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The join of split graphs whose completely regular endomorphisms form a monoid

  • Hailong Hou , Yanhua Song and Rui Gu EMAIL logo
Published/Copyright: June 22, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

In this paper, completely regular endomorphisms of the join of split graphs are investigated. We give conditions under which all completely regular endomorphisms of the join of two split graphs form a monoid.

Keywords: Endomorphism; Monoid; Join of split graphs; Completely regular
MSC 2010: 05C25; 20M20

1 Introduction and preliminaries

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 (cf. [1–4]). The endomorphism monoids of graphs have valuable applications (cf. [1]) and are related to automata theory (cf. [2, 5]). Let X be a graph. Denote by End(X) the set of all endomorphisms of X. It is known that End(X) forms a monoid with respect to composition of mappings. We call End(X) the endomorphism monoid of X. An element a of a semigroup S is said to be regular if there exists x ∈ S such that axa = a. Let f ∈ End(X). Then f is called a regular endomorphism of X if it is a regular element in End (X). Denote by rEnd(X) the set of all regular endomorphisms of X. For a monoid S, the composition of its two regular elements is not regular in general. So it is natural to ask: Under what conditions does the set rEnd(X) form a monoid for a graph X? In [6], Hou, Gu and Shang characterized the regular endomorphisms of the join of split graphs and the conditions under which the regular endomorphisms of the join of split graphs form a monoid were given. The endomorphism monoids of split graphs and the joins of split graphs were studied by several authors (cf. [7–13]).

An element a of a semigroup S is said to be completely regular if a = axa and xa = ax hold for some x ∈ S. Let f ∈ End(X) for a graph X. Then f is called a completely regular endomorphism of X if it is a completely regular element in End(X). Denote by cEnd(X) the set of all completely regular endomorphisms of graph X. In general, the composition of its two completely regular elements is also not completely regular for a monoid S. So it is natural to ask: Under what conditions does the set cEnd(X) form a monoid for X? However, it seems difficult to obtain a general answer to this question. Therefore a natural strategy for work towards answering this question is to find various kinds of conditions for various kinds of graphs. In [14], completely regular endomorphisms of split graphs were characterized and the conditions under which the completely regular endomorphisms of split graphs form a monoid were given. In this paper we give an answer to this question in the range of the joins of split graphs.

The graphs considered in this paper are finite undirected graphs without loops and multiple edges. The vertex set of X is denoted by V(X) and the edge set of X is denoted by E(X). If two vertices x1 and x2 are adjacent in X, the edge joining them is denoted by {x1, x2}. For a vertex v of X, denote by NX(v) (or just by N(v)) the set {x ∈ V(X)|{x, v} ∈ E(X)}. The cardinality of NX(v) is called the degree of v in X and is denoted by dX(v) (or just d(v)). A subgraph H is called an induced subgraph of X if for any a, b ∈ H, {a, b} ∈ E(H) if and only if {a, b} ∈ E(X). A graph X is called complete if {a, b} ∈ E(X) for any a, b ∈ V(X). We denote by Kn (or just K) a complete graph with n vertices. A clique of a graph X is a maximal complete subgraph of X. A subset K ⊆ V(X) is said to be complete if {a, b} ∈ E(X) for any two vertices a, b ∈ K with a ≠ b. A subset S ⊆ V(X) is said to be independent if {a, b} ∉ E(X) for any two vertices a, b ∈ S. A graph X is called a split graph if its vertex set V(X) can be partitioned into disjoint (non-empty) sets K and S such that K is a complete set and S is an independent set. In this paper, we always assume that a split graph X has a fixed partition V(X) = K ∪ S, where K is a maximum complete set and S is an independent set. Since K is a maximum complete set of X, 0 ≤ dX(y) ≤ n−1 for any y ∈ S. Let X and Y be two graphs. The join of X and Y, denoted by X + Y, is a graph such that V(X + Y) = V(X)∪ V(Y) and E(X + Y) = E(X)∪ E(Y) ∪ {{x, y} | x ∈ V(X), y ∈ V(Y)}.

Let X and Y be two graphs. A mapping f from V(X) to V(Y) is called a homomorphism (from X to Y) if {x1, x2} ∈ E(X) implies that {f(x1), f(x2)} ∈ E(Y). A homomorphism f from X to itself is called an endomorphism of X. A endomorphism f is said to be half-strong if {f(a), f(b)} ∈ E(X) implies that there exist x1, x2 ∈ V(X) with f(x1) = f(a) and f(x2) = f(b) such that {x1, x2} ∈ E(X). Denote by hEnd(X) the set of all half-strong endomorphisms of X.

