Is a Blaschke product/rational function a covering map for a $n$-sheeted covering of $S^{1}$? We have a Blaschke product $B(z)$ of order $n$ (you can think of it as a rational function with $n$ zeros and $n$ poles), the zeros are obviously inside $\mathbb{D}$. Why is $B(z) \colon S^{1} \to S^{1}$ a covering map for an $n$-sheeted covering of $S^{1}$ ?

$\textbf{Hint:}$ A Blaschke Product on $\mathbb{S^1}$ is locally conformal, i.e, its derivative doesn't vanish anywhere on $\mathbb{S^1}$. Any local homeomorphism from a compact space to a connected space is a covering map.