[Retracted] Idempotent Elements of the Endomorphism Semiring of a Finite Chain

Abstract

Idempotents yield much insight in the structure of finite semigroups and semirings. In this article, we obtain some results on (multiplicatively) idempotents of the endomorphism semiring of a finite chain. We prove that the set of all idempotents with certain fixed points is a semiring and find its order. We further show that this semiring is an ideal in a well-known semiring. The construction of an equivalence relation such that any equivalence class contains just one idempotent is proposed. In our main result we prove that such an equivalence class is a semiring and find its order. We prove that the set of all idempotents with certain jump points is a semiring.

1. Introduction

The idempotents play an essential role in the theory of finite semigroups and semirings. It is well known that in a finite semigroup some power of each element is an idempotent, so the idempotents can be taken to be like a generating system of the semigroup or the semiring. For deep results, using idempotents in the representation theory of finite semigroups, we refer the reader to [1, 2].

Let us briefly survey the contents of our paper. After the preliminaries, in Section 3 we show some facts about the fixed points of idempotent endomorphisms. The central result here is Theorem 9 where we prove that the set of all idempotents with 𝑠 fixed points 𝑘₁,…,𝑘_𝑠, 1≤𝑠≤𝑛−1, is a semiring of order ∏^𝑠−1_𝑚=1(𝑘_𝑚+1−𝑘_𝑚). Moreover, this semiring is an ideal of the semiring of all endomorphisms having at least 𝑘₁,…,𝑘_𝑠 as fixed points. In the next section we consider an equivalence relation on some finite semigroup 𝑆 such that for any 𝑥,𝑦∈𝑆 follows 𝑥∼𝑦 if and only if 𝑥^𝑘=𝑦^𝑚=𝑒, where 𝑘,𝑚∈𝐍 and 𝑒 is an idempotent of 𝑆. Then we consider the equivalence classes of semigroup 𝑆=(ℰ_𝒞𝑛,⋅) which is, see [3], one subsemigroup of 𝒫𝒯_𝑛. Here we investigate the so-called jump points of the endomorphism and prove that between any two fixed points 𝑘_𝑖 and 𝑘_𝑖+1 of an endomorphism, where 𝑘_𝑖+1≠𝑘_𝑖, there is a unique jump point.

The main result of the paper is Theorem 19 where we prove that such an equivalence class is a semiring of order 𝐶_𝑘1,1^ℓ−1_𝑖=1𝐶_𝑡𝑖𝐶_𝑠𝑖𝐶_{𝑛−1−𝑘ℓℓ,𝑚}, (1) where 𝐶_𝑝 is the 𝑝th Catalan number.

In the last section of the paper we consider idempotent endomorphisms with arbitrary fixed points but with certain jump points. Here we prove that the set of idempotent endomorphisms with identical jump points is a semiring.

2. Preliminaries

We consider some basic definitions and facts concerning finite semigroups and that can be found in any of [1, 2, 4, 5]. As the terminology for semirings is not completely standardized, we say what our conventions are.

An algebra 𝑅=(𝑅,+,⋅) with two binary operations + and  ⋅  on 𝑅 is called a semiring if(i)(𝑅,+) is a commutative semigroup,(ii)(𝑅,⋅) is a semigroup,(iii) both distributive laws hold 𝑥⋅(𝑦+𝑧)=𝑥⋅𝑦+𝑥⋅𝑧 and (𝑥+𝑦)⋅𝑧=𝑥⋅𝑧+𝑦⋅𝑧 for any 𝑥,𝑦,𝑧∈𝑅.

Let 𝑅=(𝑅,+,⋅) be a semiring. If a neutral element 0 of semigroup (𝑅,+) exists and it satisfies 0⋅𝑥=𝑥⋅0=0 for all 𝑥∈𝑅, then it is called zero. If a neutral element 1 of semigroup (𝑅,⋅) exists, it is called identity element.

For a semilattice ℳ the set ℰ_ℳ of the endomorphisms of ℳ is a semiring with respect to the addition and multiplication defined by(i) ℎ=𝑓+𝑔  when  ℎ(𝑥)=𝑓(𝑥)∨𝑔(𝑥)  for  all  𝑥∈ℳ,(ii)ℎ=𝑓⋅𝑔  when  ℎ(𝑥)=𝑓(𝑔(𝑥))  for  all  𝑥∈ℳ.

This semiring is called the endomorphism semiring of ℳ. In this paper all semilattices are finite chains. Following [6, 7], we fix a finite chain 𝒞_𝑛=({0,1,…,𝑛−1},∨) and denote the endomorphism semiring of this chain with ℰ_𝒞𝑛. We do not assume that 𝛼(0)=0 for arbitrary 𝛼∈ℰ_𝒞𝑛. So, there is not a zero in endomorphism semiring ℰ_𝒞𝑛. Subsemiring ℰ_𝒞𝑛=ℰ⁰_𝒞𝑛 of ℰ_𝒞𝑛 consisting of all endomorphisms 𝛼 with property 𝛼(0)=0 has zero and is considered in [6–8].

If 𝛼∈ℰ_𝒞𝑛 such that 𝑓(𝑘)=𝑖_𝑘 for any 𝑘∈𝒞_𝑛 we denote 𝛼 as an ordered 𝑛-tuple ≀𝑖₀,𝑖₁,𝑖₂,…,𝑖_𝑛−1≀. Note that mappings will be composed accordingly, although we will usually give preference to writing mappings on the right, so that 𝛼⋅𝛽 means “first 𝛼, then 𝛽.” The identity 𝐢=≀0,1,…,𝑛−1≀ and all constant endomorphisms 𝜅_𝑖=≀𝑖,…,𝑖≀ are obviously (multiplicatively) idempotents.

An element 𝑎∈𝒞_𝑛 satisfying 𝛼(𝑎)=𝑎 is called a fixed point of the endomorphism 𝛼. Any 𝛼∈ℰ_𝒞𝑛 has at most 𝑛 fixed points, and only identity 𝐢 has just 𝑛 fixed points.

For other properties of the endomorphism semiring we refer the reader to [3, 6–8].

In the following sections we use some terms from [4] having in mind that in [3] we show that some subsemigroups of the partial transformation semigroup are indeed endomorphism semirings.

3. Idempotent Endomorphisms and Their Fixed Points

The set of all idempotents (or idempotent elements) of semiring ℰ_𝒞𝑛 is not a semiring namely; endomorphisms 𝛼=≀0,0,2≀ and 𝛽=≀0,1,1≀ are the idempotents of semiring ℰ_𝒞3 but 𝛼𝛽=≀0,0,1≀ is not an idempotent. The following several facts are well known or are consequences of well-known facts.

Proposition 1. The endomorphism 𝛼∈ℰ_𝒞𝑛 is an idempotent if and only if for any 𝑘∈𝒞_𝑛, which is not a fixed point of 𝛼, image 𝛼(𝑘) is a fixed point of 𝛼.

Note that in [9] maps 𝛼_𝑘 are considered such that 𝛼_𝑘(𝑥)=𝑘 for all 𝑥∈ℳ, where ℳ is not necessarily a finite semilattice. These maps are called constant endomorphisms.

For any 𝛼∈ℰ_𝒞𝑛 with one fixed point the cardinality of set Im(𝛼) is an arbitrary number from 2 to 𝑛−1. Indeed, for 𝛼=≀𝑛−2,𝑛−1,…,𝑛−1≀ with unique fixed point 𝑛−1 we have |Im(𝛼)|=2 and for 𝛽=≀1,2,…,𝑛−2,𝑛−1,𝑛−1≀ with the same unique fixed point, |Im(𝛽)|=𝑛−1. So, the following consequence of Proposition 1 is important.

Corollary 2. An endomorphism with only one fixed point is an idempotent if and only if it is a constant.

More generally follows Corollary 3.

Corollary 3. An endomorphism 𝛼∈ℰ_𝒞𝑛 with 𝑠 fixed points 𝑘₁,…,𝑘_𝑠, 1≤𝑠≤𝑛−1, is an idempotent if and only if Im(𝛼)={𝑘₁,…,𝑘_𝑠}.

Let 𝛼∈ℰ_𝒞𝑛 have just 𝑛−1 fixed points. Then 𝛼≠𝐢 and 𝑛−1≤|Im(𝛼)|<𝑛. So, |Im(𝛼)|=𝑛−1 and from Corollary 3 follows.

