A construction of uniquely n-colorable digraphs with arbitrarily large digirth

A natural digraph analogue of the graph-theoretic concept of an ‘independent set’ is that of an ‘acyclic set’, namely a set of vertices not spanning a directed cycle. Hence a digraph analogue of a graph coloring is a decomposition of the vertex set into acyclic sets and we say a digraph is uniquely n-colorable when this decomposition is unique up to relabeling. It was shown probabilistically in [A. Harutyunyan et al., Uniquely D-colorable digraphs with large girth, Canad. J. Math., 64(6): 1310– 1328, 2012] that there exist uniquely n-colorable digraphs with arbitrarily large girth. Here we give a construction of such digraphs and prove that they have circular chromatic number n. The graph-theoretic notion of ‘homomorphism’ also gives rise to a digraph analogue. An acyclic homomorphism from a digraph D to a digraph H is a mapping φ : V (D) → V (H) such that uv ∈ A(D) implies that either φ(u)φ(v) ∈ A(H) or φ(u) = φ(v), and all the ‘fibers’ φ−1(v), for v ∈ V (H), of φ are acyclic. In this language, a core is a digraph D for which there does not exist an acyclic homomorphism from D to a proper subdigraph of itself. Here we prove some basic results about digraph cores and construct highly chromatic cores without short cycles.


Introduction
The author previously constructed digraphs with arbitrarily large digirth and chromatic number in [13].In fact, the construction strengthens the probabilistic result in [2] because it produces a digraph with digirth k and chromatic number n for each pair k, n of integers exceeding one.It is also of interest that unlike the analogous graph constructions in [7], [8], and [11], the construction is primitively recursive in n.The main result appearing here constructs uniquely n-colorable digraphs with digirth k for every reasonable pair n, k of integers.In Section 3 we show that the digraphs constructed in [13] are 'cores'.Finally, we give a simple construction of highly chromatic digraphs without two cycles in Section 4.
Although basic terminology can be found in [1], we include our main definitions for completeness.The digirth of a digraph D, denoted → g (D), is the length of its shortest directed cycle.Following [2] we define the chromatic number χ(D) of D to be the minimum number of parts in a partition of V (D) into acyclic sets, and we say that D is n-chromatic if χ(D) = n.It should be noted that this chromatic number of a digraph was first defined by Neumann-Lara in [12], where it was called the 'dichromatic number'.We would now like to relate digraph colorings to homomorphisms.An acyclic homomorphism from a digraph D to a digraph H is a mapping ϕ : V (D) → V (H) such that uv ∈ A(D) implies that either ϕ(u)ϕ(v) ∈ A(H) or ϕ(u) = ϕ(v), and all the fibers of ϕ are acyclic.If ϕ is an acyclic homomorphism such that all uv ∈ A(D) satisfy ϕ(u) = ϕ(v) then ϕ is a non-contracting homomorphism.As with graphs, if there exists an acyclic homomorphism from a digraph D to a digraph H we say that D is homomorphic to H, denoted by D → H, and D is H-colorable.Since we deal almost exclusively with acyclic homomorphisms when considering digraphs, we often write 'homomorphism' when it is clear from the context that we mean 'acyclic homomorphism'.As in the case of the graph coloring analogue, an equivalent definition of the chromatic number is χ(D) = min{n | D → ↔ K n } (where ↔ K n denotes the complete biorientation, see [1], of K n ).One nice property of the chromatic number of a digraph is that it generalizes the chromatic number of a graph because χ(G) = χ( ↔ G) for every finite simple graph G.
We define a digraph D to be uniquely n-colorable if D is n-chromatic and any two n-colorings of D induce the same partition of V (D).A digraph D is uniquely H-colorable if it is surjectively H-colorable, and for any two H-colorings φ, ψ of D, the functions φ and ψ differ by an automorphism of H, and a digraph D is a core if it is uniquely D-colorable.It is well known that a digraph D is uniquely n-colorable if and only if it is uniquely ↔ K n -colorable.In order to confirm the correctness of our theorems, we will also need the fact that D → H implies that → g (D) ≥ → g (H), which is a direct consequence of Propositions 1.2 and 1.3 in [2].It is worth noticing the subtle difference between the last statement and its graph analogue which is true only for odd girth.

Uniquely n-colorable digraphs without short cycles
The proof of Theorem 8 constructs uniquely n-colorable digraphs with digirth k for any pair n, k of suitable integers.This result is a constructive version of (an important case of) the probabilistic proof appearing in [5] and is analogous to the undirected construction appearing in [14].For the proof of Theorem 8 we first need to prove a few lemmas and to construct a few digraphs.The first of which, denoted D n , was constructed in [13] in order to prove the following theorem.
Theorem 1.For any given integers k and n exceeding one, there exists an n-chromatic digraph D with → g (D) = k.
