Homomorphically Full Oriented Graphs

Homomorphically full graphs are those for which every homomorphic image is isomorphic to a subgraph. We extend the definition of homomorphically full to oriented graphs in two different ways. For the first of these, we show that homomorphically full oriented graphs arise as quasi-transitive orientations of homomorphically full graphs. This in turn yields an efficient recognition and construction algorithms for these homomorphically full oriented graphs. For the second one, we show that the related recognition problem is GI-hard, and that the problem of deciding if a graph admits a homomorphically full orientation is NP-complete. In doing so we show the problem of deciding if two given oriented cliques are isomorphic is GI-complete.


Introduction and Background
Oriented graphs can be considered to arise in one of two ways: as a special class of digraphs defined by a restriction on the existence of directed 2-cycles, or from simple graphs by assigning a direction to each edge.The distinction between these perspectives has consequences in the study of homomorphisms of oriented graphs.For example, each leads to a different definition of vertex colouring of oriented graphs (see the work of Feder et al. (2003) and Sopena (2016)).We use the term antisymmetric digraph to specifically refer to digraphs with the restriction that there are no directed 2-cycles, and the term oriented graph otherwise, including when the distinction between perspectives is unimportant.
For standard graph theoretic notation we refer the reader to the seminal text by Bondy and Murty (2008).We generally use Greek capital letters to refer to graphs and Latin capital letters to refer to antisymmetric digraphs and oriented graphs.
A homomorphism of a graph Γ to a graph Λ is a function ϕ : V (Γ) → V (Λ) such that xy ∈ E(Γ) implies ϕ(x)ϕ(y) ∈ E(Λ).When ϕ is a homomorphism of Γ to Λ we write ϕ : Γ → Λ.When the existence of a homomorphism (rather than a particular homomorphism) is of interest we write Γ → Λ.The definition of a homomorphism of a digraph G to a digraph H is identical.

Bellitto, Duffy and MacGillivray
A homomorphism of Γ to Λ induces a mapping of A(Γ) to A(Λ).A homomorphism ϕ : Γ → Λ is called complete when both ϕ and the induced mapping of A(Γ) to A(Λ) are surjective.If there is a complete homomorphism of Γ to Λ, then Λ is called a homomorphic image of Γ.The corresponding definitions are the same for digraphs.
If Λ is a subgraph of Γ, then a homomorphism ϕ of Γ to Λ is called a retraction when ϕ(h) = h for every vertex h ∈ V (Λ).If there is a retraction of Γ to Λ, then Λ is called a retract of Γ.A retract of Γ is necessarily an induced subgraph of Γ, but the converse is false.The corresponding definitions for digraphs are identical.
Vertices x and y of a graph G are called neighbourhood comparable when N (x) ⊆ N (y) or N (y) ⊆ N (x).Notice that if N (x) ⊆ N (y), then there is a retraction of G to G − x.A graph Γ is called homomorphically full when every homomorphic image of Γ is isomorphic to a subgraph of Γ.The homomorphically full graphs were first characterized by Brewster and MacGillivray (1996).
Theorem 1. 1 Brewster and MacGillivray (1996) Let Γ be a graph.The following statements are equivalent: (e) Γ contains neither 2K 2 nor P 4 as an induced subgraph.
(f) Γ is the comparability graph of an up-branching.
Homomorphic images of simple graphs are implicitly understood to be simple graphs.If a homomorphic image of a simple graph could be a graph with loops, then there would be no homomorphically full simple graphs: a single vertex with a loop would be a homomorphic image of every graph.
Analogous to the definition of graphs, we say an oriented graph or antisymmetric digraph G is homomorphically full when every homomorphic image of G is isomorphic to a subgraph of G.The two perspectives on how an oriented graph G arises are germane in understanding its homomorphic images and the meaning of this definition.When G is an antisymmetric digraph, a homomorphic image of G is a digraph H (which may have directed 2-cycles) for which there is a complete homomorphism G → H.When G is an oriented graph, a homomorphic image of G is an oriented graph H for which there is a complete homomorphism G → H.In this case, two vertices joined by a directed path of length 2 (a 2-dipath) must map to different vertices of H.
