Finite fiber of universal covering

Question

Suppose that $Z$ admits a universal covering $\pi:Z'\to Z$. Show that $\forall z\in Z$, $\pi^{-1}(z)$ is a finite set.

Answer 1

This is not necessarily true. The map $\pi : \mathbb R \to S^1$ defined by $\pi(t) = (\cos(2\pi t), \sin(2\pi t))$ is a universal covering of the circle, yet $\pi^{-1}(\{(0, 0)\}) = \mathbb Z$ is not finite.