#### Abstract

A graph is said to be End-completely-regular (resp., End-inverse) if its endomorphism monoid End() is completely regular (resp., inverse). In this paper, we will show that if [] is End-completely-regular (resp., End-inverse), then both and are End-completely-regular (resp., End-inverse). We give several approaches to construct new End-completely-regular graphs by means of the lexicographic products of two graphs with certain conditions. In particular, we determine the End-completely-regular and End-inverse lexicographic products of bipartite graphs.

#### 1. Introduction and Preliminary Concepts

Endomorphism monoids of graphs are generalizations of automorphism groups of graphs. In recent years, much attention has been paid to endomorphism monoids of graphs and many interesting results concerning graphs and their endomorphism monoids have been obtained. The aim of this research is to develop further relationship between graph theory and algebraic theory of semigroups and to apply the theory of semigroups to graph theory. The bipartite graph is a class of famous graphs. Their endomorphism monoids are studied by several authors. In [1], the connected bipartite graphs whose endomorphism monoids are regular were explicitly found. In [2], Fan gave a characterization of connected bipartite graphs with an orthodox monoid. The joins of bipartite graphs with regular endomorphism monoids were characterized in [3]. The joins of bipartite graphs with completely regular endomorphism monoids were characterized in [4]. The endomorphism monoids and endomorphism regularity of graphs were considered by several authors (see [5–8]). In this paper, we will characterize the End-completely-regular and End-inverse lexicographic products of two graphs. We give several approaches to construct new End-completely-regular graphs by means of the lexicographic products of two graphs with certain conditions. In particular, we will determine the End-completely-regular and End-inverse lexicographic products of bipartite graphs.

The graphs considered in this paper are undirected finite simple graphs. The vertex set of is denoted by and the edge set of is denoted by . If two vertices and are adjacent in , the edge connecting and is denoted by and we write . A subgraph is called an* induced subgraph* of if for any , if and only if . A graph is called* bipartite* if has no odd cycle. It is known that if a graph is a bipartite graph, then its vertex set can be partitioned into two disjoint nonempty subsets such that no edge joins two vertices in the same set.

Let and be two graphs. The* join* of and , denoted by , is a graph such that and . The* lexicographic product* of and , denoted by , is a graph with vertex set , and with edge set , or and . Denote for any .

Let and be graphs. A mapping from to is called a* homomorphism* (from to ) if implies that . A homomorphism is called an* isomorphism* if is bijective and is a homomorphism. A homomorphism (resp., isomorphism) from to itself is called an* endomorphism* (resp.,* automorphism*) of (see [9]). The sets of all endomorphisms and automorphisms of are denoted by and , respectively. A graph is said to be* unretractive* if . For any , it is easy to see that if and only if is injective.

A* retraction* of a graph is a homomorphism from to a subgraph of such that the restriction of to is the identity mapping on . In this case, is called a* retract* of . It is known that the idempotents of are retractions of . Denote by the set of all idempotents of . Let be an endomorphism of a graph . A subgraph of is called the* endomorphic image* of under , denoted by , if and if and only if there exist and such that . By we denote the equivalence relation on induced by ; that is, for , if and only if . Denote by the equivalence class containing with respect to .

An element of a semigroup is called* regular* if there exists such that . An element of a semigroup is called* completely regular* if and hold for some . A semigroup is called* regular* (resp.,* completely regular*) if all its elements are regular (resp., completely regular). An* inverse semigroup* is a regular semigroup in which the idempotents commute. A graph is said to be End-regular (resp., End-completely-regular, End-inverse) if its endomorphism monoid is regular (resp., completely regular, inverse). Clearly, End-completely-regular graphs as well as End-inverse graphs are End-regular.

For undefined notation and terminology in this paper, the reader is referred to [9–14]. We list some known results which will be used in the sequel.

