Hedetniemi s conjecture and dense boolean lattices

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1 Hedetniemi s conjecture and dense boolean lattices Claude Tardif Royal Military College of Canada PO Box Stn Forces, Kingston, ON Canada, K7K 7B4 December 4, 2009 Keywords: Graph colourings, graph products, Hedetniemi s conjecture AMS 1991 Subject Classification: 05C15. Abstract The category D of finite directed graphs is cartesian closed, hence it has a product and exponential objects. For a fixed K, let K D be the class of all directed graphs of the form K G, preordered by the existence of homomorphisms, and quotiented by homomorphic equivalence. It has loong been known that K D, is always boolean lattice. In this paper we prove that for any complete graph K n with n 3, K D n is dense, hence up to isomorphism it is the unique countable dense boolean lattice. In graph theory, the structure of K D n is connected to the conjecture of Hedetniemi on the chromatic number of a categorical product of graphs. 1 Introduction 1.1 Hedetniemi s conjecture The chromatic number χ(g) of a directed graph G is the minimum number of colours needed to colour its vertices so that the endpoints of every arc get different colours. The categorical product G H of G and H has vertex-set V (G H) = V (G) V (H), and its arcs are the couples ((u,u ),(v,v )) such that (u,v) is an arc of G and (u,v ) is an arc of H. Colourings and chromatic numbers are usually considered in the context of symmetric directed graphs, which are usually called undirected graphs. 1

2 Conjecture 1 (Hedetniemi [5], 1966) If G and H are undirected graphs, then χ(g H) = min{χ(g),χ(h)}. Hedetniemi s conjecture is a fundamental problem in chromatic graph theory. There are many positive partial results on it, presented in the three surveys [10, 11, 13]. The present paper is inspired by the negative approach to the conjecture, that is, the work started in [8] surrounding minimalistic versions of the conjecture. Conjecture 2 (The undirected weak Hedetniemi conjecture) For every integer n, there exists an integer f(n) such that for every undirected graphs G and H satisfying min{χ(g),χ(h)} > f(n), we have χ(g H) > n. It is embarassing to have no proof of Conjecture 2 and at the same time no counterexample to Conjecture 1. The directed version of Conjecture 1 is known to be false, with fairly small counterexamples. The directed version of Conjecture 2 is still open. Conjecture 3 (The directed weak Hedetniemi conjecture) For every integer n, there exists an integer g(n) such that for every directed graphs G and H satisfying min{χ(g),χ(h)} > g(n), we have χ(g H) > n. Proposition 4 ([12]) The undirected weak Hedetniemi conjecture is equivalent to the directed weak Hedetniemi conjecture. Proposition 5 ([9]) The directed weak Hedetniemi conjecture holds if and only if g(3) is well defined (i.e., finite). So here directed graphs enter the picture, and results on the directed weak Hedetniemi conjecture impacts on its undirected counterpart. We will now present order-theoretic interpretations of these conjectures, and then state and interpret our main theorem. 1.2 The homomorphism order A homomorphism between (directed or undirected) graphs G and H is an arc-preserving map from the vertex-set of G to that of H. We write G H if there exists a homomorphism from G to H, and G H otherwise. Let D and U denote the classes of finite directed and undirected graphs respectively. It is well known (see [6]) that induces distributive lattice orders on D/ and U/, where is the equivalence relation defined by G H G H and H G. 2

3 The meet operation in D/ is induced by the categorical product, and the join is induced by the disjoint union +. (U/ is a sublattice of D/.) Distributivity is then a consequence of the straightforward identity G (H 1 + H 2 ) = (G H 1 ) + (G H 1 ). The complete graph K n is the graph with vertex set {1,...,n} and arc set {(i,j) i j}. Thus χ(g) n if and only if G K n. By adding copies of K n as summands in the factors, we get the following equivalence. Remark 6 For an integer n, the following are equivalent. For any undirected graphs G,H such that χ(g H) = n, we have min{χ(g),χ(h)} = n. K n / is meet-irreducible in the lattice U/. Thus Conjecture 1 admits a lattice-theoretic formulation. There is no such direct lattice-theoretic formulations of Conjectures 2 and 3 in terms of U/ and D/, because the chromatic number is a graphtheoretic property of the factors. We get a better interpretation in terms of exponentiation and exponential lattices introduced next. 1.3 Exponential lattices Let G, K be directed graphs. The directed graph K G is defined as follows. The vertices of K D n are all functions f : V (G) V (K), and the arcs of K G are the pairs (f,g) such that for every arc (u,v) of G, (f(u),g(v)) is an arc of K. Note that a function f : V (G) V (K) is a homomorphism from G to K if and only if (f,f) is an arc of K G. When G K, K G has no loops. However K G always contains a copy of K induced by the constant maps. The fundamental property of exponentiation is the equivalence G H K G K H, which has several immediate consequences: 1. G K KG, 2. G H K H K G, 3. K G+H K G K H, 4. G = K H for some H G K KG. 3