For a graph Γ and non-adjacent vertices u, v ∈ V (Γ), let Γ uv denote the graph produced by identifying u and v into a single vertex named u v .The homomorphism ϕ : Γ → Γ uv that sends u and v to u v and fixes all other vertices is a complete homomorphism.As such, Γ uv is a homomorphic image of Γ.We call such a homomorphism elementary.Every homomorphism can be expressed as a composition of elementary homomorphisms.These statements remain true when considered for oriented graphs and antisymmetric digraphs.However for oriented graphs we are restricted to considering pairs of vertices that are neither adjacent nor the ends of a 2-dipath.In any case, to check whether a graph, oriented graph, or antisymmetric digraph is homomorphically full it is enough to check that the target of each elementary homomorphism is isomorphic to a subgraph.
Let H be the directed path on three vertices.With Theorem 1.1 one can verify that the underlying graph (i.e., the path on three vertices) is homomorphically full.Interpreting H as an antisymmetric digraph, there is an elementary homomorphism that identifies the two ends of the directed path.The resulting homomorphic image, the directed 2-cycle, is not isomorphic to any subgraph of H and so H is not homomorphically full.On the other hand, interpreting H as an oriented graph, the vertices at the end of the path cannot be identified by an elementary homomorphism.In this case the only homomorphic image of H is H itself.And so we conclude H is homomorphically full.
Let G be a directed graph.We say that an ordered pair of non-adjacent vertices, u, v ∈ V (G) are neighbourhood comparable when As with graphs, we observe that if u and v are neighbourhood comparable, then, we have The implied isomorphism here is, in some sense, trivial.Each vertex other than v maps to itself and v maps to u v .This occurs as there is a retraction G → G uv where u v in G uv is relabelled as v.
An oriented graph or antisymmetric digraph G is homomorphically full if and only if for all u and v that can be identified by an elementary homomorphism we have that G uv is isomorphic to a subgraph of G.And so, similar to the case for graphs, if for an oriented graph or antisymmetric digraph G every pair of vertices that can be identified by an elementary homomorphism is neighbourhood comparable, then G is necessarily homomorphically full.We will see that the converse of this statement holds for homomorphically full antisymmetric digraphs but not for homomorphically full oriented graphs.
Our remaining work proceeds as follows.In Section 2 we fully classify homomorphically full antisymmetric digraphs as those that are quasi-transitive and whose underlying graph is homomorphically full.In doing so we provide a theorem for homomorphically full antisymmetric digraphs analogous to Theorem 1.1.These results imply that homomorphically full antisymmetric digraphs can be identified in polynomial time.And also that one may decide in polynomial time if a graph is an underlying graph of a homomorphically full antisymmetric digraph.In Section 3 we show for oriented graphs that neighbourhood comparability does not fully characterize homomorphic fullness.This work leads us to study, in Section 4, the problem of deciding if an oriented graph is homomorphically full.Though the analogous problems for graphs and antisymmetric digraphs are Polynomial, we show the problem to be GI-hard for oriented graphs.We continue studying the complexity of problems related to homomorphic fullness of oriented graphs in Section 5. We find that deciding if a graph admits a homomorphically full orientation is NP-complete.We conclude in Section 6 with discussion and further remarks.
Since the presence or absence of multiple edges in Γ or Λ does not matter in the definition of a homomorphism of Γ to Λ, we consider only graphs in which there are no multiple edges.By contrast, the presence of loops matters.A homomorphism of a graph can map adjacent vertices to a vertex with a loop.These statements remain true when we replace graphs with oriented graphs (under either interpretation).To simplify matters, herein we assume that all graphs and oriented graphs are irreflexive.We comment more on the nature of this problem for graphs that may have loops as part of our further remarks in Section 6.

