Neighborhood complexes, homotopy test graphs and a contribution to a conjecture of Hedetniemi
aa r X i v : . [ m a t h . C O ] O c t Neighborhood complexes, homotopy test graphs and acontribution to a conjecture of Hedetniemi
Samir Shukla ∗† Abstract
The neighborhood complex N ( G ) of a graph G were introduced by L. Lovász in hisproof of Kneser conjecture. He proved that for any graph G , χ ( G ) ≥ conn ( N ( G )) + 3 . (1)In this article we show that for a class of exponential graphs the bound given in(1) is tight. Further, we show that the neighborhood complexes of these exponentialgraphs are spheres up to homotopy. We were also able to find a class of exponentialgraphs, which are homotopy test graphs.Hedetniemi’s conjecture states that the chromatic number of the categorical productof two graphs is the minimum of the chromatic number of the factors. Let M ( G ) denotesthe Mycielskian of a graph G . We show that, for any graph G containing M ( M ( K n )) asa subgraph and for any graph H , if χ ( G × H ) = n + 1 , then min { χ ( G ) , χ ( H ) } = n + 1 .Therefore, we enrich the family of graphs satisfying the Hedetniemi’s conjecture. Keywords : Neighborhood complexes, exponential graphs, chromatic number.2010
Mathematics Subject Classification: primary 05C15, secondary 57M15
In 1978, L. Lovász ([22]) proved the famous Kneser conjecture ([18]) by using Borsuk-Ulam theorem. In his proof he introduced the concept of a simplicial complex N ( G ) calledneighborhood complex for a graph G and then related the topological connectivity of N ( G ) to the chromatic number of G . Theorem 1.1. (Lovász) For any graph G , χ ( G ) ≥ conn ( N ( G )) + 3 . (2)Here for a space X , conn ( X ) is the largest integer n such that X is n -connected. Lovászproved that in the case of Kneser graph the bound given in (2) is tight. In [5], Björnerand Longueville showed that the neighborhood complexes of a family of vertex critical ∗ Department of Mathematics, Indian Institute of Technology Bombay, India. [email protected] † The author was financially supported by Indian Institute of Technology Bombay, India. K m is homotopyequivalent to the ( m − -sphere S m − . In [25], Nilakantan and the author have studiedthe neighborhood complexes of the exponential graphs K K n m . We have shown that N ( K K n m ) is homotopy equivalent to S m − for m < n . For all the above mentioned classes of graphsthe bound given in (2) is sharp. It is a natural question to ask whether we can classify thegraphs for which the inequality (2) is sharp. In this article we show that the bound givenin (2) is sharp for a class of exponential graphs. Further, we show that the neighborhoodcomplexes of these graphs are spheres up to homotopy.For r ≥ , let L r denotes the path of length r , i.e., it is a graph with vertex set V ( L r ) = { , . . . , r } and edge set E ( L r ) = { ( i, i + 1) | ≤ i ≤ r − } . For a subset A ⊆ { , . . . , r } , L r ( A ) is the graph obtained from L r by adding a loop at x for each x ∈ A .For a graph G , we define the graph G rA = ( G × L r ( A )) / ∼ r , where ∼ r is the equivalencewhich identifies all vertices whose second coordinate is r . For a graph G and r ≥ , thegraph G r { } = ( G × L r ( { } )) / ∼ r is called the r - th generalised Mycielskian of G and wedenote it by M r ( G ) . The graph M ( G ) is called the Mycielskian of G . In this article, ifno confusion arises we denote M ( G ) simply by M ( G ) . We say that a simple graph G satisfies the property P , if the following is true: for anytwo disjoint edges e = ( v , w ) and e = ( v , w ) of G if ( v , v ) is an edge in G then either ( w , v ) or ( w , w ) is an edge in G .Here, by disjointness of edges e = ( v , w ) and e = ( v , w ) , we mean that { v , w } ∩{ v , w } = ∅ . Clearly, all the complete graphs satisfies the property P . We prove thefollowing. Theorem 1.2.
Let ≤ m < χ ( T ) be a positive integer and let T be a connected graphsatisfying the property P . Then, for K Tm the bound given in (2) is sharp. Moreover, N ( K Tm ) ≃ S m − . Theorem 1.3.
Let n ≥ , r ≥ , ≤ i ≤ r − and A = { , . . . , i } . Let T be a connectedgraph satisfying the property P . Then for any m ≤ χ ( T ) , N ( K T rA m ) ≃ S m − . In particular, N ( K M r ( K n ) m ) ≃ S m − for all m ≤ n . The following corollary is a direct application of Theorem 1.3.
Corollary 1.4.
Let T be a connected graph satisfying the property P . Let ≤ m ≤ χ ( T ) , r ≥ , ≤ i ≤ r − and A = { , . . . , i } . Then, for K T rA m the bound given in (2) is sharp. Theorem 1.5.
Let n ≥ and m ≤ n be positive integers. Then, inequality (2) is sharp for K M ( M ( K n )) m . Moreover, N ( K M ( M ( K n )) m ) ≃ S m − . The Hom complex were originally defined by Lovász to give a topological obstruction tograph coloring and were first mainly investigated by Babson and Kozlov in [2]. The Hom The generalized Mycielskians (also known as cones over graphs) of a graph were introduced by Tardif([29]), which are the natural generalization of Mycielskian. The Mycielskian of a graph was introduced by Mycielski ([24]) in 1955. ( G, H ) for graphs G and H is a polyhedral complex, whose -dimensional cellsare the graph homomorphisms from G to H . It is known that Hom ( K , G ) and N ( G ) arehomotopy equivalent ([2]). Definition 1.6. [21] A graph T is called a homotopy test graph if the inequality χ ( G ) ≥ χ ( T ) + conn ( Hom(T,G) ) + 1 , (3)holds for all graph G .Since Hom ( K , G ) ≃ N ( G ) , it follows from Theorem 1.1 that K is a homotopy testgraph. In [2], Babson and Kozlov generalize this result and showed that all complete graphsare homotopy test graphs. Some other examples of homotopy test graphs are given by oddcycles C r +1 [3], bipartite graphs [23] and some of the stable Kneser graphs [28]. For furtherdevelopment and related topics, we refer to [10, 17, 20, 21].In this article, we prove the following. Theorem 1.7.
Let n ≥ and let G be a graph. If M ( M ( K n )) is a subgraph of G , then χ ( K Gm ) = m for m ≤ n + 1 . The following corollary is an application of Theorem 1.7.
Corollary 1.8.