Corollary 4. Every endomorphism 𝛼∈ℰ_𝒞𝑛 with 𝑛−1 fixed points is an idempotent.

Let 𝛼∈ℰ_𝒞𝑛 have just 𝑛−2 fixed points. Let 𝛼(𝑘)≠𝑘 and 𝛼(ℓ)≠ℓ, where 𝑘,ℓ∈𝒞_𝑛. If we assume that 𝑘 and ℓ are not consecutive, then 𝛼(𝑘)=𝑘−1 or 𝛼(𝑘)=𝑘+1. Similarly, 𝛼(ℓ)=ℓ−1 or 𝛼(ℓ)=ℓ+1. Since 𝑘−1, 𝑘+1, ℓ−1, and ℓ+1 are fixed points, from Proposition 1 it follows that 𝛼 is an idempotent. Let us assume that ℓ=𝑘+1. If (1)𝛼(𝑘)=𝑘−1,𝛼(𝑘+1)=𝑘−1, (2)𝛼(𝑘)=𝑘−1,𝛼(𝑘+1)=𝑘+2, (3)𝛼(𝑘)=𝑘+2,𝛼(𝑘+1)=𝑘+2,

then it easily follows that 𝛼 is an idempotent. Let us assume that 𝛼(𝑘)=𝑘+1 and 𝛼(𝑘+1)=𝑘+2. Then from Proposition 1 it follows that 𝛼 is not an idempotent endomorphism. But it is clear that 𝛼²(𝑘)=𝑘+2=𝛼³(𝑘) and 𝛼²(𝑘+1)=𝑘+2=𝛼³(𝑘+1).

Thus we prove the following.

Corollary 5. Every endomorphism 𝛼∈ℰ_𝒞𝑛 with 𝑛−2 fixed points is either an idempotent or 𝛼³=𝛼².

By similar reasonings, we may prove the following more general fact.

Corollary 6. Let 𝛼∈ℰ_𝒞𝑛 have 𝑠 fixed points 𝑘₁,…,𝑘_𝑠, where [(𝑛+1)/2]≤𝑠≤𝑛−1 and all other two points 𝑘 and ℓ, 𝑘<ℓ, are not consecutive; that is 𝑘≠𝑘_𝑖, ℓ≠𝑘_𝑖, where 𝑖=1,…,𝑠, and ℓ≠𝑘+1. Then 𝛼 is an idempotent.

Let 𝛼∈ℰ_𝒞𝑛 have just two fixed points 𝑘 and ℓ, where 𝑘<ℓ. From Corollary 3 it follows that 𝛼=≀𝑘,…,𝑘,𝑖_𝑘+1,…,𝑖_ℓ−1,ℓ,…,ℓ≀, (2) where images 𝑖_𝑠, 𝑠=𝑘+1,…,ℓ−1, are equal either to 𝑘 or to ℓ.

Thus the endomorphisms of such type are 𝛼₀𝑘↑𝑘ℓ↑ℓ⋯𝛼=≀𝑘,…,,…,𝑘,,…,ℓ≀_{ℓ−𝑘−2}𝑘↑𝑘ℓ↑ℓ𝛼=≀𝑘,…,,𝑘,ℓ…,,…ℓ≀_{ℓ−𝑘−1}𝑘↑𝑘ℓ↑ℓ=≀𝑘,…,,ℓ…,,…,ℓ≀. (3) For these maps we have 𝛼₀<⋯<𝛼_{𝑙−𝑘−2}<𝛼_{𝑙−𝑘−1}. (4) Hence 𝛼_𝑖+𝛼_𝑗=𝛼_𝑗+𝛼_𝑖=𝛼_𝑗, where 𝑗≥𝑖. It is easy to see that 𝛼_𝑖⋅𝛼_𝑗=𝛼_𝑖 for all 𝑖,𝑗∈{0,…𝑙−𝑘−2}. So, we prove the following.

Proposition 7. The set of idempotent endomorphisms with two fixed points 𝑘 and ℓ, 𝑘<ℓ, is a semiring of order  𝑙−𝑘.

Example 8. The idempotent endomorphisms of semiring  ℰ_𝒞7 with fixed points 1 and 5 are 𝜑₁𝜑=≀1111155≀,₂𝜑=≀1111555≀,₃𝜑=≀1115555≀,₄=≀1155555≀. (5) The semiring consisting of these maps has the following addition and multiplication tables: +𝜑₁𝜑₂𝜑₃𝜑₄𝜑₁𝜑₁𝜑₂𝜑₃𝜑₄𝜑₂𝜑₂𝜑₂𝜑₃𝜑₄𝜑₃𝜑₃𝜑₃𝜑₃𝜑₄𝜑₄𝜑₄𝜑₄𝜑₄𝜑₄⋅𝜑₁𝜑₂𝜑₃𝜑₄𝜑₁𝜑₁𝜑₁𝜑₁𝜑₁𝜑₂𝜑₂𝜑₂𝜑₂𝜑₂𝜑₃𝜑₃𝜑₃𝜑₃𝜑₃𝜑₄𝜑₄𝜑₄𝜑₄𝜑₄ (6)

Theorem 9. The subset of ℰ_𝒞𝑛,  𝑛≥3, of all idempotent endomorphisms with 𝑠 fixed points 𝑘₁,…,𝑘_𝑠, 1≤𝑠≤𝑛−1, is a semiring of order ∏^𝑠−1_𝑚=1(𝑘_𝑚+1−𝑘_𝑚).

Proof. Let 𝑘₁ be the least fixed point of idempotent endomorphism 𝛼. If 𝑘<𝑘₁, then 𝛼(𝑘)≤𝑘₁. We assume that 𝛼(𝑘)=𝑗<𝑘₁. Then 𝛼(𝑗)=𝑗 is a contradiction to the minimal choice of 𝑘₁. Hence 𝛼(𝑘)=𝑘₁, that is in the first 𝑘₁+1 positions of the ordered 𝑛-tuple which represent endomorphism 𝛼 occurs only 𝑘₁.
Let 𝑘_𝑠 be the biggest fixed point of idempotent endomorphism 𝛼. If 𝑘>𝑘_𝑠, then 𝛼(𝑘)≥𝑘_𝑠. We assume that 𝛼(𝑘)=𝑗>𝑘_𝑠.
Now follows 𝛼(𝑗)=𝑗 which is a contradiction to the maximal choice of 𝑘_𝑠. Hence 𝛼(𝑘)=𝑘_𝑠, that is the last 𝑛−𝑘_𝑠 positions of the ordered 𝑛-tuple which represent endomorphism 𝛼 occurs only 𝑘_𝑠.
We denote by 𝑝_𝑘𝑚, where 1≤𝑚≤𝑠, the number of coordinates in the ordered 𝑛-tuple which represents endomorphism 𝛼 equal to 𝑘_𝑚. For 𝑝_𝑘𝑚 it follows 𝑘₁+1≤𝑝_𝑘1≤𝑘₂1≤𝑝_𝑘2≤𝑘₃−𝑘₂⋮1≤𝑝_𝑘𝑚≤𝑘_𝑚+1−𝑘_𝑚⋮𝑛−𝑘_𝑠≤𝑝_𝑘𝑠≤𝑛−𝑘_𝑠−1−1. (7)
So every idempotent endomorphism with 𝑠 fixed points 𝑘₁,…,𝑘_𝑠 has the following type: 𝛼=≀𝑘₁,…,𝑘_1𝑝𝑘1,𝑘₂,…,𝑘_2𝑝𝑘2,…,𝑘_𝑠,…,𝑘_{𝑠𝑝𝑘𝑠}≀. (8)
Let another endomorphism 𝛽 have the same type: 𝛽=≀𝑘₁,…,𝑘_{1𝑝′𝑘1},𝑘₂,…,𝑘_{2𝑝′𝑘2},…,𝑘_𝑠,…,𝑘_{𝑠𝑝′𝑘𝑠}≀. (9)
Since (𝛼+𝛽)(𝑘_𝑖)=𝛼(𝑘_𝑖)+𝛽(𝑘_𝑖)=𝑘_𝑖+𝑘_𝑖=𝑘_𝑖, where 𝑖=1,…,𝑠, it follows that 𝛼+𝛽 has the type of 𝛼 and 𝛽.
For these 𝛼 and 𝛽 it follows that 𝛼⋅𝛽=𝛼. (So, the multiplication of idempotent endomorphisms with fixed points  𝑘₁,…,𝑘_𝑠 is like the multiplication of constant endomorphisms—every idempotent is a right identity).
Thus we prove that the set of all idempotent endomorphisms with 𝑠 fixed points 𝑘₁,…,𝑘_𝑠, 1≤𝑠≤𝑛−1, is a semiring.
From the inequalities of numbers 𝑝_𝑘𝑚, where 1≤𝑚≤𝑠, it follows that all possibilities to have 𝑘_𝑚  in  𝑛-tuple are 𝑘_𝑚+1−𝑘_𝑚.
The order of this semiring is equal to the product of all possibilities for the fixed points 𝑘_𝑚, where 1≤𝑚≤𝑠. This number is equal to product ∏^𝑠−1_𝑚=1(𝑘_𝑚+1−𝑘_𝑚), and the proof is completed.