Homomorphically Full Antisymmetric Digraphs
We begin our study of homomorphically full antisymmetric digraphs by examining their underlying graphs.
Lemma 2.1 If G is a homomorphically full antisymmetric digraph, then its underlying graph U (G) is homomorphically full.
Proof: Let G be a homomorphically full antisymmetric digraph.Notice that each of G and U (G) have the same set of elementary homomorphisms.For uv / We note that the converse of this lemma is false.As discussed in our introductory remarks, an undirected path on three vertices is homomorphically full, but the directed path on three vertices is not a homomorphically full antisymmetric digraph.
Recall that a directed graph is quasi-transitive when for each vertex w, there is complete adjacency between the in-neighbours and the out-neighbours of w.Quasi-transitive directed graphs were first studied and subsequently fully classified by Ghouila-Houri (1962).A graph is an underlying graph of a quasitransitive digraph if and only if it is a comparability graph.Further, every comparability graph is the underlying graph of some quasi-transitive digraph.
Theorem 2.2 An antisymmetric digraph is homomorphically full if and only if it is quasi-transitive and its underlying graph is homomorphically full.Further, every homomorphically full graph is the underlying graph of a homomorphically full antisymmetric digraph Proof: Let G be an antisymmetric directed graph whose underlying graph is homomorphically full.
If G is not quasi-transitive, then there exists a vertex w such that there is not complete adjacency between the in-neighbours and the out-neighbours of w.That is, there exists u, v ∈ V (G) such that uwv is an directed path of length 2 (a 2-dipath).The elementary homomorphism that identifies u and v is complete.As such G uv is a homomorphic image of G. However G uv is not isomorphic to any subgraph of G; it contains a directed 2-cycle.Therefore G is not a homomorphically full antisymmetric digraph.
Assume now that G is quasi-transitive.Consider a pair of non-adjacent vertices u, v ∈ V (G).Since U (G) is homomorphically full, we have, without loss of generality, that Let Γ be a homomorphically full graph.By Theorem 1.1, Γ has no induced P 4 , and hence is a cograph.Every cograph is a comparability graph.As proven by Ghouila-Houri (1962), every comparability graph is the underlying graph of a quasi-transitive digraph. 2 As with graphs, homomorphically full antisymmetric digraphs can be classified using neighbourhood comparability, homomorphic images and retracts.We note, however, that such a classification does not exist for oriented graphs.We explore this further in Section 3. Proof: Let G be a homomorphically full antisymmetric digraph.
Assume G is a homomorphically full antisymmetric digraph.By Lemma 2.1, we have that U (G) is homomorphically full.By Theorem 2.2 it follows that G is quasi-transitive.That is, (a) ⇒ (b).
Assume G is quasi-transitive and its underlying graph is homomorphically full.By Theorem 1.1 it follows that every pair of non-adjacent vertices of U (G) is neighbourhood comparable.Since G is quasitransitive, it follows that G has no induced 2-dipath.Therefore every pair of non-adjacent vertices of G are neighbourhood comparable.That is, (b) ⇒ (c).
Assume every pair of non-adjacent vertices of G are neighbourhood comparable.Let H be a homomorphic image of G. Let ϕ : G → H be a complete homomorphism.Since ϕ is complete it suffices to assume it is an elementary homomorphism.Since ϕ is an elementary homomorphism we have that H ∼ = G uv for some pair of non-adjacent vertices u and v. Since u and v are neighbourhood comparable, it follows without loss of generality that G uv ∼ = G − u.Thus there exists a retraction G → G − u.And so it follows that every homomorphic image of G is isomorphic to a retract of G.That is, (c) ⇒ (d).By definition, every retract of G is isomorphic to an induced subgraph of G. Thus (d) ⇒ (e).
