ON THE COMPLETE DIGRAPHS WHICH ARE SIMPLY DISCONNECTED Abstract

DAVIDE C. DEMARIA AND ,I . CARLOS S . KHHL Homotopic methods are employed for the characterization of the complete digraphs which are the composition of non-trivial highly regular tournaments.


Introduction
It is known that one can also construct a homotopy theory for categories of spaces having a structure weaker than a topology.For example, one can take the category of prespaces or Cech closure spaces .
To every digraph D one can associate, in a natural way, two finite prespaces P(D) and P* (D) ; and vice versa, to every finite prespace one can associate two digraphs G and G*, dually oriented.Hence one can relate the category of the digraphs with that of the finite prespaces.Therefore a homotopy theory can be defined for digraphs, by setting the regular homotopy group Q,(D) of D to be the homotopy group 7r,(P(D)) of the associated prespace (see [6]).
In [3] Burzio and Demaria proved that the groups Q,(D) are isomorphic to the classical homotopy groups 7r,(IKDI), where jKD1 is the polyhedron of a suitable simplicial compeex KD associated with the digraph D .Then in [5] they obtained, as an application of the regular homotopy of digraphs, a structural characterization of tournaments T, called simply disconnected tour- naments, whose fundamental group QI(T) is non trivial .In [4], they have obtained another characterization for the simply disconnected tournaments by using coned 3-cycles.
In this paper we extend those results to the case of digraphs D which are complete, and we get analogous results if QI(D) ~¿ 0. First of all, we must generalize the concept of simple quotient for every type of digraph and we prove the following theorem : In this way we obtain the following theorems:

. Some Definitions and Notations
Definition 2.1.Let V be a finite non-empty set and E a set of ordered pairs (u, v) E V x V, such that u =,A v.We call the pair D = (V, E) a directed graph or digraph.The elements of V are the vertices of D, the cardinality of V the order of D, and the elements of E the ares of D .Moreover, we write u -> v instead of (u, v), and we call u a predecessor of v and v a successor of u.
Remark 1.Given two distinct vertices u and v, we have a priori four possibilities, and four types of are : (1) there is no oriented are between u and v -we denote by uI v the null are; (2) there is the oriented are (u, v), but not the are (v, u) -we denote the simple are by u -> v; (3) there is the oriented are (v, u), but not the are (u, v) -we denote the simple are by u -v ; (4) there are both oriented ares (u, v) and (v, u) -we denote the double are by u <--> v. (A double are is also called a symmetric pair.)Definition 2.2.A digraph is called oriented if, between two distinct vertices, there is at most one ordered are -that is, the possible ares are either simple ares or null ares.A digraph is called a non-oriented graph if, between two distinct vertices, there is either a double are or a null are.A digraph is called complete if, between two distinct vertices, there is at least one ordered are ; the possible ares in this case are either simple or double ares .Definition 2.3.A digraph T is a tournament if, between every pair of distinct vertices, there is one and only one are.A tournament T is called hamiltonian if it contains a spanning cycle -that is, a cycle passing through all the vertices of T.
Definition 2.4.Let D = (V, E) and D' = (V', E') be digraphs.A function f : V -> V' is a homomorphism between D and D' if, for every u, v E Remark 2. We can consider two kinds of dualities for a given digraph D: (a) the first one is the dually oriented digraph D, which is obtained by changing the orientation of the ares; in this case, both digraphs are of the same type.
(b) the second kind is the digraph D, which is obtained by maintaining the simple ares and by changing the null ares into double ares, and vice-versa.In this case, oriented digraphs become complete digraphs, and vice-versa .On the other hand, tournaments and non-oriented graphs do not change .
It is known that, given a tournament T, we can associate an algebraic structure to T in a natural way.In fact we have (see [71) : Proposition 2.5 .A tournament T becomes a commutative groupoid A(T) if we define the following binary operation * : Remark 3 .Similarly we can associate with T the dual commutative groupoid A'(T), by defining : u, ifv-->uoru=v ; forallu,vETu * v=v * u={ v, if u --~v .

Remark 4. Every homomorphism between two tournaments T and T' is also an algebraic homomorphism between the commutative groupoids A(T) and A(T') (orA'(T) and A'(T')), and vice-versa.
The same definitions can be applied to the case of a digraph D of any type, and then we have the associated groupoids A(D) and A'(D), which are dual.In this case, we set for A(D) : on the other hand from uIv we get: Remark 5.In general the two groupoids are not commutative .In the first case, from u <--> v we get u * v=uandv * u=v ; u*v=uandv*u=u.Remark 6.Whereas the homomorphisms between two tournaments coincide with the algebraic homomorphisms between the associated groupoids, this is not true in the general case, for there are homomorphisms between digraphs which are not algebraic homomorphisms between the associated groupoids .u, ifu-~voru=v ; for all u, v E D, u * v v, if v 74 u; and for A'(D) : For example, given the 3-cycle C : u , v -> w -; u and the symmetric pair D : x <--> y, if we define f : C --+ D by f (u) = x, f (v) = f (w) = y, then in A(C) and A(D) we have Remark 7. We can still associate other two groupoids tl(D) and fh(D) to the digraph D, in the following way : We observe that, if we change the orientation of the arcs, then A(D) becomes A(D) and A'(D) becomes Á'(D) ; in changing the double arcs into null arcs and the null arcs into double arcs, A(D) becomes th(D), and A'(D) becomes A(D) .