The semiring of the idempotent endomorphisms of ℰ_𝒞𝑛 with 𝑠 fixed points 𝑘₁,…,𝑘_𝑠 is denoted by ℐ𝒟(𝑘₁,…,𝑘_𝑠). Since the semiring of all the endomorphisms with fixed points 𝑘₁,…,𝑘_𝑠 is ⋂^𝑠_𝑟=1ℰ^(𝑘𝑟)_𝒞𝑛, it follows that ℐ𝒟(𝑘₁,…,𝑘_𝑠) is a subsemiring of ⋂^𝑠_𝑟=1ℰ^(𝑘𝑟)_𝒞𝑛.

Remark 10. (a) From Theorem 9 semiring ℐ𝒟(0,…,𝑘−1,𝑘+1,…,𝑛−1) has two elements: idempotent endomorphisms 𝑘⁺=≀0,…,𝑘−1,𝑘+1,𝑘+1,…𝑛−1≀ and 𝑘⁻=≀0,…,𝑘−1,𝑘−1,𝑘+1,…𝑛−1≀. This semiring has the following addition and multiplication tables: +𝑘⁻𝑘⁺𝑘⁻𝑘⁻𝑘⁺𝑘⁺𝑘⁺𝑘⁺⋅𝑘⁻𝑘⁺𝑘⁻𝑘⁻𝑘⁻𝑘⁺𝑘⁺𝑘⁺ (10)
(b) The product of endomorphisms from different semirings ℐ𝒟(𝑘₁,…,𝑘_𝑠) is not, in general, an idempotent. Indeed, in semiring ℰ_𝒞8 for (),()𝛼=≀11555555≀∈ℐ𝒟1,5𝑔=≀22255555≀∈ℐ𝒟2,5 (11) it follows that 𝛼⋅𝛽=≀22555555≀, but this endomorphism is not an idempotent.
(c) The sum of two idempotent endomorphisms such that the fixed points of the first one are part of the fixed points of another endomorphism is not surely an idempotent. For instance, ≀0033≀+≀0223≀=≀0233≀.

Corollary 11. Let 𝑛≥3 and 𝑘₁,…,𝑘_𝑠∈𝒞_𝑛, where 𝑠=1,…,𝑛−2. Semiring ℐ𝒟(𝑘₁,…,𝑘_𝑠) is an ideal of ⋂^𝑠_𝑟=1ℰ^(𝑘𝑟)_𝒞𝑛.

Proof. Using Theorem 9, it will be enough to show that ℐ𝒟(𝑘₁,…,𝑘_𝑠) is closed under the left and right multiplications by elements of ⋂^𝑠_𝑟=1ℰ^(𝑘𝑟)_𝒞𝑛.
Let 𝛼=≀𝑘₁,…,𝑘_1𝑝𝑘1,𝑘₂,…,𝑘_2𝑝𝑘2𝑘,…,_𝑠,…,𝑘_{𝑠𝑝𝑘𝑠}≀∈ℐ𝒟𝑘₁,…,𝑘_𝑠,𝛽=≀𝑖₀,𝑖₁,…,𝑖_𝑘1−1,𝑘₁,𝑖_𝑘1+1𝑖,…,_𝑘2−1,𝑘₂,…,𝑘_𝑠≀∈^𝑠_𝑟=1ℰ^(𝑘𝑟)_𝒞𝑛. (12) Hence 𝛼⋅𝛽=𝛼.
Now we calculate 𝛽⋅𝛼=≀𝑖₀,𝑖₁,…,𝑖_𝑘1−1,𝑘₁,𝑖_𝑘1+1,…,𝑖_𝑘2−1,𝑘₂,…,𝑘_𝑠≀⋅≀𝑘₁,…,𝑘₁,𝑘₂,…,𝑘₂,…,𝑘_𝑠,…,𝑘_𝑠≀. (13)
Since 𝑖_𝑠≤𝑘₁, where 𝑠=0,…,𝑘₁−1, then 𝛼(𝑖_𝑠)≤𝛼(𝑘₁)=𝑘₁, and from the proof of Theorem 9 it follows that 𝛼(𝑖_𝑠)=𝑘₁.
Since 𝑘₁≤𝑖_𝑘1+1≤⋯≤𝑖_𝑘2−1≤𝑘₂, it follows that 𝑘₁=𝛼𝑘₁≤𝛼𝑖_𝑘1+1𝑖≤⋯≤𝛼_𝑘2−1≤𝛼𝑘₂=𝑘₂. (14)
But endomorphism 𝛼 maps all the elements between 𝑘₁ and 𝑘₂ either in 𝑘₁, or in 𝑘₂. Then 𝛼(𝑖_𝑘1+1),…,𝛼(𝑖_𝑘2−1) are either 𝑘₁, or 𝑘₂. We use such arguments for the next elements of Im(𝛽) which are between the other fixed points. Thus 𝛽⋅𝛼=𝛾=≀𝑘₁,…,𝑘_{1𝑝′𝑘1},𝑘₂,…,𝑘_{2𝑝′𝑘2}𝑘,…,_𝑠,…,𝑘_{𝑠𝑝′𝑘𝑠}≀∈ℐ𝒟𝑘₁,…,𝑘_𝑠. (15)

4. Roots of Idempotent Endomorphisms

Let (𝑆,⋅) be a finite semigroup. It is well known, see [10], that for any 𝑥∈𝑆 there is a positive integer 𝑘=𝑘(𝑥) such that 𝑎^𝑘 is an idempotent element of 𝑆.

Now we consider the following relation: for any 𝑥,𝑦∈𝑆 we define 𝑥𝑥∼𝑦⟺∃𝑘,𝑚∈𝐍,_𝑘=𝑦^𝑚=𝑒, (16) where 𝑒 is an idempotent element of 𝑆.

Obviously, the relation ∼ is reflexive and symmetric.

Let 𝑥∼𝑦 and 𝑦∼𝑧. Then 𝑥^𝑘=𝑦^𝑚=𝑒₁ and 𝑦^𝑟=𝑧^𝑠=𝑒₂ where 𝑒₁ and 𝑒₂ are idempotents. Now it follows that (𝑦^𝑚)^𝑟=𝑒^𝑟₁=𝑒₁ and (𝑦^𝑟)^𝑚=𝑒^𝑚₂=𝑒₂; thus 𝑒₁=𝑒₂. Hence 𝑥^𝑘=𝑧^𝑠=𝑒₁; that is, 𝑥∼𝑧. So, we prove that ∼ is an equivalence relation on 𝑆.

Note that two different idempotents belong to different equivalence classes modulo ∼. If 𝑒 is an idempotent, the elements of the equivalence class containing 𝑒 are called roots of idempotent 𝑒.

The following natural question arises: are there any finite noncommutative semigroups such that the equivalence classes (modulo the previous relation) are semigroups?

Of course, we must avoid trivial examples of commutative semigroups, nilpotent semigroups, and others.

To answer the previous question, we consider the semigroup (ℰ_𝒞𝑛,⋅) and the already defined equivalence relation.

Let 𝛼∈ℰ_𝒞𝑛. Element 𝐣∈𝒞_𝑛 is called a jump point of 𝛼 if 𝐣≠0, and one of the following conditions holds:(1)𝛼(𝐣−1)≤𝐣−1 and 𝛼(𝐣)>𝐣,(2)𝛼(𝐣−1)<𝐣−1 and 𝛼(𝐣)≥𝐣.