Let n ≥ and let G be a graph. If M ( M ( K n )) is a subgraph of G , then K Gm is a homotopy test graph for m ≤ n + 1 . One of the famous open problem in graph theory is the following conjecture of Hedet-niemi about the chromatic number of categorical product of graphs (for categorical productof graphs, see Definition 2.1).
Conjecture 1.1. [15] For any two graphs G and H , χ ( G × H ) = min { χ ( G ) , χ ( H ) } .Since we have projections G × H → G and G × H → H , it is straightforward to verifythat χ ( G × H ) ≤ min { χ ( G ) , χ ( H ) } . The difficulty lies in deriving the other inequality i.e.,to prove that χ ( G × H ) ≥ min { χ ( G ) , χ ( H ) } . It is easy to prove that Conjecture 1.1 is truefor graphs of chromatic numbers and . In [12], El-Zahar and Sauer, showed that thechromatic number of the product of two -chromatic graph is . The following are the someof the results in the support of the Conjecture 1.1. Theorem 1.9. [7] Let G be graph such that every vertex of G is in an n -clique. For everygraph H , if χ ( G × H ) = n , then min { χ ( G ) , χ ( H ) } = n . Theorem 1.10. [11][33] Let G and H be connected graphs containing n -cliques. If χ ( G × H ) = n , then min { χ ( G ) , χ ( H ) } = n . Theorem 1.11. [6, Theorem ] Let G be be graph such that the clique number of G is n ≥ . Let A be the set of vertices of G that are not contained in an n -clique. Let thereexists a coloring with n + 1 colors , . . . , n + 1 of G such that each vertex of A is colored byeither n or n + 1 . For every graph H , if χ ( G × H ) = n , then min { χ ( G ) , χ ( H ) } = n .
3n all above results, for a pair of graph G and H , the Hedetniemi’s conjecture is true,when χ ( G ) = n + 1 and G contains an n -clique. It is natural to ask, what can we say, if χ ( G ) = n + 2 and the size of the maximal clique in G is n . In this article we prove thefollowing, which is an immediate consequence of Theorem 1.7. Theorem 1.12.
Let n ≥ be positive integer. Let G be a graph containing M ( M ( K n )) asa subgraph. For every graph H , if χ ( G × H ) = n + 1 , then min { χ ( G ) , χ ( H ) } = n + 1 . Some other special cases of the Conjecture 1.1 have also been proved to be true [1, 6,7, 11, 27, 31, 32, 33]. For further development and related research, we refer to the surveyarticles [26, 30, 34].
A graph G is a pair ( V ( G ) , E ( G )) , where V ( G ) is the set of vertices of G and E ( G ) ⊆ V ( G ) × V ( G ) denotes the set of edges. If ( x, y ) ∈ E ( G ) , it is also denoted by x ∼ y and x is said to be adjacent to y . All the graphs in this article are finite, undirected, i.e., ( x, y ) ∈ E ( G ) implies ( y, x ) ∈ E ( G ) and without multiple edges.A graph G is said to be simple if ( x, y ) ∈ E ( G ) implies that x = y . A subgraph H of G is a graph with V ( H ) ⊆ V ( G ) and E ( H ) ⊆ E ( G ) . For a subset U ⊆ V ( G ) , the inducedsubgraph G [ U ] is the subgraph whose set of vertices V ( G [ U ]) = U and the set of edges E ( G [ U ]) = { ( a, b ) ∈ E ( G ) | a, b ∈ U } .For a positive integer r , cyclic graph C r is the graph with V ( C r ) = { , . . . , r } and E ( C r ) = { ( i, i + 1) | ≤ i ≤ r − } ∪ { (1 , r ) } .The neighborhood of v ∈ V ( G ) is defined as N G ( v ) = { w ∈ V ( G ) | ( v, w ) ∈ E ( G ) } . If A ⊆ V ( G ) , the neighborhood of A is defined as N G ( A ) = { x ∈ V ( G ) | ( x, a ) ∈ E ( G ) ∀ a ∈ A } .A graph homomorphism from G to H is a set map φ : V ( G ) → V ( H ) which preserve theedges, i.e, ( x, y ) ∈ E ( G ) = ⇒ ( φ ( x ) , φ ( y )) ∈ E ( H ) . A graph homomorphism f is called an isomorphism if f is bijective and f − is also a graph homomorphism. Two graphs are called isomorphic , if there exists an isomorphism between them. If G and H are isomorphic, wewrite G ∼ = H .The complete graph K n is a graph with V ( K n ) = { , . . . , n } and E ( K n ) = { ( i, j ) | i = j } .A graph isomorphic to K n is called an n -clique or a clique of size n . The chromatic number of a simple graph is defined by χ ( G ) := min { n | ∃ a graph homomorphism from G to K n } .The clique number ω ( G ) of a graph G is the largest integer k such that G contains an k -clique as a subgraph. Definition 2.1 (Categorical product) . The categorical product of two graphs G and H ,denoted by G × H is the graph where V ( G × H ) = V ( G ) × V ( H ) and ( g, h ) ∼ ( g ′ , h ′ ) in G × H if g ∼ g ′ and h ∼ h ′ in G and H respectively. Definition 2.2 (Exponential graph) . Let G and H be graphs. The exponential graph H G is a graph where V ( H G ) contains all the set maps from V ( G ) to V ( H ) and any two vertices f and f ′ in V ( H G ) are said to be adjacent, if v ∼ v ′ in G implies that f ( v ) ∼ f ′ ( v ′ ) in H .4ore details about product of graphs and exponential graphs can be found in [13, 16].A topological space X is said to be k - connected if every continuous map from a m -dimensional sphere S m → X can be extended to a continuous map from the ( m + 1) -dimensional ball B m +1 → X for m = 0 , , . . . , k. The connectivity of X is the largest k suchthat X is k -connected and it is denoted by conn ( X ) . If X is a non empty disconnectedspace, it is said to be − connected.A finite abstract simplicial complex X is a collection of finite sets where τ ∈ X and σ ⊂ τ , implies σ ∈ X . The elements of X are called the simplices of X . If σ ∈ X and | σ | = k + 1 , then σ is said to be k dimensional . Definition 2.3.
