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 $p^{1}(K)$?

$\begingroup$ 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
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.

$\begingroup$ 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