There are endomorphisms without jump points, namely, identity 𝐢 and constant endomorphisms 𝜅_𝑘=≀𝑘𝑘⋯𝑘≀, 𝑘=0,1,…,𝑛−1. For 𝑘,ℓ∈𝒞_𝑛, 𝑘<ℓ endomorphism 𝛼_𝑗(𝑖)=𝑘,𝑖≤𝑗−1,ℓ,𝑖≥𝑗 (17) has no jump points if 𝑗>ℓ. The endomorphisms 𝑘⁺ and 𝑘⁻, see Remark 10, have jump points: 𝑘 and 𝑘+1 are jump points of 𝑘⁺ and 𝑘⁻, respectively.

Theorem 12. Let 𝛼∈ℰ_𝒞𝑛, 𝑛≥3, be an endomorphism with 𝑠 fixed points 𝑘₁,…,𝑘_𝑠, 1≤𝑠≤𝑛−2. Let for some 𝑖, 𝑖=1,…,𝑠−1, fixed points 𝑘_𝑖 and 𝑘_𝑖+1 be non consecutive; that is, 𝑘_𝑖+1≠𝑘_𝑖+1. Then there is a unique jump point 𝐣_𝑖 of 𝛼 such that 𝑘_𝑖+1≤𝐣_𝑖≤𝑘_𝑖+1.

Proof. For 𝑘_𝑖+1 there are two possibilities:(1)𝛼(𝑘_𝑖+1)>𝑘_𝑖+1, then 𝐣_𝑖=𝑘_𝑖+1 is the searched jump point of 𝛼, or(2)𝛼(𝑘_𝑖+1)<𝑘_𝑖+1.
Let 𝛼(𝑘_𝑖+𝑚)<𝑘_𝑖+𝑚 for any 𝑚=1,2,…,ℓ−1. If 𝛼(𝑘_𝑖+ℓ)≥𝑘_𝑖+ℓ, then 𝐣_𝑖=𝑘_𝑖+ℓ is the searched jump point of 𝛼.
We assume that 𝛼(𝑘_𝑖+𝑚)<𝑘_𝑖+𝑚 for all 𝑚=1,2,…,𝑘_𝑖+1−𝑘_𝑖−1. Now 𝛼(𝑘_𝑖+1−1)<𝑘_𝑖+1−1. But 𝛼(𝑘_𝑖+1)=𝑘_𝑖+1, so, 𝐣_𝑖=𝑘_𝑖+1 is the searched jump point of 𝛼.
Suppose that 𝐣 and 𝐣^∘ are two jump points of 𝛼 such that 𝑘_𝑖+1≤𝐣≤𝑘_𝑖+1 and 𝑘_𝑖+1≤𝐣^∘≤𝑘_𝑖+1. Let 𝐣<𝐣^∘. Now 𝛼(𝐣)>𝐣. Let 𝑚∈𝒞_𝑛, and 𝐣≤𝑚<𝐣^∘ is the maximal element such that 𝛼(𝑚)>𝑚. Then 𝛼(𝑚+1)<𝑚+1, and it follows that 𝛼(𝑚)≥𝑚+1>𝛼(𝑚+1) which is a contradiction. By the same arguments we show that 𝐣^∘<𝐣 is impossible. So 𝐣=𝐣^∘ is the unique jump point of 𝛼 such that 𝑘_𝑖+1≤𝐣≤𝑘_𝑖+1.

Let 𝑘₁ be the least fixed point of endomorphism 𝛼 and 𝑖∈𝒞_𝑛, 𝑖<𝑘₁. If we suppose that 𝛼(𝑖)<𝑖, using 𝛼(0)>0, it follows that for some maximal element 𝑚, where 0≤𝑚<𝑖, we have 𝛼(𝑚)>𝑚 and then 𝛼(𝑚+1)<𝑚+1, which is impossible. So, for every 𝑖<𝑘₁ it follows that 𝛼(𝑖)>𝑖. Similarly for every 𝑖>𝑘_𝑠, where 𝑘_𝑠 is the biggest fixed point of 𝛼, it follows that 𝛼(𝑖)<𝑖. So, we prove the following.

Corollary 13. Every endomorphism 𝛼∈ℰ_𝒞𝑛, 𝑛≥3, with just ℓ fixed points 𝑘₁,…,𝑘_ℓ, which are not consecutive, that is, 𝑘_𝑖+1≠𝑘_𝑖+1 for 𝑖=1,…,ℓ−1, has just ℓ−1 jump points 𝐣_𝑖 such that 𝑘_𝑖+1≤𝐣_𝑖≤𝑘_𝑖+1.

Remark 14. From Theorem 9 it follows that the number of idempotents with fixed points 𝑘₁,…,𝑘_𝑠 is a semiring of order ∏^𝑠−1_𝑖=1(𝑘_𝑖+1−𝑘_𝑖). But if two fixed points 𝑘_𝑖 and 𝑘_𝑖+1 are consecutive, the difference 𝑘_𝑖+1−𝑘_𝑖 is equal to 1, and they do not appear in the product. So, the order of the semiring is equal to ∏^ℓ−1_𝑖=1(𝑘_𝑖+1−𝑘_𝑖), where 𝑘₁,…,𝑘_ℓ are (after suitable renumbering) non consecutive fixed points. But for any 𝑖 the difference 𝑘_𝑖+1−𝑘_𝑖 is just the number of possible jump points 𝐣_𝑖,𝑡=𝑘_𝑖+𝑡, where 𝑡=1,…,𝑘_𝑖+1−𝑘_𝑖. So, for given fixed points 𝑘₁,…,𝑘_𝑠 the order of semiring of idempotents is equal to the number of all ℓ−1-tuples (𝐣_1,𝑡1,𝐣_2,𝑡2,…,𝐣_{ℓ−1,𝑡ℓ−1}), where 𝐣_{𝑖,𝑡𝑖}=𝑘_𝑖+𝑡_𝑖 for 𝑡_𝑖=1,…,𝑘_𝑖+1−𝑘_𝑖 and 𝑖=1,…,ℓ−1. Hence, any ℓ−1-tuple (𝐣_1,𝑡1,…,𝐣_{ℓ−1,𝑡ℓ−1}) defines just one idempotent from the semiring considered.
To describe precisely all the fixed points of an arbitrary endomorphism 𝛼∈ℰ_𝒞𝑛 we will use new indices. Let first fixed point of 𝛼 be 𝑘_1,1 and let some fixed points after 𝑘_1,1 be consecutive; that is, 𝑘_1,2,…,𝑘_1,𝑚1 are fixed points such that 𝑘_1,𝑖+1−𝑘_1,𝑖=1 where 𝑖=1,…,𝑚₁−1. Let the next fixed point be 𝑘_2,1 such that 𝑘_2,1−𝑘_1,𝑚1>1. So we construct the first pair of two fixed points which are not consecutive. Let the following fixed points be 𝑘_2,2,…,𝑘_2,𝑚2 such that 𝑘_2,𝑖+1−𝑘_2,𝑖=1 where 𝑖=1,…,𝑚₂−1. The next fixed point is 𝑘_3,1 and 𝑘_3,1−𝑘_2,𝑚2>1. Let the last pair of two non consecutive fixed points be 𝑘_{ℓ−1,𝑚ℓ−1} and 𝑘_ℓ,1. Then the last fixed points are 𝑘_ℓ,2,…𝑘_ℓ,𝑚ℓ such that 𝑘_ℓ,𝑖+1−𝑘_ℓ,𝑖=1 where 𝑖=1,…,𝑚_ℓ−1. So, we construct a partition of a set of fixed points of 𝛼 such that we can distinguish the fixed points which are not consecutive.
Let 𝐣_{𝑖,𝑡𝑖} be the jump points of 𝛼 such that 𝐣_{𝑖,𝑡𝑖}=𝑘_{𝑖,𝑚𝑖}+𝑡_𝑖, where 𝑡_𝑖=1,…,𝑘_𝑖+1,1−𝑘_{𝑖,𝑚𝑖} and 𝑖=1,…,ℓ−1.
An endomorphism 𝛼 with fixed points 𝑘_1,1,…,𝑘_ℓ,𝑚ℓ and jump points 𝐣_{𝑖,𝑡𝑖} from the previous definitions is called a endomorphism of type[𝑘_1,1,…,𝑘_1,𝑚1,𝐣_1,𝑡1,𝑘_2,1,…,𝑘_{ℓ−1,𝑚ℓ−1},𝐣_{ℓ−1,𝑡ℓ−1},𝑘_ℓ,1,…,𝑘_ℓ,𝑚ℓ]. (18)
Let us consider endomorphism 𝜀 of this type such that(i)𝜀(𝑥)=𝑘_1,1 for any 0≤𝑥≤𝑘_1,1,(ii)𝜀(𝑥)=𝑘_{𝑖,𝑚𝑖} for any 𝑘_{𝑖,𝑚𝑖}≤𝑥≤𝐣_{𝑖,𝑡𝑖}−1, where 𝑖=1,…,ℓ−1,(iii)𝜀(𝑥)=𝑘_𝑖+1,1 for any 𝐣_{𝑖,𝑡𝑖}≤𝑥≤𝑘_𝑖+1,1, where 𝑖=1,…,ℓ−1,(iv)𝜀(𝑥)=𝑘_ℓ,𝑚ℓ for any 𝑘_ℓ,𝑚ℓ≤𝑥≤𝑛−1.