The neighborhood complex N ( G ) of a graph G is the abstract simplicialcomplex whose simplices are those subsets of vertices of G , which have a common neighbor,i.e., N ( G ) = { A ⊆ V ( G ) | N G ( A ) = ∅} .In this article we consider N ( G ) as a topological space, namely its geometric realization.For definition of geometric realization and details about simplicial complexes, we refer tobook [21] by Kozlov. Definition 2.4. [2, Definition . ] For any two graphs G and H , Hom complex Hom ( G, H ) is the polyhedral complex whose cells are indexed by all functions η : V ( G ) → V ( H ) \ {∅} ,such that if ( v, w ) ∈ E ( G ) then η ( v ) × η ( w ) ⊆ E ( H ) .Let G be a graph and N G ( u ) ⊆ N G ( v ) for two distinct vertices u and v of G . Thegraph G \ { u } is called a fold of G . Here, V ( G \ { u } ) = V ( G ) \ { u } and the edges inthe subgraph G \ { u } are all those edges of G which do not contain u . In this case, themap V ( G ) → V ( G ) \ { u } , which sends u to v and fixes all other vertices, is a graphhomomorphism. Thus, we conclude that χ ( G \ { u } ) = χ ( G ) .From [19, Theorem . ] we have the following result which allows us to replace a graphby a subgraph in the Hom complex. Proposition 2.5.
Let G \ { v } be a fold of G and let H be some graph. Then Hom ( G, H ) ≃ Hom ( G \ { v } , H ) and Hom ( H, G ) ≃ Hom ( H, G \ { v } ) . From [9, Proposition 3.5] we have a relationship between the exponential graph and thecategorical product in the Hom complex.
Proposition 2.6.
Let G , H and K be graphs. ThenHom ( G × H, K ) ≃ Hom ( G, K H ) . Throughout this article, for any positive integer n , we denote the set { , . . . , n } by [ n ] .For a map f , we let Im f denote the image set of f .A simple graph G is called perfect if, for every induced subgraph H of G , the number ofvertices in a maximum clique is χ ( H ) . The compliment graph G c of G is the graph on samevertex set as of G and E ( G c ) = { ( x, y ) | ( x, y ) / ∈ E ( G ) } . The well known Strong Perfect5raph Theorem ([8]) says that G is perfect if and only if no induced subgraph of G or G c is an odd cycle of length greater than three. Using this characterization of perfect graphswe show that any graph satisfying the property P (defined in section 1.1) is perfect. Proposition 3.1.
Every graph satisfying the property P is perfect. Proof.
Let T be a graph satisfying the property P . It is clear from the definition ofproperty P that no induced subgraph of T is isomorphic to an odd cycle of length greaterthan . Suppose T c has an induced subgraph H isomorphic to an odd cycle of length r ≥ .Without loss of generality we assume that V ( H ) = { , . . . , r } and E ( H ) = { ( i, i +1) | ≤ i ≤ r − } ∪ { (1 , r ) } . Then, clearly (1 , , (2 , and (3 , are edges in T , but neither (2 , nor (2 , is an edge in T , which is a contradiction to the fact that T is satisfying the property P . We first prove the following lemma, which plays a crucial role in the proof of Theorems1.2 and 1.3.
Lemma 3.2.
Let m be a positive integer and let T be a connected graph having the property P . Let f ∈ V ( K Tm ) such that f ( v ) = f ( w ) = a for some edge ( v, w ) ∈ E ( G ) . Then, thereexists ˜ f such that N K Tm ( f ) ⊆ N K Tm ( ˜ f ) and ˜ f ( x ) = a for all x ∈ V ( T ) . Proof.
Define f : V ( T ) → [ m ] by f ( x ) = ( a, if x ∈ N T ( v ) ,f ( x ) , otherwise . Let g ∈ N K Tm ( f ) . We first show that g ∈ N K Tm ( f ) . Since f ( x ) = f ( x ) for all x / ∈ N T ( v ) ,observe that to prove g ∼ f , it is enough to show that for each x ∈ N T ( v ) and y ∼ x , g ( x ) ∼ f ( y ) and g ( y ) ∼ f ( x ) in K m . Let x ∈ N T ( v ) and y ∼ x in T . If y ∈ N T ( v ) , then f ( y ) = a . In this case, since x ∼ v in T and f ∼ g , we see that g ( x ) ∼ f ( v ) = a = f ( y ) .Now, if y / ∈ N T ( v ) , then f ( y ) = f ( y ) and therefore g ( x ) ∼ f ( y ) = f ( y ) . Hence, g ( x ) ∼ f ( y ) in K m .We now show that g ( y ) ∼ f ( x ) in K m . Since, f ( x ) = a , it is sufficient to show that g ( y ) = a . If x = w , then g ∼ f implies that g ( y ) ∼ f ( w ) (in K m ). Hence, g ( y ) = f ( w ) = a .So assume that x = w . If y ∈ { v, w } , then since ( v, w ) ∈ E ( T ) and f ( v ) = f ( w ) = a , wesee that g ∼ f = ⇒ g ( y ) = a . If y / ∈ { v, w } , then since ( v, x ) , ( x, y ) belong to E ( T ) and T is satisfying the property P , we conclude that either ( y, v ) or ( y, w ) is an edge in T . Now, f ∼ g implies that g ( y ) = a .Thus, g ∼ f and therefore N K Tm ( f ) ⊆ N K Tm ( f ) . Now, we apply the above argumentfor f and get a f ∈ V ( K Tm ) such that N K Tm ( f ) ⊆ N K Tm ( f ) and f ( x ) = a for all x ∈ N T ( v ) S y ∈ N T ( v ) N T ( y ) . Since T is a finite connected graph, after finite number of steps, weget a f k ∈ V ( K Tm ) such that N K Tm ( f ) ⊆ N K Tm ( f ) , N K Tm ( f ) ⊆ N K Tm ( f ) , . . . , N K Tm ( f k − ) ⊆ N K Tm ( f k ) and f k ( x ) = a for all x ∈ V ( T ) . We take ˜ f = f k .6 roof of Theorem 1.2. Since, m < χ ( T ) , there exists no graph homomorphism from T to K m . Hence, for any f ∈ V ( K Tm ) , there exists and edge ( v, w ) ∈ E ( T ) such that f ( v ) = f ( w ) and therefore from Lemma 3.2, K Tm is folded onto a subgraph T where each vertex of T is a constant map from V ( T ) to [ m ] . Observe that T is isomorphic to the complete graph K m . Since, folding preserve the chromatic number, χ ( K Tm ) = χ ( T ) = m .Since, N ( G ) ≃ Hom ( K , G ) , from Proposition 2.5, we conclude that N ( K Tm ) ≃ N ( T ) ≃N ( K m ) ≃ S m − . Now, since conn ( S m − ) = m − result follows. Remark 3.3.
