices) to refer to anchors, and u and v (also with possible indices) to refer to words from G + .Next, we list some general terms and notations that we will use throughout this paper.These terms identify special words from G + and their characteristics that will be important to us.We use the same terminology of [6] and the notions are very much alike as the reader can verify.The main difference, as mentioned already, is on what we consider now a landscape.
and anchors a j ∈ A such that all triplet subwords g i−1 a i g i are anchored.We denote by L the set of all landscapes of G + .Note also that single letters g ∈ G ′ are particular landscapes; whence G ′ ⊆ L. We make notice that some of the terms below refer to properties only of the letters g i of a landscape (and not to its anchors a i ).Ridge: Hence, a ridge is never the first nor the last letter of a landscape.Peak: highest ridge of a landscape u.Note that u can have several peaks, but all of them have the same height.If u has only one peak, we denote it by κ(u).The height υ(u) of the landscape u is the maximum between the height of its peaks, the height of σ(u), and the height of τ (u).Note that this definition of height agrees with the notion of height for elements of G introduced earlier.
As with ridges, rivers are never endpoints of landscapes.
In the former case we have an uphill since υ(g i ) = υ(g i−1 ) + 1, while on the latter case we have a downhill since υ(g i ) = υ(g i−1 )−1.We denote by L 1 , L + 1 and L − 1 the set of all hills, uphills and downhills, respectively.Further, L + 2 will denote the set of all landscapes composed by an uphill followed by a downhill.Valley: landscape composed by a downhill followed by an uphill.Thus, a valley has always one (and only one) river at its lower height letter.We denote by L − 2 the set of all valleys of G + , and by L 2 the set L + 2 ∪ L − 2 .Mountain range: landscape u with σ(u) = τ (u) = 1.Thus, the height of a nontrivial mountain range is the height of its peaks.Note that all mountain ranges have length 4k + 1 for some k ∈ N 0 , and u = 1 is the only mountain range of length 1.If u = 1, we denote by λ l (u) the maximal prefix of u which is also an uphill, and by λ r (u) the maximal suffix of u which is also a downhill.We call λ l (u) and λ r (u) the left hill and the right hill of u, respectively.For technical reasons, we define also λ l (1) = λ r (1) = 1.We denote by MR the set of all mountain ranges of G + .Mountain: mountain range u with no rivers.Thus, it is either the trivial mountain range u = 1, or it is composed by the left hill λ l (u) followed by the right hill λ r (u).Nontrivial mountains have only one ridge.We denote by M 1 and M the sets of all mountains and nontrivial mountains, respectively.So M = MR ∩ L + 2 .
We shall use the previous terminology also to refer to subwords of a given word u ∈ G + .For example, a hill of u ∈ L is a subword of u that is also a hill.A hill of u ∈ L is called maximal if it is not properly contained in another hill of u, while a valley of u is called maximal if it is not properly contained in another valley of u.Thus, each nontrivial mountain range is uniquely decomposable into its left hill, followed by a possible empty sequence of maximal valleys between its consecutive ridges, and ending with its right hill.However, we must be careful since u 1 u 2 is not a landscape if u 1 and u 2 are two landscapes such that g = τ (u 1 ) = σ(u 2 ): it appears a double gg at the junction of u 1 with u 2 .Therefore, we will use the notation u 1 * u 2 to represent the landscape obtained from u 1 u 2 by replacing gg with g.Note that u 1 u 2 ≈ u 1 * u 2 since [g] is an idempotent of F 1  1 .We will represent landscapes as line graphs where the anchors label the edges and the other letters label the vertices.We will include information about the height of the letters from G ′ in this line graphs as follows: we draw them such that all vertices labeled by letters with the same height belong to the same imaginary horizontal line; and labels of vertices in higher imaginary horizontal lines have higher height.We try to illustrate this idea in Figure 1.For example, looking carefully to that picture, we can conclude that (g 1 , a 2 ) = (g s 2 , g sa 2 ) and (a 7 , g 7 ) = ((g ta 6 ) ′ , g t 6 ) for some s, t ∈ {l, r}.Using these line graphs, we can see that the terminology introduced above has a natural interpretation.We draw again the line graph of Figure 1 in Figure 2 (omitting some unnecessary information) to illustrate some of those terms.
Next, we associate a mountain to each element of G.We begin by observing that 11g xx ′ x1, 1x ′ g xx ′ 11 and 11g xx ′ 11 are mountains.It may seem strange the appearance of double ones (11).This is so because one of them is consider as a letter from G ′ while the other is an anchor.Of course, since we know that 1 will represent the identity element of F 1  1 , we could have replaced these double ones by a single one or even omit them.However, this would mean that we would have to treat this particular letter different from the other letters from G ′ .For technical reason it is better not to do so, and consider the letter 1 from G ′ as any other letter.Therefore, we will see very often these double ones (or even longer sequences of ones) in our landscapes.
