Prove that $f:\mathbb{R}\rightarrow S^1$ is a fiber bundle of fiber $\mathbb{Z}$

Question

I'm asked to prove that $f:\mathbb{R}\rightarrow S^1$, given by $f(t)=(\cos t,\sin t )$ is a fiber bundle of fiber $\mathbb{Z}$. I know that for every $p\in S^1$ I have to find an open set $U\subseteq S^1$, with $p\in U$, and a diffeomorphism $\rho_U: f^{-1}(U)\rightarrow U\times \mathbb{Z}$ such that

$\require{AMScd}$ \begin{CD} f^{-1}(U) @>{{\rho_U}}>> U\times \mathbb{Z}\\ @V{f}VV @V{\pi_1}VV\\ U @>{id}>> U \end{CD}

commutes, where $\pi_1$ is the first projection. (I don't know how to do triangular diagrams). However, I don't really know how to start. Can you give me a hint or an idea of how to approach this? Thank you very much!!

Answer 1

Here are some basic facts about $f$.

1. $f$ is peridioc with period $2\pi$.

2. $f$ is an open surjection.

3. $f$ is injective on each open interval $(a,a+2\pi)$.

4. $f((a,a+2\pi)) = S^1 \setminus \{f(a) \}$.

Facts 2. - 4. show that $f$ maps $(a,a+2\pi)$ homeomorphically onto $S^1 \setminus \{f(a) \}$.

Now consider $p \in S^1$. The set $U = S^1 \setminus \{-p\}$ is an open neigborhood of $p$ on $S^1$. Pick $t \in \mathbb R$ such that $f(t) = -p$. From 1. and 3. we see that $f^{-1}(-p) = \{t +2k\pi \mid k \in \mathbb Z\}$. Hence

$$f^{-1}(U) = \mathbb R \setminus f^{-1}(-p) = \bigcup_{k \in \mathbb Z} (t+2k\pi, t+2(k+1)\pi) = \bigcup_{k \in \mathbb Z}V_k$$ with $V_k = (t+2k\pi, t+2(k+1)\pi)$. By definition $f(V_k) = U$.

We know that each $f_k : V_k \stackrel{f}{\to} U$ is a homeomorphism. Therefore

$$\rho_U : f^{-1}(U) \to U \times \mathbb Z, \rho_U(x) = (f(x), k) \text{ for } x \in V_k$$ is a homeomorphism.