Recall that End-regular bipartite graphs are characterized in Lemma 1.4 and End-regular lexicographic products of bipartite graphs are determined in Lemma 1.6. In this section, we characterize the End-regular generalized lexicographic products of connected bipartite graphs. The following theorem is our main result.

**Theorem**2.1. *Let G be a connected bipartite graph and let* {*B*_{i}| *i* ∈ *V*(*G*)} *be a family of connected bipartite graphs. Then G*[*B*_{i}]_{i∈V(G)} *is End-regular if and only if one of the following conditions hold*:

(1) *G = K*_{2} *with V*(*G*) *=* {1,2}, *B*_{1} *is End-regular and B*_{2} *= K*_{2}*;*

(2) *G = K*_{2} *with V*(*G*) *=* {1, 2}, *B*_{1} *and B*_{2} *are trees of diameter* 2*;*

(3) *G = K*_{2} *with V*(*G*) *=* {1,2}, *B*_{1} *and B*_{2} *are complete bipartite graphs;*

(4) *G = K*_{2} *with V*(*G*) *=* {1_{;}2}, *B*_{1} *is a tree of diameter* 2 *and B*_{2} *is a complete bipartite graph;*

(6) *G* ≠ *K*_{2} *is End-regular and B*_{i} = K_{2}*for any i* ∈ *V*(*G*).

To prove our main result, we need the following characterizations of the regular endomorphisms of lexicographic products of bipartite graphs.

**Lemma 2.2***Let G be a bipartite graph and let* {*B*_{i}|*i* ∈ *V*(*G*)} *be a family of connected bipartite graphs. If G*[*B*_{i}]_{i∈V(G)} *is End-regular*, *then G is End-regular*

*Proof* Since *B*_{i} is bipartite, its vertex set *V*(*B*_{i}) can be partitioned into two disjoint non-empty subsets *A*_{i} and *C*_{i}, such that no edge joins two vertices in the same set. Take *y*_{i}_{0} ∈ *A*_{i} and z_{i0} ∈ *C*_{i} such that {*y*_{i}_{0}*, z*_{i}_{0}} ∈ *E*(*B*_{i}) for any *i* ∈ *V*(*G*). Define the mapping on the vertex set *V*(*B*_{i}) as follows:

$${g}_{i}(x)=\{\begin{array}{c}{y}_{i0},x\in {A}_{i},\\ {z}_{i0},x\in {C}_{i}.\end{array}$$Let *f* ∈ *End*(*G*). Define a mapping *F* from *V*(*G*[*B*_{i}]_{i∈V(G)}) to itself by

$$F\left(\left(i,x\right)\right)=\{\begin{array}{c}\left(f\left(i\right),{y}_{f\left(i\right)0}\right),x\in {A}_{i},\\ \left(f\left(i\right),{z}_{f\left(i\right)0}\right),x\in {C}_{i}.\end{array}$$Let (*i*, *x*_{i}), (*j*, *x*_{2}) ∈ *V*(*G*[*B*_{i}]_{i∈V(G)}) with {(*i*, *x*_{i}), (*j*, *x*_{2})} e *E*(*G*[*Bi*]_{isViG)}). If {*i*, *j*} ∈ *E*(*G*), then {*f*(*i*), *f*(*j*)} ∈ *E*(*G*) and so {*F*((*i*, *x*_{1})), *F*((*j*, *x*_{2}))} ∈ *E*(*G*[*B*_{i}]_{i∈V(G)})*. If i = j* and{*x*_{1}, *x*_{2}} ∈ *E*(*B*_{i}), then {*F*((*i*, *x*_{1})), *F*((*j*, *x*_{2}))} *=* {(*f* (*i*), *y*_{f}_{(i)0}),(*f*(*i*), *z*_{f}_{(i)0})} ∈ *E*(*G*[*B*_{i}]_{i∈V(G)}). Hence *F* ∈ *End*(*G*[*B*_{i}]_{i∈V(G)}). Note that *F* is regular. Thus there exists pseudoinverse *K* ∈ *End*(*G*[*B*_{i}]_{i∈V(G)}) such that *FKF D F*.