We define We define also β 1 (1) = 1, the trivial mountain.The definition of β 1 (g) for a 5-tuple g of  height greater than 1 is more complex.These words are long and, therefore, we will define first their left and right hills.Let g ∈ G 5 such that υ(g) = 2n or υ(g) = 2n + 1 for some n ∈ N. Set The 'brackets' under these words highlight the four element blocks used as patterns in their construction.To build β 1,l (g), we add to the left of each g c i the block g c i+1 g c i laa g c i l g c i la ; while, to build β 1,r (g), we add to the right of each g c i the block (g c i ra ) ′ g c i r (g c i raa ) ′ g c i+1 .Consequently, β 1,l (g) is an uphill from g c n to g, while β 1,r (g) is a downhill from g to g c n .Furthermore, , and We now define . If υ(g) = 2n + 1, then β 1 (g) is also a mountain since both 11g xx ′ and g xx ′ 11 are hills.We depict the mountain β 1 (g) in Figure 3 to help visualizing it.We denote by λ l (g) and λ r (g) the left hill and the right hill of β 1 (g), respectively, for each g ∈ G 5 .Hence Then β 1 (u) ∈ MR.In the following result we prove that u ≈ β 1 (u) for all words u ∈ G + .
Proof.Note that we just need to prove that β 1 (h) ≈ h for all h ∈ G. Indeed, if β 1 (h) ≈ h for all h ∈ G, then We have already observed that β 1 (a) ≈ a for all a ∈ A. Clearly by definition of ρ s .Since we have also 11g xx ′ ≈ g xx ′ , we conclude that λ l (g) ≈ β 1,l (g) ≈ g for all g ∈ G 5 .Dually, we can conclude also that λ r (g) ≈ β 1,r (g) ≈ g.Consequently We have proved that β 1 (h) ≈ h for all h ∈ G as wanted.
3.5.Uplifting of rivers.Let u = g 0 a 1 g 1 • • • a n g n be a landscape and assume that g i is a river of u.Then g i−1 a i g i a i+1 g i+1 is a subword of u such that g i−1 a i g i is a right anchored triplet and g i a i+1 g i+1 is a left anchored triplet.In other words, (g i , a i ) = g s i−1 , (g sa i−1 ) ′ and (g i , a i+1 ) = (g t i+1 , g ta i+1 ) , for some s, t ∈ {l, r}.
Let h i be the 5-tuple (g i+1 , a ′ i+1 , g i , a i , g i−1 ).If If h i ∈ G 5 , then the triplet g i−1 a i h i is left anchored, while the triplet h i a i+1 g i+1 is right anchored; whence v is also landscape.If h i ∈ G 5 , then g i−1 a i+2 g i+2 = g i+1 a i+2 g i+2 is an anchored triplet and again v is a landscape too.We have shown that v is always a landscape.We say that v is obtained from u by uplifting a river and write u → v (or u g i − → v if one needs to identify the river uplifted).Thus → is a binary relation defined on the set L of all landscapes.We illustrate the two distinct instances of this operation in Figure 4.
The word v either keeps the length of u or decreases it by 4 units.If h i ∈ G 5 , the uplifting of g i eliminates the river g i , but may create a new river g i−1 if neither g i−1 nor g i+1 are ridges of u.If h i ∈ G 5 , then the uplifting of g i replaces the river g i with a ridge h i .However, g i−1 and g i+1 become rivers of v unless they were ridges of u, respectively.
Let us also analyze the impact of this operation onto the 'anchors subsequence' of the landscape u.Continuing to use the notation above, if h i ∈ G 5 , then this anchors subsequence remains unchanged.However, if h i ∈ G 5 , then we obtain a new anchors subsequence by deleting a subword of the form xx ′ , x ′ x or 11 from the original one.
The next result tells us that the relation → is contained in ρ.
Lemma 3.3.With the notation introduced above, u ≈ v. where Figure 4. Illustration of the uplifting of a river. Proof.
By definition of ρ s , we conclude that g i−1 a i h i a i+1 g i+1 ≈ g i−1 a i g i a i+1 g i+1 ; whence u ≈ v also in this case.
We will denote by * − → the reflexive and transitive closure of →.Due to the previous result, we know that If we assume that u has length 2n + 1 and height i, then the landscapes from U (u) all have length at most 2n + 1 and height at most j = ⌊i + n/2⌋.They all have in common with u its first and last letters.If we look to the number of rivers, we conclude they all must have at most ⌊n/2⌋ rivers each.Hence, for each v ∈ U (u), we record the number of rivers of v in a j-tuple r(v) = (r , where r k (v) denotes the number of rivers of v of height k − 1.The sum of all entries of r(v) gives us the number of rivers of v and so is at most The set R(u) is obviously finite.We order the elements of R(u) using the lexicographic order.If v ∈ U (u) and v → v 1 by uplifting a river of height k, then v 1 ∈ U (u), for all other k 1 ; whence r(v 1 ) < r(v).Therefore, we cannot apply upliftings of rivers indefinitely to u.We must stop only when we get a landscape with no river, that is, a landscape from L + 2 ∪ L 1 ∪ G ′ .Note that these landscapes give us j-tuples with all entries equal to 0. Thus R(u) must always contain the j-tuple with all entries equal to 0, and this j-tuple is obviously the smallest element of R(u).
It is also easy to see that the uplifting of rivers is a commutative operation.In other words, if g i and g j are two rivers of u ∈ L, then the uplifting of g i followed by the uplifting of g j gives the same landscape as the uplifting of g j followed by the uplifting of g i .Hence, the uplifting of rivers is a system of rules commonly known as a noetherian locally confluent system of rules.An important property of this kind of systems is that, independently of the order of the rules that we choose to apply to an element, we must always stop after a finite number of steps with the same 'reduced' element (see [5]).In the case considered here, this means that, independently of the order of upliftings of rivers that we choose to apply to u ∈ L, we will always end up with the same landscape from U (u) with no rivers.In particular, U (u) has a unique landscape with no rivers.We designate it by β 2 (u).Note further that Proof.This corollary is an obvious consequence of Lemma 3.3.