Finally, assume every homomorphic image of G is isomorphic to an induced subgraph of G. From the definition of homomorphically full, it follows that G is homomorphically full.That is, (e) ⇒ (a).This completes the proof. 2 Let G be a homomorphically full antisymmetric digraph and let u and v be a pair of non-adjacent vertices.By Theorem 2.3 we can assume, without loss of generality, that G uv ∼ = G − u.In some sense, this isomorphism is trivial -vertices other than u and v may be mapped to themselves.Vertex u v may be mapped to v. By changing the label of u v to v we arrive at a homomorphic image that is a subgraph of G.The statement of Theorem 2.3 suggests that for antisymmetric digraphs we may remove the word isomorphic from within the definition of homomorphically full.The same observation holds for homomorphically full graphs.And in fact, the word isomorphism does not appear in the original definition of homomorphically full given by Brewster and MacGillivray (1996).In the following section, we will see that the existence of these trivial isomorphisms is not guaranteed for homomorphically full oriented graphs.

Homomorphically Full Oriented Graphs
We turn now to the interpretation of oriented graphs as arising from simple graphs and the subsequent definition of homomorphically full.Consider the oriented graphs in Figure 1.Since v and v ′ are neighbourhood comparable we have One can construct such an isomorphism by first noting that the sole vertex of degree 2 in G − v ′ − xu ′ must map to the sole vertex of degree 2 in G uu ′ .As these are the only elementary homomorphisms of G, we conclude G is homomorphically full.The example shows that a theorem for homomorphically full oriented graphs akin to Theorems 1.1 and 2.3 is not possible.That Fig. 1: A homomorphically full oriented graph with a pair of vertices that are not neighbourhood comparable is, this example shows that the property of homomorphic fullness is not necessarily related to neighbourhood comparability, nor is it necessarily related to the existence of retracts or homomorphisms to induced subgraphs.
Recall that an oriented clique is an oriented graph in which any two non-adjacent vertices are at directed distance 2. The name oriented clique arises from the fact an oriented colouring (i.e., a homomorphism) of such a graph assigns a different colour (i.e., image) for each vertex.Observe that oriented cliques have no elementary homomorphisms.From this observation the following two facts follow.
Theorem 3.1 Every oriented clique is a homomorphically full oriented graph.Theorem 3.2 Let G be a homomorphically full oriented graph.The core of G is an oriented clique.
Proof: Let G be a homomorphically full oriented graph and let H be the oriented graph with the fewest vertices such that there is a homomorphism ϕ : G → H that is onto with respect to A(H).Observe that H is necessarily an oriented clique as if otherwise, H has a pair of vertices u and v such that G → H uv .By the definition of homomorphically full, H is isomorphic to a subgraph of G. Therefore H is isomorphic to the core of G. 2 Let G be an homomorphically full oriented graph.As our definition of homomorphically full oriented graph restricts homomorphism to targets that are oriented graphs, an elementary homomorphism of G necessarily identifies a pair of non-adjacent vertices that are not the ends of a 2-dipath.That is, an elementary homomorphism necessarily identifies vertices that are not at directed distance either 1 or 2. We define the undirected closure of G, denoted cl(G), to be the graph formed from G by adding an edge between any pair of vertices at the end of 2-dipath and then considering all arcs as edges.That is, we have uv ∈ E(cl(G)) when u and v are at directed distance at most 2 in G.Note that for any u, v ∈ V (G) we have that there exists an elementary homomorphism G → G uv if and only if there exists an elementary homomorphism cl(G) → cl(G) uv .
Lemma 3.3 Let G be an oriented graph and let u and v be a pair of vertices that are neither adjacent nor at directed distance 2. We have cl(G) uv ⊆ cl(G uv ).
Proof: Let G be an oriented graph.Let u v denote the vertex in both cl(G uv ) and cl(G) uv formed by identifying u and v.As cl(G uv ) and cl(G) uv have the same vertex set, to show cl(G) uv ⊆ cl(G uv ) it suffices to show that for all xy ∈ E(cl(G) uv ) we have xy ∈ E(cl(G uv )).
