Lifting property of a covering space

Question

A common example of a covering space is $p:\mathbb R \rightarrow S^1 $, $p(t)= e^{2 \pi it}$, which can be looked at as an infinitely long spiral placed over a circle. Consider the double loop $\gamma: [0,1] \rightarrow S^1$, i.e. the path that starts at $x_0$, goes twice around the circle, and ends at $x_0$. I am having troubles visualising (in an "infinite spiral" fashion) a continious path $ \tilde \gamma: [0,1] \rightarrow \mathbb R$ that will lift the double loop: $p \tilde \gamma = \gamma$. Could you please help me?

Answer 1

Take the path $\tilde\gamma(t)=2t$.

Think of the covering $\mathbb R\to S^1$ as that infinitely long spiral over the circle. Choose a point over $x_0$ on this spiral, then trace the spiral upwards until you've gone around the circle twice, ending at another point over $x_0$. The segment of the spiral that you've traced is the path $\tilde\gamma$.