k-to-1 Functions on an Arc

Katsuura, Hidefumi; Kellum, Kenneth R.

doi:10.2307/2046660

Cited by 7 publications

(6 citation statements)

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“…In 1986, Jo Heath proved that, for any 2‐to‐1 function from [0, 1] (i.e., a K 2 ) onto any Hausdorff space, the set of discontinuities is infinite (of course, all graphs are Hausdorff spaces). In 1987, Katsuura and Kellum showed that, for any

k \geq 2

, there is no k ‐to‐1 function from K 2 onto K 2 with only finitely many discontinuities. Finally, building on these earlier results, Jo Heath clarified the whole issue of whether or not there is a k ‐to‐1 finitely discontinuous function between graphs, as stated in Theorem 1.4.…”

Section: Finitely Discontinuous K‐to‐1 Functions Between Graphsmentioning

confidence: 99%

Wiggles and Finitely Discontinuous k‐to‐1 Functions Between Graphs

Gauci

Hilton

Stark

2013

Journal of Graph Theory

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A function between graphs is k‐to‐1 if each point in the codomain has precisely k preimages in the domain. In this article, we approach the topic of continuous, or finitely discontinuous, k‐to‐1 functions between graphs from three different points of view. Harrold (Duke Math J 5 (1939), 789–793) showed that there is no 2‐to‐1 continuous function from a closed interval onto a circle (i.e., from K2 onto C3). In the first part of this article, we describe all 3‐to‐1 continuous functions from an edge onto a cycle. Such a description is just one step away from a description of all 3‐to‐1 continuous functions from double-struckR onto double-struckR, which is in fact our main initial emphasis. Second, given two graphs, G and H, and an integer k≥1, and considering G and H as subsets of double-struckR3, Jo Heath gave a simple criterion for the existence of a finitely discontinuous k‐to‐1 function from G onto H. Such functions often involve a limiting construction which we call a wiggle. In the second part of this article, we give a simple formula (related to Jo Heath's construction) which counts the number of wiggles. The question of whether there is a continuous k‐to‐1 function (i.e., a k‐to‐1 map in the usual topological sense) from G onto H is more complicated. In the third part of this article, we consider complete graphs Kn and Km. In the cases where n and m have the same parity, and n≤m, then we determine exactly when there is a k‐to‐1 continuous function from Kn onto Km. Other cases are considered elsewhere (J. K. Dugdale, S. Fiorini, A. J. W. Hilton, J. B. Gauci, Discrete Math 310 (2010), 330–346 and J. B. Gauci, A. J. W. Hilton, J Graph Theory 65 (2010), 35–60).

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k \geq 2

Section: Finitely Discontinuous K‐to‐1 Functions Between Graphsmentioning

confidence: 99%

Wiggles and Finitely Discontinuous k‐to‐1 Functions Between Graphs

Gauci

Hilton

Stark

2013

Journal of Graph Theory

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“…In the proof of Lemma 1.21, we need one result which was proved by Katsuura and Kellum in [13], stated below. The proof in the cases when the domain is ]0, 1[ or [0, 1[ is very similar to the proof in the case when the domain is [0, 1] due to Katsuura and Kellum [13], and we give a proof of all three at the same time.…”

Section: Lemma 110 ([3])mentioning

confidence: 99%

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

Dugdale

Fiorini

Hilton

et al. 2010

Discrete Mathematics

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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.

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“…Indeed, it relies on the existence of continuous functions with a given assignment of the fibers. Many authors have investigated various aspects of this problem (see, e.g., [2,4,6,7,10] and references therein), giving answers to questions as "Does there exist a continuous k-to-1 function defined on a given topological space?". Actually, as we will see, the possibility to rearrange a map to a continuous one, depends only on the existence of a continuous function with a given assignment of the cardinality of the fibers.…”

Section: Introductionmentioning

confidence: 99%

“…This is a purely combinatorial classification of Y X . On the other hand, it discloses a variety of topological problems, which have been widely investigated in recent literature (see, for instance, [2,4,6,7,10]).…”

Section: Definition 22 Introduce the Setmentioning

confidence: 99%