We will provide the reader with the construction but we refer the reader to [13] for the proof of Theorem 1.For n = 2, the directed k-cycle will suffice.For n ≥ 2, we proceed by induction on n and suppose that we have already constructed a digraph D n with chromatic number n, digirth k, and , and the arc sets are Finally, we define D n+1 to be the digraph with and In order to illustrate this construction, we include in Figure 1 The following lemma provides us with an important result about the arcs of D n that will be useful both in this section and Section 3.
Proof.We notice that the assertion is true for n = 2 because D 2 ∼ = → C k and proceed by induction.Next assume its truth for D n and let uv be an arc in D n+1 .If, for an i ∈ [m], uv is an arc in D i n , which is isomorphic to D n , then we may use the inductive hypothesis to see that uv is in a k-cycle.Another easy case is when uv is an arc of some P d i since uv is in a k-cycle for all i ∈ [m] by our construction.Similarly, for all i ∈ [m], our construction implies that uv is in a k-cycle when either u ∈ with i = j.In this case, by our construction, uv is an arc in D n+1 if and only if d i d j is an arc in D n .Thus d i d j is in a k-cycle of D n by the induction hypothesis.Finally this in turn implies that uv is in a k-cycle of D n+1 and the proof is complete.
Another of these digraphs, denoted B n , is a spanning subdigraph of D n .We will define B n inductively and start by setting B 2 to be the path of length k − 1.We now define , and the arc sets are Finally, we define B n+1 to be the digraph with and It may be helpful for the reader to view Figure 2 for an example of this construction.
Proof.We proceed by induction and notice first that B 2 is acyclic as it is just a directed path.Now, assuming that B n is acyclic, we see that each subdigraph B i n of B n+1 is acyclic by our induction hypothesis.Thus there does not exist a cycle in B n+1 containing a vertex from any B i n since there are no arcs from any vertex of the P d i 's to any vertex of the B i n 's.Since the subdigraph Σ of B n+1 induced by the vertices of the P d i 's is homomorphic to B n via projection onto the first coordinate, we see that Proof.Proceeding by induction, again we see that the statement is true for B 2 .Thus by our induction hypotheses the statement is true for any shortest cycle contained in any D i n .Since all shortest cycles containing a vertex (d, i) from a D i n and a vertex from a P d j have the form (d, i), (d i , p 1 ), .., (d i , p k−1 ) , the unique arc in A(D n+1 ) A(B n+1 ) for such cycles is (d i , p k−1 )(d, i).Thus it remains to show that the statement is true for shortest cycles contained in Σ (defined on p. 3).So let us now suppose that the cycle α 0 , α 1 , ..., α k−1 is contained in Σ; as we have seen in Theorem 1 we may assume that α j = (d j , p r j ), where (d 0 , d 1 , ..., d k−1 ) is a shortest cycle in D n−1 .Thus the induction hypothesis yields that there exists a unique such that d d +1 is in A(D n−1 ) A(B n−1 ).Therefore α α +1 is the unique arc of α 0 , α 1 , ..., α k−1 in A(D n ) A(B n ), and the proof is complete.
Next we construct digraphs D n with digirth k.For n = 2, the directed k-cycle will suffice.For n ≥ 2, we proceed by induction on n and suppose that we have already constructed a digraph D n with digirth k and V (D n ) = {d 1 , d 2 , ..., d m }.We now define D n+1 .
, and the arc sets are Finally, we define D n+1 to be the digraph with and Proof.We observe that D 2 has digirth k and proceed by induction.Using the induction hypothesis, we see that Thus we need only consider cycles which contain vertices α and β where α ∈ V (H (n, i)), β ∈ V (H (n, j)) and i = j.But this implies, from the construction of D n+1 , that there exists a path in D n from d i to d j and from d j to d i .Thus the induction hypothesis also implies that the cycle containing α and β has length at least k.Combining these observations, we see that → g (D n+1 ) = k, and induction gives the lemma.
Finally we define B n inductively and start by letting B 2 be the path of length k − 1.We now define , and the arc sets are Finally, we define B n+1 to be the digraph with and, for all s, t ∈ [m], It is clear from the construction that B n is a spanning subdigraph of B n .Figure 4 may help bring to light some of the nuances of this construction.
Proof.It is easy to see that B 2 is acyclic and thus we continue by induction.Once again the induction hypothesis and the construction of Since there is no arc from any vertex of a P d i to any vertex of a B j n , it suffices to consider the subdigraph of B n+1 induced by the vertices of the P d i 's and the subdigraph induced by the B i n 's.Both of these subdigraph s are homomorphic to B n and thus are acyclic by induction.Therefore B n+1 is acyclic.