Let (*i, x*) ∈ *V*(*G*[*B*_{i}]_{i∈V(G)}) and *K*((*i, x*)) *=* (*j, x*′) for some *j* ∈ *V*(*G*) and *x*′ ∈ *V*(*B*_{j}). If *x* ∈ *A*_{i}, then *x*′ ∈ *A*_{j}; If *x* ∈ *C*_{i}, then *x*′ ∈ *C*_{j} Define a mapping *k* from *V*(*G*) to itself by *k*(*i*) *= j*, where *j* satisfies *K*((*i*, *x*)) = (*j*, *x*′) for some *x*′ ∈ *V*(*B*_{j}). We now prove that *k* ∈ *End*(*G*). Let *i*, *j* ∈ *V*(*G*) be such that {*i*, *j*} ∈ *E*(*G*)*. Then* {(*i*, *x*_{1}), (*j*, *x*_{2})} ∈ *E*(*G*[*B*_{i}]_{i∈V(G)}) for any *x*_{1} ∈ *A*_{i} and *x*_{2} ∈ *A*_{j} Since *K* ∈ *End*{*G*[*B*_{i}]_{i∈EV(G)}), there exists ${x}_{1}^{\text{'}}$ ∈ *V*(*B*_{k}_{(i)}) and ${x}_{2}^{\text{'}}$ ∈ *V*(*B*_{k}_{(j)}) such that {*K*((*i*, *x*_{1})), *K*((*j*, *x*_{2}))} *=* {{*k*{*i*), ${x}_{1}^{\text{'}}$), (*k*(*j*), ${x}_{2}^{\text{'}}$)} ∈ *E*(*G*[*B*_{i}]_{i∈V(G)})*. If* {*k*(*i*), *k*(*j*)} ∉ *E*(*G*), then *k*(*i*) *= k*(*j*) and {${x}_{1}^{\text{'}}$, ${x}_{2}^{\text{'}}$} ∈ *E*(*B*_{k}_{(i)}). Note that ${x}_{1}^{\text{'}}$, ${x}_{2}^{\text{'}}$ ∈ *A*_{k}_{(i)}. This is a contradiction. Hence {k(i), *k*(*j*)} ∉*E*(*G*) and so *k* ∈ *End*(*G*). And by the definition of *k*, we get that there exists *x*′ ∈ *V*(*B*_{fkf}_{(i)}) such that *FKF*((*i, x*)) = (*fkf*(*i*), *x*′). Hence *fkf = f* This means that *k* is a pseudoinverse of *f* and so *f* is regular. Therefore *G* is End-regular. □

**Lemma 2.3***Let G be a bipartite graph and let* {*B*_{i}|*i* ∈ *V*(*G*)} *be a family of connected bipartite graphs. If G*_{1} *is a retract of G and* {*H*_{i}|*i* ∈ *V*(*G*)} *are retracts of* {*B*_{i}|*i* ∈ *V*(*G*)}, *then* *G*_{1}[*H*_{i}]_{i∈V(G1)} *is a retract of G*[*B*_{i}]_{i∈V(G)}.

*Proof*. We only need to show that there exists a retraction from *G*[*B*_{i}]_{i∈V(G)} to *G*_{1}[*H*_{i}]_{i∈V(G1)}. Since *G*_{1} and {*H*_{i}|*i* ∈ *V*(*G*)} are retracts of *G* and {*Bi*| *i* ∈ *V*(*G*)} respectively, there exist retractions f from *G* to *G*_{1} and *h*_{i} from *B*_{i} to *H*_{i} For any *j* ∈ *V*(*G*_{1}), take *y*_{j}_{1} ∈ *A*_{j} ⋂ *I*_{hj} and *Zj*_{1} ∈ *C*_{j} ⋂ *I*_{hj}.

