Let $S$ be the dyadic solenoid.

Let $x\in S$, and let $X$ be the union of all arcs (homeomorphic copies of $[0,1]$) in $S$ containing $x$.

$X$ is called a composant of $S$.

It is well-known that $X$ is a dense first category one-to-one continuous image of the reals, and that $S$ is a homogeneous continuum.

Now let $y$ and $z$ be any two points in $S\setminus X$.

Is there a self-homeomorphism $h:S\to S$ such that $h[X]=X$ and $h(y)=z$?


1 Answer 1


In this paper of J.Kwapisz I have found the following

Theorem 1. Any homeomorphism $h$ of the dyadic solenoid $S$ is isotopic to the "affine" homeomorphism of the form $g:x\mapsto \pm(2^n x+b)$ for some $n\in\mathbb Z$ and some $b\in S$.

If $h$ preserves the path-connected component $X$ of the neutral element, then so does the affine homeomorphism $g$, which implies that $b\in X$. It follows that for any $y\in S\setminus X$ the image $h(y)$ belongs to $g(y)+X\subset \pm 2^{\mathbb Z}y+X$, which is contained in a countable union of path-connected components and hence cannot be an arbitrary element $z\in S\setminus X$.

So, the answer to the original question is negative.

The same negative answer holds for homeomorphisms of any (not necessarily dyadic) solenoid.

  • $\begingroup$ That is very surprising to me! $\endgroup$ Aug 15, 2018 at 15:49
  • $\begingroup$ One typo: I think $h(x)$ should be $h(y)$ $\endgroup$ Aug 15, 2018 at 15:57
  • $\begingroup$ @ForeverMozart The typo is corrected. Thanks for noticing. $\endgroup$ Aug 15, 2018 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.