Lemma 1 (see [2]). *If is End-regular, then both and are End-regular.*

Lemma 2 (see [15]). *Let be a graph and let . Then is completely regular if and only if .*

Lemma 3 (see [15]). *Let be a bipartite graph. Then is End-completely-regular if and only if is one of , , , , , and .*

Lemma 4 (see [4]). *Let and be two bipartite graphs. Then is End-completely-regular if and only if one of them is End-completely-regular and the other is or .*

Lemma 5 (see [16]). *Let be a graph and . Then is completely regular if and only if there exists such that and .*

Lemma 6 (see [4]). *Let be a bipartite graph. Then is End-inverse if and only if or .*

Lemma 7 (see [17]). *Let and be two graphs. Then if and only if for any and , there exists such that .*

#### 2. Main Results

In this section, we will characterize the End-completely-regular and End-inverse lexicographic products of two graphs. We first show that if is End-completely-regular, then both and are End-completely-regular.

Theorem 8. *Let and be two graphs. If is End-completely-regular, then both and are End-completely-regular.*

*Proof. *By Lemma 2, to show that is End-completely-regular, it is only necessary to verify that is an automorphism of for each . Define a mapping from to itself by
Then . Since is End-completely-regular, by Lemma 2, is an automorphism of . It is easy to see that . For any distinct and , and hold. Since is an automorphism of , . Hence and so is an automorphism of .

Let . Define a mapping from to itself by
Then . Since is End-completely-regular, by Lemma 2, is an automorphism of . It is easy to see that . For any and , and . Since is an automorphism of , , we get that and so is an automorphism of , as required.

The following example shows that and being End-completely-regular does not yield that is End-completely-regular.

*Example 9. *Let and be two graphs with , , , and . By Lemma 3, and are End-completely-regular. It is easy to see that . Also by Lemma 3, this is not End-completely-regular.

In the following, we give some sufficient conditions for to be End-completely-regular. To this aim, we need the following result due to Fan [17].

Lemma 10 (see [17]). *Let and be two -free connected graphs. If or is odd, then , where is the wreath product of the monoids and .*

Let and be two -free connected graphs such that or is odd. In [2], Fan proved that if both of and are End-regular and one of them is unretractive, then is End-regular. Here we prove that if is an End-completely-regular graph and is an unretractive graph, then is End-completely-regular.

Theorem 11. *Let and be two -free connected graphs with or being odd, and assume that*(1)*is End-completely-regular,*(2)* is unretractive.**Then is End-completely-regular.*

*Proof. *Let and be two graphs satisfying the assumptions. To show that is End-completely-regular, we prove that for any , there exists an idempotent endomorphism such that and .

Let . Since , for some and . Thus, for any , there exists . Let and be -free connected graphs with or being odd. By Lemma 7, for any , for some . Note that is unretractive. Then . Since is End-completely-regular and , by Lemma 5, there exists such that and . Clearly, is an induced subgraph of . Hence is an induced subgraph of .

Since is End-completely-regular, is an automorphism of . Thus for any , there exists only one vertex such that . Then . Now for any , there exists only one vertex such that . Define a mapping from to itself in the following way. If , then ; if , then for some . Now let , where is the only vertex in such that . Then it is easy to see that is well-defined. Let . If , then . Thus . Hence . If , then . Hence . Therefore, for any .

Let be such that . If , then . If and , then and . Thus and . Since and , . Hence . If and , there are two cases.*Case **1.* Assume that . Then and . Since and , . Hence we have .*Case **2.* Assume that and . Then we have and . Since is an isomorphism from to , . Therefore, .

If , then . If , then . Thus . Hence . Clearly, .

Suppose for some . In fact, it is easy to prove that . Let . Then, by the definition of , we have for some with . So and thus . Hence .

Next we start to seek the conditions for bipartite graphs and under which is End-completely-regular.

Lemma 12. *Let be a graph and be a retract of . If is not End-completely-regular, then is not End-completely-regular.*

*Proof. *Let be a retract of . Then there exists such that . Let . Since is not End-completely-regular, there exists such that is not completely regular. By Lemma 2, is not an automorphism of . Thus there exist with such that . Define a mapping from to itself by
Then and . Now it is easy to see that and . It follows from that is not an automorphism of . Hence is not End-completely-regular.