Let *F* be the mapping from *G*[*B*_{i}]_{i∈V(G)} to itself defined as follows:

$$F\left(\left(k,y\right)\right)=\{\begin{array}{ll}\left(k,{h}_{k}\left(y\right)\right),\hfill & if\text{\hspace{0.17em}\hspace{0.17em}}k\in V\left({G}_{1}\right),\hfill \\ \left(f\left(k\right),{y}_{f(k)1}\right),\hfill & if\text{\hspace{0.17em}\hspace{0.17em}}k\notin V\left({G}_{1}\right)and\text{\hspace{0.17em}\hspace{0.17em}}y\in {A}_{k},\hfill \\ \left(f\left(k\right),{z}_{f(k)1}\right),\hfill & if\text{\hspace{0.17em}\hspace{0.17em}}k\notin V\left({G}_{1}\right)and\text{\hspace{0.17em}\hspace{0.17em}}y\in {C}_{k}.\hfill \end{array}$$*Let* (*j*, *x*_{1}), (*k*, *x*_{2}) ∈ *V*(*G*[*B*_{i}]_{i∈V(G)}) be such that {(*j*, *x*_{1}), (*k*, *x*_{2})} ∈ *E*(*G*[*B*_{i}]_{i∈V(G)})*. Then* {*j*, *k*} ∈ or *j* = *k* and {*x*_{1}, *x*_{2}} ∈ *E*(*B*_{j}). If {*j*, *k*} ∈ *E*(i), then there exist ${x}_{1}^{\text{'}}$ ∈ *V*(*B*_{f}_{(j)}) and ${x}_{2}^{\text{'}}$ ∈ *V*(*B*_{f}_{(k)}) such that {*F*((*j*, *x*_{1})), *F*((*k*, *x*_{2}))} = {(*f*(*j*), ${x}_{1}^{\text{'}}$, (*f*(*k*), ${x}_{2}^{\text{'}}$} ∈ *E*(*G*[*B*_{i}]_{i∈V(G)}). If *j* = *k* and {*x*_{1},*x*_{2}} ∈ *E*(*B*_{j}), without loss of generality, suppose *x*_{1} ∈ *A*_{j} and *x*_{2} ∈ *C*_{j}. If *j* ∈ *V*(G_{1}), then {*F*((*j*, *x*_{1})), *F*((*k*, *x*_{2}))} = {(*j*, *h*_{j} (*x*_{1})), (*j*, *h*_{j} (*x*_{2}))}. Since *h*_{j} is a retraction of *B*_{j}, {*h*_{j} (*x*_{1}), *h*_{j} (*x*_{2})} ∈ *E*(*B*_{j}). Hence {(*j*, *h*_{j} (*x*_{1})), (*j*, *h*_{j} (*x*_{2}))} ∈ *E*(*G*[*B*_{i}]_{i∈V(G)}). If *j* ∉ *V*(*G*_{1}), then {*F*((*j*, *x*_{1})), *F*((*k*, *x*_{2}))} = {(*f*(*j*), *y*_{f}_{(j)1}), (*f*(*j*), *z*_{f}_{(j)1})} ∈ *E*(*G*[*B*_{i}] _{i∈V(G)}). *Therefore F* ∈ *End*(*G*[*B*_{i}]_{i∈V(G)}). It is easy to check that *I*_{F} = *G*_{1}[*H*_{i}]_{i∈V(G1)}. Let (*j*, *x*) ∈ *V*(*G*_{1}[*H*_{i}]_{i∈V(G1)}. Then *F*((*j*, *x*)) = (*j*, *h*_{j}(*x*)) = (*j*, *x*) and so *F* is a retraction from *G*[*B*_{i}]_{i∈V(G)} to *G*_{1}[*H*_{i}]_{i∈V(G1)}. □

Next we start to seek the conditions for a generalized lexicographic product of bipartite graphs under which *G*[*B*_{i}]_{i∈V(G)} is End-regular.