Note that the set of all mountain ranges is closed for
Since the only mountain ranges with no rivers are the mountains, the landscape β 2 (u) must be a mountain if u is a mountain range.Thus, if we set β(v) = β 2 (β 1 (v)) for any word v ∈ G + , then β(v) ∈ M 1 .The previous corollary and Lemma 3.2 also allow us to conclude that β(v) ≈ v.We leave this observation registered in the next corollary for future reference.
Let us take a pause and look more carefully to the structure of the mountains.First, they are words u = g 0 a 1 g 1 • • • g 2n−1 a 2n g 2n of length 4n + 1 (n ∈ N 0 ), g 0 = g 2n = 1, and peak g n .If n ≥ 1, then also g 1 = g 2n−1 = g xx ′ ∈ G e .In fact, if g i is the highest letter of λ l (u) belonging to G e , then the subsequence In contrast, the subsequence a 1 • • • a i may vary: a k can be any anchor from {1, x} if k even, or from {1, x ′ } if k odd.
If we look now to the upper part of the uphill λ l (u) (above g i ), quite the opposite occurs.Each g k−1 is either the left or the right entry of g k , for i < k ≤ n.However, the side of g k−1 inside g k completely determines a k : a k is the anchor of g k corresponding to the side of g k−1 .Therefore, while the (possible empty) subsequence g i • • • g n may vary (although the peak g n is fixed), the subsequence d if it is not empty.As we saw, the uphill λ l (u) is divided into two sections with very distinct properties.The same conclusions can be drawn for the downhill λ r (u).If g j is the first letter from G e in λ r (u), then g k = g l k−j j = g r k−j k ∈ G e and g j fixes the subsequence g j • • • g 2n ∈ G + e .In contrast, for each k > j, a k ∈ {1, x} if k even, or a k ∈ {1, x ′ } if k odd; whence the subsequence a j+1 • • • a 2n does not depend of g j .In the upper part of λ r (u), each g k can be either the left or the right entry of g k−1 , for n < k ≤ j.However, for n < k ≤ j, a k is the inverse anchor of the anchor of g k determined by the side of g k−1 .
From the conclusions described above, it is now obvious that the number of uphills and downhills between 1 and some g ∈ G n is 2 n in both cases.We leave this conclusion registered in the following corollary for future reference.
Corollary 3.6.If g ∈ G n , then there are 2 n uphills from 1 to g and 2 n downhills from g to 1.
3.6.The word problem for G, R .In the next sequence of results we show that β(u) completely characterizes the ρ-class of u ∈ G + .We will prove that [u] = [v] if and only if β(u) = β(v), for all u, v ∈ G + .This result solves the word problem for G, R : we just need to compute both β(u) and β(v) and check if we get the same mountain to know if We want to show that ρ = ρ 1 , but we already know that ρ 1 ⊆ ρ by Corollary 3.5.The next couple of results show us that ρ e ⊆ ρ 1 .Lemma 3.7.Let g ∈ G ′ .Then:

Proof. (i). Clearly
(ii).The first part of (ii) is proved by induction on υ(g).Clearly λ r (1) * λ l (1) = 1 and λ r ( and assume that λ r (h) * λ l (h) * − → h for all h ∈ G j with j < i.Since λ r (g) * λ l (g) = g(g ra ) ′ g r (g raa ) ′ λ r (g c ) * λ l (g c ) g laa g l g la g , we conclude that λ r (g) * λ l (g) * − → g(g ra ) ′ g r (g raa ) ′ g c g laa g l g la g by the induction hypothesis.But g(g ra ) ′ g r (g raa ) ′ g c g laa g l g la g g c − → g(g ra ) ′ g r g ra g(g la ) ′ g l g la g g r , g l − −− → g .
Let g ∈ G i with i > 2, and assume that λ r (h t ) * β 1 (h ta ) * λ l (h) * − → h t h ta h and λ r (h) * β 1 ((h ta ) ′ ) * λ l (h t ) * − → h(h ta ) ′ h t for all h ∈ G j with 2 ≤ j < i and t ∈ {l, r}.Let s ∈ {l, r} and recall that g sa = (g saa ) ′ and g sa = g c .Then where * − → follows by the induction hypothesis.Note now that if s = l, then g s (g saa ) ′ g c g laa g l g la g = g l (g laa ) ′ g c g laa g l g la g g c − → g l g la g = g s g sa g ; and if s = r, then g s (g saa ) ′ g c g laa g l g la g = g r (g raa ) ′ g c g laa g l g la g g c − → g r (g raa ) ′ gg laa g l g la g = g r g ra g(g la ) ′ g l g la g g l − → g r g ra g = g s g sa g .
(ii).For i ∈ {0, • • • , n}, let u i = g 0 a 1 g 1 • • • a i g i .Thus all u i are prefixes of u and u = u n .Consequently, (ii) becomes proved once we show that ).By definition of landscape and (i), we have also Now, using the induction hypothesis, we obtain We have finished the proof of (ii).
(iii) follows easily from (ii) since g 0 = 1 = g n if u ∈ MR: and (iv) is obvious from (iii).