A retraction of a graph X is an endomorphism f from X to a subgraph Y of X such that the restriction f |Y to V(Y) is the identity mapping on V(Y). It is well known that the idempotents of End(X) are retractions of X. Denote by Idpt(X) the set of all idempotents of End(X). Let f ∈ End(X). A subgraph of X is called the endomorphic image of X under f, denoted by If, if V(If) = f(V(X)) and {f(a), f(b)} ∈ E(If) if and only if there exist c ∈ f−1(f(a)) and d ∈ f−1(f(b)) such that {c, d} ∈ E(X). By ρf we denote the equivalence relation on V(X) induced by f, i.e., for a, b ∈ V(X), (a, b) ∈ ρf if and only if f(a) = f(b). Denote by [a]ρf the equivalence class containing a ∈ V(X) with respect to ρf.

We use the standard terminology and notation of semigroup theory as in [2, 15] and of graph theory as in [16]. We list some known results which are used in this paper.

Lemma 1.1

([17]). Let X be a graph and f ∈ End(X). Then

  1. f ∈ hEnd(X) if and only if If is an induced subgraph of X.

  2. If f is regular, then f ∈ hEnd(X).

Lemma 1.2

([18]). Let G be a graph and f ∈ End(G). Then f is completely regular if and only if there exists g ∈ I dpt(G) such that ρg = ρf and Ig = If.

Lemma 1.3

([6]). Let X + Y be a join of split graphs and f ∈ End(X + Y). Then the following statements are equivalent.

  1. There exists h ∈ I dpt(X + Y) such that Ih = If.

  2. If is an induced subgraph of X + Y and {x, y} ∉ E(X + Y) for any x ∈ Ki \ If and y ∈ Si ∩ If (where i = 1, 2).

Lemma 1.4

([6]). Let X + Y be a join of split graphs and f ∈ End(X + Y). Then there exists g ∈ I dpt(X + Y) such that ρg = ρf if and only if there exists b ∈ [a]ρf such that N(b) = ∪x∈[a]ρf N(x) for any a ∈ V(X + Y).

2 Main results

Let X be a split graph with V(X) = K1 ∪ S1, where K1 = {k1, k2, ⋯, kn} is a maximal complete set and S1 = {x1, x2, ⋯, xs} is an independent set. Let Y be another split graph with V(Y) = K2 ∪ S2, where K2 = {r1, r2, ⋯, rm} is a maximal complete set and S2 = {y1, y2, ⋯, yt} is an independent set. It is easy to see that the vertex set V(X + Y) of X + Y can be partitioned into three parts K, S1 and S2, i.e., V(X + Y) = K ∪ S1 ∪ S2, where K = K1 ∪ K2 is a complete set, S1 and S2 are independent sets. Obviously, the subgraph of X + Y induced by K is complete and the subgraph of X + Y induced by S = S1 ∪ S2 is complete bipartite. Hence in graph X + Y, N(xi) = NX(xi) ∪ V(Y) for xi ∈ S1 and N(yi) = NY(yi)∪ V(X) for yi ∈ S2. Clearly, X + Y is a split graph extended by the edge set {{xi, yj} | xi ∈ S1, yj ∈ S2}. In this section, we investigate the completely regular endomorphisms of X + Y and give the conditions under which the completely regular endomorphisms of X + Y form a monoid.

First, we give a characterization of the completely regular endomorphisms for X + Y.

Theorem 2.1

Let X + Y be a join of split graphs and f ∈ End(X + Y). Then f is completely regular if and only if the following conditions hold:

  1. If is an induced subgraph of X + Y and {x, y} ∉ E(X + Y) for any x ∈ K \ If and y ∈ S ∩ If.

  2. N(b) = ∪x∈[b]ρf N(x) for any b ∈ V(If) with [b]ρf⊆ S.

  3. f(a)≠ f(b) for any a, b ∈ V(If) with a ≠ b.

Proof

Necessity. Since f is completely regular, by Lemma 1.2, there exists g ∈ I dpt(X + Y) such that ρg = ρf and Ig = If. By Lemma 1.3, If is an induced subgraph of X + Y and {x, y} ∉ E(X + Y) for any x ∈ K \ If and y ∈ S ∩ If. Hence (1) holds. Let a ∈ V(If). Note that g ∈ I dpt(X + Y) and Ig = If, then g(a) = a. It follows from ρg = ρf that g(x) = a for any x ∈ [a]ρf. Let b ∈ V(If) be such that [b]ρf⊆ S. Then b ∈ S and N(b)∩ K ⊆ V(If) by (1). Thus g(x) = x for any x ∈ N(b)∩ K and so N(b)∩ K ⊆ g(N(b)). We claim that N(b) = ∪x∈[b]ρf N(x). Otherwise, there exists y ∈ [b]ρf such that N(y)⊈ N(b). Then there exists k ∈ K such that k ∈ N(y) and k ∉ N(b). Note that {k, t} ∈ E for any t ∈ N(b). Then {g(k), g(t)} ∈ E and so g(k)∉ g(N(b)). In particular, g(k) ∉ N(b)∩ K. If g(k)∈ N(b)∩ S, then g2 (k) = g(k) since g is idempotent. Since {k, g(k)} ∈ E(X + Y), g(k) forms a loop in X + Y. This is a contradiction. Hence g(k)∉ N(b). Now we get that {g(y), g(k)} = {b, g(k)}∉ E(X + Y). This contradicts {y, k} ∈ E. Hence (2) holds. If f(a)= f(b) for some a, b ∈ V(If), then [a]ρf = [b]ρf. Note that g ∈ I dpt(X + Y) and ρg = ρf, then g(a) = g(b). This means that {a, b} ⊈ V(Ig) and so Ig ≠ If, which yields a contradiction. Hence (3) holds.

Sufficiency. Let X + Y be a join of split graphs and f ∈ End(X + Y) be such that (1), (2) and (3). Note that f(a) ≠ f(b) for any a, b ∈ V(If). Then for any a ∈ V(X + Y), there exists only one vertex in [a]ρf∩ V(If). Denote it by [a]ρf. Define a mapping g from V(X + Y) to itself by

g(a)=[a]ρf¯foralla∈V(X+Y).

Then g is well-defined. In the following, we show that g ∈ End(X + Y). Let {a, b} ∈ E(X + Y) for some a, b ∈ V(X + Y). Since K is a maximum complete set of X + Y, f(K) is a clique of size n + m. We have K ⊆ If, |K ∩ If| = n + m − 1 or |K ∩ If| = n + m − 2.

Assume that K ⊆ If. Then g(k) = k for any k ∈ K. If a, b ∈ V(If), then g(a) = a and g(b) = b. Thus {g(a), g(b)} = {a, b} ∈ E(X + Y). If a ∈ V(If) and b ∉ V(If), then b ∈ S. If [b]ρf⊆ S, then g(b) = [b]ρf. Note that N([b]ρf) = ∪x ∈ [b]ρf N(x). Then a ∈ N([b]ρf). Hence {g(a), g(b)} = {a, [b]ρf} ∈ E(X + Y); If [b]ρf ⊈ S, then there exists k ∈ K ∩[b]ρf. Obviously, {a, k} ∈ E(X + Y). Hence {g(a), g(b)} = {a, k} ∈ E(X + Y). If a ∉ V(If) and b ∉ V(If), then a, b ∈ S. Without loss of generality, we may assume that a ∈ S1 and b ∈ S2. If g(a)∈ S, then g(a) = [a]ρf for some [a]ρf ∈ S1; If g(a)∈ K, then g(a) = k1 for some k1 ∈ K1. Similarly, if g(b)∈ S, then g(a) = [b]ρf for some [b]ρf ∈ S2; If g(b)∈ K, then g(b) = k2 for some k2 ∈ K2. It is a routine manner to check that {g(a), g(b)} ∈ E(X + Y) for every case. Consequently, g ∈ End(X + Y).