**Lemma 2.4***Let G = K*_{2} *with V*(*G*) *=* {1,2} *and let* {*B*_{i}|*i* ∈ *V*(*G*)} *be two connected bipartite graphs. Then G*[*B*_{i}]_{i∈V(G)} *is End-regular if and only if one of the following conditions hold*:

(1) *B*_{1} *is End-regular and B*_{2} *= K*_{2},

(2) *Both B*_{1} *and B*_{2} *are trees of diameter* 2,

(3) *Both B*_{1} *and B*_{2} *are complete bipartite graphs*,

(4) *B*_{1} *is a tree of diameter* 2 *and B*_{2} *is a complete bipartite graph*

*Proof* Note that *G*[*B*_{i}]_{i∈V(G)} is isomorphic to *B*_{1} *+ B*_{2}. Then the result follows directly from Lemma 1.5. □

**Lemma 2.5***Both P*_{3} [*P*_{3}*, K*_{2}*, K*_{2}] *and P*_{3} [*K*_{2}*, P*_{3}*, K*_{2}] *are not End-regular*

Fig. 1 *Graph P*_{3}[*P*_{3}*, K*_{2}*, K*_{2}]

Fig. 2 *Graph P*_{3}[*K*_{2}*, P*_{3}*, K*_{2}]

*Proof* Let

$$f=\left(\begin{array}{lllllll}{x}_{1}\hfill & {x}_{2}\hfill & {x}_{3}\hfill & {y}_{1}\hfill & {y}_{2}\hfill & {z}_{1}\hfill & {z}_{2}\hfill \\ {x}_{1}\hfill & {y}_{1}\hfill & {x}_{1}\hfill & {y}_{2}\hfill & {x}_{2}\hfill & {x}_{3}\hfill & {y}_{1}\hfill \end{array}\right).$$Then *f* ∈ *End*(*P*_{3}[*P*_{3}, *K*_{2}, *K*_{2}]). It is easy to see that *ρ*_{f} = {[*x*_{1}, *x*_{3}], [*y*_{2}], [*z*_{1}], [*x*_{2}, *z*_{2}], [*y*_{1}]}. Suppose that there exists an idempotent endomorphism *g* of *P*_{3}[*P*_{3}*, K*_{2}*, K*_{2}] such that *ρ*_{g} = ρ_{f} Then *g*(*y*_{1}*) = y*_{1}, *g*(*y*_{2}) *= y*_{2} and *g*(*z*_{1}) *= z*_{1}. Since *z*_{2} is adjacent to every vertices of {*y*_{1}*, y*_{2}*, z*_{1}}, *g*(*z*_{2}) is adjacent to every vertex of *g*({*y*_{1}*, y*_{2}*, z*_{1}}) *=* {*y*_{1}, *y*_{2}, *z*_{1}}. Note that *z*_{2} is the only vertex in *V*(*P*_{3}[*P*_{3}, *K*_{2}, *K*_{2}]) adjacent to every vertices of {*y*_{1}, *y*_{2}, *z*_{1}}. Then *g*(*x*_{2}) *= g*(*z*_{2}) *= z*_{2}. Since {*x*_{1}*, x*_{2}} ∈ *E*(*P*_{3}[*P*_{3}, *K*_{2}, *K*_{2}]), {*g*(*x*_{1}) *g*(*x*_{2})} *=* {*g*(*x*_{1}), *z*_{2})} ∈ *E*(*P*_{3}[*P*_{3}, *K*_{2}, *K*_{2}]). Thus *g*(*x*_{1}) ∈ {*y*_{1}, *y*_{2}*, z*_{1})}. This is a contradiction. By Lemma 1.1 *f* is not regular and so *P*_{3}[*P*_{3}, *K*_{2}, *K*_{2}] is not End-regular.

