Why is $p^{-1}(1)=\mathbb{Z}$?

Question

I encountered this proof of fundamental group of the circle by Allen Hatcher. There is a part where I have some doubt.

There is this fact that: For each path $f:I\to X$ starting at a point $x_0\in X$ and each $\tilde{x_0}\in p^{-1}(x_0)$ there is a unique lift $\tilde{f}:I\to\tilde{X}$ starting at $\tilde{x_0}$.

Let $f:I\to S^1$ be a loop at the basepoint $x_0=(0,1)$, representing a given element of $\pi_1(S^1,x_0)$. By the fact above there is a life $\tilde{f}$ starting at $0$. This path $\tilde{f}$ ends at some integer $n$ since $p\tilde{f}(1)=f(1)=x_0$ and $p^{-1}(x_0)=\mathbb{Z}\subset\mathbb{R}$.

My doubt is why $p^{-1}(x_0)=\mathbb{Z}$? How could I make sense of this? Could somebody help clarify my doubt?

Thanks.

Answer 1

The map $p : \Bbb R\to \mathcal S^1$ may be defined as $t\mapsto e^{2\text{i}\pi t}$. Thus, the fiber over a point $e^{2\text{i}\pi \theta}\in\mathcal S^1$ is $\{\theta+k : k\in\Bbb Z\}\simeq \Bbb Z$. That's all.

Answer 2

$$p^{-1}(0,1)=p^{-1}(1)=\{ p \in \mathbb{R} | p(t)=1\}$$

$$=\{ p \in \mathbb{R} | e^{2\pi t}=1+0i\}$$

$$=\{ p \in \mathbb{R} | \cos2 \pi t +i\sin2 \pi t =1 +0\}$$

$$=\{ p \in \mathbb{R} | 2 \pi t=2\pi n , n \in \mathbb{Z}\}$$

$$=\{ p \in \mathbb{R} | t=n , n \in \mathbb{Z}\}=\mathbb{Z}$$

$$p^{-1}(1)=\mathbb{Z}$$