Consider first an arc xy ∈ E(cl(G) uv ) such that x, y ̸ = u v .Since xy ∈ E(cl(G) uv ) and x, y ̸ = u v , we have that x and y are adjacent in U (G) or there is a 2-dipath from x to y in G. Therefore x and y are adjacent in cl(G) and so xy ∈ E(cl(G uv )).
Assume without loss of generality that x = u v .If one of uy or vy is an arc in G, then u v y is an arc in G uv .Thus xy is an edge in cl(G uv .Otherwise, neither of uy or vy is an arc in G. Thus xy as an edge in cl(G) uv could only have arisen from an edge of the form uy or vy in cl(G).Therefore, there is a 2-dipath in G with one end at y and the other at one of u or v. Since u and v are not adjacent, this 2-dipath exists in G uv .Therefore xy is an edge in cl(G uv ). 2 Using Lemma 3.3 we find a connection between homomorphically full graphs and homomorphically full oriented graphs.
Theorem 3.4 If G is a homomorphically full oriented graph, then cl(G) is homomorphically full.
Proof: Let G be an homomorphically full oriented graph.Consider cl(G) and u, v ∈ V (G) such that u and v are not adjacent in cl(G).Since G is homomorphically full, it follows that G uv isomorphic to a subgraph of G.That is, there exists a subgraph H of G such that G uv ∼ = H.Since H is a subgraph of G, it follows that cl(H) is a subgraph of cl(G).And so by Lemma 3.3 we have Thus, cl(G) uv is isomorphic to a subgraph of cl(G).Therefore cl(G) is homomorphically full. 2 Corollary 3.5 A homomorphically full oriented graph has at most one nontrivial component.
Proof: Let G be a homomorphically full oriented graph.Notice that U (G) and cl(G) have the same number of nontrivial components.The result now follows from Theorems 1.1 and 3.4.2 Statement (e) in Theorem 1.1 implies that homomorphically full graphs admit a forbidden subgraph characterization.Similarly, statement (b) in Theorem 2.3 implies that homomorphically full antisymmetric digraphs admit a forbidden subgraph characterization.We find this to not be the case for homomorphically full oriented graphs.
Theorem 3.6 Every oriented graph is an induced subgraph of an homomorphically full oriented graph.
Proof: By Lemma 3.1, it suffices to show that every oriented graph appears as an induced subgraph of some oriented clique.Let G be an oriented graph with n vertices.
Let B n be the oriented complete bipartite graph with bipartition ({a 1 , a 2 , . . ., a n }, {b 1 , b 2 , . . ., b n }) obtained by orienting the edge a i b j from a i to b j if i ≤ j and from b j to a i if i > j.One can verify that B n is an oriented clique.
Denote by B ′ n the oriented clique obtained from the oriented clique B n by adding arcs such that the subgraph induced by {a 1 , a 2 , . . ., a n } is isomorphic to G.
By construction, the oriented graph G is an induced subgraph of B ′ n .The result now follows by our previous remarks.As homomorphically full oriented graphs do not admit a forbidden subgraph orientation, one can wonder if there is an efficient recognition algorithm for homomorphically full oriented graphs.We study this problem in the following section.

Deciding if G is Homomorphically Full
We consider the problem of deciding if an oriented graph is in fact homomorphically full.The corresponding decision problem for graphs and antisymmetric digraphs are Polynomial -it is enough to compare the neighbourhoods of pairs of non-adjacent vertices.However the example in Figure 2 shows that such a procedure does not suffice for homomorphically full oriented graphs.We see that u and u ′ are not neighbourhood comparable, yet G uu ′ ∼ = G − v.We use the example in Figure 2 to show that the problem of deciding if an oriented graph is homomorphically full is GI-hard.We do this by first showing that the problem of deciding if a pair of oriented cliques are isomorphic is GI-complete.For these ends we define the following decision problems.