Let

$$h=\left(\begin{array}{lllllll}{a}_{1}\hfill & {a}_{2}\hfill & {b}_{1}\hfill & {b}_{2}\hfill & {b}_{3}\hfill & {c}_{1}\hfill & {c}_{2}\hfill \\ {a}_{1}\hfill & {b}_{1}\hfill & {a}_{2}\hfill & {b}_{2}\hfill & {a}_{2}\hfill & {b}_{3}\hfill & {a}_{1}\hfill \end{array}\right).$$Then *h* ∈ *End*(*P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}]). It is easy to see that *ρ*_{h} = {[*a*_{1}, c_{2}], [*a*_{2}], [*b*_{1}, *b*_{3}], [*b*_{2}], [*c*_{1}]}. Suppose that there exists an idempotent endomorphism *k* of *P*_{3}[*K*_{2}, *P*_{3}*, K*_{2}] such that *ρ*_{h} = ρ_{h} Then *k*(*a*_{2}) *= a*_{2}, *k*(*b*_{2}) = *b*_{2} and *k*(*c*_{1}) *= c*_{1}. Since *b*_{1} is adjacent to every vertex of {*a*_{2}, *b*_{2}, *c*_{1}}, *k*(*b*_{1}) is adjacent to every vertex of *k*({*a*_{2}, *b*_{2},*c*_{1}}) *=* {*a*_{2}, *b*_{2}, *c*_{1}}. Note that {*a*_{1}, *a*_{2}} ∈ *E*(*P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}]), {*c*_{2}, *b*_{2}} ∈ *E*(*P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}]) and {*c*_{2}, *c*_{1}} ∈ *E*(*P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}]). Then *k*(*a*_{1}) = *k*(*c*_{2}) is adjacent to every vertex of *k*({*a*_{2}, *b*_{2}, *c*_{1}}) *=* {*a*_{2}, *b*_{2}, *c*_{1}}. It follows from {*a*_{1}, *b*_{1}} ∈ *E*(*P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}]) that {*f*(*a*_{1}), *f*(*b*_{1})} ∈ *E*(*P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}]). Note that there are no two adjacent vertices in *V*(*P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}]) adjacent to every vertex of {*a*_{2}, *b*_{2}, *c*_{1}}. This is a contradiction. By Lemma 1.1 *h* is not regular and so *P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}] is not End-regular. □

**Lemma 2.6***Let G* ≠ *K*_{2} *be a connected End-regular bipartite graph and let* {*B*_{i}|*i* ∈ *V*(*G*)} *be a family of connected bipartite graphs. Then G*[*B*_{i}]_{i∈V(G)} *is End-regular if and only if B*_{i} = K_{2} *for any i* ∈ *V*(*G*).

*Proof*. Necessity. Suppose that *G*[*B*_{i}]_{i∈V(G)} is End-regular. Then *G* is End-regular by Lemma 2.2 and *B*_{i} is End-regular for any *i* ∈ *V*(*G*) by Lemma 1.8. Note that *G* is a connected bipartite graph and *G* ≠ *K*_{2}. Then *G* contains at least two edges. Thus *P*_{3} is a retract of *G* Let {*B*_{i}|*i* ∈ *V*(*G*)} be a family of connected bipartite graphs. If there exists *j* ∈ *V*(*G*) such that *B*_{j} ≠ *K*_{2}, then *P*_{3} is a retract of *B*_{j} Note that *K*_{2} is a retract of any bipartite graphs. By Lemma 2.3, *P*_{3}[*P*_{3}, *K*_{2}, *K*_{2}] or *P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}] is a retract of *G*[*Bi*]_{i∈V(G)}. Since both *P*_{3}[*P*_{3}, *K*_{2}, *K*_{2}] and *P*_{3}[*K*_{2}, *P*_{3}, *K*_{2}] are not End-regular, by Lemma 1.3, *G*[*B*_{i}]_{i∈V(G)} is not End-regular. This is a contradiction.

Sufficiency. If *B*_{i} = K_{2} for any *i* ∈ *V*(*G*), then *G*[*B*_{i}]_{i∈V(G)} *= G*[*K*_{2}], the lexicographic products of *G* and *K*_{2}. By Lemma 1.6, *G*[*B*_{i}]_{i∈V(G)} is End-regular. □

With these preparations, the proof of our main result is straightforward.

*Proof of Theorem 2.1* This follows directly from Lemmas 2.4 and 2.6. □

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