# The preimage of continuum on Torus

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 $$p^{-1}(K)$$?

• 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 ($\mathbb{R}^2 \to \mathbb{T}^2$) – 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 $$X=\{ (x,y)\in R^2: y=1/x,0 is disconnected, while its image $$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.
• 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? – Yee Neil Nov 2 '18 at 3:58