We begin by showing OCLIQUEISO is GI-complete.For our reduction, we require the following result from Zemlyachenko et al. Theorem 4.1 Zemlyachenko et al. (1985) DAGISO is GI-complete.
Given an instance G, H of DAGISO we construct oriented cliques G ⋆ and H ⋆ such that G ∼ = H if and only if G ⋆ ∼ = H ⋆ .
We construct G ⋆ using a pair of disjoint copies of G, say G L and G R , and a fixed regular tournament Construct G ⋆ by adding the following arcs to the oriented graph formed from the disjoint union of G L , G R and T : • u L i t for all t ∈ V (T ) and all 1 ≤ i ≤ n; and • tu R i for all t ∈ V (T ) and all 1 ≤ i ≤ n.
Lemma 4.2 Let G and H be directed acyclic graphs.We have Proof: Let G and H be directed acyclic graphs with n vertices.Let V (G) = {u 1 , u 2 , . . .u n } and V (H) = {v 1 , v 2 , . . ., v n } Let ϕ : G → H be an isomorphism.Using ϕ we construct an isomorphism ϕ ⋆ : G ⋆ → H ⋆ as follows: • ϕ ⋆ (t) = t for all t ∈ V (T ); Proof: The reduction is from DAGISO.Given an instance G, H of DAGISO construct the instance G ⋆ , H ⋆ of OCLIQUEISO.Such a construction can be carried out in polynomial time.The result follows from Lemma 4.2 and Theorem 4.1.2 Using Theorem 4.3, we show HOMFULL is GI-hard.Given an instance G 1 , G 2 of OCLIQUEISO we construct an instance Ĝ of HOMFULL such that G 1 ∼ = G 2 if and only if Ĝ is homomorphically full.We construct Ĝ from a copy of J (as labelled as in Figure 2), a copy of G 1 and two copies of G 2 , say G 2 and G ′ 2 , by adding a new vertex q and the following arcs: • qx for all x ∈ V (J); and arcs yq for • y 1 y 2 , y 2 y ′ 2 and y 2 y 1 for all • wy 1 for all y 1 ∈ V (G 1 ).
Proof: Let G 1 and G 2 be oriented cliques.By construction Ĝ has two elementary homomorphisms: Let ϕ : G 2 → G 1 be an isomorphism We extend ϕ as follows such that ϕ Assume now Ĝ is homomorphically full.Consider Ĉ, the core of Ĝ.By Theorem 3.2, Ĉ is an oriented clique.By observation we produce a copy of Ĉ by removing v ′ and u ′ from Ĝ.We see then that Ĉ is an oriented clique with |V (G 1 )| + 2|V (G 2 )| + 4 vertices.We also observe that neither u ′ nor v ′ are contained within a copy of Ĉ.Consider now Ĝr uu ′ .Identifying a pair of vertices if Ĝ cannot change the core of Ĝ and so Ĉ is the core of Ĝuu ′ .This implies that Ĝuu ′ has as a subgraph an oriented clique with |V (G 1 )| + 2|V (G 2 )| + 4 vertices.