Now it is easy to show that this endomorphism is an idempotent.

Let 𝜀 be another idempotent of the same type. Then from the reasonings just before Corollary 13 it follows that(i)𝜀(𝑥)=𝑘_1,1 for any 0≤𝑥≤𝑘_1,1, (ii)𝜀(𝑥)=𝑘_ℓ,𝑚ℓ for any 𝑘_ℓ,𝑚ℓ≤𝑥≤𝑛−1.

Since 𝜀 is an idempotent, we conclude that for some 𝑥, where 𝑘_{𝑖,𝑚𝑖}≤𝑥≤𝑘_𝑖+1,1, it follows either 𝜀(𝑥)=𝑘_{𝑖,𝑚𝑖}, or 𝜀(𝑥)=𝑘_𝑖+1,1. By using that  𝜀  is of type (18) it follows that 𝜀(𝑥)=𝜀(𝑥) for all 𝑥, where 𝑘_{𝑖,𝑚𝑖}≤𝑥≤𝑘_𝑖+1,1 and 𝑖=1,…,ℓ−1. Hence, there is only one idempotent of the given type (18). This endomorphism is 𝜀=≀𝑘_1,1,…,𝑘_1,1↑𝑘_1,1,…,𝑘_1,𝑚1↑𝑘_1,𝑚1,…,𝑘_1,𝑚1,𝑘_2,1↑𝐣_1,𝑡1𝑘,…,_{ℓ−1,𝑚ℓ−1},𝑘_ℓ,1↑𝐣_{ℓ−1,𝑡ℓ−1},…,𝑘_ℓ,𝑚ℓ↑𝑘_ℓ,𝑚ℓ,…,𝑘_ℓ,𝑚ℓ≀. (19)

Lemma 15. Let 𝜀∈ℰ_𝒞𝑛 be an idempotent endomorphism. If 𝛼∈ℰ_𝒞𝑛 is a root of 𝜀, then both endomorphisms 𝛼 and 𝜀 have the same fixed points.

Proof. Let 𝑘 be a fixed point of 𝜀. If 𝛼(𝑘)<𝑘, then 𝛼²(𝑘)≤𝛼(𝑘)<𝑘, and by the same arguments 𝛼^𝑚(𝑘)<𝑘 for arbitrary natural number 𝑚, but this is a contradiction to 𝛼^𝑚=𝜀 for some 𝑚∈𝐍. Analogously, assuming that 𝛼(𝑘)>𝑘, we obtain a contradiction to equality 𝛼^𝑚=𝜀. Hence 𝛼(𝑘)=𝑘.
Let 𝑘 be a fixed point of 𝛼. Since it is a fixed point of 𝛼^𝑚 for arbitrary natural 𝑚, it follows that it is a fixed point of 𝜀.

Lemma 16. Let 𝜀∈ℰ_𝒞𝑛 be an idempotent endomorphism. If 𝛼∈ℰ_𝒞𝑛 is a root of 𝜀, then both endomorphisms 𝛼 and 𝜀 have the same jump points.

Proof. Let 𝐣 be a jump point of 𝜀 and 𝜀(𝐣−1)≤𝐣−1, 𝜀(𝐣)>𝐣. If we assume that 𝛼(𝐣−1)>𝐣−1, then 𝛼^𝑚(𝐣−1)>𝐣−1 for arbitrary natural 𝑚 which is a contradiction to 𝛼^𝑚=𝜀 for some 𝑚∈𝐍.
Analogously, if we assume that 𝛼(𝐣)≤𝐣, it follows that 𝛼^𝑚(𝐣)≤𝐣 for arbitrary natural 𝑚 which is also a contradiction. Hence 𝐣 is a jump point of 𝛼 such that 𝛼(𝐣−1)≤𝐣−1 and 𝛼(𝐣)>𝐣. From similar reasonings, it follows that if 𝐣 is a jump point of 𝜀 such that 𝜀(𝐣−1)<𝐣−1 and 𝜀(𝐣)≥𝐣 for every endomorphism 𝛼, which is a root of 𝜀, we have that 𝐣 is a jump point of 𝛼, 𝛼(𝐣−1)<𝐣−1, and 𝛼(𝐣)≥𝐣.
Let 𝐣 be a jump point of 𝛼 such that 𝛼(𝐣−1)≤𝐣−1 and 𝛼(𝐣)>𝐣. Then for arbitrary natural 𝑚 it follows 𝛼^𝑚(𝐣−1)≤𝐣−1 and 𝛼^𝑚(𝐣)>𝐣. So, 𝐣 is a jump point of 𝜀, 𝜀(𝐣−1)≤𝐣−1 and 𝜀(𝐣)>𝐣. Analogously, 𝛼(𝐣−1)<𝐣−1 and 𝛼(𝐣)≥𝐣 imply 𝛼^𝑚(𝐣−1)<𝐣−1 and 𝛼^𝑚(𝐣)≥𝐣 for every natural number 𝑚, so 𝐣 is a jump point of 𝜀, 𝜀(𝐣−1)<𝐣−1 and 𝜀(𝐣)≥𝐣.

Immediately from Lemmas 15 and 16 follows Proposition 17.

Proposition 17. All the endomorphisms of one equivalence class modulo ∼ are of the same type.

Let the idempotent endomorphism 𝜀 from (19) be an element of equivalence class 𝐸. Then 𝐸 is called an equivalence class of type (18).

Lemma 18. Let 𝐸 be an equivalence class of type (18). Then any 𝛼∈𝐸 satisfies the following conditions:(A)𝛼(𝑥)>𝑥, where 0≤𝑥<𝑘_1,1, 𝐣_1,𝑡1≤𝑥<𝑘_2,1, 𝐣_2,𝑡2≤𝑥<𝑘_3,1, …, or 𝐣_{ℓ−1,𝑡ℓ−1}≤𝑥<𝑘_ℓ,1;(B)𝛼(𝑥)<𝑥, where 𝑘_1,𝑚1<𝑥<𝐣_1,𝑡1, 𝑘_2,𝑚2<𝑥<𝐣_2,𝑡2, …, 𝑘_{ℓ−1,𝑚ℓ−1}<𝑥<𝐣_{ℓ−1,𝑡ℓ−1}, or 𝑘_ℓ,𝑚ℓ<𝑥≤𝑛−1.   

