1
$\begingroup$

Let $p:\mathbb{R}^2\rightarrow\mathbb{R}^2/\mathbb{Z}^2$ be the natural projection, obviously $\mathbb{R}^2/\mathbb{Z}^2$ is the torus $\mathbb{T}^2$, if $K$ is a connected and compact subset of $\mathbb{T}^2$, and $Q$ is the component of $p^{-1}(K)$, then whether $p(Q)=K$? What is the relation between $Q$ and other components of $p^{-1}(K)$?

$\endgroup$
  • $\begingroup$ If I remember my topology correctly Decktransformationsshould help you with the relations between the components of $Q$. Also since $K$ is connected, $p(Q)$ has to be $K$. you should be able to proof that by using the connectedness of $K$ and lifting of paths to the universal cover ($\mathbb{R}^2 \to \mathbb{T}^2$) $\endgroup$ – Enkidu Nov 1 '18 at 11:42
1
$\begingroup$

The answer is no. Take a cylinder, $C=R\times T$, where $T$ is the unit circle. And consider the universal covering $f:R^2\to C$ given by the formula $(x,y)\mapsto (x,e^{iy})$. Now the set $$X=\{ (x,y)\in R^2: y=1/x,0<x\leq 1\}\cup \{(x,y):x=0\}$$ is disconnected, while its image $K=f(X)$ is closed and connected, it consists of a circle and a spiral accumuating on this circle. No component of $f^{-1}(K)$ is mapped surjectively: one has the circle as the image, and another the spiral.

To make a torus glue the ends of this cylinder together.

$\endgroup$
  • $\begingroup$ I agree with your example, thanks a lot ! I wonder if I assume that the component $Q$ of $p^{-1}(K)$ is bounded, is it true? $\endgroup$ – Yee Neil Nov 2 '18 at 3:58

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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