4 These and other general properties of exponentiation are discussed in [1], where they are considered folklore. We shall be concerned with exponential lattices, that is, objects of the type K U = {(K G )/ G U} and K D = {(K G )/ G D}. Hedetniemi s conjecture 1 is equivalent to some of these objects being trivial. Remark 7 For an integer n, the following are equivalent. For any undirected graphs G,H such that χ(g H) = n, we have min{χ(g),χ(h)} = n. K U n is a two-element lattice. Therefore it appears that determining the cardinality of exponential lattices is often difficult. On the other hand, the algebraic structure of these lattices is well known in the folklore (see [1]). For C = U or D, K/ is the minimal element in K C, and the maximal element is L/, where L is the one-vertex graph with a loop. By 3, K C is closed under. However K C is not a sublattice of C/. Indeed for G/,H/ K C we usually do not have (G + H)/ K C. The join of G/ and H/ in K C is then G/ H/ = ( K KG+H) /, which is generally larger (with respect to the order ) than (G + H)/. However still distributes over. By 2, the complementation mapping G K C to G = K G is well-defined. We then have G G K and G G L/. Thus K C is a boolean lattice. 1.4 Dense boolean lattices. If Hedetniemi s conjecture 1 fails, then for some n the boolean lattice Kn U has more than two elements. If Kn U is finite or at least contains an atom G/, then the value f(n) = max{χ(g),χ(kn G )} satisfies the property described in Conjecture 2. Thus if Conjecture 2 fails, then Kn U is dense. A similar argument holds for directed graphs. Therefore by Propositions 4 and 5, if K3 D contains an atom, then Conjectures 3 and 2 are true. However our main result rules out this eventuality. Theorem 8 For every n 3, K D n is dense. It is well known (see [4]) that there is only one isomorphism type of dense countable boolean lattice, namely the lattice of finite unions of intervals of the type (a,b] (0,1], with rational endpoints. Thus Theorem 8 completely characterises the structure of the lattices Kn D, n 3. (KD 1 and KD 2 are twoelement boolean lattices.) Theorem 8 has no direct impact on conjectures 1, 4

5 2 and 3. At first sight it seems to provide evidence against these conjectures. On the other hand, Conjecture 1 asks for a complete characterisation of the lattices K U n. Our result shows that their superlattices K D n can be completely characterised. Perhaps the tools used and the insight gained will be helpful in settling Hedetniemi s conjecture one way or the other. In the next section, we will see that Theorem 8 is an easy consequence of our Theorem 10 on exponentiation with the iterated shift graphs of [6] as powers. The remainder of the paper will be devoted to the proof of Theorem 10, using tree duality as a main tool. 2 Proof of Theorem Density It is well known that a boolean lattice is dense if and only if it contains no atoms. Therefore to prove Theorem 8, it suffices to show that if G/ K D n and G K n, then there exists a digraph H such that H/ K D n, K H G and G H K. The standard construction, used in particular to prove the density of U (see [6]) uses a graph of the form H = K n +(G M), where M is a graph with girth larger than V (G) and chromatic number larger than χ(k G n ). It can be shown (see below) that such a choice of H is indeed strictly between K n and G. The difficulty lies in selecting M such that H/ is in K D n. For this purpose, we introduce the shift graphs. 2.2 Shift graphs The arc graph δ(g) of a directed graph G is the digraph whose vertices are the arcs of G, and whose arcs are the couples ((u,v),(v,w)) of consecutive arcs in G. The transitive tournament T m is the digraph with vertex-set {1,...,m} and arcs (i,j) : i < j. Iterated arc-graphs of the form δ k (T m ) are called shift graphs. The vertices of δ k (T m ) can be represented by increasing sequences (i 0,i 1,...,i k ) of elements in {1,...,m}, with arcs joining consecutive sequences. Lemma 9 ([6], Section 2.5) (i) For k 1, χ(δ k (T n )) log 2 (χ(δ k 1 (T n )) (and χ(δ 0 (T n )) = χ(t n ) = n). (ii) The odd girth of δ k (T n ) is at least 2k + 1. Therefore the family of shift graphs provide simple examples of graphs with large chromatic number and large odd girth. 5