Proof. We assume that for some 𝑥 such that 0≤𝑥<𝑘_1,1,or𝐣_1,𝑡1≤𝑥<𝑘_2,1,or𝐣_2,𝑡2≤𝑥<𝑘_3,1,…,or𝐣_{ℓ−1,𝑡ℓ−1}≤𝑥<𝑘_ℓ,1 (20) it follows that 𝛼(𝑥)≤𝑥. Then 𝛼²(𝑥)≤𝛼(𝑥)≤𝑥 and by induction 𝛼^𝑚(𝑥)≤𝑥 for arbitrary natural 𝑚. It is a contradiction to the assumption that 𝛼^𝑚=𝜀 for some natural 𝑚, where 𝜀 has the form (19). Thus we prove that 𝛼(𝑥)>𝑥.
Analogously, we assume that for some 𝑥 such that 𝑘_1,𝑚1<𝑥<𝐣_1,𝑡1,or𝑘_2,𝑚2<𝑥<𝐣_2,𝑡2,…,or𝑘_{ℓ−1,𝑚ℓ−1}<𝑥<𝐣_{ℓ−1,𝑡ℓ−1},or𝑘_ℓ,𝑚ℓ<𝑥≤𝑛−1 (21) it follows 𝛼(𝑥)≥𝑥. Then by induction 𝛼^𝑚(𝑥)≥𝑥 for arbitrary natural 𝑚. This contradicts the assumption that 𝛼^𝑚=𝜀 for some natural 𝑚, where 𝜀 has the form (19). Thus we prove that 𝛼(𝑥)<𝑥.

Note that if two endomorphisms have the same fixed points but different jump points, their product can posses a new fixed point. For instance, endomorphisms 𝛼=≀2,2,2,4,5,5,5≀∼𝜀₁=≀2,2,2,5,5,5,5≀ and 𝛽=≀2,2,2,2,3,5,5≀∼𝜀₂=≀2,2,2,2,2,5,5≀ have fixed points 2 and 5 but 𝛼⋅𝛽=≀2,2,2,3,5,5,5≀ and 𝛽⋅𝛼=≀2,2,2,2,4,5,5≀ have new fixed points 3 and 4, respectively.

Let us remind that the jump points, which we use in (18), are defined by equality 𝐣_{𝑖,𝑡𝑖}=𝑘_{𝑖,𝑚𝑖}+𝑡_𝑖. So it follows that 𝐣_{𝑖,𝑡𝑖}−𝑘_{𝑖,𝑚𝑖}=𝑡_𝑖. Now we denote 𝑘_𝑖+1,1−𝐣_{𝑖,𝑡𝑖}=𝑠_𝑖. The indices 𝑡_𝑖 and 𝑠_𝑖 will be used in the next theorem.

Theorem 19. Every equivalence class modulo ∼ of type (18) is a subsemiring of ℰ_𝒞𝑛, 𝑛≥2. The order of this semiring is 𝐶_𝑘1,1^ℓ−1_𝑖=1𝐶_𝑡𝑖𝐶_𝑠𝑖𝐶_{𝑛−1−𝑘ℓℓ,𝑚}, (22) where 𝐶_𝑝 is the 𝑝th Catalan number.

Proof. Let 𝐸 be an equivalence class of type (18) and 𝛼,𝛽∈𝐸. Using Lemma 18 it follows that(A)𝛼(𝑥)>𝑥 and 𝛽(𝑥)>𝑥, where 0≤𝑥<𝑘_1,1, 𝐣_1,𝑡1≤𝑥<𝑘_2,1, 𝐣_2,𝑡2≤𝑥<𝑘_3,1,…, or 𝐣_{ℓ−1,𝑡ℓ−1}≤𝑥<𝑘_ℓ,1. Then (𝛼+𝛽)(𝑥)=𝛼(𝑥)∨𝛽(𝑥)>𝑥∨𝑥=𝑥 and (𝛼⋅𝛽)(𝑥)=𝛽(𝛼(𝑥))≥𝛽(𝑥)>𝑥.(B)𝛼(𝑥)<𝑥 and 𝛽(𝑥)<𝑥, where 𝑘_1,𝑚1<𝑥<𝐣_1,𝑡1, 𝑘_2,𝑚2<𝑥<𝐣_2,𝑡2, …, or 𝑘_{ℓ−1,𝑚ℓ−1}<𝑥<𝐣_{ℓ−1,𝑡ℓ−1}, or 𝑘_ℓ,𝑚ℓ<𝑥≤𝑛−1. Then (𝛼+𝛽)(𝑥)=𝛼(𝑥)∨𝛽(𝑥)<𝑥∨𝑥=𝑥 and (𝛼⋅𝛽)(𝑥)=𝛽(𝛼(𝑥))≤𝛽(𝑥)<𝑥.

Thus we prove that equivalence class 𝐸 is a subsemiring of ℰ_𝒞𝑛.

For the second part of the proof of the theorem we needed two combinatorial lemmas.

Lemma 20. The number of the ordered 𝑝-tuples (𝑖₀,…,𝑖_𝑝−1), where (1)𝑖_𝑟∈{0,…,𝑝−1} for 𝑟=0,…,𝑝−1,(2)𝑖_𝑟≤𝑖_𝑟+1 for 𝑟=0,…,𝑝−1,(3)𝑖_𝑟<𝑟 for 𝑟=1,…,𝑝−1,

is the 𝑝th Catalan number 𝐶_𝑝=(1/𝑝)^2𝑝−1_𝑝−1.

The proof of this lemma is a part of the proof of Proposition 3.4 of [7]. This result (using the so-called Dyck paths) and many other applications of Catalan numbers can be found in [11].

Lemma 21. The number of the ordered 𝑝-tuples (𝑖₀,…,𝑖_𝑝−1), where (1)𝑖_𝑟∈{1,…,𝑝} for 𝑟=0,…,𝑝−1, 𝑖_𝑝=𝑝,(2)𝑖_𝑟≤𝑖_𝑟+1 for 𝑟=0,…,𝑝−1,(3)𝑖_𝑟>𝑟 for 𝑟=0,…,𝑝−1

is the 𝑝th Catalan number 𝐶_𝑝=(1/𝑝)^2𝑝−1_𝑝−1.

Proof of Lemma 21. Let (𝑖₀,…,𝑖_𝑝−1) be the ordered 𝑝-tuple. Since 𝑖_𝑟>𝑟 for any 𝑟=0,…,𝑝−1, follows that 𝑝=𝑖_𝑝≥𝑖_𝑝−1>𝑝−1; that is, 𝑖_𝑝−1=𝑝. Instead of the 𝑝-tuple (𝑖₀,…,𝑖_𝑝−1) we consider the 𝑝-tuple (𝑗₀,…,𝑗_𝑝−1), where 𝑗_𝑚=𝑝−𝑖_{𝑝−𝑚−1}, 𝑚=0,…,𝑝−1. Then 𝑗_𝑚<𝑚 and 𝑗_𝑚≤𝑗_𝑚+1. From Lemma 20 it follows that the number of all 𝑘-tuples of this kind is 𝐶_𝑘.
Now we continue the proof of the theorem.
For all the intervals considered in part (A) (from the beginning of this proof) we apply Lemma 21. Then, for all the intervals considered in part (B) we apply Lemma 20. Hence we find that the number of the endomorphisms in the equivalence class modulo ∼ of type (1) is 𝐶_𝑘1,1(∏^ℓ−1_𝑖=1𝐶_𝑡𝑖𝐶_𝑠𝑖)𝐶_{𝑛−1−𝑘ℓℓ,𝑚}.

Remark 22. The equivalence relation ∼ is not a congruence on ℰ_𝒞𝑛. For 𝑛=8 we consider the endomorphisms 𝛼=≀2,2,2,2,2,6,7,7≀∼𝜀₁=≀2,2,2,2,2,7,7,7≀,𝛽=≀2,2,2,2,3,7,7,7≀∼𝜀₁=≀2,2,2,2,2,7,7,7≀,𝛾=≀1,1,1,3,3,4,5,6≀∼𝜀₂=≀1,1,1,3,3,3,3,3≀. (23) Now we compute 𝛼⋅𝛾=≀1,1,1,1,1,5,6,6≀ and 𝛽⋅𝛾=≀1,1,1,1,3,6,6,6≀. So, it follows that 𝛼∼𝛽, but 𝛼⋅𝛾≁𝛽⋅𝛾.

5. The Crucial Role of Jump Points

Here we consider the idempotent endomorphisms with arbitrary fixed points but we assume that each endomorphism has 𝐣₁,…,𝐣_𝑟 for jump points.

Two jump points 𝐣_𝑠 and 𝐣_𝑠+1 of the idempotent 𝜀 are called consecutive if 𝐣_𝑠+1=𝐣_𝑠+1. First, let us answer the question: are there any consecutive jump points of the idempotent endomorphism?

