Topology Atlas Document # qaaa-11 | Production Editor: Harriet Lord

© 1997 Copyright by E. Santillan. All rights reserved.


Question

Peano Continua which fix the n+1-dimensional triodes

E. Santillan

Let I be the closed unit interval of the real line and let T be a (simple) triode with vertex point t. (T is the union of three arcs which are pairwise disjoint, except for the common point t. Both sets I and T are endowed with the standard topology).

Definition The topological space X is said to fix the n+1-dimensional triodes (n is a natural number), if for every mapping f : T ×In --> X ×In which is a homeomorphism from Y ×In onto its image, there exists a point x in X such that f(t ×In) 'subset or equal' {x} ×In.

If a topological space X fixes the n+1-dimensional triodes, it will contains no copies of the real plane R2. I think that the converse implication holds when X is a Peano continuum, but I have not be able to show this fact.

Open Problem Will each Peano continuum X, which contains no copies of the real plane R2, fix the n+1-dimensional triodes?

Nowadays, I am working on cancellation laws of topological products, and I got this open problem when I was showing the following theorem (I am preparing a paper with this theorem).

The spaces S1, R, and R0+ = [0, \infty) 'subset or equal' R are respectively the closed unit circle, the real line and the semi-closed interval of the real line. (They are all endowed with the standard topology).

Theorem Let H be a metrisable connected n-manifold with boundary (n is a natural number) such that the following four products are not homeomorphic by pairs:

  1. S1 ×H,
  2. I ×H,

  3. R0+ ×H,

  4. R ×H.

Then the cancellation law:

A ×H \cong B ×H iff A \cong B
holds for all Peano continua A and B which both fix the n+1-dimensional triodes.


Received by the editors: November 7, 1997
Revised: November 10, 1997