We may now define directed graphs D m * ↔ K n which will be shown to be uniquely n-colorable Proof.First we show that there exists an m-coloring φ It is easy to see that φ 2 exists and thus we may proceed by induction.Define a mapping φ m : Now suppose that αβ ∈ A(D m ) A(B m ) which implies that either α, β are vertices of some D i m−1 , or α is a vertex of some P d i and β is a vertex of D i m−1 , or α is a vertex of some P d i and β is a vertex of some P d j .In the case where α, β are vertices of some D i m−1 , we may assume that α = (d j , i) and β = (d h , i), where d j d h is in A(D m−1 ) A(B m−1 ).Thus using the induction hypothesis we see that In the second case, where α is a vertex of some P d i and β is a vertex of D i m−1 , it is clear from the definition of φ m that φ m (α) = φ m (β).In the last case, in which α is a vertex of some P d i and β is a vertex of some P d j , we may assume that

. Thus the inductive hypotheses and the definition of
We now define Γ to be the subdigraph of n ) and notice that Γ is a subdigraph of Ω n (n, 1) and the digraph induced by D i (m − 1, 1) is isomorphic to the digraph induced by D i (m − 1, j) for all j ∈ {1, 2, ..., |V (D m−1 )|}.Consider the mapping ρ : is well-defined and bijective.We now suppose that αβ ∈ A(D n ) in order to show that ρ is in fact a homomorphism.The first case is when αβ is an arc in B n which implies that (α, 1)(β, 1) ∈ A(B n+1 ) and thus (α, 1), φ n (α) (β, 1), φ n (β) ∈ A(Γ) whether or not φ n (α) = φ n (β) (because of the definition of our * -product).The second case is when αβ is an arc in D n but not B n .From the preceding paragraph we know that this implies that φ n (α) = φ n (β).Also the construction of D n+1 implies that (α, 1)(β, 1) is an arc in D n+1 .Hence (α, 1), φ n (α) (β, 1), φ n (β) is an arc in Γ and ρ is a homomorphism.This now implies that χ(Γ) ≥ n and in fact χ(Γ) = n since we saw above that Finally we have all the necessary tools to prove the deepest result of this paper.
Proof.We have seen that the canonical projection π : In other words we need only show that the fibers of ψ are a relabeling of the fibers of π.The preceding lemma has the direct consequence that for all j ∈ {1, 2, ..., |V (D m−1 )|} and For each such j and γ s , let α s j be such that α s j ∈ V (Ω n (m − 1, j)) and ψ(α s j ) = γ s .
Consider two vertices d 0 and d 1 of D m−1 such that there is an arc from Lemmas 2 and 4 imply that there exists a cycle (d 0 , d 1 , ..., d k−1 ) in D m−1 and there exists a unique . Hence for all s ∈ [n] and some i ∈ [n] (which depends on s) the vertices α s and α s +1 lie in D i (m − 1, ) and D i (m − 1, + 1) respectively, for otherwise (α s 0 , α s 1 , ..., α s k−1 ) would be a monochromatic cycle with respect to ψ.Similarly, supposing that for otherwise (α s 0 , α s 1 , ..., α s , ν, α s +2 , ..., α s k−1 ) and (α s 0 , α s 1 , . . ., α s −1 , µ, α s +1 , α s +2 , ..., α s k−1 ) would be monochromatic cycles with respect to ψ.Now consider the set {i ∈ [n]| there exists an s such that α s ∈ D i (m−1, )} and suppose that the size of this set is less than n.This implies that for all s ∈ [n] there does not exist an α s ∈ D 1 (m − 1, ), where the 1 is taken without loss of generality.This implies further that no α s +1 , for s ∈ [n], lies in D 1 (m−1, +1) as we've established that for each fixed s ∈ [n], the vertices α s and α s +1 lie in D i (m−1, ) and D i (m−1, +1) respectively (i.e. share the same superscript i here).However there exists some s 1 ∈ [n] and β ∈ D 1 (m − 1, ) such that ψ(β) = γ s 1 because every vertex is sent to some color, and we just concluded that α s 1 +1 cannot be in D 1 (m − 1, + 1).Hence we reach a contradiction because this leads to the cycle (α s 1 0 , α s 1 1 , ..., α s 1 −1 , β, α s 1 +1 , ..., α s 1 k−1 ) being monochromatic with respect to ψ.Thus we may conclude that for all i ∈ [n] there exists an α s ∈ D i (m − 1, ).Suppose that α s i ∈ D i (m − 1, ).Appealing to (1), we now see that for every i ∈ [n], when r = i, there does not exist a vertex in D r (m − 1, + 1) nor a vertex in H r (m − 1, ) either of which is colored γ s i .Therefore H i (m − 1, ) and D i (m − 1, + 1) are both monochromatic of the same color