The topological fixed point property –- an elementary continuum-theoretic approach

Mańka, Roman

doi:10.4064/bc77-0-14

Cited by 5 publications

(1 citation statement)

References 11 publications

Supporting

Mentioning

Contrasting

Order By: Relevance

“…Questions 1 and 2 are also open for tree-like plane continua. For more information and unsolved fixed-point problems, see [2], [7], and [9].…”

Section: Resultsmentioning

confidence: 99%

Fixed points in indecomposable 𝑘-junctioned tree-like continua

Hagopian¹

2009

Proc. Amer. Math. Soc.

View full text Add to dashboard Cite

Abstract. Let M be an indecomposable k-junctioned tree-like continuum. Let f be a map of M that sends each composant of M into itself. Using an argument of O. H. Hamilton, we prove that f has a fixed point.D. P. Bellamy [1] in 1979 defined a tree-like continuum that does not have the fixed-point property (also see [4], [10], [11], and [13]). In 1998, the author [6] proved that every tree-like continuum has the fixed-point property for deformations. This was accomplished by showing that every map of a tree-like continuum M that sends each arc-component of M into itself has a fixed point. The collection of arc-components is a partition of a continuum. Hence a map that sends each arc-component into itself is sometimes called an arc-component-preserving map. Bellamy's continuum is indecomposable and its arc-components are composants. The collection of composants of an indecomposable continuum is a partition that is refined by the collection of arc-components. The difference between these two partitions is sometimes extreme. Each arc-component consists of one point in P. Minc's example [11] of a hereditarily indecomposable tree-like continuum without the fixed-point property. Must every composant-preserving map of an indecomposable tree-like continuum M have a fixed point? We prove the answer is yes if M is k-junctioned. PreliminariesA space X has the fixed-point property if for each map f of X into X there is a point x of X such that f (x) = x.A chain is a finite collection A collection E of sets is coherent if for each nonempty proper subcollection F of E there is an element of F that intersects an element of E \ F.

show abstract

“…Questions 1 and 2 are also open for tree-like plane continua. For more information and unsolved fixed-point problems, see [2], [7], and [9].…”

Section: Resultsmentioning

confidence: 99%