TITLE: Do solenoids embed into Möbius strips?
QUESTION [11 upvotes]: I found a strange attractor which looks a lot like a solenoid. The attractor continuum is the closure of a continuous line which limits onto itself, and it is locally homeomorphic to Cantor set times Reals. It sits in the Möbius strip.

Does the Dyadic solenoid embed into the Möbius strip?  What about solenoids in general?  Could the strange attractor above actually be a solenoid?

REPLY [8 votes]: No solenoid can be embedded into the Mobius strip. To derive a contradiction, assume that some solenoid $S$ embeds into the Mobius strip $M$. Let $\pi:C\to M$ be a 2-fold covering map of the cylinder $C$ onto the Mobius strip. It is well-known that the solenoid $S$ contains a dense subset $D$ which is the image of the real line under a continuous map $\phi:\mathbb R\to D$ (and this image of $\mathbb R$ is called a composant of the solenoid). By the lifting property of the covering map $\pi$, there exists a continuous map $\varphi:\mathbb R\to C$ such that $\pi\circ\varphi=\phi$. Then the closure $K$ of the connected set $\phi(\mathbb R)$ in $C$ is a continuum such that $\pi(K)=\bar D=S$. Taking into account that the cylinder $C$ embeds into the plane, we conclude that $K$ is a planar continuum and hence the solenoid $S$ is a continuous image of a planar continuum.
On the other hand, by a result of Krasinkiewicz in his paper Mappings onto circle-like continua, no solenoid is a continuous image of a planar continuum (as solenoids have infinitely divisible first cohomology group whereas the first cohomology group of any planar continuum is finitely divisible). This is a desired contradiction.