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Let $B$ be the closed unit ball in $\mathbb R^n$ and $f\colon B\to B$ a continuous map.

Consider the infinite product $B^{\mathbb Z}$ equipped with the product topology. Consider the solenoid $$ S_f=\{\{x_n; n\in\mathbb Z\}: x_{n+1}=f(x_n)\} $$ equipped with the induced topology.

Question: Is $S_f$ contractible? If yes, is there a deformation retraction?

Motivation is here: Two commuting mappings in the disk

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up vote 8 down vote accepted

No, it is not even path-connected in general, already for $n=1$.

Consider the folding map $f:[0,1]\to[0,1]$, namely $f(t)=2t$ for $t\le 1/2$ and $f(t)=2(1-t)$ for $t\ge 1/2$. There is no path connecting the orbits of the two fixed points: 0 and 2/3.

Indeed, suppose there is a continuous path $t\mapsto \{x_n(t):n\in\mathbb Z\}$ in $S_f$ such that $x_n(0)=0$ and $x_n(1)=2/3$ for all $n\in\mathbb Z$. Consider $x_0(t)$ and $x_{-n}(t)$ where $n>0$. While $x_{-n}(t)$ travels from 0 to 1/2, $x_0(t)$ will take the value 0 at $t_0=0$, then the value 1 at some $t_1>t_0$, then 0 at some $t_2>t_1$, then 1 again, and so on, $2^n$ times. Since $t\mapsto x_0(t)$ is a continuous function, it can have only finite number of alternating 0 and 1 values, but $n$ can be arbitrarily large, a contradiction.

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Indeed, a nasty space. Thank you. – Andrey Gogolev Jun 15 '10 at 23:05
This construction is somewhat like the Smale horseshoe, just modifying the generating map a little to map the ball into the ball itself. – X.M. Du Jun 16 '10 at 13:06

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