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](https://mathoverflow.net/questions/3332/two-commuting-mappings-in-the-disk/28003#28003)



Solenoid of a continuous map of a ball, is it contractible?
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.