As Ĝ is homomorphically full, Ĝuu ′ is isomorphic to some subgraph of Ĝ − s for some vertex s ∈ V ( Ĝ).Since Ĝ − u ′ has a vertex of degree 2 (and Ĝuu ′ does not), it cannot be that s 2 ) ∪ {q}, then Ĝ − s does not contain a copy of Ĉ. Therefore s = v ′ .Notice that Ĝ − v ′ and Ĝuu ′ have the same number of arcs and the same number of vertices.And so there is an isomorphism Each of Ĝuu ′ and Ĝ − v ′ has a single vertex of degree three.Therefore β(v ′ ) = u ′ .In Ĝuu ′ , the vertex v ′ has a single out-neighbour: w.Therefore β(w) = v, as v is the lone out-neighbour of the image of v ′ under β.In Ĝuu ′ , the vertex v ′ has a pair of in-neighbours: q and uu ′ .By considering the direction of the arc between q and uu ′ we see β(q) = q and β(uu ′ ) = w.Since β(q) = q and q has only four out-neighbours (three of which we have already considered), it follows that β(v) = u.Since β is an isomorphism, and all other vertices are accounted for, it must be that 2 ).As β(w) = v it must be that β(y) ∈ V (G 2 ) for all y ∈ V (G 1 ).And so restricting β to V (G 1 ) gives an isomorphism G 1 → G 2 . 2 Given an instance Γ, of OCLIQUE we construct an instance Γ of FULLORIENT such that Γ admits an orientation as an oriented clique if and only if Γ admits an orientation that is homomorphically full.Let V (Γ) = {v 1 , v 2 , . . ., v n }.Let Λ be a complete graph on n + 2 vertices with vertex set V (Λ) = {v ′ 1 , v ′ 2 , . . ., v ′ n , s, t}.We construct Γ from the disjoint union of Γ and a complete graph by the following edges i , for all 1 ≤ i ≤ n, and • tv i for each 1 ≤ i ≤ n.
Lemma 5.4 Let Γ be a graph.The graph Γ admits an orientation as an oriented clique if and only if Γ admits an orientation that is homomorphically full.
Proof: Let Γ be a graph.Notice that no two vertices of Γ are neighbourhood comparable.Therefore Γ has an orientation that is a homomorphically full oriented graph if and only if it has an orientation that is an oriented clique.
Suppose Γ has an orientation as an oriented clique, G. Extend this orientation to Γ by orienting the edge between v i and v ′ i from v i to v ′ i , 1 ≤ i ≤ n; orienting Λ to be a transitive tournament, H, in which s has in-degree 0 and t has out-degree 0; and orienting the edges between t and v 1 , v 2 , . . ., v n from t to v i , 1 ≤ i ≤ n.
Since G and H are oriented cliques, it remains to verify that there is a directed path of length at most 2 between each vertex of G and each vertex of H − t.For x ∈ V (H) − {t} and 1 ≤ i ≤ n there is a 2-dipath x, t, v i .Thus this orientation of Γ is an oriented clique, and hence is a homomorphically full oriented graph.
Now suppose Γ has an orientation G that is a homomorphically full oriented graph.As noted above, this orientation G is an oriented clique.Let G and H respectively be the subgraphs of G induced by the vertex sets of Γ and Λ.Since there is no path of length 2 in Γ joining a vertex of Γ and a vertex of Λ, no 2-dipath joining vertices of G contains a vertex of H. Therefore G is an oriented clique.And so Γ admits an orientation as an oriented clique.
(a) Γ is homomorphically full.(b) If x and y are non-adjacent vertices of Γ, then x and y are neighbourhood comparable (c) Every homomorphic image of Γ is isomorphic to a retract of Γ.(d) Every homomorphic image of Γ is isomorphic to an induced subgraph of Γ.
Theorem 2.3 Let G be an antisymmetric digraph.The following statements are equivalent: (a) G is a homomorphically full antisymmetric digraph.(b) G is quasi-transitive and its underlying graph is homomorphically full.(c) Every pair of non-adjacent vertices of G are neighbourhood comparable.(d) Every homomorphic image of G is isomorphic to a retract of G.(e) Every homomorphic image of G is isomorphic to an induced subgraph of G.

Fig. 2 :
Fig.2: A homomorphically full oriented graph with a pair of vertices that are not neighbourhood comparable

HOMFULL
Instance: An oriented graph G. Question: Is G homomorphically full?OCLIQUEISO Instance: A pair of oriented cliques G and H. Question: Is G ∼ = H? DAGISO Instance: A pair of directed acyclic graphs G and H Question: