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Polyhedra for which every homotopy domination over itself is a homotopy equivalence.
- Source :
-
Topology & Its Applications . Nov2023:Part A, Vol. 339, pN.PAG-N.PAG. 1p. - Publication Year :
- 2023
-
Abstract
- In this paper we study an open problem: Is it true that each homotopy domination of a polyhedron over itself is a homotopy equivalence? Since every ANR has the homotopy type of a polyhedron, this question is related in an obvious way to the longstanding Borsuk's problem (1967) Is it true that two ANR's homotopy dominating each other have the same homotopy type? The answer is positive for all polyhedra (or ANR's) with polycyclic-by-finite fundamental groups, manifolds, and clearly for 1-dimensional polyhedra. In some previous paper, we proved that this also holds for all 2-dimensional polyhedra with weakly Hopfian (hence Hopfian and co-Hopfian) fundamental groups (a group is weakly Hopfian if it is not isomorphic to a proper retract of itself). In this paper we extend this result to all (G , n) -complexes (where n ≥ 2). A (G , n) -complex, is an n -dimensional finite CW -complex with fundamental group G and all the homotopy groups π r in dimensions 1 < r < n trivial. We also discuss fundamental groups of potential counterexamples. As a corollary we obtain that for the same classes of spaces, the Borsuk's question: Is it true that the homotopy types of two quasi-homeomorphic ANR's are equal? , has positive answer (this question was also asked by S. Ferry in the recent edition of Scottish Book (1981)). [ABSTRACT FROM AUTHOR]
- Subjects :
- *POLYHEDRA
*HOMOTOPY groups
*HOMOTOPY equivalences
Subjects
Details
- Language :
- English
- ISSN :
- 01668641
- Volume :
- 339
- Database :
- Academic Search Index
- Journal :
- Topology & Its Applications
- Publication Type :
- Academic Journal
- Accession number :
- 172291984
- Full Text :
- https://doi.org/10.1016/j.topol.2023.108572