Lemma 13. *Let and be two graphs. If at least one of and is not End-completely-regular, then is not End-completely-regular (where is the disjoint union of and ).*

*Proof. *Without loss of generality, we may suppose that is not End-completely-regular. By Lemma 2, there exists such that is not an automorphism of . Define a mapping from to itself by
Then . Now it is easy to see that and for any . Since is not an automorphism of , is not an automorphism of . Hence is not End-completely-regular.

Theorem 14. *Let and be two bipartite graphs. Then is End-completely-regular if and only if*(1)* and is End-completely-regular or*(2)* is End-completely-regular and or .*

*Proof **Sufficiency.* Since and , we have immediately that is End-completely-regular if and only if is End-completely-regular. If , then . Thus is a group. Since any group is a completely regular semigroup, is End-completely-regular. If and , then . By Lemma 3, it is End-completely-regular. In the following, we show that is End-completely-regular for the following cases (see Figure 1).*Case **1. * and . Let . If , then . Otherwise, or . Without loss of generality, we can suppose . Since is adjacent to every vertex of and , is adjacent to every vertex of . Note that there is no vertex in adjacent to vertices. This is a contradiction. Hence and so is completely regular. If , then or . Without loss of generality, we can suppose . Then . Otherwise, a similar argument as above will show that is adjacent to every vertex of , which yield a contradiction. Thus . Since any endomorphism maps a clique to a clique of the same size, . By Lemma 2, is completely regular. Hence is End-completely-regular.*Case **2. * and . Let . If , then . Otherwise, . Without loss of generality, we can suppose . Since any endomorphism maps a clique to a clique of the same size and there is only one clique of size in , . Note that . Then . Thus , which is a contradiction. Clearly, for any . Hence and so is completely regular. If , then for some . Without loss of generality, we can suppose . Then . Thus . Hence and is completely regular. Consequently, is End-completely-regular.*Case **3. * and . Let . If for any , then and so is completely regular. If for some with , without loss of generality, we can suppose . Since , , , is a clique of size in , , , , is also a clique of size in . Note that , , are adjacent to . Then , , are adjacent to . Thus and . Hence and is completely regular. Consequently, is End-completely-regular. *Necessity*. We only need to show that is not End-completely-regular in the following cases.*Case **1* *.* Then . By Lemma 4, is not End-completely-regular for the corresponding .*Case **2* *.* Then is a retract of . Since is not End-completely-regular for , , by Lemma 12, is not End-completely-regular for the corresponding .*Case **3* *.* Then . If is bipartite, then is also bipartite. By Lemma 3, is not End-completely-regular for the corresponding .*Case **4* *.* Then . Since is not End-completely-regular for , , by Lemma 13, is not End-completely-regular for the corresponding .*Case **5* *.* Then . Since is not End-completely-regular for , , by Lemma 13, is not End-completely-regular for the corresponding .

Next we start to seek the conditions for a lexicographic product of bipartite graphs and under which is End-inverse.

Theorem 15. *Let and be two graphs. If is End-inverse, then both and are End-inverse.*

*Proof. *Since is End-inverse, is End-regular. By Lemma 1, both and are End-regular. To show that is End-inverse, we only need to prove that the idempotents of commute.

Let and be two idempotents in . Define two mappings and from to itself by
Then and are two idempotents of and so , since is End-inverse. For any , we have
Clearly, . Hence is End-inverse.

Similarly, let and be two idempotents in . Define two mappings and from to itself by
Then and are two idempotents of and so , since is End-inverse. For any , we have
Clearly, . Hence is End-inverse, as required.

The next theorem characterizes the End-inverse lexicographic products of bipartite graphs.

Theorem 16. *Let and be two bipartite graphs. Then is End-inverse if and only if is one of , , , and .*

*Proof **Necessity.* This follows directly from Lemma 6 and Theorem 15. *Sufficiency*. It is easy to see that , , and are End-inverse, since they are unretractive.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors want to express their gratitude to the referees for their helpful suggestions and comments. This research was partially supported by the National Natural Science Foundation of China (nos. 11301151 and 11226047), the Key Project of the Education Department of Henan Province (no. 13A110249), and the Project of Science and Technology Department of Henan Province (no. 132300410411).