Proposition 3.10.Let g ∈ G 5 with υ(g) ≥ 2 and u, v Proof.(i).By Lemma 3.9.(ii), Thus β(g c g L g) = β(g).We have also β(gg R g c ) = β(g) by duality.Again by Lemma 3.9.(ii),we have Thereby (ii).We already know that [u] = [v] if β(u) = β(v) by Corollary 3.5.From Corollary 3.8, we have also ρ e ⊆ ρ 1 .Now, we proved in (i) that ρ s ⊆ ρ 1 ; whence R ⊆ ρ 1 .To finish the proof of (ii), we just need to conclude that ρ 1 is a congruence on G + .But this is obvious since β(u Although G and ρ e ∪ ρ s are infinite sets, Proposition 3.10.(ii)gives us a solution for the word problem for F 1  1 as explained before: to see if u ≈ v, we just need to compute both β(u) and β(v), and check if we get the same mountain.Thus: Corollary 3.11.The word problem for F 1  1 is decidable.Proposition 3.10.(ii)tells us also that each ρ-class [u] has a unique mountain, namely β(u).We will call β(u) the canonical form of u ∈ G + .We introduce the operation ⊙ on both M 1 and M as follows: Proposition 3.12.(M 1 , ⊙) and (M, ⊙) are models for F 1  1 and Proof.The first part is just a consequence of Proposition 3.10.(ii),while (ii) follows obviously from (i).Hence, we only need to prove (i).But notice that, by definition of uplifting of rivers, if u 1 → u 2 for a mountain range u 1 , then λ l (u 1 ) is a prefix of λ l (u 2 ) and λ r (u 1 ) is a suffix of λ r (u 2 ).Applying several times the previous observation, we conclude that λ l (u) = λ l (u * v) The ground ǫ(g) of a letter g ∈ G ′ is defined recursively as follows: if υ(g) ≥ 1.Thus υ(g 1 ) < υ(g) for any g 1 ∈ ǫ(g)\{g}.We define the relation on G ′ by setting h g if h ∈ ǫ(g).Due to statement (i) of the following result, is a partial order on G ′ .We extend the notion of ground to any landscape by setting, for u = g 0 a Again by statement (i) of the following result, ǫ(u) is the union of the grounds of its ridges, and the highest letters of ǫ(u) are the peaks of u.In particular,
(ii).Let h ∈ ǫ(g) \ {g}.We prove (ii) by induction on n = υ(g) − υ(h).If n = 1, then h = g s for some s ∈ {l, r}.Hence u = hg sa g is an uphill satisfying the conditions of (ii).Assume now that n ≥ 2. Then h ∈ ǫ(g s ) for some s ∈ {l, r}.Without loss of generality, we assume that h ∈ ǫ(g l ).Since υ(g l ) − υ(h) = n − 1, there exists 1 such that all g i ∈ ǫ(g l ) for 0 ≤ i ≤ n − 1, g 0 = h and g n−1 = g l , by the induction hypothesis.Clearly u = g 0 a 1 g 1 • • • a n−1 g n−1 g la g is now an uphill satisfying the conditions stated in (ii).
We describe next the partial orders ≤ R , ≤ L and ≤ J on (M 1 , ⊙).
Proof.We prove only (i) and (iii) since (ii) is the dual of (i).(i).Assume that u ≤ R v. Thus u = v ⊙ w for some w ∈ M 1 .By Proposition 3.12.(i),λ l (v) is a prefix of λ l (u).Assume now that λ l (v) is a prefix λ l (u).Then u = λ l (v) * u 1 for some landscape κ(u).Conversely, if κ(v) κ(u), then let u 1 be an uphill such that σ(u 1 ) = κ(v) and τ (u 1 ) = κ(u), whose existence is guaranteed by Lemma 4.3.(ii).Observe The next two results are corollaries of the previous proposition.The first one is an obvious consequence.For the second one, we need to say something more about the H and the D relations.(i) v covers u for ≤ R if and only if λ l (u) = λ l (v)aκ(u) for some a ∈ A.
(ii) v covers u for ≤ L if and only if λ r (u) = κ(u)aλ r (v) for some a ∈ A.
Corollary 4.6.Let u, v ∈ M 1 .Then: Proof.The first three statements follow from the corresponding statements of Proposition 4.4.Then (iv) is a consequence of (i) and (ii).So, we only need to prove that J ⊆D. Assume that u J v. Then w = λ l (u) * λ r (v) ∈ M 1 since κ(u) = κ(v) by (iii).Now, by (i) and (ii), we conclude that u R w L v, whence u D v.
The statements (iii) and (v) of the previous corollary tell us that the D=J -classes of F 1  1 are in one-to-one correspondence with the elements of G ′ , and that the set {[g] | g ∈ G ′ } is a transversal (or cross-section) for the set of D-classes of F 1  1 .The next result gives us the size of each R, L and Proof.We only need to prove that follows by duality and the statements about D follow from the statements about R and L and from Corollary 4.6.(iv).But Corollary 4.6.(i)tells us that the size of R [g] is equal to the number of downhills from g to 1, and this number is 2 n due to Corollary 3.6.Hence, Let v be the mountain λ l (g c )g laa λ r (g l ).By Lemma 3.9.(ii),β 1 (g c g laa g l ) * − → v, and so g c g laa g l ≈ v. Hence and so u = v ⊙ β 1 (g la ) ⊙ u .By Proposition 3.12.(i),λ l (v) = λ l (g c )g laa g l is a prefix of λ l (u).In a similar way, we conclude also that g r (g raa ) ′ λ r (g c ) is a suffix of λ r (u).