In the proof of Theorem 1.2, we have shown that χ ( K Tm ) = m for m < χ ( T ) .But, this is directly follows from Theorem 1.9, by using the following argument. FromProposition 3.1, T is a perfect graph and therefore it contains a clique of order χ ( T ) .Hence, Theorem 1.9 implies that χ ( T × H ) = min { χ ( T ) , χ ( H ) } for any graph H . Inparticular χ ( T × K Tm ) = min { χ ( T ) , χ ( K Tm ) } . Since the map φ : T × K Tm → K m definedby φ ( x, f ) = f ( x ) is a graph homomorphism, we see that χ ( T × K Tm ) ≤ m . But, since χ ( K Tm ) ≥ m and χ ( T ) > m , χ ( T × K Tm ) = min { χ ( T ) , χ ( K Tm ) } implies that χ ( K Tm ) = m .In the rest of the section, we assume that T is a connected graph satisfying the property P , m and r are positive integers, A ⊆ { , . . . , r − } and T = T rA . For any f ∈ V ( K T m ) and ≤ l ≤ r − , we define f l : V ( T ) → [ m ] by f l ( x ) = f (( x, l )) for all x ∈ V ( T ) . Lemma 3.4.
Let f ∈ V ( K T m ) such that f l ( v ) = f l ( w ) = a for some ( v, w ) ∈ E ( T ) andsome ≤ l ≤ r − . There exists ˜ f ∈ V ( K T m ) such that ˜ f l ( x ) = a for all x ∈ V ( T ) and N K T m ( f ) ⊆ N K T m ( ˜ f ) . Proof.
The proof is similar to that of proof of the Lemma 3.2. Define f : V ( T ) → [ m ] by f ( x ) = ( a, if x ∈ N T ( v ) × { l } ,f ( x ) , otherwise . Let h ∈ N K T m ( f ) . We first show that h ∼ f in K T m . Here, it is enough to show thatfor each x ∈ N T ( v ) × { l } and y ∈ N T ( x ) , h ( x ) ∼ f ( y ) and h ( y ) ∼ f ( x ) (in K m ). Let x = ( x , l ) and y = ( y , j ) for some x , y ∈ V ( T ) , ≤ j ≤ r such that x ∈ N T ( v ) × { l } and y ∈ N T ( x ) . Since, ( x, y ) ∈ E ( T ) , we see that ( j, l ) ∈ E ( L r ( A )) and ( x , v ) ∈ E ( T ) .If y ∈ N T ( v ) × { l } , then f ( y ) = a . In this case, y ∼ x in T and x ∼ v in T impliesthat x ∼ ( v, l ) in T . Hence, h ∼ f implies that h ( x ) ∼ f (( v, l )) = f l ( v ) = a = f ( y ) .Now, if y / ∈ N T ( v ) × { l } , then f ( y ) = f ( y ) and therefore h ( x ) ∼ f ( y ) = f ( y ) . Hence, h ( x ) ∼ f ( y ) in K m .Here, f ( x ) = a . Hence, to show that h ( y ) ∼ f ( x ) , it is enough to prove that h ( y ) = a .If x = w , then x = ( w, l ) and therefore h ∼ f implies that h ( y ) = f ( x ) = f (( w, l )) = f l ( w ) = a . If y ∈ { v, w } , then either (( y , j ) , ( v, l )) ∈ E ( T ) or (( y , j ) , ( w, l )) ∈ E ( T ) .Since, f (( v, l )) = f (( w, l )) = a and h ∼ f , we see that h (( y , j )) = a . So, assume that x = w and y / ∈ { v, w } . Since ( w, v ) , ( v, x ) , ( x , y ) are edges in T and T is satisfyingthe property P , we conclude that either ( y , v ) or ( y , w ) is an edge in T . Hence, either (( y , j ) , ( v, l )) ∈ E ( T ) or (( y , j ) , ( w, l )) ∈ E ( T ) . Now, f ∼ h implies that h (( y , j )) = a .Thus N K T m ( f ) ⊆ N K T m ( f ) . We now apply the above argument on f and get a f such that N K T m ( f ) ⊆ N K T m ( f ) and f (( x, l )) = a for all x ∈ N T ( v ) S y ∈ N T ( v ) N T ( y ) . Since7 is finite connected graph, after finite number of steps, we get a f k ∈ K T m such that N K T m ( f ) ⊆ N K T m ( f ) , N K T m ( f ) ⊆ N K T m ( f ) , . . . , N K T m ( f k − ) ⊆ N K T m ( f k ) and f k (( x, l )) = a for all x ∈ V ( T ) . We take ˜ f = f k . Lemma 3.5.
Let ≤ l ≤ r − and let f ∈ V ( K T m ) such that the map f l is a graphhomomorphism from T to K m . If m ≤ χ ( T ) , then N K T m ( f ) = ∅ . Proof.
Let n = χ ( T ) . From Proposition 3.1, T is a perfect graph and therefore it containsa subgraph isomorphic to the complete graph K n . Without loss of generality we can assumethat K n is a subgraph of T , i.e., V ( K n ) = [ n ] ⊆ V ( T ) .Suppose N K T m ( f ) = ∅ and let h ∈ N K T m ( f ) . For ≤ i ≤ r − , let ˜ f i and ˜ h i arerestrictions of f i and h i on K n respectively. Since f l is a graph homomorphism from T to K m , ˜ f l is also a graph homomorphism from K n to K m and therefore ˜ f l ( i ) = ˜ f l ( j ) for all ≤ i = j ≤ n .Since m ≤ χ ( T ) = n and ˜ f l is a graph homomorphism, we conclude that m = n and Im ˜ f l = [ m ] . For any ≤ i = j ≤ n , since ( i, l + 1) ∼ ( j, l ) in T and f ∼ h in K T m , we have h (( i, l + 1)) = f (( j, l )) . Hence, ˜ h l +1 ( i ) = ˜ f l ( j ) for i = j and therefore ˜ h l +1 ( i ) = ˜ f l ( i ) for all ≤ i ≤ n . Thus, ˜ h l +1 = ˜ f l and Im ˜ h l +1 = [ m ] .For any i = j , since ( i, l + 2) ∼ ( j, l + 1) , we see that f (( i, l + 2)) = h (( j, l + 1)) . Usingthe fact that ˜ h l +1 = ˜ f l , we conclude that ˜ f l +2 = ˜ h l +1 = ˜ f l and therefore Im ˜ f l +2 = [ m ] . Byapplying the similar argument, since ( i, l + 2) ∼ ( j, l + 3) for all ≤ i = j ≤ n , we concludethat Im ˜ h l +3 = [ m ] and ˜ h l +3 = ˜ f l +2 = ˜ h l +1 = ˜ f l .By repeating the above argument, we observe that Im ˜ f l + t = [ m ] for all even t suchthat l + t ≤ r − and Im ˜ h l + s = [ m ] , for all odd s such that l + s ≤ r − . Hence, eitherIm ˜ f r − = [ m ] or Im ˜ h r − = [ m ] .Let w = ( ∗ , r ) be the vertex of T rA , obtained by identifying all vertices whose secondcoordinate is r . Since, w ∼ ( i, r − for all ≤ i ≤ n and either Im ˜ f r − = [ m ] orIm ˜ h r − = [ m ] , we see that f ∼ h = ⇒ either h ( w ) / ∈ [ m ] or f ( w ) / ∈ [ m ] , which is acontradiction. Thus N K T m ( f ) = ∅ .Using the above two lemmas we prove the following more general result, from whichproof of Theorem 1.3 follows easily. Theorem 3.6.
Let r be a positive integer and let T be a connected graph having the property P . Then for any A ⊆ { , . . . , r − } and ≤ m ≤ χ ( T ) , N ( K T rA m ) ≃ N ( K L r ( A ) m ) . Proof.
By using Lemma 3.4, K T rA m is folded onto an induced subgraph T where f ∈ V ( T ) if and only if, for each ≤ l ≤ r − , the map f l : V ( T ) → [ m ] defined by f l ( x ) = f (( x, l )) is either a constant map or a graph homomorphism from T to K m . For any f ∈ V ( T ) such that f l is non-constant for some ≤ l ≤ r − , N T ( f ) = ∅ from Lemma 3.5. Hence, N ( T ) ≃ N ( T ) , where T is the induced subgraph of T on the set of vertices V ( T ) = { f ∈ V ( T ) | f l is a constant map ∀ ≤ l ≤ r − } .8ince χ ( T ) ≥ , T contains at least one edge and using this fact it is easy to seethat T ∼ = K L r ( A ) m . Since N ( G ) ≃ Hom ( K , G ) , using Proposition 2.5, we conclude that N ( K T rA m ) ≃ N ( T ) ≃ N ( T ) ≃ N ( K L r ( A ) m ) . Proof of Theorem 1.3.
From Theorem 3.6, N ( K T rA m ) ≃ N ( K L r ( A ) m ) . Since A = { , . . . , i } ,we observe that K × L r ( A ) is folded onto a subgraph isomorphic to K . Hence, fromPropositions 2.5 and 2.6, we conclude that N ( K T rA m ) ≃ N ( K L r ( A ) m ) ≃ Hom ( K , K L r ( A ) m ) ≃ Hom ( K × L r ( A ) , K m ) ≃ Hom ( K , K m ) ≃ N ( K m ) ≃ S m − . Proof of Corollary 1.4.
Let χ ( T ) = n . We first show that χ ( K T rA m ) = m . Since T is a perfectgraph, it has a subgraph isomorphic to K n . Without loss of generality we assume that K n isa subgraph of T . From Lemma 3.4, K T rA m is folded onto a subgraph G , where f ∈ V ( G ) if andonly if, for each ≤ l ≤ r − , the map f l is either a constant map or a graph homomorphismfrom T to K m . Choose ≤ l ≤ r − such that l ∈ A . Define φ : V ( G ) → [ m ] by φ ( f ) = ( f l (1) if f l is a constant map , , otherwise . Clearly φ is well defined. Let f, h ∈ V ( G ) such that f ∼ h . For any g ∈ V ( G ) suchthat g i is non-constant for some ≤ i ≤ r − , N K TrAm ( g ) = 0 from Lemma 3.5 and therefore N G ( g ) ⊆ N K TrAm ( g ) implies that N G ( g ) = ∅ . Hence, f l and h l must be constant maps. Since n ≥ , and (1 , l ) ∼ (2 , l ) in T rA , f ((1 , l )) = h ((2 , l )) = h ((1 , l )) . Hence φ ( f ) = φ ( h ) . Itfollows that φ is a graph homomorphism.Thus, χ ( G ) ≤ m . Since, G contains a subgraph isomorphic to K m , namely induced byall constant maps, we conclude that χ ( G ) = m . Thus, χ ( K T rA m ) = χ ( G ) = m .Since, conn ( S m − ) = m − , result follows from Theorem 1.3. Remark 3.7.
In the proof of Corollary 1.4, we have shown that χ ( K T rA m ) = m for m ≤ χ ( T ) .But, this is directly follows from Theorem 1.11. For, first observe that T rA contains M r ( K n ) as a subgraph, where n = χ ( T ) . Since, M r ( K n ) satisfies the condition of Theorem 1.11, weconclude that χ ( K M r ( K n ) m ) = m for all m ≤ n . Now, since i ∗ : K T rA m → K M r ( K n ) m defined by i ∗ ( f )( x ) = f ( x ) is a graph homomorphism, it follows that χ ( K T rA m ) = m .We now fix some notations. Let m and n be positive integers such that m ≤ n + 1 .For any f ∈ V ( K M ( M ( K n )) m ) and i, j ∈ { , } , we let f i,j : [ n ] → [ m ] defined by f i,j ( x ) = f ( x, i, j ) for all x ∈ [ n ] . For i ∈ { , } , we let w i = ( ∗ , i ) ∈ V ( M ( M ( K n ))) , where ∗ ∈ V ( M ( K n )) obtained by identifying all the vertices whose second coordinate is . Let w ∈ V ( M ( M ( K n ))) obtained by identifying all the vertices whose second coordinate is , i.e.,by identifying all the vertices of the form ( x, where x ∈ V ( M ( K n )) . Lemma 3.8.
Let f ∈ V ( K M ( M ( K n )) m ) . Let there exist p, q ∈ { , } and i , i ∈ [ n ] , i = i such that f p,q ( i ) = f p,q ( i ) . Then there exists ˜ f ∈ V ( K M ( M ( K n )) m ) such that ˜ f p,q ( j ) = f p,q ( i ) for all j ∈ [ n ] and N K M ( M ( Kn )) m ( f ) ⊆ N K M ( M ( Kn )) m ( ˜ f ) . roof. Define ˜ f : V ( M ( M ( K n ))) → [ m ] by ˜ f ( x ) = ( f p,q ( i ) , if x = ( j, p, q ) for some j ∈ [ n ] ,f ( x ) , otherwise . Let h ∈ N K M ( M ( Kn )) m ( f ) . We show that h ∼ ˜ f . Let x = ( j , p, q ) for some j ∈ [ n ] and let y ∼ x . Observe that, to prove h ∼ ˜ f , it is enough to show that h ( x ) = ˜ f ( y ) and h ( y ) = ˜ f ( x ) = f p,q ( i ) . If y = ( j, p, q ) for some j ∈ [ n ] , then ˜ f ( y ) = f p,q ( i ) . Since, y ∼ x , we conclude that x is adjacent to either ( i , p, q ) or ( i , p, q ) . Hence, h ∼ f and f (( i , p, q )) = f (( i , p, q )) implies that h ( x ) = f (( i , p, q )) = f p,q ( i ) = ˜ f ( y ) . If y = ( j, p, q ) for all j ∈ [ n ] , then ˜ f ( y ) = f ( y ) and therefore h ( x ) = ˜ f ( y ) . Thus, h ( x ) = ˜ f ( y ) .If y = w i for some i ∈ { , , } , the since y ∼ x , we see that y ∼ ( i , p, q ) . Hence, f ∼ h implies that h ( y ) = f (( i , p, q )) = f p,q ( i ) . So, assume that y = w i for all i ∈ { , , } .For any i = j , since i ∼ j in K n , we observe that either y ∼ ( i , p, q ) or y ∼ ( i , p, q ) .Since, f ∼ h and f (( i , p, q )) = f (( i , p, q )) , we conclude that h ( y ) = f p,q ( i ) . Thus, h ( y ) = ˜ f ( x ) . Lemma 3.9.
Let m ≤ n and let f ∈ V ( K M ( M ( K n )) m ) . If there exist p, q ∈ { , } such that f p,q is a graph homomorphism from K n to K m , then N K M ( M ( Kn )) m ( f ) = ∅ . Proof.
Since f p,q is a graph homomorphism from K n to K m and m ≤ n , we conclude that m = n and Im f p,q = [ m ] . Suppose N K M ( M ( Kn )) m ( f ) = ∅ and let h ∈ N K M ( M ( Kn )) m ( f ) . Letus first assume that q = 1 . Then, w ∼ ( j, p, q ) for all j ∈ [ n ] . Hence, f ∼ h implies that h ( w ) / ∈ Im f p,q = [ m ] , which is a contradiction.Now, let q = 0 . If p = 1 , then since w ∼ ( j, , for all j ∈ [ n ] , we conclude that h ( w ) / ∈ Im f p,q = [ m ] , a contradiction. So, assume p = 0 . For any i = j , since ( i, , ∼ ( j, , and f ∼ h , we have h , ( i ) = f , ( j ) . Hence, h , ( i ) = f , ( i ) for all i . Now, since w ∼ ( i, , for all i , we see that f ( w ) / ∈ Im h , = Im f , = [ m ] , which is a contradiction. Proof of Theorem 1.5.
Using Lemma 3.8, K M ( M ( K n )) m is folded onto an induced subgraph G , where f ∈ V ( G ) if and only if, for each p, q ∈ { , } the map f p,q is either a constantmap or a graph homomorphism from K n to K m . Let G be the induced subgraph of G ,where f ∈ V ( G ) if and only if f p,q is a constant map for all p, q ∈ { , } . From Lemma3.9, N G ( f ) = ∅ for all f such that f p,q is non-constant for some p, q ∈ { , } . Hence, N ( G ) ≃ N ( G ) . It is easy to see that G ∼ = K M ( L ( { } )) m . Observe that M ( L ( { } )) isfolded onto L ( { } ) and therefore K × M ( L ( { } )) is folded onto K × L ( { } ) . Since, K × L ( { } ) is folded onto a subgraph isomorphic to K , we have, N ( K M ( M ( K n )) m ) ≃ N ( G ) ≃ N ( G ) ≃ N ( K M ( L ( { } )) m ) ≃ Hom ( K , K M ( L ( { } )) m ) ≃ Hom ( K × M ( L ( { } )) , K m ) ≃ Hom ( K × L ( { } ) , K m ) ≃ Hom ( K , K m ) ≃ N ( K m ) ≃ S m − .Result follows from Theorem 1.7. As in the previous Section, let m and n be positive integers such that m ≤ n + 1 . Forany f ∈ V ( K M ( M ( K n )) m ) and i, j ∈ { , } , we define f i,j : [ n ] → [ m ] by f i,j ( x ) = f ( x, i, j ) for all x ∈ [ n ] . For i ∈ { , } , w i = ( ∗ , i ) ∈ V ( M ( M ( K n ))) , where ∗ ∈ V ( M ( K n )) obtained10y identifying all the vertices whose second coordinate is and w ∈ V ( M ( M ( K n ))) isobtained by identifying all the vertices of the form ( x, , where x ∈ V ( M ( K n )) .Let G be the induced subgraph of K M ( M ( K n )) m where f ∈ V ( G ) if and only if, for any i, j ∈ { , } the map f i,j is either a constant map or a graph homomorphism from K n to K m .We first prove few lemmas, which we need in the proof of Theorem 1.7. Lemma 3.10.
Let f ∈ V ( G ) such that f , and f , are graph homomorphisms.(i) If f , ( i ) = f , ( j ) for some i , j ∈ [ n ] , i = j , then there exists ˜ f ∈ V ( G ) suchthat Im ˜ f , = { f , ( i ) } and N G ( f ) ⊆ N G ( ˜ f ) .(ii) If Im f , = Im f , and f , ( i ) = f , ( i ) for some i ∈ [ n ] , then there exists ˜ f ∈ V ( G ) such that Im ˜ f , = { f , ( i ) } and N G ( f ) ⊆ N G ( ˜ f ) .(iii) If Im f , = Im f , and f , ( i ) = f , ( i ) , f , ( j ) = f , ( j ) for some i, j ∈ [ n ] , i = j ,then there exists ˜ f ∈ V ( G ) such that ˜ f , is a constant map and N G ( f ) ⊆ N G ( ˜ f ) . Proof. (i) Define ˜ f : V ( M ( M ( K n ))) → [ m ] by ˜ f ( x ) = ( f , ( i ) , if x = ( j, , for some j ∈ [ n ] ,f ( x ) , otherwise . Let h ∼ f . Let x = ( j, , for some j ∈ [ n ] and let y ∼ x . Let y = ( y , p, q ) . Observethat p ∈ { , } and q ∈ { , } . If p = 2 , then y = w q . In this case, y ∼ ( j , , and therefore f ∼ h implies that h ( y ) = f (( j , , f , ( j ) = f , ( i ) . Assumethat p = 0 . In this case, since i = j , we see that either ( y , , q ) ∼ ( i , , (if y = i ) or ( y , , q ) ∼ ( j , , (if y = j ). Since, f ( i , ,
0) = f ( j , , , we derivethat h ( y ) = f , ( i ) . Hence, h ( y ) = ˜ f ( x ) .Since, y ∼ x , y = ( i, , for any i ∈ [ n ] and therefore ˜ f ( y ) = f ( y ) . Hence, h ( x ) =˜ f ( y ) . Now, from the definition of ˜ f , we conclude that ˜ f ∼ h . Thus, N G ( f ) ⊆ N G ( ˜ f ) .(ii) Since Im f , = Im f , and f , ( i ) = f , ( i ) , we see that there exists j = i such that f , ( j ) = f , ( i ) . Proof follows from ( i ) .(iii) Since f , is a graph homomorphism from K n to K m , we see that f , ( i ) = f , ( j ) .Further, since f , is a graph homomorphism, | Im f | = n . Now, m ≤ n + 1 impliesthat { f , ( i ) , f , ( j ) } ∩ Im f , = ∅ .If f , ( i ) ∈ Im f , , then f , ( p ) = f , ( i ) for some p . Since, f , ( i ) = f , ( i ) , we seethat p = i . Similarly, if f , ( j ) ∈ Im f , and f , ( j ) = f , ( q ) , then q = j . Prooffollows from ( i ) . Lemma 3.11.
Let f ∈ V ( G ) such that f , is a graph homomorphism.(i) Let f , is a graph homomorphism such that Im f , = Im f , . If f ( w ) / ∈ Im f , ,then N G ( f ) = ∅ . ii) Let f , is a graph homomorphism, Im f , = Im f , and f ( w ) ∈ Im f , . If f , and f , differ at only one vertex, then N G ( f ) = ∅ .(iii) Let f , is a constant map such that f , (1) / ∈ Im f , . If f ( w ) ∈ Im f , , then N G ( f ) = ∅ . Proof. If m ≤ n , then since f , is a graph homomorphism, N G ( f ) = ∅ from Lemma 3.9.So, assume that m = n + 1 . Suppose N G ( f ) = ∅ and let h ∈ N G ( f ) . Observe that, since w ∼ ( i, j, for all ≤ i, j ≤ n and h ∼ f , f ( w ) / ∈ Im h , ∪ Im h , ∪ h ( w ) . Therefore,Im h , ∪ Im h , ∪ h ( w ) = [ m ] .(i) Since f ( w ) / ∈ Im f , , | Im f , | = n and m = n +1 , we see that { f ( w ) } = [ m ] \ Im f , .Since, ( i, , ∼ ( j, , for i = j and h ∼ f , we see that h (( i, , ∈ { f , ( i ) , f ( w ) } .For any i ∈ [ n ] , since ( i, , ∼ w , we conclude that f ( w ) = h (( i, , . Hence, h (( i, , f (( i, , for all i , i.e., h , = f , . Since w ∼ ( i, , for all i ∈ [ n ] , h ∼ f implies that h ( w ) / ∈ Im f , = Im f , = Im h , . Thus, Im h , ∪ h ( w ) = [ m ] ,which implies that Im h , ∪ Im h , ∪ h ( w ) = [ m ] , a contradiction. Therefore, N G ( f ) = ∅ .(ii) Let { a } = [ m ] \ Im f , . Since Im f , = Im f , and | Im f , | = n , there exists i ∈ [ n ] such that f , ( i ) = a . Further, since f , and f , differs at only one vertex, weconclude that f , ( j ) = f , ( j ) for all j ∈ [ n ] , j = i .Since ( i, , ∼ ( j, , for all i = j and h ∼ f , we see that h (( i, , ∈ { f , ( i ) , a } .If i = i , then since ( i, , ∼ ( i , , and f (( i , , f , ( i ) = a , we seethat h (( i, , = f (( i , , a . Hence, h (( i, , f , ( i ) for all i = i and h (( i , , ∈ { f , ( i ) , a } . Thus, h , is either f , or f , .Let h , = f , . Since w ∼ ( i, , for all i , we see that h ( w ) / ∈ Im f , and thereforeIm h , ∪ h ( w ) = [ m ] , which is a contradiction.Now, let h , = f , . If h , is constant map, then since ( i, , ∼ ( j, , for all i = j , h ∼ f implies that Im h , = a . Hence Im h , ∪ Im h , = [ m ] , a contradiction. Soassume that h , is non-constant. Since, h ∈ V ( G ) , h , is a graph homomorphism.For any i ∈ [ n ] , w ∼ ( i, , and therefore f ( w ) / ∈ Im h , . Since, f ( w ) ∈ Im f , ,we see that f ( w ) = a . Now, | Im h , | = n implies that a ∈ Im h , and therefore Im h , ∪ Im h , = [ m ] , which is not possible. Hence, N G ( f ) = ∅ .(iii) Let { a } = Im f , . Then a / ∈ Im f , . Since ( i, , ∼ ( j, , for all i = j , h ∼ f implies that a / ∈ Im h , . Further, since ( i, , ∼ ( j, , for all i = j , we concludethat h , ( k ) = f , ( k ) for all k ∈ [ n ] . Hence, h , = f , .Let us first assume that h , is a constant map and Im h , = { b } . Observe that b / ∈ Im f , . Thus, h , = f , implies that Im h , ∪ { b } = [ m ] , which is a contradiction.Now, assume that h , is a graph homomorphism. Observe that f ( w ) / ∈ Im h , . Since f ( w ) ∈ Im f , , there exists i such that f , ( i ) = f ( w ) . Since, ( i , , ∼ ( j, , , h , ( i ) = f , ( j ) for all j = i . Hence, either h , ( i ) = f , ( i ) or h , ( i ) = a . But,since, f , ( i ) = f ( w ) / ∈ Im h , , we conclude that h , ( i ) = a i.e., a ∈ Im h , . Thus,Im h , ∪ Im h , = Im f , ∪ Im h , = [ m ] , which is not possible. Hence, N G ( f ) = ∅ .12 roof of Theorem 1.7. We first prove that, for any m ≤ n + 1 , χ ( K M ( M ( K n )) m ) = m . FromLemma 3.8, K M ( M ( K n )) m is folded onto a subgraph G , where f ∈ V ( G ) if and only if for any i, j ∈ { , } the map f i,j is either a constant map or a graph homomorphism from K n to K m . Using Lemma 3.10, G is folded onto the induced subgraph on vertex set V ( G ) , where f ∈ V ( G ) if and only if the following are true:(A) if Im f , = Im f , , then f , = f , (from Lemma 3.10 ( ii ) ).(B) if f , , f , are graph homomorphisms and Im f , = Im f , , then f , and f , differat only one vertex (from Lemma 3.10 ( iii ) ).We define the subsets U , U , U ⊆ V ( G ) by U = { f ∈ V ( G ) | f , , f , are graph homomorphisms , Im f , = Im f , and f ( w ) / ∈ Im f , } , U = { f ∈ V ( G ) | f , , f , are graph homomorphisms , Im f , = Im f , , f ( w ) ∈ Im f , and , f , and f , differ at only one vertex } and U = { f ∈ V ( G ) | f , is a graph homomorphism , f , is a constanta map , f , (1) / ∈ Im f , and f ( w ) ∈ Im f , } .Let G be the induced subgraph of G on vertex set V ( G ) \ ( U ∪ U ∪ U ) . From Lemma3.11, N G ( f ) = ∅ for all f ∈ U ∪ U ∪ U . Since N G ( f ) ⊆ N G ( f ) , we see that N G ( f ) = ∅ for all f ∈ U ∪ U ∪ U . Thus, we conclude that χ ( G ) = χ ( G ) .We define φ : V ( G ) → [ m ] by φ ( f ) = f , (1) , if f , is a constant map ,f , (1) , if f , is a graph homomorphism and f , is a constant map ,f ( w ) , if f , and f , are graph homomorphisms . Clearly, φ is well defined. We show that φ is a graph homomorphism from G to K m .Let ( f, h ) ∈ E ( G ) . Let us first assume that m ≤ n . In this case, from Lemma 3.9, N G ( g ) = ∅ for all g such that g , is a graph homomorphism. Hence, f , and h , mustbe constant maps. Here, φ ( f ) = f , (1) and φ ( h ) = h , (1) . Since, (1 , , ∼ (2 , , , f , (1) = f ((1 , , = h ((2 , , h ((1 , , h , (1) . Hence φ ( f ) = φ ( h ) .So, assume that m = n + 1 . We consider the following cases. Case 1. f , is a constant map.In this case φ ( f ) = f , (1) . If h , is a constant map, then φ ( h ) = h , (1) . Since (1 , , ∼ (2 , , , f ((1 , , = h ((2 , , h ((1 , , . Hence φ ( f ) = φ ( h ) . If h , is a graph homomorphism and h , is a constant map, then φ ( h ) = h , (1) . In this case,since (1 , , ∼ (2 , , , we see that f ((1 , , = h ((2 , , h ((1 , , and therefore φ ( f ) = φ ( h ) . Now, assume that both h , and h , are graph homomorphisms. Here, φ ( h ) = h ( w ) . Subcase 1.1. Im h , = Im h , .In this case from the condition ( A ) given above, we conclude that h , = h , . Since, h / ∈ U , we see that h ( w ) ∈ Im h , . Since f , is a constant map and h ∼ f , observe that f , (1) / ∈ Im h , . Thus φ ( f ) = f , (1) = h ( w ) = φ ( h ) . Subcase 1.2. Im h , = Im h , . 13ince, h , and h , are graph homomorphisms, m = n + 1 and Im h , = Im h , ,we see that Im h , ∪ Im h , = [ m ] . From the condition ( B ) given above, we concludethat h , and h , differ at only one vertex. Since, h / ∈ U , we see that h ( w ) / ∈ Im h , .Now, Im h , ∪ Im h , = [ m ] implies that h ( w ) ∈ Im h , . Since, f , is a constant mapand ( i, , ∼ ( j, , for all i = j , we see that f ∼ h = ⇒ f , (1) / ∈ Im h , . Hence, φ ( f ) = φ ( h ) . Case 2. f , is a graph homomorphism and f , is a constant map.In this case φ ( f ) = f , (1) . If h , is a constant map then φ ( h ) = h , (1) . Since ( i, , ∼ ( j, , for all i = j and f ∼ h , we conclude that h , (1) = f , (1) .Let h , is a graph homomorphism and h , is a constant map. Here, φ ( h ) = h , (1) .Since ( i, , ∼ ( j, , for all i = j and f ∼ h , we see that h , (1) / ∈ Im f , . If f , (1) ∈ Im f , , then φ ( h ) = φ ( f ) . So assume that f , (1) / ∈ Im f , . Since f / ∈ U , f ( w ) / ∈ Im f , .Using the fact that | Im f , | = n and m = n + 1 , we conclude that f ( w ) = f , (1) = φ ( f ) .Since w ∼ ( i, , for all i , we see that h , (1) = f ( w ) . Thus φ ( f ) = φ ( h ) .Now, let both h , and h , are graph homomorphisms. Here, φ ( h ) = h ( w ) . Since w ∼ ( i, , for all i , we have f , (1) = h ( w ) and therefore φ ( f ) = φ ( h ) . Case 3.
Both f , and f , are graph homomorphisms.In this case φ ( f ) = f ( w ) . If h , is a constant map or h , is a constant map, then byreversing the roles of h and f , we conclude from Case and Case that φ ( f ) = φ ( h ) . Soassume that both h , and h , are graph homomorphisms. Here, φ ( h ) = h ( w ) . Subcase 3.1. Im f , = Im f , .From ( A ) , f , = f , . Since f / ∈ U , f ( w ) ∈ Im f , . Further, since w ∼ ( i, , for all i , we see that h ( w ) / ∈ Im f , . Since f , = f , , f ( w ) ∈ Im f , . Hence φ ( f ) = f ( w ) = h ( w ) = φ ( h ) . Subcase 3.2. Im f , = Im f , .From ( B ) , f , and f , differ at only one vertex. Since, f / ∈ U , f ( w ) / ∈ Im f , . SinceIm f , = Im f , , we see that f ( w ) ∈ Im f , . Observe that h ( w ) / ∈ Im f , . Hence φ ( f ) = φ ( h ) .Thus φ ( f ) ∼ φ ( h ) in K m . Therefore φ is a graph homomorphism. Hence, χ ( G ) ≤ m .Since G contains a clique of order m , namely induced by constant maps, we conclude that χ ( G ) = m . Since folding preserve chromatic number, we have χ ( K M ( M ( K n )) m ) = χ ( G ) = χ ( G ) = χ ( G ) = m .Now, result follows from the fact that we have graph homomorphism i ∗ : K Gm → K M ( M ( K n )) m defined by φ ( f )( x ) = f ( x ) and K Gm has a subgraph isomorphic to the com-plete graph K m , namely induced by all constant maps.In [23], Matsushita proved the following sufficient condition for a graph to be a homotopytest graph. Theorem 3.12. [23, Theorem ] Let A be a graph of chromatic number n . If A has asubgraph isomorphic to K n , then A is a homotopy test graph.Proof of Corollary 1.8. Since, K Gm has a subgraph isomorphic to K m , result follows fromTheorem 1.7 and Theorem 3.12. 14 .3 Proof of the result of Section 1.3 Proof of Theorem 1.12.
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