Quotient Digraphs
We say that a subset X of vertices of a digraph D is a set of equivalent vertices if for any vertex u in D -X the oriented arcs from u to any vertex v in X are all of the same type.Of course the type of oriented arc can change if we vary the vertex u in D -X.
If p : A(D) -+ A(Q) is a surjective algebraic homomorphism between the groupoids which are associated to the two digraphs D and Q,,,, of order n and m, respectively, then we see that the groupoid A(Q) is isomorphic to the quotient groupoid A(D)/p.For, if we consider the m pre-images of the vertices v,, . . ., v,, in Q,,, and we set S(i) = p-' (vi), i = 1, . . ., m, then we can subdivide the n vertices of D in m disjoint subdigraphs S(I), . . ., S(-) of equivalent vertices, because the type of arc which joins the vertex vi to vj is of the same type as the arcs which join every vertex in S(i) to every vertex in SO) .
If the above conditions hold, then we write D,, = Q,(S('), . . ., S(m)), and we say that the digraph D,, is the composition of the m digraphs S(I), . . ., S(-).The subdigraphs S(') , . . ., S(n) are called the componente of the digraph D, and Q, is the quotient of the digraph D,n .
We say that a digraph is simple if the composition D,, = Q,,,(S~I) , . . ., S(m)) implies that m = 1 or m = n -that is, if the quotient Q,, or the componente S(2) coincide with the trivial digraph of order 1 .

Properties of the Quotient Digraphs
Proposition 4.1.Let D be a digraph, Q be one of the quotient digraphs of D, and S be one of the components of D with respect to Q; then S is a set of equivalent vertices .Conversely, if X is a set of equivalent vertices, then X is a component of D .
Proof. .The first statement is obvious .For the second, we consider the partition of D in the subset X and the singular subsets of D -X Proposition 4 .2.Let X and Y be two sets of equivalent vertices of a digraph D .If X fl Y =,A 0 and X U Y =~D, then X U Y is a set of equivalent vertices .
Proof. .Let u be a vertex in X n Y and let v be a vertex in D -(X U Y) ; for each vertex w in X U Y, the oriented arc between v and w is of the same type as the oriented arc between v and u Proposition 4.3.Let D be a digraph and Q one of its quotient digraphs; then Q is isomorphic to a subdigraph E of D.