If g ∈ G d , then g l = g r .Hence, for this case, λ l (g) is the only uphill from 1 to g with prefix λ l (g c )g laa g l and λ r (g) is the only downhill from g to 1 with suffix g r (g raa ) ′ λ r (g c ).We have concluded that λ l (g) = λ l (u) and λ r (g) = λ r (u) since κ(u) = g.Thus, u = β 1 (g) and The case g ∈ G e is more complex since g l = g r , and thus there are two distinct uphills from 1 to g, namely λ l (g) and λ l (g c )g laa g l g ra g.There are also two distinct downhills form g to 1, namely λ r (g) and g(g la ) ′ g r (g raa ) ′ λ r (g c ).
We have concluded that , which is a contradiction.Therefore a 2 must be equal to g ra .Similarly, we conclude also that a 1 = g la .Consequently λ l (u) = λ l (g) and λ r (u) = λ r (g), that is, u = β 1 (g) and [u] = [g] if g ∈ G e also.We have proved the statement of this proposition.
We have now all the ingredients necessary to show that F 1 is weakly generated by [x].
Proof.Let S be a regular subsemigroup of F } by induction on υ(g).
Let g ∈ G i for i ≥ 3, and assume that [h] belongs to S for all h ∈ G j with j < i.Then [g l ], [g r ] and [g c ] belong to S by the induction hypothesis (note that g c = 1 because υ(g) ≥ 3).Since [x] and [x ′ ] also belong to S, we conclude that [g R g c ] ∈ S and [g c g L ] ∈ S. Again, we know that Since this latter set generates F 1 , we must have S = F 1 .Therefore, F 1 is weakly generated by [x].
In this section we have already described the Green's relations on F 1 .We end it describing the idempotents, the inverses and the natural partial order on F 1 .A canyon is a valley w with σ(w) = τ (w) and a gorge is a canyon such that w * − → σ(w) = τ (w).Thus β 2 (w) = σ(w) for any gorge w, and w 1 * − → σ(w) for any w 1 ∈ G + such that w * − → w 1 .Note however that w 1 may not be a gorge since we cannot guarantee that w 1 is even a valley.We will use gorges to characterize the idempotents, the inverses and the natural partial order, but this characterization will be just theoretical.For practical purposes, it is hard to identify the gorges, and consequently also the idempotents, the inverses and the natural partial order.(i) u is an idempotent if and only if the canyon λ r (u) * λ l (u) is a gorge.
(ii) v is an inverse of u if and only if λ r (u) * λ l (v) and λ r (v) * λ l (u) are both gorges.
for some and u is an idempotent.Conversely, if u is an idempotent, then u 2 * − → u.By definition of uplifting of rivers, it is obvious that we must have λ r (u) * λ l (u) * − → κ(u).We have proved (i).The proof of (ii) is similar and as obvious as the proof of (i).Let us prove (iii) now.
Assume first that v < u.
and a 1 , a 2 ∈ A, by Proposition 4.4.But, there exists also another mountain w ∈ M 1 such that We can now conclude that u 2 a 2 κ(u) a 1 u 1 is a gorge because λ r (w) * λ l (w) is a gorge too (note that u 2 a 2 κ(u) a 1 u 1 is a canyon).
Assume now that λ l (v) = λ l (u) a 1 u 1 and λ r (v) = u 2 a 2 λ r (u) for some and a 1 , a 2 ∈ A such that u 2 a 2 κ(u) a 1 u 1 is a gorge.In particular u = v.Note also that are well defined mountains.In fact, they are both idempotents of M 1 .For example, λ r (u) a 1 u 1 is a gorge and so w r is an idempotent.Finally, observe that u ⊙ w r = v and w l ⊙ u = v, whence v < u.

The universal property of F 1
In this section we prove that all regular semigroups weakly generated by x are homomorphic images of F 1 under a homomorphism that sends [x] into x.A crucial concept is the notion of skeleton.A skeleton mapping for a regular semigroup S is a mapping φ (iii) gφ ∈ S g φ,r , g φ,l for all g ∈ G i with i ≥ 2, where g φ,l = (g c φ)((g la ) ′ φ)(g l φ)(g la φ) and g φ,r = ((g ra ) ′ φ)(g r φ)(g ra φ)(g c φ) .
In this definition we are using the non-idempotent version of the definition of sandwich set since we cannot guarantee at this point that g φ,l and g φ,r are idempotents of S.However, after the following observations, we will prove that these elements are indeed idempotents of S. From that point on, we will consider here also the usual definition of sandwich set for idempotents.
In this paper, a skeleton of S is the image of G \{1} under some skeleton mapping.Thus, except for xφ and x ′ φ, which are mutually inverse elements with g xx ′ φ = (xφ)(x ′ φ), all other elements of a skeleton are idempotents of S.
Lemma 5.1.Let S be a regular semigroup and φ : G → S 1 be a skeleton mapping.Then g φ,l and g φ,r are idempotents of S for all g ∈ G i with i ≥ 2.
Proof.We need to prove that g φ,l and g φ,r are idempotents of S simultaneously by induction on i.For i = 2, note that g l = g xx ′ = g r and g c = 1.Thus, we can easily check that g φ,l and g φ,r belong to the set {aa ′ , a ′ a} for a = xφ and a ′ = x ′ φ; whence they are idempotents.Assume now that i > 2. By definition of skeleton mapping, g l φ ∈ S (g l ) φ,r , (g l ) φ,l , and we may assume that (g l ) φ,l and (g l ) φ,r are idempotents by the induction hypothesis.Thus (g l ) φ,l (g l φ) = g l φ and (g l φ)(g l ) φ,r = g l φ.Note also that g c = g l 2 and g la = (g l 2 a ) ′ , or g c = g lr and g la = (g lra ) ′ .If g c = g l 2 and g la = (g l 2 a ) ′ , then = (g l φ)((g l ) φ,l )(g l φ) = g l φ .
It is easy to see now that g φ,l = (g c φ)((g la ) ′ φ)(g l φ)(g la φ) is an idempotent of S. The proof that g φ,r is also an idempotent of S is similarly.
If a is an element of a regular semigroup S, we can easily construct recursively a skeleton mapping such that xφ = a.We can choose an inverse a ′ of a in S and define x ′ φ = a ′ , g xx ′ φ = aa ′ and 1φ = 1 ∈ S 1 .Then, assuming that g ∈ G i with i ≥ 2 and φ is defined for all h ∈ G j with j < i, we choose y ∈ S g φ,r , g φ,l , and set gφ = y.Note that S g φ,r , g φ,l = ∅ since S is regular.We cannot however guarantee that the skeleton mapping φ is unique.In terms of skeletons, each element a ∈ S belongs to at least one skeleton of S, but it may belong to many distinct skeletons.
In the next result we show that every skeleton mapping for a regular semigroup S can be uniquely extended into a homomorphism from F 1  1 to S 1 .Proposition 5.2.If φ : G → S 1 is a skeleton mapping, then there is a unique (semigroup) homomorphism ϕ : F 1  1 → S 1 extending φ (that is, such that gφ = [g]ϕ for all g ∈ G).Furthermore, G φ = F 1 1 ϕ is a regular subsemigroup of S 1 and a monoid with identity element 1φ.
Proof.Since G + is freely generated by G, let φ 1 : G + → S 1 be the only homomorphism that extends φ.By definition of skeleton mapping, is contained in the kernel of φ 1 .To conclude that ρ is contained in the kernel of φ 1 , it is now enough to show that (g r •g•g l , g r •g c •g l ) belongs to that kernel for all g ∈ G i with i ≥ 2.
Due to space constrains, we denote by w and z the expressions (g l φ)(g la φ)(g c φ)((g la ) ′ φ)(g l φ) and (g r φ)(g ra φ)(g c φ)((g ra ) ′ φ)(g r φ) , respectively.In the proof of Proposition 5.1 we showed that g l φ = w.Similar arguments allow us to conclude that g r φ = z.We need also the equality gφ = (g c φ)(gφ)(g c φ) that follows from an observation made prior to Lemma 5.1.So, where the fourth "=" sign is due to gφ ∈ S g φ,r , g φ,l .So, we have shown that ρ is contained in the kernel of φ 1 .Thus, if ϕ 1 : G + → F 1  1 denotes the natural quotient homomorphism, then there exists a homomorphism ϕ : F 1  1 → S 1 such that φ 1 = ϕ 1 ϕ.Hence ϕ extends φ.If ϕ ′ : F 1 1 → S 1 is another homomorphism extending φ, then ϕ 1 ϕ ′ = φ 1 = ϕ 1 ϕ.Consequently ϕ ′ = ϕ since ϕ 1 is surjective.We proved there exists a unique homomorphism ϕ : F 1  1 → S 1 extending φ.Finally, the second part of this result follows immediately from the fact that F 1  1 is a regular monoid with identity element [1] (this property is inherited by homomorphic images) and it is generated by {[g] | g ∈ G}.
Note that, in the previous proposition, F 1  1 ϕ may not be a submonoid of S 1 since [1]ϕ may not be the identity of element of S 1 .Given a skeleton mapping φ, we can only guarantee that 1φ is an identity element for the subsemigroup F 1  1 ϕ.Also by the previous result, a skeleton of S generates always a regular subsemigroup of S. We leave this conclusion registered in the following corollary for future reference.
Corollary 5.3.If A is a skeleton of S, then A is a regular subsemigroup of S.
Let T be a regular subsemigroup of a (not necessarily regular) semigroup S. Next, we prove that if T is weakly generated by an element a, then there exists a homomorphism ϕ : F 1 → S such that [x]ϕ = a and F 1 ϕ = T .Corollary 5.4.Let T be a regular subsemigroup of a semigroup S. If T is weakly generated by a, then T is the image of F 1 under some homomorphism ϕ : F 1 → S such that [x]ϕ = a.
Proof.Let φ : G → T 1 be a skeleton mapping such that xφ = a.We have already observed that φ exists.Let ϕ : F 1  1 → T 1 be the unique homomorphism extending φ given by Proposition 5.2.Note that ϕ | F 1 is a homomorphism from F 1 to T .Thus F 1 ϕ is a regular subsemigroup of T containing a. Since T is weakly generated by a, we conclude that F 1 ϕ = T .Finally, note that we can consider ϕ as a homomorphism from F 1 to S, and [x]ϕ = a as wanted.
The main result of this paper is now an obvious consequence of the previous corollary.