with respect to ψ.Now suppose, in order to reach a contradiction, that for some color γ s 1 there exists a vertex This implies that there does not exist a vertex β ∈ V (P t 2 +1 ) such that ψ(β) = γ s 1 , for otherwise the cycle (α s 1 0 , α s 1 1 , ..., α s 1 , β, µ, α s 1 +3 ..., α s 1 k−1 ) would be monochromatic with respect to ψ. Hence there exist vertices β 2 , β 3 ∈ V (P t 2 +1 ) such that ψ(β 2 ) = γ s 2 , ψ(β 3 ) = γ s 3 , and s 1 , s 2 and s 3 are distinct because every vertex in is in a cycle with the vertices in P t 2 +1 .This implies that there does not exist a vertex ) would be monochromatic with respect to ψ. However this implies that Ω Thus for all i ∈ [n], the sets D i (m − 1, + 1) and D i (m − 1, + 2) are monochromatic of the same color which in turn implies that P i +1 is monochromatic of the same color.Therefore we may inductively argue that all H i (m − 1, j), for j ∈ [k], are monochromatically colored the same.As D m is strongly connected, this implies that ψ((α, i)) = ψ((β, i)) for all vertices α, β of D m and i ∈ V ( ↔ K n ), and as we noted in the first paragraph of this proof, this is enough to Recall from p. 8 and Lemma 5 that We also note that we were able to construct D m with → g (D m ) = k for any pair m, k of integers exceeding one.Therefore our proof of Theorem 8 constructs a uniquely n-colorable digraph with digirth k for every pair n, k of integers both exceeding one.Thus we now have a constructive and more precise version of (an important case of) the main theorem appearing in [5].
We remark here on the number of vertices in the D n 's (and consequently the number of vertices in D m * ↔ K n ).We see from the construction that we have the recurrence relation: It is easy to confirm that |V (D n )| = 2 2 n−1 −1 for k = 3.For a general k we do not have a closed form.However we may inductively argue that since D m is a spanning subdigraph of D m .In order to put the size of these digraphs into context we point to a construction of uniquely 3-colorable graphs with girth 4 by Nešetřil [10] in which the smallest example has over 500 million vertices.The digraph constructed in the proof of Theorem 8 with the least vertices while still being uniquely 3-colorable with digirth In [2] the authors define the circular chromatic number χ c of a digraph This digraph invariant can be seen to generalize the graph version of this invariant.For p ≥ q, we define the directed complete rational graph → K p/q to be the digraph with vertex set {0, 1, . . ., p − 1} and an arc from i to j if j − i ∈ {q, q + 1, . . ., p − 1} (with arithmetic modulo p).It is easy to check that → K p/1 ∼ = ↔ K p for every positive integer p.It was shown in [2] that we may define Next we would like to show that χ c (D m * ↔ K n ) = n in order to construct digraphs with digirth k, chromatic number n, and circular chromatic number n for every pair k, n of integers exceeding one.To this end we will prove the following proposition.Proof.By way of contradiction we assume D is uniquely n-colorable, χ c (D) = p/q < n, ) is an acyclic homomorphism.It has been proved that every digraph D satisfies χ(D) − 1 < χ c (D) ≤ χ(D), see e.g.[2], allowing us to assume that n − 1 < p/q.Now consider the mapping ϕ : V ( In order to show that ϕ is an acyclic homomorphism it suffices to show that the fibers of ϕ are acyclic because the target digraph is complete.It is easy to check that for j ∈ {0, 1, . . ., n − 2} ϕ −1 (j) = {jq, jq + 1, . . ., (j + 1)q − 1} and ϕ −1 (n − 1) = {(n − 1)q, (n − 1)q + 1, . . ., p − 1}.Thus for every j ∈ V (K n ) and pair of vertices s, t ∈ ϕ −1 (j) we have 0 < t − s < q whenever s < t implying that st is not an arc in K p/q .Therefore the fibers of ϕ are acyclic and ϕ is an acyclic homomorphism.Next consider the mapping σ : V ( → K p/q ) → V ( ↔ K n ) defined by σ(i) = (i + 1)/q for i = 0, 1, . . ., p − 2 and σ(p − 1) = 0.A similar argument to that for ϕ yields that σ is an acyclic homomorphism.Thus we see that ϕ Thus we see that Proposition 9 implies that χ c (D m * ↔ K n ) = n and hence we have a construction of digraphs with digirth k, chromatic number n, and circular chromatic number n for every pair k, n of integers exceeding one.