Yes, for instance, 𝜀=≀1,1,1,3,5,5≀∈ℰ_𝒞6 is an idempotent and jump points 3 and 4 are consecutive. Note that 3 is also a fixed point of 𝜀. So, we modify the question: If the first jump point is not a fixed point, is it possible fort the next point to be a jump point?

The answer is negative. Indeed, if 𝜀 is an idempotent and 𝜀(𝐣)=𝑘>𝑗, then from Proposition 1 it follows that 𝜀(𝑘)=𝑘, and, since 𝐣+1≤𝑘, we have 𝜀(𝐣+1)≤𝑘. So, 𝐣+1 is not a jump point of 𝜀.

Lemma 23. Let 𝜀 be an idempotent and 𝐣_𝑠 and let 𝐣_𝑠+1 be non consecutive jump points of 𝜀. Then in interval [𝐣_𝑠,𝐣_𝑠+1−1] one of the following holds:(1)𝜀 is a constant endomorphism.(2)𝜀 is an identity.(3)𝜀 is an identity in interval [𝐣_𝑠,𝑘] and a constant endomorphism in interval [𝑘,𝐣_𝑠+1−1].(4)𝜀 is a constant endomorphism in interval [𝐣_𝑠,𝑘], and an identity in interval [𝑘,𝐣_𝑠+1−1].(5)𝜀 is a constant endomorphism in interval [𝐣_𝑠,𝑘], an identity in interval [𝑘,ℓ] and a constant endomorphism in interval [ℓ,𝐣_𝑠+1−1].

Proof. Let 𝜀(𝐣_𝑠)=𝐣_𝑠. Since 𝐣_𝑠+1 is not a jump point, there are two possibilities: 𝜀(𝐣_𝑠+1)=𝐣_𝑠 or 𝜀(𝐣_𝑠+1)=𝐣_𝑠+1. In the first case the equality 𝜀(𝐣_𝑠+2)=𝐣_𝑠+1 is impossible; see Proposition 1. So, either 𝜀(𝐣_𝑠+2)≥𝐣_𝑠+2 and 𝐣_𝑠+2 is the next jump point or 𝜀(𝐣_𝑠+2)=𝐣_𝑠. From these reasonings, it follows that 𝜀 is a constant endomorphism in interval [𝐣_𝑠,𝐣_𝑠+1−1]. When 𝜀(𝐣_𝑠+1)=𝐣_𝑠+1 for point 𝐣_𝑠+2, there are three possibilities. If 𝜀(𝐣_𝑠+2)>𝐣_𝑠+2, then 𝐣_𝑠+2 is the next jump point and 𝜀 is an identity in the interval between the considered jump points. If 𝜀(𝐣_𝑠+2)=𝐣_𝑠+1, by the previous reasonings, we find that 𝜀 is an identity in interval [𝐣_𝑠,𝐣_𝑠+1] and a constant endomorphism in the second interval. Let 𝜀(𝐣_𝑠+2)=𝐣_𝑠+2. Now if for any 𝑘≤𝐣_𝑠+1−1, it follows that 𝜀(𝐣_𝑠+𝑘)=𝐣_𝑠+𝑘. Then 𝜀 is an identity in the whole interval. If for some 𝑘 we have 𝜀(𝐣_𝑠+𝑘+1)=𝜀(𝐣_𝑠+𝑘)=𝐣_𝑠+𝑘, then 𝜀 is an identity in interval [𝐣_𝑠,𝑘] and a constant endomorphism in interval [𝑘,𝐣_𝑠+1−1].
Let 𝜀(𝐣_𝑠)=𝑘>𝐣_𝑠. Since 𝜀 is an idempotent, it follows that 𝜀(𝑘)=𝑘; that is 𝜀 is a constant endomorphism in interval [𝐣_𝑠,𝑘]. If 𝜀(𝑘+1)=𝑘 then 𝜀 is a constant endomorphism in the whole interval. Let 𝜀(𝑘+1)=𝑘+1, and for any ℓ, where ℓ=𝑘+1,…,𝐣_𝑠+1−1, we have 𝜀(ℓ)=ℓ. So, 𝜀 is a constant endomorphism in interval [𝐣_𝑠,𝑘] and an identity in interval [𝑘,𝐣_𝑠+1−1]. In the end, if there is some ℓ, ℓ=𝑘+1,…,𝐣_𝑠+1−1, such that 𝜀(ℓ+1)=𝜀(ℓ)=ℓ, then 𝜀 is a constant endomorphism in interval [𝐣_𝑠,𝑘], an identity in interval [𝑘,ℓ], and a constant endomorphism in interval [ℓ,𝐣_𝑠+1−1].

It is straightforward to show that if one of the conditions (1)–(5) of Lemma 23 holds for endomorphism 𝜀, then 𝜀 is an idempotent.

From Lemma 23 it follows that the graph of the arbitrary idempotent endomorphism has the shape displayed on Figure 1. Note that each of the three segments of this graph can have length zero.

Figure 1 

Let 𝐣_𝑠 and 𝐣_𝑠+1 be consecutive jump points of the idempotent 𝜀. Then the graph of 𝜀 is a particular case of those displayed in Figure 1 containing one point (𝐣_𝑠,𝜀(𝐣_𝑠)), where 𝜀(𝐣_𝑠)=𝐣_𝑠.

Lemma 24. Let 𝜀 and 𝜀 be idempotent endomorphisms of ℰ_𝒞𝑛 with the same jump points 𝐣₁,…,𝐣_𝑟. Then 𝜀+𝜀 is also an idempotent endomorphism with the same jump points.

Proof. First we will prove that 𝐣₁,…,𝐣_𝑟 are jump points of endomorphism 𝜀+𝜀. Let 𝐣 be any common jump point of 𝜀 and 𝜀. If 𝐣 is also a fixed point of 𝜀 and 𝜀, that is, both idempotents satisfied the inequalities 𝜀(𝐣−1)<𝐣−1 and 𝜀(𝐣)≥𝐣, 𝜀(𝐣−1)<𝐣−1 and 𝜀(𝐣)≥𝐣, then (𝜀+𝜀)(𝐣−1)<𝐣−1 and (𝜀+𝜀)(𝐣)≥𝐣. When for 𝜀 it follows that 𝜀(𝐣−1)≤𝐣−1 and 𝜀(𝐣)>𝐣 or for 𝜀 it follows that 𝜀(𝐣−1)≤𝐣−1 and 𝜀(𝐣)>𝐣, then (𝜀+𝜀)(𝐣−1)≤𝐣−1 and (𝜀+𝜀)(𝐣)>𝐣.
Now we will prove that sum 𝜀+𝜀 is an idempotent endomorphism. Let us, like in Lemma 23, fix two nonconsecutive jump points 𝐣_𝑠 and 𝐣_𝑠+1 of idempotents 𝜀 and 𝜀 and the interval [𝐣_𝑠,𝐣_𝑠+1−1].
When the idempotent 𝜀 satisfies some the of conditions (1)–(5) of Lemma 23, we denote this by 𝜀₍₁₎,…,𝜀₍₅₎.
When 𝜀 satisfies (1) and 𝜀 satisfies (5), we denote this by 𝜀₍₁₎&𝜀₍₅₎. Let, in this case, 𝜀 be equal to constant 𝐶 in interval [𝐣_𝑠,𝐣_𝑠+1−1], and 𝜀 is consecutively equal to constant 𝑘, identity 𝐢, and constant ℓ. If 𝐶≤𝑘 or 𝐶≥ℓ, then 𝜀+𝜀=𝐶. If 𝑘<𝐶<ℓ, then 𝜀+𝜀 satisfy condition (5). This reasoning we denote by 𝜀₍₁₎&𝜀₍₅₎⇒(𝜀+𝜀)₍₁₎  or  (𝜀+𝜀)₍₅₎.
The verification of all fifteen possibilities is easy: 𝜀₍₁₎&𝜀₍₁₎⟹(𝜀+𝜀)₍₁₎,𝜀₍₁₎&𝜀₍₂₎⟹(𝜀+𝜀)₍₁₎or(𝜀+𝜀)₍₂₎or(𝜀+𝜀)₍₄₎,𝜀₍₁₎&𝜀₍₃₎⟹(𝜀+𝜀)₍₁₎or(𝜀+𝜀)₍₃₎or(𝜀+𝜀)₍₅₎,𝜀₍₁₎&𝜀₍₄₎⟹(𝜀+𝜀)₍₁₎or(𝜀+𝜀)₍₄₎or(𝜀+𝜀)₍₅₎,𝜀₍₁₎&𝜀₍₅₎⟹(𝜀+𝜀)₍₁₎or(𝜀+𝜀)₍₅₎,𝜀₍₂₎&𝜀₍₂₎⟹(𝜀+𝜀)₍₂₎,𝜀₍₂₎&𝜀₍₃₎⟹(𝜀+𝜀)₍₂₎,𝜀₍₂₎&𝜀₍₄₎⟹(𝜀+𝜀)₍₄₎,𝜀₍₂₎&𝜀₍₅₎⟹(𝜀+𝜀)₍₄₎,𝜀₍₃₎&𝜀₍₃₎⟹(𝜀+𝜀)₍₃₎,𝜀₍₃₎&𝜀₍₄₎⟹(𝜀+𝜀)₍₄₎,𝜀₍₃₎&𝜀₍₅₎⟹(𝜀+𝜀)₍₅₎,𝜀₍₄₎&𝜀₍₄₎⟹(𝜀+𝜀)₍₄₎,𝜀₍₄₎&𝜀₍₅₎⟹(𝜀+𝜀)₍₄₎,𝜀₍₅₎&𝜀₍₅₎⟹(𝜀+𝜀)₍₅₎. (24)
Thus we prove that the sum of two idempotent endomorphisms in interval [𝐣_𝑠,𝐣_𝑠+1−1] is also an idempotent in this interval. So, it follows that 𝜀+𝜀 is an idempotent endomorphism in interval [𝐣₁,𝐣_𝑟]. For 0≤𝑥<𝐣₁ each of idempotents 𝜀 and 𝜀 satisfies one of conditions (1)–(5) of Lemma 23. So, from the previous reasonings it follows that sum 𝜀+𝜀 is an idempotent endomorphism in [0,𝐣₁−1]. In a similar way we show that 𝜀+𝜀 is an idempotent endomorphism in [𝐣_𝑟,𝑛−1]. Hence, 𝜀+𝜀 is an idempotent endomorphism in the whole chain 𝒞_𝑛.

