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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

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

Hailong Hou
• School of Mathematics and Statistics, Henan University of Science and Technology, Luoyang, Henan, 471023, China
• Other articles by this author:
/ Yanhua Song
• School of Mathematics and Statistics, Henan University of Science and Technology, Luoyang, Henan, 471023, China
• Other articles by this author:
/ Rui Gu
• Corresponding author
• School of Mathematics and Statistics, Henan University of Science and Technology, Luoyang, Henan, 471023, China
• Email
• Other articles by this author:
Published Online: 2017-06-22 | DOI: https://doi.org/10.1515/math-2017-0071

## 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.

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. [14]). 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 xS such that axa = a. Let fEnd(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. [713]).

An element a of a semigroup S is said to be completely regular if a = axa and xa = ax hold for some xS. Let fEnd(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 {xV(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, bH, {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, bV(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 KV(X) is said to be complete if {a, b} ∈ E(X) for any two vertices a, bK with ab. A subset SV(X) is said to be independent if {a, b} ∉ E(X) for any two vertices a, bS. 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) = KS, 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 yS. 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} | xV(X), yV(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, x2V(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 fEnd(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 cf−1(f(a)) and df−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, bV(X), (a, b) ∈ ρf if and only if f(a) = f(b). Denote by [a]ρf the equivalence class containing aV(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 fEnd(X). Then

1. fhEnd(X) if and only if If is an induced subgraph of X.

2. If f is regular, then fhEnd(X).

#### Lemma 1.2

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

#### Lemma 1.3

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

1. There exists hI dpt(X + Y) such that Ih = If.

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

#### Lemma 1.4

([6]). Let X + Y be a join of split graphs and fEnd(X + Y). Then there exists gI 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 aV(X + Y).

## 2 Main results

Let X be a split graph with V(X) = K1S1, 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) = K2S2, 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) = KS1S2, where K = K1K2 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 = S1S2 is complete bipartite. Hence in graph X + Y, N(xi) = NX(xi) ∪ V(Y) for xiS1 and N(yi) = NY(yi)∪ V(X) for yiS2. Clearly, X + Y is a split graph extended by the edge set {{xi, yj} | xiS1, yjS2}. 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 fEnd(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 xK \ If and ySIf.

2. N(b) = ∪x∈[b]ρf N(x) for any bV(If) with [b]ρfS.

3. f(a)≠ f(b) for any a, bV(If) with ab.

#### Proof

Necessity. Since f is completely regular, by Lemma 1.2, there exists gI 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 xK \ If and ySIf. Hence (1) holds. Let aV(If). Note that gI 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 bV(If) be such that [b]ρfS. Then bS and N(b)∩ KV(If) by (1). Thus g(x) = x for any xN(b)∩ K and so N(b)∩ Kg(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 kK such that kN(y) and kN(b). Note that {k, t} ∈ E for any tN(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, bV(If), then [a]ρf = [b]ρf. Note that gI dpt(X + Y) and ρg = ρf, then g(a) = g(b). This means that {a, b} ⊈ V(Ig) and so IgIf, which yields a contradiction. Hence (3) holds.

Sufficiency. Let X + Y be a join of split graphs and fEnd(X + Y) be such that (1), (2) and (3). Note that f(a) ≠ f(b) for any a, bV(If). Then for any aV(X + Y), there exists only one vertex in [a]ρfV(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 gEnd(X + Y). Let {a, b} ∈ E(X + Y) for some a, bV(X + Y). Since K is a maximum complete set of X + Y, f(K) is a clique of size n + m. We have KIf, |KIf| = n + m − 1 or |KIf| = n + m − 2.

