Locally trivial fibration $S^1\to S^1$

Question

I have read the answers to this question.

I am trying to understand why the map $f_n:S^1\to S^1$ defined by $f(z)=z^n$ is a locally trivial fibration.

It is clear that $f_n$ is surjective, and I think I know how to construct a homeomorphism $\psi:f_n^{-1}(U_z)\to U_z\times F$, where $U_z$ is an open set around $z\in S^1$ and $F$ is the set of $n$th roots of unity, such that the diagram below commutes: $\require{AMScd}$ \begin{CD} f_n^{-1}(U_z) @>{\psi}>>U_z\times F\\ @V{f_n}VV @VV{\text{projection}}V\\ U_z @>>{\text{identity}}> U_z \end{CD}

I know that the $U_z$ component of $\psi(\alpha)$ must be $\alpha^n$ in order to ensure that the diagram commutes.

I know that the preimage of $U_z\times F$ under $f_n$ is a set of $n$ copies of $U_z$. I think therefore that $\psi$ should be given by $\psi(\alpha):=(\alpha^n,\zeta^k)$, where $\zeta$ is the $n$th root of unity and $k\in\{1,\ldots,n\}$ is the copy of $U_z$ in $f_n^{-1}(U_z)$ to which $\alpha$ belongs. The inverse should then be $\psi^{-1}((\beta,\zeta^k))=\beta^{1/n}$ where $\beta^{1/n}$ is in the $k$th copy of $U_z$ in $f^{-1}(U_z)$.

Does this work? Is there a clearer way of writing down the maps $\psi$ and $\psi^{-1}$ explicitly?

Answer 1

... and $k\in \{1,…,n\}$ is the copy of $U_z$ in $f^{−1}_n(U_z)$ to which $\alpha$ belongs.

This doesn't define $k$ until you have some method of identifying those copies and assigning a unique $k$ to each of them. This is what you have to figure out. To do this, you first will want to restrict $U_z$ to be sufficiently small so that its copies in $f_n^{-1}(U_z)$ do not overlap each other. Then find a way of uniquely identifying one of those copies to be the "base" copy (for $k = 0$). Then show that each of the other copies is $\zeta^k$ times the base copy for some $k$.

I trust it is obvious how small $U_z$ has to be. The are any number of ways you can identify a particular copy of $U_z$ to be the base copy. It doesn't matter which copy gets picked. While you cannot depend on any copy to contain some particular value such as $1$ (it might fall in the interstitial regions between copies), you can take the first copy encountered when travelling counter-clockwise around from $1$. From there, the final step is just algebra.