Lemma 25. Let 𝜀 and 𝜀 be idempotent endomorphisms of ℰ_𝒞𝑛 with the same jump points 𝐣₁,…,𝐣_𝑟. Then 𝜀⋅𝜀 is also an idempotent endomorphism with the same jump points.

Proof. First we will prove that 𝐣₁,…,𝐣_𝑟 are the jump points of endomorphism 𝜀⋅𝜀. Let 𝐣 be any common jump point of 𝜀 and 𝜀. Let 𝐣 be also a fixed point of 𝜀 or of 𝜀; that is, 𝜀(𝐣−1)<𝐣−1 and 𝜀(𝐣)≥𝐣 or 𝜀(𝐣−1)<𝐣−1 and 𝜀(𝐣)≥𝐣. Then it follows that in all cases (𝜀⋅𝜀)(𝐣−1)=𝜀(𝜀(𝐣−1))<𝐣−1 and (𝜀⋅𝜀)(𝐣)=𝜀(𝜀(𝐣))≥𝐣. Let 𝐣 be not a fixed point of both idempotents; that is, 𝜀(𝐣−1)≤𝐣−1 and 𝜀(𝐣)>𝐣 and also 𝜀(𝐣−1)≤𝐣−1 and 𝜀(𝐣)>𝐣. At last, we obtain that (𝜀⋅𝜀)(𝐣−1)=𝜀(𝜀(𝐣−1))≤𝜀(𝐣−1)≤𝜀(𝐣−1) and (𝜀⋅𝜀)(𝐣)=𝜀(𝜀(𝐣−1))>𝜀(𝐣−1)>𝜀(𝐣−1).
Now we will prove that 𝜀⋅𝜀 is an idempotent endomorphism. Like in the proof of the previous lemma, we fix two nonconsecutive jump points 𝐣_𝑠 and 𝐣_𝑠+1 of idempotents 𝜀 and 𝜀 and interval [𝐣_𝑠,𝐣_𝑠+1−1].
Using the notations from the proof of Lemma 24 we easily verify all twenty-five possibilities: 𝜀₍₁₎⋅𝜀₍₁₎=𝜀₍₁₎,𝜀₍₁₎⋅𝜀₍₂₎=𝜀₍₁₎,𝜀₍₁₎&𝜀₍₃₎⟹(𝜀⋅𝜀)₍₁₎,𝜀₍₁₎&𝜀₍₄₎⟹(𝜀⋅𝜀)₍₁₎,𝜀₍₁₎&𝜀₍₅₎⟹(𝜀⋅𝜀)₍₁₎,𝜀₍₂₎⋅𝜀₍₁₎=𝜀₍₁₎,𝜀₍₂₎⋅𝜀₍₂₎=𝜀₍₂₎,𝜀₍₂₎⋅𝜀₍₃₎=𝜀₍₃₎,𝜀₍₂₎⋅𝜀₍₄₎=𝜀₍₄₎,𝜀₍₂₎⋅𝜀₍₅₎=𝜀₍₅₎,𝜀₍₃₎⋅𝜀₍₁₎=𝜀₍₁₎,𝜀₍₃₎⋅𝜀₍₂₎=𝜀₍₃₎,𝜀₍₃₎&𝜀₍₃₎⟹(𝜀⋅𝜀)₍₃₎,𝜀₍₃₎&𝜀₍₄₎⟹(𝜀⋅𝜀)₍₁₎or(𝜀⋅𝜀)₍₅₎,𝜀₍₃₎&𝜀₍₅₎⟹(𝜀⋅𝜀)₍₁₎or(𝜀⋅𝜀)₍₅₎,𝜀₍₄₎⋅𝜀₍₁₎=𝜀₍₁₎,𝜀₍₄₎⋅𝜀₍₂₎=𝜀₍₄₎,𝜀₍₄₎&𝜀₍₃₎⟹(𝜀⋅𝜀)₍₁₎or(𝜀⋅𝜀)₍₅₎,𝜀₍₄₎&𝜀₍₄₎⟹(𝜀⋅𝜀)₍₄₎,𝜀₍₄₎&𝜀₍₅₎⟹(𝜀⋅𝜀)₍₁₎or(𝜀⋅𝜀)₍₅₎,𝜀₍₅₎⋅𝜀₍₁₎=𝜀₍₁₎,𝜀₍₅₎⋅𝜀₍₂₎=𝜀₍₅₎,𝜀₍₅₎&𝜀₍₃₎⟹(𝜀⋅𝜀)₍₁₎or(𝜀⋅𝜀)₍₅₎,𝜀₍₅₎&𝜀₍₄₎⟹(𝜀⋅𝜀)₍₁₎or(𝜀⋅𝜀)₍₅₎,𝜀₍₅₎&𝜀₍₅₎⟹(𝜀⋅𝜀)₍₁₎or(𝜀⋅𝜀)₍₅₎. (25)
Thus we prove that the product of both idempotent endomorphisms in interval [𝐣_𝑠,𝐣_𝑠+1−1] is also an idempotent in this interval. It easily implies that 𝜀⋅𝜀 is an idempotent endomorphism in the whole interval [𝐣₁,𝐣_𝑟]. In interval [0,𝐣₁−1] each one of the idempotents 𝜀 and 𝜀 satisfies one of the conditions (1)–(5), of Lemma 23. So, from the previous reasonings it follows that the product 𝜀⋅𝜀 is an idempotent endomorphism in [0,𝐣₁−1]. In a similar way we show that 𝜀⋅𝜀 is an idempotent endomorphism in [𝐣_𝑟,𝑛−1]. Hence, 𝜀⋅𝜀 is an idempotent endomorphism in the whole 𝒞_𝑛.

Immediately from Lemmas 24 and 25 follows Theorem 26.

Theorem 26. The set of the idempotent endomorphisms of ℰ_𝒞𝑛 with the same jump points is a subsemiring of ℰ_𝒞𝑛.

From Lemmas 24 and 25, and Figure 1 it is easy to prove the following.

Proposition 27. The set of the idempotent endomorphisms of ℰ_𝒞𝑛 without jump points is a subsemiring of ℰ_𝒞𝑛 of order 2𝑛+1.