Assume that |K ∩ If| = n + m − 1. Then there exists x1 ∈ K \ If. Since any endomorphism f maps a clique to a clique of the same size, f(K) is a clique of size n + m in X + Y. Thus there exist y1 ∈ S ∩ If such that y1 is adjacent to every vertex of K \ {x1}. Now g(x1) = y1. If a, b ∈ V(If), then g(a) = a and g(b) = b. Thus {g(a), g(b)} = {a, b} ∈ E(X + Y). If a = x1 and b ∈ K\ {x1}, then g(b) = b. Thus {g(a), g(b)} = {y1, b} ∈ E(X + Y). If a = x1 and b ∈ S, then b ∉ V(If) by (1). Now [b]ρf⊈ S. Otherwise, there exists [b]ρf ∈ V(If)∩ S such that N([b]ρf) = ∪x∈[b]ρf N(x). Since {x1, b} ∈ E(X + Y), {x1, [b]ρf} ∈ E(X). Note that g([b]ρf) = [b]ρf ∈ Ig = If. This contradicts (1). Then there exists k1 ∈ K \ {x1} such that k1 ∈ [b]ρf. Thus {g(a), g(b)} = {y1, k1} ∈ E(X + Y). If a ∈ K \ {x1} and b ∈ S \ If, then g(a) = a. If [b]ρf⊆ S, then g(b) = [b]ρf. Note that N([b]ρf) = ∪x∈[b]ρf N(x). Then a ∈ N([b]ρf). Thus {g(a), g(b)} = {a, [b]ρf} ∈ E(X + Y); If [b]ρf ⊈ S, then there exists k ∈ K ∩[b]ρf. Thus {g(a), g(b)} = {a, g(k)}. If k ∈ K \ {x1}, then g(k) = k and so {a, g(k)} = {a, k} ∈ E(X + Y); If k = x1, then g(k) = y1 and so {a, g(k)} = {a, y1} ∈ E(X + Y). If a ∈ S \ V(If) and b ∈ S \ V(If), without loss of generality, we may assume that a ∈ S1 and b ∈ S2. If g(a)∈ S, then g(a) = [a]ρf for some [a]ρf ∈ S1; If g(a)∈ K, then g(a) = k1 for some k1 ∈ K1. Similarly, if g(b)∈ S, then g(a) = [b]ρf for some [b]ρf ∈ S2; If g(b)∈ K, then g(b) = k2 for some k2 ∈ K2. It is a routine manner to check that {g(a), g(b)} ∈ E(X + Y) for each cases. Consequently, g ∈ End(X + Y).

Assume that |K ∩ If| = n + m − 2. Then there exist x1∈ K1 \ If and x2∈ K2 \ If. Note that f(K) is a clique of size n + m in X + Y. Then there exist y1, y2 ∈ S ∩ If such that y1 is adjacent to every vertex of K \ {x1} and y2 is adjacent to every vertex of K \ {x2}. Obviously, g(x1) = y1 and g(x2) = y2. If a, b ∈ V(If), then g(a) = a and g(b) = b. Thus {g(a), g(b)} = {a, b} ∈ E(X + Y). If a ∈ {x1, x2} and b ∈ K \ {x1, x2}, then g(b) = b. Without loss of generality, we may assume that a = x1. Thus {g(a), g(b)} = {y1, b} ∈ E(X + Y). If a ∈ {x1, x2} and b ∈ S, then b ∉ V(If) by (1). Without loss of generality, we may assume that a = x1. We claim that [b]ρf⊈ S. Otherwise, there exists [b]ρf ∈ V(If)∩ S such that N([b]ρf) = ∪x∈[b]ρf N(x). Since {x1, b} ∈ E(X + Y), {x1, [b]ρf} ∈ E(X + Y). Note that g([b]ρf) = [b]ρf ∈ Ig = If. This contradicts (1). Then there exists k1 ∈ K \ {x1} such that k1 ∈ [b]ρf. If k1≠ x2, then {g(a), g(b)} = {y1, k1} ∈ E(X + Y); If k1 = x2, then {g(a), g(b)} = {y1, y2} ∈ E(X + Y). If a ∈ K \ {x1, x2} and b ∈ S \ If, then g(a) = a. If [b]ρf⊆ S, then g(b) = [b]ρf. Note that N([b]ρf) = ∪x ∈ [b]ρf N(x). Then a ∈ N([b]ρf). Thus {g(a), g(b)} = {a, [b]ρf} ∈ E(X + Y); If [b]ρf ⊈ S, then there exists k ∈ K ∩[b]ρf. Thus {g(a), g(b)} = {a, g(k)}. If k ∈ K \ {x1, x2}, then g(k) = k and so {a, g(k)} = {a, k} ∈ E(X + Y); If k ∈ {x1, x2}, then g(k)∈ {y1, y2} and so {a, g(k)} = {a, yi} (where i = 1 or 2). Note that a ∉ {x1, x2}, then {a, yi} ∈ E(X + Y). If a = x1 and b = x2, then {g(a), g(b)} = {y1, y2} ∈ E(X + Y). If a ∈ S \ V(If) and b ∈ S \ V(If), without loss of generality, we may assume that a ∈ S1 and b ∈ S2. If g(a)∈ S, then g(a) = [a]ρf for some [a]ρf ∈ S1; If g(a)∈ K, then g(a) = k1 for some k1 ∈ K1. Similarly, if g(b)∈ S, then g(a) = [b]ρf for some [b]ρf ∈ S2; If g(b)∈ K, then g(b) = k2 for some k2 ∈ K2. It is routine manner to check that {g(a), g(b)} ∈ E(X + Y) for each case. Consequently, g ∈ End(X + Y).

