Continuous k-to-1 functions between complete graphs of even order

Abstract

a b s t r a c tA function between graphs is k-to-1 if each point in the co-domain has precisely k preimages in the domain. Given two graphs, G and H, and an integer k ≥ 1, and considering G and H as subsets of R 3 , there may or may not be a k-to-1 continuous function (i.e. a k-to-1 map in the usual topological sense) from G onto H. In this paper we review and complete the determination of whether there are finitely discontinuous, or just infinitely discontinuous k-to-1 functions between two intervals, each of which is one of the following: ]0, 1[, [0, 1[ and [0, 1]. We also show that for k even and 1 ≤ r < 2s, (r, s) = (1, 1) and (r, s) = (3, 2), there is a k-to-1 map from K 2r onto K 2s if and only if k ≥ 2s.

Continuous k‐to‐1 functions between complete graphs whose orders are of a different parity

“…Let k = 2s. We consider all the 2r +1 vertices x i , 1≤ i ≤ 2r +1, in G and map them to one vertex v 1 of H. Then we again obtain r(2r +1) "loops" on v 1 Also, the off-diagonal elements of kA 0 −B 0 are equal to either (k −2) or k, which, in either case, is even and non-negative since k = 2s. Thus Condition (2) (3) is also satisfied.…”

“…Since 2s−1>r(2r +1), the valency of any vertex in the co-domain graph H is greater than the number of edges in the domain graph. We consider all the 2r +1 vertices x i , 1≤ i ≤ 2r +1, in G and map them to one vertex v 1 of H. Then, since in G there are r(2r +1) edges, we obtain r(2r +1) "loops" on v 1 . Mapping one loop on each of r(2r +1) edges of H incident to v 1 , we have (2s−1)−r(2r +1) edges left.…”

“…In this paper we describe how these folds can be composed together (in Definition 2.5), but first we present again the definitions given in [1]. It is useful to view these functions as the actual "folding" of the edges as illustrated in The three folds F(m+1, 2m+1, m+1), F(m+1, 2m+1, 1) and F(1, 2m+1, 1) are the only ones we have used in our studies up to now.…”

“…Mapping one loop on each of r(2r +1) edges of H incident to v 1 , we have (2s−1)−r(2r +1) edges left. We "invent" this number of further loops by taking one of the existing loops which is already mapped to an edge [v 1 , z] of H, and extending it to cover the edges incident to v 1 that still have no pre-images. By doing this, the number of pre-images of v 1 is increased by (2s−1−r(2r +1)), to become |f −1 (v 1 )| = r +2s−2r 2 .…”

“…In [1] we presented our first results on k-to-1 continuous functions (or maps) between complete graphs with the ultimate objective of determining all triples (n, k, m) for which there is k-to-1 (continuous) map from K n onto K m . We showed that for k even and 1 ≤ r<2s, (r, s) = (1, 1) and (r, s) = (3, 2), there is a k-to-1 map from K 2r onto K 2s if and only if k ≥ 2s.…”