HOMOMORPHISMS AND RELATED CONTRACTIONS OF GRAPHS

For every homomorphism ϕ of a graph G there exists a contraction θϕ on G¯, the complement of G. Here we study the graph equation ϕ(G)=θϕ(G¯). In the course of our work we show that Hadwiger's Conjecture is true for every self-complementary graph.


I. INTRODUCTION.
By a graph of order n we mean a set V(G) of n vertices together with a set E(G) of unordered pairs of distinct vertices in V(G) called edges.A graph G is isomorphic to a graph H if there is a bijection from V(G) onto V(H) which preserves both adjacency and non-adjacency, in which case we will write G H. A graph with the property that G , where denotes the complement of G, is called self-complementary.
An elementary homomorphism of a graph G is the identification of two non-adjacent vertices of G, and a homomorphism is a sequence of elementary homomorphisms.Thus a homomorphism of a graph G onto a graph H is a function from V(G) onto V(H) such that whenever u and v are adjacent in G, (u) and (v) are adjacent in H.Likewise, an elementary contraction of a graph G is the identification of two adjacent vertices of G, and a contraction is a sequence of elementary contractions.Thus for every homomorphism of G there is a contraction of that we may construct as follows: is a sequence of elementary homomorphisms el,e 2 en each of which identifies two non-adjacent vertices in G, so we let be the sequence of elementary contrac- tions 1'O2'''''en such that i identifies the same vertices in G that e i identifies in G.
Recently [I] the graph equation (G)   (G) was studied.Here we consider a similar equation, namely, We will employ the following notation as the need arises: 0G(U) will denote the valency of the vertex u in G and AG(U) will be the set of all vertices in G that are to u.Thus OG(U) lAG(U) where IAG(U) is the cardlnality of the set adjacent AG(U ).As usual (G) will denote the chromatic number of G.

SOME GENERAL RESULTS.
THEOREM i.There is no graph G of order n > such that (I.I) is satisfied for every homomorphism of G.
If G is not self-complementary then the identity homomophism and its related null contraction suffice to satisfy the theorem.We will postpone the remainder of the proof until section 3, where we restrict our attention to results on self- complementary graphs.
THEOREM 2. If there exists a homomorphism of G that satisfies (I.I) then: (a) G must be connected.
(b) G cannot be a tree of order n 5.
PROOF.(a) If G is not connected, then no contraction of G is connected.
However, G not connected implies G is connected, thus every homomorphic image of G will be connected.
(b) If G is a tree of order n 5 then G and every homomophic image of G contains K 3, the complete graph on three vertices, as a subgraph.But every contraction of G will be a tree and so cannot contain K 3 as a subgraph.THEOREM

If x(G)
2 and is connected then there exists a homomorphism such that (G) PROOF.Since is connected, the image of under any contraction will be connected.Thus every contraction of G onto two points must have K 2 as its image.
From [3] we know that there exists a homomorphism # of G such that #(G) Kx(G... Hence, using this homomorphism and its related contraction, we have (G) K 2 0(G).
Since every homomorphism is a sequence of elementary homomorphisms, we now turn our attention to the equation where e is an elementary homomorphism.
PROOF.In [3] Harary et.al. proved the following inequalities: X(G) + g Since (2.1) holds we have (e(G)) X(0 (G)) and so from the first inequality g X(0 (G)) equals either X(G) or X(G) + Putting these values into the second g inequality above yields: which completes the proof.
The following result will be needed in the next section.
LEMMA I. Let G be of order n and an elementary homomorphism that identifies since when an elementary homomorphism e is applied to a graph G, e(G) will lose one edge for each vertex which is adjacent to both vertices that were identified, and all other edges will remain in e(G).Likewise, by the same reasoning as above, IE( e())l IE()I IA(Ul)N A(u2) i, where the extra edge removed is the one which was contracted.Now if (2.1) holds both e(G) and e (G) must contain the same number of edges.Thus IE(G) I-IAG(UI) AG(U2) IE()I-IA(uI) A(u2) I, where IE(8 which proves the lemma 2 3. SELF-COMPLEMENTARY GRAPHS. We first summarize some results from [4], [5] and [6] that will be needed.Let S be a self-complementary graph.If S is of order n then n E 0,1(mod 4).Suppose f is an isomorphism such that f(S)