To conclude this section, we leave the reader with a conjecture.Noting that ↔ K n is a core, and with an eye to the nonconstructive results of [5], we would like to construct digraphs with arbitrarily large girth which are uniquely H-colorable for any core H.In fact we feel confident that the construction is done, but the proof still eludes us.Consider the following conjecture concerning the digraphs D m constructed for Theorem 8, and notice that Conjecture 11.For all cores H and some constant c, the digraph D m * H is uniquely

D n are cores
We now turn our attention to saying something substantially stronger about the digraphs constructed in the proof of Theorem 1: they are cores.This suggests that there is some sort of minimality to this construction.In order to proceed we need a few lemmas about cores.For the sake of completeness we have also decided to include a few basic lemmas about cores of digraphs that confirm that they behave similarly to their graph analogues.Recall that a digraph D is a core if it is uniquely D-colorable.The following useful lemma was proved in [5].
The condition that every acyclic homomorphism V (D) → V (D) is a bijection is equivalent to saying that D is not homomorphic to a proper subdigraph of itself.An (acyclic) retraction of a digraph D is an acyclic homomorphism φ from D to a subdigraph H of D such that the restriction φ| H is the identity map on H. Now we can state an equivalent definition of a core that will be used to prove Theorem 20.Lemma 13.A digraph D is a core if and only if it does not retract to a proper subdigraph of itself.
Proof.The necessity is clear since an acyclic retraction is an acyclic homomorphism.Now suppose that D is not a core, and let H be a proper subdigraph of D such that φ is an acyclic homomorphism from D to H and D is not homomorphic to any proper subdigraph of H.The existence of such an H is ensured by Lemma 12.We claim that H is a core, for suppose it is not and let f be an acyclic homomorphism from H to a proper subdigraph of H. Then f • φ is an acyclic homomorphism from D to a proper subdigraph of H, contradicting our choice of H.Because of the claim, any homomorphism from H to itself is an automorphism of H. Let ϕ : V (D) → V (H) be an acyclic homomorphism.Since the restriction of a homomorphism is a homomorphism, ψ := ϕ| H is an automorphism of H. Hence ψ −1 exists and ψ −1 • ϕ : V (D) → V (H) is an acyclic retraction to a proper subdigraph of D.
We now define a subdigraph H of a digraph D to be a core in D if there exists an acyclic retraction from D to H and H is a core.Lemma 14.An acyclic retract of a digraph D is an induced subdigraph of D.
Proof.Let φ be an acyclic retraction from D to a subdigraph H of D. Suppose that x, y ∈ V (H) and xy ∈ A(D).Since H is a retract, both φ(x) = x and φ(y) = y and, as φ is a homomorphism, xy is an arc in H.
We say that two digraphs D and H are homomorphically equivalent if H is homomorphic to D and D is homomorphic to H. Lemma 15.If H and K are cores then they are homomorphically equivalent if and only if they are isomorphic.
Proof.Let φ : H → K and ψ : K → H be acyclic homomorphisms.This implies that ψ • φ and φ • ψ are bijections since H and K are cores.Thus φ and ψ are both bijective and hence H ∼ = K since, e.g., φ is a bijective homomorphism.
Lemma 16.Every finite digraph D has a core, which is an induced subdigraph and is unique up to isomorphism.
Proof.Since D is finite and the identity mapping is an acyclic retraction, the family of subdigraphs of D to which D has an acyclic retraction is finite and nonempty and thus has a minimal element D • with respect to inclusion.From the definition of 'core in D' and Lemma 13, we see that D • is a core in D. Since D • is an acyclic retract, it is an induced subdigraph by Lemma 14.Now let H 1 and H 2 be cores of D, and, for i = 1, 2, let φ i be an acyclic retraction from D to H i .Then φ 1 | H 2 is an acyclic homomorphism from H 2 to H 1 and similarly there exists an acyclic homomorphism from H 1 to H 2 .Therefore, by the preceding lemma, In the remainder of this paper we will always use D • for 'core of D' as is done for the graph-theoretic analogue in [4].
Lemma 17. Cores of connected digraphs are connected.
Proof.Let D be a connected digraph and ϕ a retraction to D • .Suppose that x, y ∈ V (D • ).Then x, y are vertices of D because ϕ is a retraction.Since D is connected there exists a sequence of vertices x = u 1 , u 2 , . . ., u n = y in D such that for all i ∈ [n − 1] we have The following result displays one use of cores for testing homomorphic equivalence.
Lemma 18.Two digraphs are homomorphically equivalent if and only if their cores are isomorphic.
Proof.Clearly, a digraph and its core are homomorphically equivalent.The sufficiency of the condition follows.For necessity, let D • and H • be cores of the digraphs D and H respectively.Assuming D and H are homomorphically equivalent, we have that D • is homomorphic to D, D is homomorphic to H, and H is homomorphic to H • .Thus D • is homorphic to H • using the fact that the composition of acyclic homomorphisms is an acyclic homomorphism.Similarly H • is homomorphic to D • .Hence by Lemma 15, H • and D • are isomorphic.
