1992
DOI: 10.4064/fm-141-3-269-276
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Exactly two-to-one maps from continua onto some tree-like continua

Abstract: Abstract. It is known that no dendrite (Gottschalk 1947) and no hereditarily indecomposable tree-like continuum (J. Heath 1991) can be the image of a continuum under an exactly 2-to-1 (continuous) map. This paper enlarges the class of tree-like continua satisfying this property, namely to include those tree-like continua whose nondegenerate proper subcontinua are arcs. This includes all Knaster continua and Ingram continua. The conjecture that all tree-like continua have this property, stated by S. Nadler Jr.… Show more

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Cited by 5 publications
(7 citation statements)
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“…If f is a simple map from a continuum X onto a dendroid and f (x) = f (y), then either (1) there is an arc in X from x to y, or (2) there are disjoint twin arcs xx and yy in X such that for some indecomposable continuum I in X containing x and y , f (x ) is the center of f (I), or (3) there is an indecomposable continuum I in X containing x and y such that f (x) is the center of f (I).…”
Section: Lemma 5 Suppose F Is a Simple Map From An Indecomposable Comentioning
confidence: 99%
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“…If f is a simple map from a continuum X onto a dendroid and f (x) = f (y), then either (1) there is an arc in X from x to y, or (2) there are disjoint twin arcs xx and yy in X such that for some indecomposable continuum I in X containing x and y , f (x ) is the center of f (I), or (3) there is an indecomposable continuum I in X containing x and y such that f (x) is the center of f (I).…”
Section: Lemma 5 Suppose F Is a Simple Map From An Indecomposable Comentioning
confidence: 99%
“…Suppose |f We will extend the maximal triple T (m) to complete the contradiction. Using M as the continuum X in Lemma 6, and using x and y to denote the two inverse points of m, there are only three possibilities: (1) there is an arc in M from x to y, (2) there are disjoint twin arcs xx and yy in M and an indecomposable continuum J in M containing x and y such that f (x ) is the center of f (J), or (3) there is an indecomposable continuum J in M containing x and y such that f (x) is the center of f (J).…”
Section: A Set Is Open In the Image Iff Its Preimage Is Open In The Dmentioning
confidence: 99%
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“…That question is answered only for some cases. For instance, it is known that a tree-like continuum cannot be the image of a 2-to-1 map if it is hereditarily indecomposable [10], or if it is an indecomposable arc-continuum (every proper subcontinuum is an arc) [3]. In the decomposable case, Gottschalk [4] showed that no dendrite can be the image of a 2-to-1 map.…”
Section: Introductionmentioning
confidence: 99%