Theorem 5.5.All regular semigroups weakly generated by an element a are homomorphic images of F 1 under a homomorphism that sends [x] into a.
The reverse of this theorem is not true however, that is, there are homomorphic images of F 1 that are not weakly generated by the image of [x].Next, we present one such example.This example is adapted from the one used in [6] to show that not all homomorphic images of FI 2 are weakly generated by two idempotents.
The semigroup S 1 is the one obtained from F 1 by collapsing all D-classes into the 0 element except for D [x] , D [g 2,e,1 ] , D [g 2,e,2 ] and D [g 3,d,2 ] .In fact, the first set of relations identifies g 1 , g 2 and h with the elements g 2,e,1 , g 2,e,2 and g 3,d,2 , respectively, while the second set of relations collapses all other D-classes into 0.
But the semigroup we are interested in is the semigroup S given by the presentation X, R where R = R 1 ∪ {(g 1 , g 1 x ′ hxg 1 ), (g 2 , g 2 x ′ hxg 2 )} .Since R contains R 1 , the semigroup S is also a homomorphic image of F 1 .The semigroup S is, in fact, obtained from S 1 by merging the D-classes of g 1 and g 2 into distinct 'blocks' of the D-class of h.We illustrate the "Egg-box" diagram of the D-classes of S in Figure 5 (note that each congruence class is represented by one of its elements; we write also e = xx ′ , f = x ′ x and h 1 = x ′ hx due to space constrains).
The Semigroup S is a homomorphic image of F 1 but it is not weakly generated by x.The subsemigroup T generated by the set {x, x ′ , g 1 x ′ hxg 2 } is a proper regular subsemigroup of S. It is obtained from S by deleting the fifth and sixth rows and the third and fourth columns of D h .
The fact that we were able to adapt the example for FI 2 to an example for F 1 shouldn't be so surprising.In the next section we show that FI 2 can be embedded into F 1 in a natural way.But before, let us end this section by proving that F 1 is unique up to isomorphism.Proposition 5.6.Let T be a regular semigroup weakly generated by x such that all other regular semigroups weakly generated by x are homomorphic images of T (under a homomorphism that fixes x).Then T and F 1 are isomorphic.
Proof.By the universal property of both T and F 1 , there are surjective homomorphisms ϕ : F 1 → T and ψ : By Lemma 4.8, the restriction of ϕ • ψ to the set {[g] | g ∈ G \{1}} must be the identity mapping.Therefore, ϕ • ψ is the identity automorphism of F 1 since this semigroup is generated by the previous set.We have shown that ϕ and ψ are mutually inverse isomorphisms, whence T and F 1 are isomorphic.

Embedding FI 2 into F 1
In this section we prove that F 1 has a regular subsemigroup F It is also important that the reader recalls the notation used in Section 3 for the elements of G 2,e (page 7) and for some of the elements of G 3,d (page 8).

The construction of F •
1 .We begin by defining a subset G The next result is obvious by induction on the indexes i.We omit its proof.
Clearly (a) is satisfied, and if υ(g i−1 ) = 1, then also (b) is satisfied.If υ(g i−1 ) = 1, then υ(g i+1 ) = 1, and by condition (b) (applied to u) we know that a i−1 = a ′ i and a i+1 = a ′ i+2 .Thus a i−1 = a ′ i+2 since a i = a ′ i+1 , and v continues to satisfy (b) even in this case.In conclusion, v ∈ MR In this case, v clearly satisfies (b) since the anchors subsequence does not change and υ(h i+1 by condition (b) (applied to u) and so . Thus, we have ended the proof of this result.We denote by M • the set of all mountains from MR • .If u ∈ M • , then (a) is equivalent to κ(u) = g n ∈ G • by Lemma 6.1; while (b) is equivalent to the anchors subsequence a 1 a 2 • • • a 2n−1 a 2n having a prefix and a suffix from the set {11, x ′ x}.In particular, the only two mountains of M • of height 1 are β 1 (xx ′ ) = 11g xx ′ 11 and β 1 (x ′ x) = 1x ′ g xx ′ x1.In the next result we prove that M • is a regular subsemigroup of M. Proposition 6.3.M • is a regular subsemigroup of M.
Proof.Let u, v ∈ M • .By Lemma 6.2, u ⊙ v is a mountain from MR • and M • is a subsemigroup of M. Now, M • is regular because u ∈ M • for every u ∈ M • , and u is an inverse of u.
We denote by F • 1 the regular subsemigroup of F 1 corresponding to M • .In the next subsection we prove that F • 1 is isomorphic to FI 2 .
6.2.An isomorphism from FI 2 to F • 1 .It is important for this subsection that the reader recalls the construction of FI 2 described in Section 2. It should be clear at this point the connection between the theory developed here and the one developed in [6].Roughly speaking, we can look to the theory developed in [6] as the simplified version for triples of the theory developed here, obtained by 'dropping' the information given by the anchors.So, the meaning of notions like i-landscape and i-mountain range (called landscape and mountain range in [6]) and of operations like uplifting of irivers should now be clear to the reader.Otherwise, we advice the reader to look for more information in [6] before she/he proceeds since a clear understanding of these notions is important for what follows.