Earlier we defined a digraph H to be a core if it is uniquely H-colorable.In fact we will see that there is a looser condition governing whether H is a core.The next result shows that if we find any digraph which is uniquely H-colorable, then H is a core.Lemma 19.If there exists a uniquely H-colorable digraph, then H must be a core.
Proof.Let D be uniquely H-colorable and φ : Since the reverse containment is always true, we conclude that H = H • .
Theorem 20.For any given integers n > 1 and k > 2 the digraph D n constructed in Theorem 1 is a core with C k so that we may proceed by induction.Assume that D n is a core with n ≥ 2 and let ϕ : V (D n+1 ) → V (D n+1 ) be a retraction.Define Γ to be the image of ϕ and thus our goal is to show that Γ = V (D n+1 ).Since ϕ is a retraction and Γ induces a subdigraph of D n+1 , the function ϕ must map k-cycles (i.e.shortest cycles) to k-cycles.Thus (2) if two vertices u, v are in the same k-cycle then ϕ(u) = ϕ(v).
We first proceed to show that for the subdigraph Σ = D n+1 m i=1 , for some i, j, and we will show that ϕ(u) = v.This and the fact that D n is strongly connected will suffice to show that ϕ(V (Σ)) ⊆ V (Σ), for we can repeat our argument below as necessary to force every such ϕ(u) into V (Σ).If i = j, then the construction of D n+1 puts u and v together in a k-cycle and hence they cannot be in the same fiber of ϕ.This proves that ϕ(u) = v for otherwise, with ϕ being the identity on Γ, we'd have ϕ(u) = ϕ • ϕ(u) = ϕ(v), contradicting (2).Notice that there exists a directed path from d i to d j for all i, j ∈ [m] because D n is strongly connected.For the case i = j, we proceed by induction on the distance, s, from d i to d j in D n .Assume that this distance is s + 1 and that for every r ∈ We now show that (3) for all i ∈ [m] there exists a j ∈ [k − 1] such that ϕ((d i , p j )) = (d i , p j ) ∈ Γ.
By way of contradiction assume that Γ is contained in V (D n+1 − P d i ).This follows from the negation of (3) because the definition of retraction implies that Notice that for all α ∈ V (D ) is an arc in D n+1 .However, in the construction of D n+1 , the only arcs from D i n to Σ are those from D i n to (d i , p 1 ).Thus ϕ(α) ∈ V (D n+1 − H(n, i)) for every α ∈ V (D i n ).However this and the fact that ϕ is a retraction imply that Γ is contained in V (D n+1 − H(n, i)) which contradicts the preceding paragraph.Thus we've established (3).This fact will now allow us to show that for every i ∈ [m] the vertices (d i , p 1 ) and (d i , p k−1 ) are sent to the same P d by ϕ.We again proceed by way of contradiction and assume that for two distinct vertices u 1 , u k−1 of D n , we have ϕ(d i , p 1 ) ∈ V (P u 1 ) and ϕ(d i , p k−1 ) ∈ V (P u k−1 ).Considering an α ∈ V (D i n ) we see that both ϕ(d i , p k−1 )ϕ(α) and ϕ(α)ϕ(d i , p 1 ) must be in the arc set of D n+1 [Γ] since ϕ is a non-contracting homomorphism.The preceding sentence and our construction of D n+1 thus imply that ϕ(α) ∈ V (Σ), say ϕ(α) ∈ V (P u 0 ), because u We may thus assume that the u j 's are distinct for otherwise u 0 , u 1 , ..., u k−1 would contain a cycle of length less than k.Now (3) shows that there exists j ∈ [k − 1] such that u j = d i .If we let = j − 1 (mod k), then the preceding sentence implies that u d i ∈ A(D n ).Thus for all r, s ∈ [k − 1] the arc (u , p r )(d i , p s ) is in D n+1 and (3) implies that ϕ(d i , p ) = (u , p t ) for some t ∈ [k − 1].Hence (u , p t ) is in the image of ϕ implying that ϕ((u , p t )) = (u , p t ) because ϕ is a retraction (ϕ is the identity on its image).This contradicts the fact that ϕ is a non-contracting homomorphism.Thus for every i ∈ [m] the vertices (d i , p 1 ) and (d i , p k−1 ) are indeed sent to the same P d by ϕ.
We next show that in fact for every i ∈ [m] all the vertices of P d i are sent to the same P d by ϕ for some d ∈ V (D n ).By way of contradiction assume that some s ∈ [k − 1] and ∈ {2, 3, ..., k − 2} satisfy ϕ(d i , p ) = (v, p s ) while ϕ(d i , p 1 ) ∈ V (P u ), where u and v are distinct vertices of D n .The preceding paragraph implies that (d i , p k−1 ) is sent to P u as well.As P d i is a directed path of length k − 2, the image under ϕ of V (P d i ) induces a directed path of length not exceeding k − 2. However since ϕ(d i , p 1 ), ϕ(d i , p k−1 ) ∈ V (P u ) and u = v, the preceding sentence implies that there is a cycle in D n containing u and v which has length less than k.This contradiction lets us deduce that for every i ∈ [m] all the vertices of P d i are sent to P d by ϕ for some d ∈ V (D n ).This implies that the restriction of ϕ to V (Σ) is the identity because of (3).For u ∈ V (D i n ), the image ϕ(u) lies in neither D j n , for j = i, nor P d i because we already know that ϕ fixes P d i .Also, ϕ(u) ∈ P d j with d j = d i implies that d j d i and d i d j are arcs in D n because u(d i , p 1 ) and (d i , p k−1 )u are arcs in D n+1 .Hence we see that for all u ∈ V (D i n ) the image ϕ(u) lies in D i n .This implies that the restriction of ϕ to each V (D i n ) is a retraction of D i n to itself.Since each D i n is isomorphic to D n , a core by induction, each D i n is also a core.Thus, the restriction of ϕ to each V (D i n ), for i ∈ [m], is the identity.Therefore, we've finally established that Γ = V (D n+1 ) and hence by induction reached the conclusion that the D n 's are cores.
Hence we have in fact constructed highly chromatic cores without short cycles.It would be of interest to construct graphs with the same properties.In the next section we give a nod to the elegant construction of highly chromatic triangle-free graphs by Mycielski, see e.g.[9], by constructing analogous digraphs in a similar manner.We note that the Mycielski Graph with chromatic number 4, M 4 , is an example of a graph core without triangles.The truth of this fact comes to light via the minimality of M 4 proved in [3].

Highly chromatic digraphs with digirth three
We conclude this paper with a construction of digraphs with arbitrarily large chromatic number and digirth at least three (i.e.without bi-cycles).Although it is not as general (it does not work for larger digirth) as the construction appearing in [13], we decided to include it for it's simplicity and the fact that it is a directed analogue of Mycielski's construction.We also ask the question: does there exist a construction of digraphs retaining the desired properties with less vertices or arcs?It should be noted that in the case where the chromatic number is 3, non-isomorphic examples with the same number of vertices and arcs as the construction which follows appear in [6].The construction will be inductive and similar to the construction appearing in [13].We will again start with the directed three cycle and call it A 2 .We now suppose A n has chromatic number n and digirth 3 and construct A n+1 from A n .We will let P 1 be the path of length one and V (P 1 ) = {p 1 , p 2 }.Let V (A n+1 ) = V (A n • → P 1 ) v and We first show that the digirth of A n+1 is three.The following lemma will be useful in attaining that goal.
Lemma 21.If M 2 is an acyclic digraph then the projection π M 1 : V (H) → V (M 1 ) is a homomorphism for every subdigraph H of M 1 • M 2 .
Proof.Let (u, v)(x, y) be an arc of H.If u = x then ux = π M 1 ((u, v))π M 1 ((x, y)) is an arc of M 1 .Thus we assume that u = x which implies that vy is an arc of M 2 and π M 1 ((u, v)) = π M 1 ((x, y)) = u.Since the subdigraph of M 1 •M 2 induced by the vertex set F u = {(u, w)|w ∈ V (M 2 )} is isomorphic to M 2 , the set π −1 M 1 (u) is acyclic in H for all u ∈ V (M 1 ).Therefore the projection π M 1 is a homomorphism.
We see from the construction that the vertex v is not a part of any bi-cycles.Thus it suffices to show that → g (A n •P 1 ) = 3. Indeed Lemma 21 gives the homomorphism π An : V (A n •P 1 ) → V (A n ) and thus → g (A n • P 1 ) ≥ 3. Therefore → g (A n • P 1 ) = 3 as it is easy to see from the construction that A n • P 1 contains a number of directed three cycles because, by induction, It is clear that χ(H n+1 ) ≥ χ(A n ) = n since A n is isomorphic to a subdigraph of A n+1 .If A n+1 is n-chromatic, then there exists an acyclic homomorphism σ : To set up the contradiction we are about to derive, fix a σ 'color' α ∈ V ( ↔ K n ) and suppose σ(v) = α.Notice that v is in a cycle with the vertices of the subdigraph induced by {a} × V (P 1 ) ∀a ∈ V (A n ), which implies that there exists a vertex u a ∈ {a} × V (P 1 ) such that σ(u a ) = α.The subdigraph Λ of A n+1 induced by {u a | a ∈ V (A n )} is isomorphic to A n .This contradicts the fact that A n has chromatic number n since σ, now seen to avoid α on V (Λ), effectively maps V (Λ) to V ( As the target digraph of φ is complete, to show that φ is an acyclic homomorphism, it will suffice to show that each fiber of φ is acyclic.Since φ −1 (n + 1) = {v} it suffices to show that the restriction of φ to A n • P 1 is an acyclic homomorphism.Notice that φ| An•P 1 = ζ • π An .As ζ and π An are acyclic homomorphisms, so too is their composition φ| An•P 1 .Therefore φ is an acyclic homomorphism which finally implies that χ(A n+1 ) = n + 1.