,T(k)) .
Suppose that h > 2 .Since the two partitions are distinct, there must exist two distinct components S and T with non empty intersection.If S is not contained in T and the other components T have non-empty intersection with S, then at least one of them cannot be contained in S, for otherwise we can replace these particular components T in the partition Q by the unique component S; this contradicts the simplicity of Qk .
Therefore there exist a component S and a component T such that S n T r,4 0, S ~t-T and T 9~-S, and we choose them to be S(1) and T (1) .We now number the components S in such a way that S(1), S(2), . . ., S(,) intersect T(1) , while the rest S(T+1), S(T+2) . . .S(h) do not intersect T(1) .
The oriented ares between a vertex u in S(1) -T (1) and any vertex in the union U = S(2) U S(3) U . . .U S(h) are all of the same type, because every component S('), i = 2, 3, . . ., h intersects the component T(1).On the other hand, Since u is a vertex in S(1), the oriented ares between any vertex in S( 1) and any vertex in U are of the same type.This means that the partition [S( 1 ), U] is a composition of D, which is a contradiction, for we have supposed that the quotient Ph is simple of order h > 2.
Finally, if h = 2, then k must be also equal to 2, for otherwise it would be sufficient to interchange the two compositions and repeat the previous argument.On the other hand, the digraphs P2 and Q2 must be isomorphic, for if P2 is the null are, then the digraph D is disconnected, if P2 is the simple are, then the digraph D is weakly (but not strongly) connected ; and if P2 is the double are, then the digraph D is strongly connected Remark 1 .It follows from the proof that, for h > 2, we have not only a unique simple quotient, but also a unique partition into components .On the other hand, for h = 2, then we may have more partitions.For example, if D = [u, v, w; u H v, u H w, v H w] we have the following partitions Remark 2. We recall that a digraph is hamiltonian if theie is a cycle passing through all the vertices .We shall consider a symmetric pair as a hamiltonian cycle.Proposition 4.5.A complete digraph is hamiltonian if and only if each of its (non-trivial) quotients is hamiltonian.
Proof.From results due to Rado (1943), Rey (1958) and Camion (1959), we know that a complete digraph D is hamiltonian if and only if the tournament T2 (the simple oriented are) is not the simple tournament related to D .We now observe that either the initial digraph or any of its quotient digraphs have the same simple quotient .The assertion follows 5. Complete Digraphs which are Simply Disconnected Definition 5.1 .A tournament T is regular if, for each vertex v E T, the numbers of predecessors and successors of v are the same (and hence the order of T is odd) .A tournament T2 .+1 is highly regular if there exists a cyclical ordering vi, . . ., v2n,,+1, Vi on the vertices of T2 ,, +1 such that vi -~vj if and only if vj is one of the first m successors of vi in the cyclical ordering of T2.+1 .Definition 5.2 .A digraph D is simply connected if its first homotopy group We have the following theorem : Theorem 5.3.A complete digraph D is simply disconnected if and only if its simple quotient is a highly regular tournament .
This theorem is a generalization of the analogous theorem for tournaments (see Theorem 3 .9, of [5]) .For the proof, we need the following lemmas .Lemma 5 .4.A complete digraph D is simply connected if and only if each of its non-trivial quotient digraphs D* is simply connected.
Proof..This is analogous to the proof of Proposition 2 .1 in [5] for tournaments Lemma 5.5.For every complete digraph Dn of order n, there exists at least one tournament of order n which is a subdigraph of Dn .
Proof. .It is sufficient to eliminate one oriented arc from each symmetric pair Lemma 5.6.Let D, and Fn be two given complete digraphs, each of order n, such that Fn, is a subdigraph of Dn .Then if Fn is simply connected, so is D, .
Proof. .Each edge-loop in the polyhedron associated to the digraph Dn is null-homotopic, since it is null-homotopic in the sub-polyhedron associated to the digraph Fn We now prove Theorem 5 .3: (a) If the simple quotient of a complete digraph D is a highly regular tournament, then D is simply disconnected .
This follows direct1y from Lemma 5 .4and the fact that a highly regular tournament is simply disconnected .
(b) Let D n be a complete digraph which is simply disconnected ; then Dn, has a highly regular tournament as a simple quotient.
A tournament T, which is a subdigraph of D, is aleo simply disconnected, and therefore is the composition of a highly regular tournament, by the analogous theorem for tournaments.
It is now sufficient to preve that the simple ares which ought to be replaced by double ares in order to pase from the tournament T to the initial complete digraph D,,, all belong to subtournaments which are componente of T,z .
FYom Lemma 5.4, we see that the replacement of simple ares by double ares in the components of T,, do not change the first homotopy group of the digraphs which are obtained one by one .
On the other hand, if we assume there exists a double are with vertices u and v, which belongs to two different components of T,, then: 1) if we construct a 3-cycle C in T, using the vertices u and v, then the edge-loop determined by C is not null-homotopic in the polyhedron associated to T,, as it is a generator of QI(T~,) (see [5,Proposition 3.6]); 2) en the other hand, if we replace the simple are between u and v by a double are, then the same loop becomes null-homotopic in the polyhedron associated with the complete digraph obtained in such a manner from T,, .
Therefore the first homotopy group of such a digraph is trivial, and hence by Lemma 5 .6 the group Ql(D~) is aleo trivial .Hence the result follows Corollary 5.7.A simply disconnected complete digraph is hamiltonian.
Proof.This follows easily from Proposition 4.5 Before we obtain a second characterization for the simply disconnected complete digraphs, we need to introduce some further definitions .Definition 5.8.A subdigraph F of a digraph D is coned if there is at least one vertex u in D -F, such that u is either a predecessor or a successor of the vertices in F; otherwise, the subdigraph E is non-coned.Definition 5.9 .A subdigraph F of a digraph D is shrinkable if there exists a proper subset of D consisting of equivalent vertices, and containing F. Proof. .By Theorem 5.3, it is sufficient to show that a complete digraph whose simply quotient is a highly regular tournament, is characterized by the conditions (a) and (b).
1) The proof that conditions (a) and (b) are necessary is analogous to that given for tournaments in [4,Th. 7] .
2) If we suppose that conditions (a) and (b) hold for the complete digraph D, then by (b) the simple quotient digraph Q of D, is a tournament, such that each of its 3-cycle is non-coned .Hence by the analogous theorem for tournaments which was mentioned above and by the simplicity of Q, we see that the tournament Q is highly regular .The result is proved Aknowledgements .This work was performed under the auspices of the Consiglio Nazionale delle Ricerche (CNR, GNSAGA) and of FAPESP (Proc .NQ 89/2042-1) .

Theorem 5 . 3 .
A complete digraph D is simply disconnected if and only if its simple quotient is a highly regular tournament.Theorem 5.10 .A complete digraph Dn is simply disconnected if and only (a) there exists in Dn a non coned 3-cycle; (b) every symmetric pair and every 3-cycle in Dn are shrinkable in Dn .

Theorem 5 .
10 .A complete digraph D. is simply disconnected if and only if (a) there exists in Dn, a non-coned 3-cycle; (b) every symmetric pair and every coned 3-cycle in D, are shrinkable in Dn, .