It is easy to check that g ∈ I dpt(X + Y), ρg = ρf and Ig = If. By Lemma 1.2, f is completely regular. □

Next, we start to seek the conditions for X + Y under which cEnd(X + Y) forms a monoid.

Lemma 2.2

If there exist yi, yj ∈ S such that N(yi) ⊂ N(yj), then cEnd(X + Y) does not form a monoid.

Proof

Suppose that there exist yi, yj ∈ S such that N(yi) ⊂ N(yj). Since K is a maximum complete set of X + Y, for any x ∈ S, there exists kx ∈ K such that {x, kx} ∉ E(X + Y). Let

f(x)=yj,x=yi,x,otherwise.andg(x)=kx,x=yj,x,otherwise.

Then f and g are idempotent endomorphisms of X + Y and so they are completely regular. It is easy to see that yj = (fg)(yi)∈ Ifg and (fg)−1(yj) = {yi}. Since |N(yi)∩ K | < |N(yj)∩ K |, Ifg is not an induced subgraph of X + Y. By Theorem 2.1, fg is not completely regular. Therefore cEnd(X + Y) does not form a monoid. □

Lemma 2.3

If there exist yi, yj ∈ S with i ≠ j such that |N(yi)∩ K |=| N(yj)∩ K |, then cEnd(X + Y) does not form a monoid.

Proof

Suppose that there exist yi, yj ∈ S with i ≠ j such that |N(yi)∩ K N(yj)∩ K |. Let p be a bijection of K such that p(N(yi)) = N(yj) and p(N(yj)) = N(yi). Since K is a maximum complete set of X + Y, for any x ∈ S, there exists kx ∈ K such that {x, kx}∉ E(X). Let

f(x)=p(x),x∈K,yj,x=yi,yi,x=yj,kx,otherwise.

It is easy to check that f is completely regular. Let

g(x)=x,x∈K,yi,x=yi,kx,otherwise.

Then g is an idempotent endomorphism of X + Y and so it is completely regular. It is easy to see that yj = (fg)(yi)∈ Ifg, (fg)(K) = K and (fg)(yj)∈ K. Thus there exists k ∈ K such that (fg)(yj) = f(k). By Theorem 2.1 fg is not completely regular. Therefore cEnd(X + Y) does not form a monoid. □

Up to now, we have obtained the following necessary conditions for cEnd(X + Y) being a monoid:

  1. N(yi) ⊄ N(yj) for any yi, yj ∈ S.

  2. |N(yi)∩ K |≠|N(yj)∩ K | for any yi, yj ∈ S with i ≠ j.

To show that (A) and (B) are also sufficient for cEnd(X + Y) being a monoid, we need the following characterization of completely regular endomorphisms of X + Y satisfying (A) and (B).

Lemma 2.4

Let X + Y be a join of split graphs satisfying (A) and (B), and let f ∈ cEnd(X + Y). If there exists y ∈ S such that [y] ρf⊆ S, then [y] ρf = {y}.

Proof

Let f ∈ cEnd(X + Y). By Theorem 2.1 (3), f(a)≠ f(b) for any a, b ∈ V(If). Thus there exists y0 ∈ V(If)∩[y]ρf. It follows from Theorem 2.1 (2) that N(y0) = ∪x∈[y] ρf N(x). Thus N(b)⊆ N(y0) for any b ∈ [y] ρf. It follows from (A) that N(b) = N(y0). In particular, N(b)∩ K = N(y0)∩ K. It follows from (B) that b = y0 = y. Hence [y] ρf = {y}. □

Theorem 2.5

Let X + Y be a join of split graphs satisfying (A) and (B), and let f ∈ End(X + Y). Then f ∈ cEnd(X + Y) if and only if one of the following conditions hold:

  1. For x ∈ K, f(x)∈ K; for y ∈ S, either f(y) = y, or f(y)∈ K.

  2. f(K)≠ K and If ≅ K.

  3. There exist x1∈ K, y1 ∈ S with N(y1) = K \ {x1} such that f(x1) = y1 and f(y1) = x1;f(K \ {x1}) = K \ {x1}; for y ∈ S with {y, y1} ∉ E(X + Y), f(y)∈ K \ {x1}; for y ∈ S with {y, y1} ∈ E(X + Y), either f(y) = y, or f(y)∈ K.

Proof

Necessity. Let X + Y be a join of split graphs satisfying (A) and (B) and let f ∈ cEnd(X + Y). We divide it into two cases to discuss:

Case 1. Assume that f(K) = K. For any y ∈ S, if f(y)∉ K, then f(y) ∈ S. Since f(K) = K, [y] ρf ⊆ S. By Lemma 2.4, [y]ρf = {y}. Since f(K) = K, |N(y)∩ K | = |N(f(y)) ∩ K |. It follows from (B) that f(y) = y. Hence f is an endomorphism of X + Y satisfying (1).

Case 2. Assume that f(K)≠ K. Then there exist xi ∈ K, y1 ∈ S such that f(xi) = y1 and |N(y1)∩ K | = n + m − 1. Now N(y1)∩ K = K \ {x1} for some x1∈ K. By (B), y1 is the only vertex in S such that |N(y1)∩ K | = n + m − 1. Thus there are exactly two cliques of order n + m in X + Y. They are induced by K and f(K) = (K \ {x1})∪{y1}, respectively. Hence f(K \ {xi}) = K \ {x1}. There are two cases:

  1. xi ≠ x1. Then x1∉ If. Otherwise, since f(K \ {xi}) = K \ {x1}, f−1(x1)⊆ S. Let y ∈ f−1(x1). By Lemma 2.4, [y]ρf = {y}. It follows from {x1, y1}∉ E(X + Y) that {xj, y}∉ E(X + Y). Thus y ≠ y1. It follows from (B) that |N(y)∩ K | ≤ n + m − 2. Thus there exists k0 ∈ K \ {xi} such that {k0, y}∉ E(X + Y). Obviously, {f(y), f(k0)} = {x1, f(k0)} ∈ E(X + Y). But {y, t}∉ E(X + Y) for any t ∈ [k0] ρf. Hence f is not half-strong. Note that f ∈ hEnd(X + Y) if and only if If is an induced subgraph of X + Y. This contradicts Theorem 2.1 (1). It follows from Theorem 2.1 (1) that y ∉ If for any y ∈ S with {y, x1} ∈ E(X + Y). In particular, y ∉ If for any y ∈ S with {y, y1} ∈ E(X + Y). Let y ∈ S with {xi, y} ∈ E(X + Y). Then {f(xj), f(y)} = {y1, f(y)} ∈ E(X + Y). Hence f(y)∈ N(y1)∩ K = K \ {x1}.

    Let y ∈ S with {xi, y}∉ E(X + Y). Then N(y)∩ K ⊆ K \ {xi}. If f(y) ∈ S, then {x1, f(y)} ∉ E(X + Y) since x1 ∉ If. Thus N(f(y))⊆ N(y1). It follows from (A) that N(f(y)) = N(y1). By (B), we have f(y) = y1. If f(y)∈ K, then f(y)≠ x1. Otherwise, [y] ρf ⊆ S. Since f ∈ cEnd(X + Y), If is an induced subgraph of X + Y and [y]ρf = {y}. Note that {x, x1} ∈ E(X + Y) for any x ∈ K \ {x1}. Then N(y)∩ K = K \ {xi}. Thus we have |N(y) ∩ K | = |N(y1)∩ K | = n + m − 1. This contradicts (B) . Hence If is a subgraph of X + Y induced by (K \ {x1}) ∪ {y1} and so If ≅ K.

  2. xi = x1. Then f(x1) = y1 and f(K \ {x1}) = K \ {x1}. Since {y1, k} ∈ E(X + Y) for any k ∈ K \ {x1}, f(y1) is adjacent to every vertex of f(K \ {x1}) = K \ {x1}. Thus f(y1) ∈ {x1, y1}.

    If f(y1) = y1, then x1∉ If by Theorem 2.1 (3). It follows from Theorem 2.1 (1) that y ∉ If for any y ∈ S with {x1, y} ∈ E(X + Y). Note that y1 is the only vertex in S, which is not adjacent to x1. Then If is a subgraph of X + Y induced by (K \ {x1})∪{y1} and so If≅ K.

    If f(y1) = x1, then x1∈ If. Let y ∈ S \ {y1} be such that {y, y1}∉ E(X + Y). It follows from (A) and (B) that {x1, y} ∈ E(X + Y). Thus {f(x1), f(y)} = {y1, f(y)} ∈ E(X + Y). If f(y)∈ S, f(y) and y1 lie in the different Si (where i ∈ {1, 2 Note that [y]ρf ⊆ S, then [y]ρf = {y}. Since f is half-strong, f((N(y)∩ K) \ {x1}) = (N(f(y))∩ K) \ {x1}. Hence |N(y)∩ K N(f(y))∩ K |. This contradicts (B). Hence f(y)∈ K \ {x1} for any y ∈ S \ {y1} with {y, y1}∉ E(X + Y). Let y ∈ S be such that {y, y1} ∈ E(X + Y). If f(y)∉ K, then f(y)∈ S. Since {x1, y} ∈ E(X + Y), {f(x1), f(y)} = {y1, f(y)} ∈ E(X + Y). Thus f(y)≠ y1. Then [y]ρf⊆ S and so [y]ρf = {y} by Lemma 2.4. Note that If is an induced subgraph of X + Y. Then |N(y)∩ K N(f(y))∩ K |. It follows from (B) that f(y) = y. Hence f is an endomorphism of X + Y satisfying (3).

Sufficiency. This follows directly from Theorem 2.1. □

Corollary 2.6

Let X + Y be a join of split graphs satisfying (A) and (B), and let f ∈ End(X + Y). If If≅ K, then f ∈ cEnd(X + Y).

Proof

This follows directly from Theorem 2.5. □

Now we prove that cEnd(X + Y) forms a monoid for X + Y satisfying (A) and (B).

Theorem 2.7

Let X + Y be a join of split graphs satisfying (A) and (B). Then cEnd(X + Y) forms a monoid.

Proof

Let X + Y be a join of split graphs satisfying (A) and (B). We only need to show that the composition of any two completely regular endomorphisms of X + Y is also completely regular in every case. Let f be an arbitrary completely regular endomorphism of X + Y. Then by Theorem 2.5, f acts in one of the following ways:

  1. For x ∈ K, f(x)∈ K; for y ∈ S, either f(y) = y, or f(y)∈ K.

  2. f(K)≠ K and If≅ K.

  3. There exist x1∈ K, y1 ∈ S with N(y1) = K \ {x1} such that f(x1) = y1 and f(y1) = x1; f(K \ {x1}) = K \ {x1}; for y ∈ S with {y, y1}∉ E(X + Y), f(y)∈ K \ {x1}; for y ∈ S with {y, y1} ∈ E(X + Y), either f(y) = y, or f(y)∈ K.

It is straightforward to see that the composition of any such two completely regular endomorphisms is still a completely regular endomorphism of X + Y. The proof is complete. □

Up to now we have

Theorem 2.8

Let X + Y be a join of split graphs. Then cEnd(X + Y) forms a monoid if and only if

  1. N(yi) ⊄ N(yj) for any yi, yj ∈ S, and

  2. | N(yi) ∩ K | ≠ | N(yj) ∩ K | for any yi, yj ∈ S with i ≠ j.

Proof

Necessity follows directly from Lemmas 2.2 and 2.3.

Sufficiency follows directly from Theorem 2.7. □

Acknowledgement

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(No.11301151), the Key Project of the Education Department of Henan Province(No.13A110249), the Subsidy Scheme of Young Teachers of Henan Province(No.2014GGJS-057) and the Innovation Team Funding of Henan University of Science and Technology(No.2015XTD010).

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Received: 2016-7-6
Accepted: 2017-5-8
Published Online: 2017-6-22

© 2017 Hou et al.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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Keywords for this article
Endomorphism; Monoid; Join of split graphs; Completely regular
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