Assume that KIf. Then g(k) = k for any kK. If a, bV(If), then g(a) = a and g(b) = b. Thus {g(a), g(b)} = {a, b} ∈ E(X + Y). If aV(If) and bV(If), then bS. If [b]ρfS, then g(b) = [b]ρf. Note that N([b]ρf) = ∪x ∈ [b]ρf N(x). Then aN([b]ρf). Hence {g(a), g(b)} = {a, [b]ρf} ∈ E(X + Y); If [b]ρfS, then there exists kK ∩[b]ρf. Obviously, {a, k} ∈ E(X + Y). Hence {g(a), g(b)} = {a, k} ∈ E(X + Y). If aV(If) and bV(If), then a, bS. Without loss of generality, we may assume that aS1 and bS2. If g(a)∈ S, then g(a) = [a]ρf for some [a]ρfS1; If g(a)∈ K, then g(a) = k1 for some k1K1. Similarly, if g(b)∈ S, then g(a) = [b]ρf for some [b]ρfS2; If g(b)∈ K, then g(b) = k2 for some k2K2. It is a routine manner to check that {g(a), g(b)} ∈ E(X + Y) for every case. Consequently, gEnd(X + Y).

Assume that |KIf| = n + m − 1. Then there exists x1K \ 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 y1SIf such that y1 is adjacent to every vertex of K \ {x1}. Now g(x1) = y1. If a, bV(If), then g(a) = a and g(b) = b. Thus {g(a), g(b)} = {a, b} ∈ E(X + Y). If a = x1 and bK\ {x1}, then g(b) = b. Thus {g(a), g(b)} = {y1, b} ∈ E(X + Y). If a = x1 and bS, then bV(If) by (1). Now [b]ρfS. Otherwise, there exists [b]ρfV(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]ρfIg = If. This contradicts (1). Then there exists k1K \ {x1} such that k1 ∈ [b]ρf. Thus {g(a), g(b)} = {y1, k1} ∈ E(X + Y). If aK \ {x1} and bS \ If, then g(a) = a. If [b]ρfS, then g(b) = [b]ρf. Note that N([b]ρf) = ∪x∈[b]ρf N(x). Then aN([b]ρf). Thus {g(a), g(b)} = {a, [b]ρf} ∈ E(X + Y); If [b]ρfS, then there exists kK ∩[b]ρf. Thus {g(a), g(b)} = {a, g(k)}. If kK \ {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 aS \ V(If) and bS \ V(If), without loss of generality, we may assume that aS1 and bS2. If g(a)∈ S, then g(a) = [a]ρf for some [a]ρfS1; If g(a)∈ K, then g(a) = k1 for some k1K1. Similarly, if g(b)∈ S, then g(a) = [b]ρf for some [b]ρfS2; If g(b)∈ K, then g(b) = k2 for some k2K2. It is a routine manner to check that {g(a), g(b)} ∈ E(X + Y) for each cases. Consequently, gEnd(X + Y).

Assume that |KIf| = n + m − 2. Then there exist x1K1 \ If and x2K2 \ If. Note that f(K) is a clique of size n + m in X + Y. Then there exist y1, y2SIf 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, bV(If), then g(a) = a and g(b) = b. Thus {g(a), g(b)} = {a, b} ∈ E(X + Y). If a ∈ {x1, x2} and bK \ {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 bS, then bV(If) by (1). Without loss of generality, we may assume that a = x1. We claim that [b]ρfS. Otherwise, there exists [b]ρfV(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]ρfIg = If. This contradicts (1). Then there exists k1K \ {x1} such that k1 ∈ [b]ρf. If k1x2, then {g(a), g(b)} = {y1, k1} ∈ E(X + Y); If k1 = x2, then {g(a), g(b)} = {y1, y2} ∈ E(X + Y). If aK \ {x1, x2} and bS \ If, then g(a) = a. If [b]ρfS, then g(b) = [b]ρf. Note that N([b]ρf) = ∪x ∈ [b]ρf N(x). Then aN([b]ρf). Thus {g(a), g(b)} = {a, [b]ρf} ∈ E(X + Y); If [b]ρfS, then there exists kK ∩[b]ρf. Thus {g(a), g(b)} = {a, g(k)}. If kK \ {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 aS \ V(If) and bS \ V(If), without loss of generality, we may assume that aS1 and bS2. If g(a)∈ S, then g(a) = [a]ρf for some [a]ρfS1; If g(a)∈ K, then g(a) = k1 for some k1K1. Similarly, if g(b)∈ S, then g(a) = [b]ρf for some [b]ρfS2; If g(b)∈ K, then g(b) = k2 for some k2K2. It is routine manner to check that {g(a), g(b)} ∈ E(X + Y) for each case. Consequently, gEnd(X + Y).

