Topological razors (ball-like spaces)

Question

Introduction

Many admire the Euclidean space, and I am not an exception. I will try to catch the topological roundness of the $n$-ball in its greatest generality. I call the resulting axiomatized space to be a razor. If my definition turns out too general then (after getting interesting counter-examples) it will be a simple matter to narrow these definitions down (e.g. by finally agreeing--if necessary--to explicit homotopic constructions) till we are left with the most possible general characterizations (as opposed to simply general). Optimistically I hope for the minimal properties though (while among others the exotic cubes have to be considered).

THE GOAL (conjecture)   is to show that for every razor $\ (X\ T)\ $ and $S\subseteq X$ (see below) there exists a non-negative integer $\ n\ $ and subspace $\ Y\subseteq X\ $ such that $\ X\setminus S\subseteq Y\ $ and pairs $\ (Y\ \,Y\!\cap\! S)\ $ and $\ (\mathbb I^n\ \,\partial(\mathbb I^n))\ $ are homeomorphic.

All topological spaces are assumed here to be $T_1$-spaces (each single-point set should be closed). Also by definition:

$$X^{(2)}\ :=\ X\times X\setminus \Delta_X$$

where $\ \Delta_X\ :=\ \{(x\ x) : x\in X\}$,   for any set $\ X$.

At this time I am not making any assumptions about compactness, separability, finite dimension ... or even about the involved subsets being closed. But if you have to do them then do them, of course.

Razors

A razor is a topological space $\ X\ $ for which there exists a razor structure. And a razor structure in $\ X\ $ is defined as a triple $\ (X\ S\ f)\ $ where $\ S\ $ is a subset of $\ X\ $, function $\ f: X^{(2)}\rightarrow\ S\ $ is continuous, f.p.p. stands for the fixed point property, and the following conditions hold:

1. $\ X\ $ has the f.p.p.
2. $\ S\ $ does not have the f.p.p.
3. $\ \forall_{x\in X}\forall_{y\in S\setminus\{x\}}\ \ f(x\ y)\ =\ y$

REMARK   Axiomat 1 is equivalent to property: $\ \ S\ $ is not a retract of $\ X\ \ $ (it's a standard observation in a situation like this).

Answer 1

A way to attack the above problem may be via non-$T_1$ spaces which otherwise satisfy the above three axioms--let's call a topological space like this, together with an appropriate function, to be a pre-razor. (In general, the finite topological spaces should play the role similar to the combinatorial simplicial complexes). Let me present the simplest pre-razor. It's defined as a 3-point space $\ (X\ T),\ $ where $\ X:= \{a\ b\ p\},\ $ such that $$ T := \{\emptyset\}\cup \{G\subseteq X: p\in G\}$$ Also $$ S:=\{a\ b\}$$

Thus $\ S\ $ is discrete, and it follows that $\ S\ $ does not have the fixed point property.

On the other hand $\ X\ $ is connected. Thus there does not exist any continuous $\ \phi:X\rightarrow X\ $ such that $\ \phi(X)=S.\ $ Thus every continuous $\ \phi:X\rightarrow X\ $ is either constant, and then it has a fixed point, or $\ p\in \phi(X).\ $ Then $\ \phi^{-1}(\{p\})\ $ is open, hence $\ \phi(p)=p.\ $ This proves that $\ X\ $ has the fixed point property.

Finally, let's define $\ f : X^{(2)}\rightarrow S\ $ as follows:

• $f^{-1}(a) := \{ (b\ p)\ \,(b\ a)\ \,(p\ a)\}$
• $f^{-1}(b) := \{ (a\ p)\ \,(a\ b)\ \,(p\ b)\}$

These sets are both open in $\ X^{(2)},\ $ say:

• $f^{-1}(a) := \{b\ p\}\!\times\!\{a\ p\}\ \cap\ X^{(2)} $
• $f^{-1}(b) := \{a\ p\}\!\times\!\{b\ p\}\ \cap\ X^{(2)} $

This shows that $\ f : X^{(2)}\rightarrow S\ $ is continuous, and it obviously satisfies the last axiom. Thus the system consisting of the topological space $\ (X\ T)\ $ and function $\ f:X^{(2)}\rightarrow S\ $ is a pre-razor.