a diagram of D 3 with k = 3.All double-tailed arrows represent numerous arcs in the diagram.The double-tailed arrows running horizontally indicate an arc from every vertex at the tail to every vertex at the head.The double-tailed arrows running up and down indicate an arc from every vertex at the tail to one vertex at the head and from one vertex at the tail to every vertex at the head respectively.The three remaining diagrams in this section follow similar schematics.It is also worth pointing out that the subdigraph σ of D n+1 induced by the vertices of the P d i 's is isomorphic to the lexicographic product D n • P k−2 .The lexicographic product D • H of two digraphs D, H is defined to be the digraph with V (D) × V (H) as its vertex set and with an arc from (d 1 , h 1 ) to (d 2 , h 2 ) if d 1 d 2 is an arc in D, or d 1 = d 2 and h 1 h 2 is an arc in H.

Figure 3 Figure 3 . D 3 with k = 3
Figure 3 is included on the following page in order to clarify the construction of D n .
The first two properties to notice about D m * ↔ K n are that it has digirth at least → g (D m ) and is n-colorable because the projections are homomorphisms.(The projection onto ↔ K n being a homomorphism relies on the fact that B m is acyclic.)We now introduce some notation for future use.Let the vertices of ↔ K n be 0, 1, ..., n − 1 and, for we will show that ψ is a composition of π with an automorphism of ↔ K n .Notice that since the target digraph is ↔ K n this amounts to showing that ψ((α, i)) = ψ((β, i)) for all vertices α, β of D m and i ∈ V ( ↔ K n ).
• ζ and σ • ζ are two homomorphisms from D to K n that induce different partitions of V (D) contradicting the assumption that D is uniquely n-colorable.Because any subdigraph H of D satisfies χ c (H) ≤ χ c (D) we have the following immediate corollary to Proposition 9. Corollary 10.If a digraph D is n-chromatic and contains a subdigraph that is uniquely n-colorable, then χ c (D) = χ(D) = n.
[m] with 0 ≤ − −−−−−− → dist(d i , d r ) ≤ s we have ϕ(u) = z for all z ∈ V (D r n ).By assumption there is a path P = (d i , d i+1 , ..., d i+s , d j ) in D n which by our construction implies that u(d i+1 , p ) is an arc in D n+1 for all ∈ [k − 1].Hence the induction hypothesis implies that ϕ(u) = ϕ((d i+1 , p )) for all ∈ [k − 1].Thus if ϕ(u) = v then vϕ(d i+1 , p ) ∈ A(D n+1 ) for all ∈ [k − 1] because ϕ is a non-contracting homomorphism.However the induction hypothesis again implies that ϕ(d i+1 , p ) / ∈ V (D j n ) which forces ϕ(d i+1 , p ) to be (d j , p 1 ) for every ∈ [k − 1].This cannot happen because k > 2 and (d i+1 , p ) and (d i+1 , p t ) are in a k-cycle together for all , t ∈ [k −1] with = t.Therefore the restriction of ϕ to V (Σ) is indeed contained in V (Σ).The next step is to show that D n+1 [Γ] cannot be a subdigraph of D n+1 − H(n, i) for any i ∈ [m].By way of contradiction we assume that there is an i such that D n+1 [Γ] is a subdigraph of D n+1 − H(n, i).Choose exactly one vertex v j from each P d j and define Λ to be the subdigraph of D n+1 induced by the v j 's which by our construction is isomorphic toD n .Define ψ : V (Σ) → V (Λ) by ψ(d , p s ) = v for all ∈ [m] and s ∈ [k − 1].Consider the homomorphism ψ| ζ • ϕ| V (Λ) : V (Λ) → V (Λ) where ζ := Im(ϕ| V (Λ) ).(Note that these restrictions compose because ψ is defined on ζ ⊂ Im(ϕ| V (Σ) ) ⊆ V (Σ).)Since we're under the assumption that the image of ϕ is contained in V (D n+1 − H(n, i)), there exists an α ∈[m] such that ϕ(v α ) = v α .Thus v α is not in the image of ψ| ζ •ϕ| V (Λ) which implies that ψ| ζ •ϕ| V (Λ)is not a bijection.Thus by Lemma 12 the digraph Λ is not a core.This contradicts the induction hypothesis that D n is a core because Λ is isomorphic to D n .Therefore D n+1 [Γ] cannot be a subdigraph of D n+1 − H(n, i) for any i ∈ [m].