Consider now the mapping ψ : H \{1, e, f } → G • \{g xx ′ } obtained recursively as follows.Set (h 2,i )ψ = g 2,e,i and (h 3,j )ψ = g 3,d,j for i ∈ {1, 2} and j ∈ {1, 2, 3, 4}.Observe that if h 3,j = (h, e 1 , h 1 ), for H 2 = {h, h 1 } and e 1 ∈ H 1 = {e, f }, then (h 3,j )ψ = (hψ, a, g xx ′ , a, h 1 ψ) , where a = 1 if e 1 = e, or a = x ′ if e 1 = f .Let now h ∈ H i with i > 3 and define recursively hψ = (h l ψ, a, h c ψ, b, h r ψ) , where a ′ = (h l ψ) sa for h ls = h c and b ′ = (h r ψ) ta for h rt = h c .By the recursive definition and the choice of a and b, the 5-tuple hψ belongs to G i and H i too.We just need to observe that the entries g l , g c and g r of g completely determine the other entries g la and g ra for all g ∈ G • i .Therefore, we can conclude by induction that ϕ | G • i and ψ | H i are mutually inverse bijections between G • i and H i for all i ≥ 2. Thus ϕ and ψ are also mutually inverse bijections between G • \{g xx ′ } and H \{1, e, f }.Note further that ϕ and ψ have the following property: (g s )ϕ = (gϕ) s and (h t )ψ = (hψ) t for any s, t ∈ {l, c, r}.
Denote by i-MR and i-M the sets of all nontrivial i-mountain ranges and i-mountains, respectively, from H + .Next, we define two mutually inverse bijections ϕ and ψ between MR • and i-MR by applying naturally the mappings ϕ and ψ to each letter.There is however a problem with this approach: both ϕ and ψ are not yet defined for letters of height less than 2, and their definition for letters of height 1 is not so obvious since G • 1 has only one letter, namely g xx ′ , while H • 1 has two letters, namely e and f .Let G •′ = (G • \{g xx ′ }) ∪ {1, 1g xx ′ 1, x ′ g xx ′ x} and extend ϕ to a bijection from G •′ onto H by setting 1ϕ = 1, (1g xx ′ 1)ϕ = e, (x ′ g xx ′ x)ϕ = f .
where h i = (a i g xx ′ a i+1 )ϕ for g i = g xx ′ , and h i = g i ϕ for the other cases.The mapping ϕ is well defined since a i a i+1 ∈ {11, x ′ x} if g i = g xx ′ by condition (b) of the definition of MR • .Also by definition of ϕ, the word uϕ is clearly a nontrivial i-landscape; whence ϕ is a mapping from MR • to i-MR.
Extend now ψ to the inverse mapping of ϕ : G •′ → H and let v = h 0 h 1 • • • h 2n−1 h 2n be a nontrivial i-mountain range.Set g i = g xx ′ if h i ∈ {e, f }, or g i = h i ψ otherwise.If h i = e [h i = f ], set also a i = 1 = a i+1 [a i = x ′ and a i+1 = x]; whence h i ψ = a i g i a i+1 if h i ∈ {e, f }.If υ(h i−1 ) ≥ 2 and υ(h i ) ≥ 2, then only one of the letters g i−1 or g i is the left or the right entry of the other (and not both entries).Hence, there is only one anchor a i that turns the triplet g i−1 a i g i anchored.Define now vψ = g 0 a 1 g 1 • • • g 2n−1 a 2n g 2n .
By the previous observations and the definition of ψ, vψ is a mountain range from MR • .A close inspection allows us to conclude also that ϕ and ψ are inverse mappings since ϕ and ψ are inverse bijections too.
Note that both u ∈ MR • and uϕ have the same line graph structure: they differ only on the labels.In particular, ϕ | M • and ψ | i-M are mutually inverse bijections between the sets M • and i-M.We leave these observations registered in the following lemma for future reference.It is not so hard to show now that ϕ | M • is an isomorphism.In fact, if we look carefully to the proof of Lemma 6.2, we see that if and u g i − → v for some river g i , then uϕ by recursion.We have proved the following result: In [6,Corollary 6.10] we proved that each regular semigroup weakly generated by a finite set of idempotents strongly divides FI 2 , that is, it is a homomorphic image of a regular subsemigroup of FI 2 .Using the Corollary 6.6 we now get the following result: Corollary 6.7.All regular semigroups weakly generated by a finite set of idempotents strongly divide F 1 .
Note that the previous corollary applies in particular to all regular semigroup generated by a finite set of idempotents and to all finite idempotent generated semigroups.In [6,Corollary 6.11] we concluded that all finite semigroups divide FI 2 , that is, are homomorphic images of (not necessarily regular) subsemigroup of FI 2 .We now have also that: Corollary 6.8.All finite semigroups divide F 1 .

Figure 1 .Figure 2 .
Figure 1.Illustration of a landscape in a line graph.

Lemma 6 . 4 .
The mappings ϕ : MR • → i-MR and ψ : i-MR → MR • are mutually inverse.Further, their restrictions ϕ | M • and ψ | i-M to the sets of (nontrivial) mountains from MR • and (nontrivial) i-mountains, respectively, give us mutually inverse bijections between these sets.
• and ψ is well defined.Clearly ϕ | G • k and ψ | H k are mutually inverse bijections between G • k and H k for k ∈ {2, 3}.If we now assume that ϕ | G • j and ψ | H j are mutually inverse bijections between G • j and H j for all j < i, we can immediately conclude that ϕ | G • i and ψ | H i are mutually inverse bijections between G •