S
If n E 0(mod 4) then f has no fixed vertex, that is f(u) # u for all u V(S).However if n l(mod 4) there exists precisely one fixed vertex for f, and moreover any such S can be constructed by appropriately n-I adding this fixed vertex, of valency -to a self-complementary graph of order n-I n-l.Lastly there are regular self-complementary graphs of degree --if and only if n E 1(mod 4).
We will assume throughout this section that S in nontrivial, that is n > I.The following sequence of four lemmas will complete the proof of Theorem I.
LEMMA 2. If S is of order n E 0(mod 4) then there exist vertices Ul,U 2 V(S) (u)= such that UlU2 E(S) and O s s(U2 PROOF.Since n E 0(mod 4), any given valency that occurs in S will occur an even number of times, [4].Thus there are vertices Ul,U 2 e V(S) such that O s(ul Os(U2).Now let f(S) , then Os(Ul) Os(U2) implies o(f(ul))=o(f(u2) and also 0s(f(ul) 0s(f(u2) ).But UlU2 e E(S) if and only if f(u I) f(u2) E(S) and so UlU2 or f(ul)f(u2) satisy the lemma.
REMARK.For any self-complementary graph S of order n, IE(S) n(n-l)4 Thus from Lemma I, if e is an elementary homomorphism of S which identifies the vertices u and u 2 and e(S) 8e ()' then IAs(Ul) N As(U2) IA(uI) A(u2) + I.
LEMMA 3. If S is of order n 0(mod4) then there is an elementary homomorphism e of S such that E(S) # 8 (S).
PROOF.Using Lemma 2 choose Ul,U 2 e V(S) such that UlU2 E(S) but 0s(UI) Os(U2).Now for any u e V(S) that is distinct from u and u 2, u can be adjacent to both u and u 2, neither u nor u 2, or adjacent to one and not the other.where k IAs(UI) N A s(u2) and k 2 IA(uI) N A(u2) I. Let g identify u and u 2 and suppose g(S) (S) so that from the proceding Remark k k 2 + I. Then g n-i Ps(Ul + p(ul) (kl +4) + (k 2 + + 4) 2k + 24 Hence n 2k + 24 + 0(mod 4), a contradiction.
We now consider those self-complementary graphs of order n E l(mod 4) and let n-I v be the fixed vertex, of valency -under the isomorphism f(S) .
LEMMA 4. Let S be of order n E l(mod 4) and g be an elementary homomorphism that identifies any u V(S) with v.If (S) (S) then S is regular.

PROOF. Let f(S)
, then for any u V(S) distinct from v, v u E(S) if and only if v f(u ) E(S) while 0s(U ) 0s(f(u)).
n-I Suppose S is not regular then there is a vertex u g V(S) such that 0s(U) 2 and, from the observation above, v u E(S).Now let g identify this vertex u and v and suppose g(S) ().We have, as in the proof of Lemma 3, --(k 2 + I), and k k 2 + since E(S) e(S).Hence (u) k + 4 k + (k 2 + l) -a contradiction, LEMMA 5.If S is a regular self-complementary graph then there is an elementary homomorphism g such that g(S) # e () n-I PROOF.First note that S v is self-complementary, with -vertices of valency and -vertices of valency --Thus there are --edges joining every n-I n-3 vertex of valency --to the vertices of valency -and these must appear in either S v of S v Since S v S v these edges must be equally divided between S v and its complement.When v is added to S v to form S it is adjacent to every n- we have vertex of valency-- Now S is regular of degree -and we let identify u, as described above, and v. Assume g(S) 8 (S) and uV(g(G)) is the image of u and v under since every vertex adjacent to both u and v in S will only account for one edge incident to u in e(S).Similarly P0 ()(u') 0(u) + O(v) [A(u) N Am(v) 2   where the 2 is substracted since the edge contracted under 0 is incident to both u and v However since e(S) e (g) we must have IA (u) As(V) IA(u) A(v) + -or 0-(U)s --2 depending on whether or not u A-(U)s A(v).Therefore 0e(s)(U') 0 ()(u) for every vertex in 0 () and hence e(S) # () since isomorphisms must preserve valencies Now that the proof of Theorem is complete we will show that for every self- complementary graph there is a homomorphism which satisfies (1.1).LEMMA 6.Let S be of order n and f(S) .Then in the set V {(uif(ui))lu i V(S), i=l n} precisely m [] are non-adjacent pairs of distinct vertices in S.
PROOF.Let u V(S) such that f(u) # u.Then u and f(u) are adjacent in S if and only if f(u) and f(f(u)) are non-adjacent in S. Now f is an isomorphism, and so f(u) and f(f(u)) must also be distinct vertices of S. Hence for every pair of distinct non-adjacent vertices in V, there is a pair of adjacent vertices in V and conversely.Thus there must be m [] pairs of distinct non-adjacent vertices in V. n THEOREM 5. Let S be of order n and m [3].There exists a homomorphism of S such that S) K e (S) n-m PROOF.For each of the m pairs of distinct non-adjacent vertices provided by Lemma 6, there exists an elementary homomorphism e.1 of S that identifies u i and f(ui).
Let i,2 em so that both (S) and @(S) will have n-m vertices.
Suppose (S) # Kn_m, then there are vertices Ul,U 2 e V((S)) such that UlU2 E((S)).Let -l(u I) {al,a2} and -l(u 2) {bl,b2}.Thus a # b I, f(a I) a 2 and f(b I) b 2 where f(S) .Now UlU2 E((S)) implies alb E(S) and so f(al)(BI) E(S).But then f(al)f(b I) alb2 e E(S) and so UlU2 E((S)), a contradiction Thus (S) K By an argument virtually identical to the one just n-m given it follows that () Kn_m, which proves the theorem.
Fr_om [3] we know that the smallest homomorphic image of any graph G is a complete graph of order x(G).Thus from Theorem 5 we have x(S) n [] for every self- complementary graph S of order n.However if a graph is contractabl to a complete graph of order t, then it has a complete contraction of order k for k t, also from [3].Hence Theorem 5 shows that every self-complementary graph satisfies Hadwiger's Conjecture [2], that is, every self-complementary graph S has a complete contraction of order x(S).
It would certainly be desirable to find some simple necessary and sufficient