6 Theorem 10 For any n 3 and m, (K n + δ k (T m ))/ K D n. The proof of Theorem 10 is the object of Section 3. We will see that Theorem 10 is all that is needed to use standard density arguments to prove Theorem Proof of Theorem 8 using Theorem 10 Since Kn D is a boolean lattice, it suffices to show that it has no atoms. Let G be a directed graph such that G/ Kn D and G K n. We need to show that G/ is not an atom. Let δ k (T m ) be a shift graph with odd girth larger than V (G) and chromatic number larger than χ(kn G ). Put H = G (K n + δ k (T m )). By Theorem 10, H/ Kn D. We have K n H G, but since χ(δ k (T m )) > χ(kn G), it follows that (K n +δ k (T m )) Kn G, whence H = G (K n + δ k (T m )) K n. Also, since G K n, G has a connected component X such that X K n. In particular X is nonbipartite and contains an odd cycle. However since the odd girth of δ k (T m ) is larger than the number of vertices in X, we also have X δ k (T m ) thus X (K n + δ k (T m )). Therefore X G (K n + δ k (T m )) = H, whence G H. This shows that G/ is not an atom. 3 Proof of Theorem A subgraph of K δk (T m) n Recall that the vertices of δ k (T m ) are arrays (i 0,i 1,...,i k ) with 1 i 0 < i 1 < < i k m. For {1,2,3} = V (K 3 ) V (K n ), we define 1 + = 2, 2 + = 3 and 3 + = 1. For a coordinate c {0,...,k}, a height h {c + 1,...,c + m k} and a label (colour) l {1,2,3}, we define the function f c,h,l K δk (T m) 3 by { l if ic h, f c,h,l (i 0,i 1,...,i k ) = l + otherwise. Note that all the functions f c,h,l with h = c + m k are equal to the constant function γ l identically equal to l. In all other cases f c,h,l has range {i,i + }. The arcs between these functions can be characterised through a straightforward application of the definition. Lemma 11 For c,c {0,...,k}, h, {c + 1,...,c + m k}, h {c + 1,...,c + m k} and l {1,2,3}, 6

7 (ll) there is an arc of K δk (T m) 3 from f c,h,l to f c,h,l if and only if h c + m k 1 and h = c + 1. (ll + ) There is an arc of K δk (T m) 3 from f c,h,l to f c,h,l + if and only if one of the following conditions holds. h c + m k 1, h < c + m k 1, c c 1 and h h + (c c + 1). (l + l) There is an arc of K δk (T m) 3 from f c,h,l + to f c,h,l if and only if one of the following conditions holds. h = c + m k, h < c + m k, c < c and h > h (c c ). Proof. (ll): In δ k (T m ), there is an arc from (1,...,k + 1) to (2,...,k + 2). Also we have f c,h,l (1,...,k + 1) = l. If (f c,h,l,f c,h,l) is an arc of K δk (T m) 3, we must have f c,h,l(2,...,k + 2) = l +, that is, h = 1 + c. Therefore, f c,h,l(m k,...,m) = l +. Since (m k 1,...,m 1) is joined by an arc to (m k,...,m) in δ k (T m ), we must have f c,h,l (m k 1,...,m 1) = l, that is, h m (k c) 1. Conversely, if h m (k c) 1 and h = 1 + c, then all non-sink elements of δ k (T m ) are mapped to l by f c,h,l, and all non-source elements of δ k (T m ) are mapped to l + by f c,h,l, therefore (f c,h,l,f c,h,l) is an arc of K δk (T m) 3. (ll + ): There is an arc from f c,h,l to f c,h,l + unless there is an arc ((i 0,...,i k ), (i 1,...,i k+1 )) of δ k (T m ) such that f c,h,l (i 0,...,i k ) = f c,h,l +(i 1,...,i k+1 ) = l +. The sequence i 0,...,i k+1 then satisfies i c > h and i c +1 h. An increasing sequence in {1,...,m} satisfying these properties exists if and only h < m (k c) 1 and either c + 1 < c, or c + 1 c and h > h + (c + 1 c). (l + l): There is an arc from f c,h,l + to f c,h,l unless there is an arc ((i 0,...,i k ), (i 1,...,i k+1 )) of δ k (T m ) such that f c,h,l +(i 0,...,i k ) = f c,h,l(i 1,...,i k+1 ) = l +. The sequence i 0,...,i k+1 then satisfies i c h and i c +1 > h. An increasing sequence in {1,...,m} satisfying these properties exists if and only h < m (k c ) and either c < c + 1, or c c + 1 and h h + c c. 7

8 Let G k,m,3 be the subgraph of K δk (T m) 3 induced by these functions f c,h,l. For n 4, the graph G k,m,n is the subgraph of K δk (T m) n obtained from G k,m,3 by adding the constant functions γ l,l = 4,...,n. Proposition 12 K G k,m,n n K n + δ k (T m ). Proof of Theorem 10 using Proposition 12. We have K n + δ k (T m ) K Kδk (Tm) n n K G k,m,n n K n + δ k (T m ). The first arrow is a basic property of exponentiation outlined in Section 1.3, the second arrow uses the fact that G k,m,n K δk (T m) n (and that exponentiation reverses arrows), and the third arrow is Proposition 12. Therefore (K n + δ k (T m )) K G k,m,n n, whence (K n + δ k (T m ))/ Kn D. We have seen that Theorem 8 is a consequence of Theorem 10, which is in turn a consequence of Proposition 12. However here the argument branches: We will split K G k,m,n n in two parts, admitting homomorphisms to K n and to δ k (T m ) respectively. Proposition 12 is a consequence of Lemma 14 and Lemma 15 given below. 3.2 Parts of K G k,m,n n G k,m,n contains a copy of K n induced by the constant functions γ 1,...,γ n. Let L n denote this copy of K n. Lemma 13 Let g : V (G k,m,n ) V (K n ) be a vertex of K G k,m,n n such that the restriction of g to L n is a bijection. Then for any (in or out) neighbour g of g and l = 1,...,n we have g (γ l ) = g(γ l ). Proof. This is actually a well-known property of exponentiation (see [2]); we repeat the short argument here for completeness. Suppose that g is an outneighbour of g. Then for any l l, γ l is adjacent to γ l, hence g (γ l ) g(γ l ). Since the restriction of g to L n is a bijection, the lone value remaining available for g (γ l ) is g(γ l ). The same argument is also valid when g is an inneighbour of g. 8

9 Let A k,m,n be the subgraph of K G k,m,n n induced by the functions g : V (G k,m,n ) V (K n ) whose restriction to L n is not bijective. By Lemma 13, no vertex of A k,m,n is adjacent to a vertex outside A k,m,n. Lemma 14 A k,m,n K n. Proof. This is again a well-known argument in the field (see [2]). Let c : A k,m,n V (K n ) be any function such that c(g) = l = l appears at least twice in g(γ 1 ),...,g(γ n ). If (g,g ) is an arc of A k,m,n, there exist l l such that c(g) = g(γ l ) g (γ l ) = c(g ), whence c is a homomorphism. The complement of A k,m,n in K G k,m,n n splits in n! components B φ k,m,n corresponding to the bijections from φ : V (L n ) V (K n ). The restriction of any g V (B φ k,m,n ) to L n is φ. By Lemma 13, no vertex of B φ k,m,n is adjacent to a vertex outside B φ k,m,n. The automorphisms of K n induce isomorphisms between the graphs B φ k,m,n, which are therefore isomorphic. Let B k,m,n be the graph B φ k,m,n, where φ is the natural bijection defined by φ(γ l) = l. Lemma 15 B k,m,n δ k (T m ). Proof of Proposition 12. K G k,m,n n is a disjoint union of A k,m,n and several copies of B k,m,n. By Lemma 14, A k,m,n K n, and B k,m,n δ k (T m ) by Lemma 15. Therefore K G k,m,n n K n + δ k (T m ). Unlike the proof of Lemma 14 which constructs an explicit homomorphism from A k,m,n to K n, the proof of Lemma 15 given in Section 3.4 uses a characterisation of the obstructions to homomorphisms to δ k (T m ), to show that none of these obstruction admits a homomorphism to B k,m,n. This is the dual approach, in a sense. 3.3 Tree duality A family F of directed graphs is a complete set of obstructions for a directed graph D if for every directed graph G the following holds. 9

10 G D if and only if there exists a F F such that F G. For every k < m, δ k (T m ) has tree duality, in the sense that there exists a family F k,m of trees that is a complete set of obstructions for δ k (T m ). To prove Lemma 15, we will use a characterisation of a complete set of obstructions for δ k (T m ). For a positive integer a, let P a be the path with vertex-set {0,1,...,a} and edges (0,1),...,(a 1,a). Every directed tree admits a homomorphism to some P a. The algebric length al(t) of a tree T is the minimum a such that T admits a homomorphism to P a. By extension, for u V (T), we write al(u) for the image of u under the unique homomorphism from T to P al(t). A tree is length-critical if al(t ) < al(t) for every proper subtree T of T. Clearly, a length-critical tree is a path P with two endpoints u,v such that al(u) = 0, al(v) = al(p), and every other vertex of the path has algebraic length strictly between 0 and al(p). Let P be a length-critical path with vertex-set {u 0,...,u b } such that al(u 0 ) = 0 and al(u b ) = al(p). We define the algebraic depth ad(u i ) of a vertex u i of P by ad(u i ) = max{al(u j ) al(u i ) j i}. By extension, we put ad(p) = max{ad(u i ) i {0,...,b}}. Thus the algebraic debth is defined only on length-critical paths. Lemma 16 ([7], Lemma 7) Let F k,m be the family of all length-critical paths with algebraic length m k and algebraic depth at most k. Then for P F k,m, we have P δ k (T m ). We will strenghten this result and prove that F k,m is a complete set of obstructions for δ k (T m ). To achieve this, we use the results of [3] characterising a complete set of obstructions for δ(h) in terms of a complete set of obstructions for H. For a tree T the family Sproink(T) consists of all the trees T constructed as follows. For every vertex u of T, T contains a subtree with parts A u, B u, where every arc goes from A u to B u. For every arc (u,v) of T, there is a vertex u (u,v) B u identified to a vertex v (u,v) A v. Note that the parts A u, B u, need not all be nonempty. If a vertex u of T has at least one outneighbour, then B u, and if u has at least one inneighbour, 10

11 then A u. If one of the parts A u, B u, is empty, then the other one must be a singleton. For a family F of trees, we put Sproink(F) T F Sproink(T). Lemma 17 ([3], Theorem 3) Let F be a family of trees which is a complete set of obstructions for a directed graph H. Then Sproink(F) is a complete set of obstructions for δ(h). Lemma 18 F k,m is a complete set of obstructions for δ k (T m ). Proof. We will use induction on k. First note that F 0,m = {P m }, which is well known to be a complete set of obstructions for T m (see e.g., [7]). Now suppose that F k,m is a complete set of obstructions for δ k (T m ). Then Sproink(F k,m ) is a complete set of obstructions for δ(δ k (T m )) = δ k+1 (T m ). Let P be a path in F k,m, with V (P) = {u 0,...,u b } where u 0,u b are the only vertices of algebraic length 0 and m k respectively. For P Sproink(P), we define the function h : V (P ) N by { al(u) if x Au, h(x) = al(u) + 1 if x B u. If x is a vertex of P corresponding to the identification of a vertex of B u to a vertex of A v, then al(v) = al(u)+1, therefore h(x) is well defined. Clearly, h is an approximation of the algebraic length on P : its range contains {1,...,al(P)} and is contained in {0,...,al(P) + 1}. The algebraic length in P is given by al(x) = h(x) min{h(y) y V (P )}. Therefore al(p ) al(p) 1 = m k 1, hence P contains a length-critical path P such that al(p ) = m k 1. The value ad(p ) is achieved as ad(x) = al(y) al(x) = h(y) h(x) for some y A ui B ui, x A uj B uj with i j. Thus ad(p ) k + 1, and P F k+1,m. Therefore, each member of Sproink(F k,m ) contains a member of F k+1,m, and by Lemma 16, no member of F k+1,m admits a homomorphism to δ k+1 (T m ). Hence F k+1,m is a complete set of obstructions for δ k+1 (T m ). 3.4 Proof of Lemma 15. We will show that for every P F k,m, we have P B k,m,n. Suppose for a contradiction that F k,m contains a path P which admits a homomorphism 11

12 to B k,m,n. We can identify each vertex of P with its image in B k,m,n. Hence V (P) = {F 0,...,F b } such that for each i, F i : G k,m,n K n is a function such that F i (γ j ) = j for all j {1,...,n}. Recall that aside from γ 1,...,γ n, the vertices of G k,m,n are functions f c,h,l : δ k (T m ) K n with range {l,l + }. Thus for j {l,l + }, γ j is both an inneighbour and an outneighbour of f c,h,l. Since F i has a neighbour F i±1, we then have F i (f c,h,l ) F i±1 (γ j ) = j. Therefore F i (f c,h,l ) {l,l + } for all i {0,...,b}, c {0,...,k}, h {c + 1,...,c + m k} and l {1,2,3}. Without loss of generality, F 0 and F b are the only vertices of P with algebraic length 0 and m k respectively. We will prove the following. Claim. For i 1, c = ad(f i ), h = al(f i ) ad(f i ) and l {1,2,3}, we have F i (f c,h,l ) = l +. Proof of Claim. We will prove the claim by induction on i, simultaneously for all l {1,2,3}. (i) We have ad(f 1 ) = 0 and al(f 1 ) = 0. For l {1,2,3}, there is an arc from the constant function γ l to f 0,1,l. Since (F 0,F 1 ) is an arc of P, we must have l = F 0 (γ l ) F 1 (f 0,1,l ) {l,l + }. Therefore F 1 (f 0,1,l ) = l +. (ii) Suppose that the claim is true for i, and (F i,f i+1 ) is an arc of P. Then al(f i+1 ) = al(f i ) + 1. If c = ad(f i ) = 0, then ad(f i+1 ) = 0. There is an arc from f 0,h,l to f 0,h+1,l +. Thus l + = F i (f 0,h,l ) F i+1 (f 0,h+1,l +) {l +,l ++ }. Therefore F i+1 (f 0,h+1,l +) = l ++. If c = ad(f i ) > 0, then ad(f i+1 ) = c 1. There is an arc from f c,h,l to f c 1,h,l +. Thus l + = F i (f c,h,l ) F i+1 (f c 1,h,l +) {l +,l ++ }. Therefore F i+1 (f c 1,h,l +) = l ++. In both cases we get F i+1 (f ad(fi+1 ),al(f i+1 )+1+ad(F i+1 ),l + ) = l ++. This being valid for all l {1,2,3}, the claim is true for i + 1. (iii) Suppose that the claim is true for i, and (F i+1,f i ) is an arc of P. Then al(f i+1 ) = al(f i ) 1, and ad(f i+1 ) = ad(f i )+1. There is an arc from f c+1,h,l + to f c,h,l. Thus {l +,l ++ } F i+1 (f c+1,h,l +) F i (f c,h,l ) = l +. Therefore F i+1 (f c+1,h,l +) = l ++. This being valid for all l {1,2,3}, the claim is true for i + 1. Thus the claim is true for all i. In particular, we have al(f b ) = m k, ad(f b ) = 0, so the claim states that F b (f 0,m k,l ) = l + for l {1,2,3}. However f 0,m k,l = γ l, and since F b is in B k,m,n, we have F b (γ l ) = l. 12

13 This contradicts the existence of the path P F k,m admitting a homomorphism to B k,m,n. Therefore by Lemma 18 B k,m,n admits a homomorphism to δ k (T m ). This completes the proof of Lemma Concluding comments There is no nonbipartite undirected graph K that is known to have the property that there exists an integer n such that if G and H are directed graphs and G H K, then min{χ(g),χ(h)} n. In particular, there is no nonbipartite undirected graph K for which K D is known not to be dense. In fact it seems likely that the methods of this paper can be adapted to prove that K D is dense for other undirected graphs K, perhaps all the nonbipartite ones. References [1] D. Duffus, N. Sauer, Lattices arising in categorial investigations of Hedetniemi s conjecture, Discrete Math. 152 (1996), [2] M.El-Zahar, N.Sauer, The chromatic number of the product of two 4-chromatic graphs is 4, Combinatorica 5 (1985), [3] J. Foniok, C. Tardif, Adjoint functors and tree duality, Discrete Mathematics and Theoretical Computer Science, Vol 11, No 2 (2009), [4] G. Grätzer Lattice Theory. First concepts and distributive lattices, W. H. Freeman and Co., San Francisco, Calif., xv+212 pp. [5] S.H.Hedetniemi, Homomorphisms of graphs and automata, University of Michigan Technical Report T, [6] P. Hell, J. Nešetřil, Graphs and homomorphisms, Oxford Lecture Series in Mathematics and its Applications, 28. Oxford University Press, Oxford, xii+244 pp. [7] J. Nešetřil, C. Tardif, Path homomorphisms, graph colourings and boolean matrices. [8] S. Poljak, V. Rödl, On the arc-chromatic number of a digraph, J. Combin. Theory Ser. B 31 (1981),

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