Article ID | Journal | Published Year | Pages | File Type |
---|---|---|---|---|
5777696 | Topology and its Applications | 2017 | 6 Pages |
Abstract
We also prove that if (X,x)âFANR, then Shn(X,x)=Sh(X,x), for some integer nâ¥2, implies that Sh(X)=1. This resolves positively Problem (5.6) from [7] (1981). Similarly as for polyhedra, first we show that if XâFANR and Sh(X)â 1, then X cannot have the shape of a proper Cartesian factor of itself.
Related Topics
Physical Sciences and Engineering
Mathematics
Geometry and Topology
Authors
Danuta KoÅodziejczyk,