It is easy to check that gI 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, yjS such that N(yi) ⊂ N(yj), then cEnd(X + Y) does not form a monoid.

#### Proof

Suppose that there exist yi, yjS such that N(yi) ⊂ N(yj). Since K is a maximum complete set of X + Y, for any xS, there exists kxK 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, yjS with ij such that |N(yi)∩ K |=| N(yj)∩ K |, then cEnd(X + Y) does not form a monoid.

#### Proof

Suppose that there exist yi, yjS with ij 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 xS, there exists kxK 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 kK 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, yjS.

2. |N(yi)∩ K |≠|N(yj)∩ K | for any yi, yjS with ij.

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 fcEnd(X + Y). If there exists yS such that [y] ρfS, then [y] ρf = {y}.

#### Proof

Let fcEnd(X + Y). By Theorem 2.1 (3), f(a)≠ f(b) for any a, bV(If). Thus there exists y0V(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 fEnd(X + Y). Then fcEnd(X + Y) if and only if one of the following conditions hold:

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

2. f(K)≠ K and IfK.

3. There exist x1K, y1S with N(y1) = K \ {x1} such that f(x1) = y1 and f(y1) = x1;f(K \ {x1}) = K \ {x1}; for yS with {y, y1} ∉ E(X + Y), f(y)∈ K \ {x1}; for yS 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 fcEnd(X + Y). We divide it into two cases to discuss:

Case 1. Assume that f(K) = K. For any yS, if f(y)∉ K, then f(y) ∈ S. Since f(K) = K, [y] ρfS. 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 xiK, y1S such that f(xi) = y1 and |N(y1)∩ K | = n + m − 1. Now N(y1)∩ K = K \ {x1} for some x1K. 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. xix1. Then x1If. Otherwise, since f(K \ {xi}) = K \ {x1}, f−1(x1)⊆ S. Let yf−1(x1). By Lemma 2.4, [y]ρf = {y}. It follows from {x1, y1}∉ E(X + Y) that {xj, y}∉ E(X + Y). Thus yy1. It follows from (B) that |N(y)∩ K | ≤ n + m − 2. Thus there exists k0K \ {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 fhEnd(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 yIf for any yS with {y, x1} ∈ E(X + Y). In particular, yIf for any yS with {y, y1} ∈ E(X + Y). Let yS 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 yS with {xi, y}∉ E(X + Y). Then N(y)∩ KK \ {xi}. If f(y) ∈ S, then {x1, f(y)} ∉ E(X + Y) since x1If. 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] ρfS. Since fcEnd(X + Y), If is an induced subgraph of X + Y and [y]ρf = {y}. Note that {x, x1} ∈ E(X + Y) for any xK \ {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 IfK.

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

If f(y1) = y1, then x1If by Theorem 2.1 (3). It follows from Theorem 2.1 (1) that yIf for any yS 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 IfK.

If f(y1) = x1, then x1If. Let yS \ {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]ρfS, 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 yS \ {y1} with {y, y1}∉ E(X + Y). Let yS 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]ρfS 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 fEnd(X + Y). If IfK, then fcEnd(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 xK, f(x)∈ K; for yS, either f(y) = y, or f(y)∈ K.

2. f(K)≠ K and IfK.

3. There exist x1K, y1S with N(y1) = K \ {x1} such that f(x1) = y1 and f(y1) = x1; f(K \ {x1}) = K \ {x1}; for yS with {y, y1}∉ E(X + Y), f(y)∈ K \ {x1}; for yS 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, yjS, and

2. | N(yi) ∩ K | ≠ | N(yj) ∩ K | for any yi, yjS with ij.

#### 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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Accepted: 2017-05-08

Published Online: 2017-06-22

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 833–839, ISSN (Online) 2391-5455,

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