If the fundamental group is finite, any path connected cover has finite fibres Let $B$ be a space with a path connected cover $p: E \to B$. Let $b_0 \in B$ and assume $\pi_1(B, b_0)$ is finite. Then $p^{-1}(b_0)$ is finite.
Assume that $p^{-1}(b_0)$ is infinite and write $\pi_1(B, b_0) = \left\{[\gamma_1], [\gamma_2], \dots, [\gamma_n]\right\}$. For each $[\gamma_j] \in \pi_1(B, b_0)$, we can pick infinitely many points $e_0, e_1, \dots$ in disjoint open sets of $E$ such that $p(e_i) = b_0$ for all $i$ and lift the homotopy equivalence class $[\gamma_j]$ to a homotopy equivalence class $[\gamma_j]_{e_i}$. I am lost as to proceed from here.
 A: Here are the important assumptions we have:

*

*$E$ is path-connected

*$\pi_1(B,b_0)$ is finite

Let's try to use these two facts as directly as possible.
First, if $p^{-1}(b_0)$ is empty, then we are done. Otherwise, fix some point $e_0 \in p^{-1}(b_0)$. Now for each $e \in p^{-1}(b_0)$, pick some path $\gamma_e : [0,1] \to E$ such that $\gamma_e(0) = e_0$ and $\gamma_e(1) = e$.
(We have now used the assumption that $E$ is path-connected!)
Next, consider the set $X := \{[p \circ \gamma_e] : e \in p^{-1}(b_0)\}$. This is a subset of $\pi_1(B,b_0)$, and thus is finite. So, there exists a natural number $n$ and points $e_1, \dots, e_n \in p^{-1}(b_0)$ such that $$X = \{[p \circ \gamma_{e_0}], \dots, [p \circ \gamma_{e_n}]\}.$$
(We have now used the assumption that $\pi_1(B,b_0)$ is finite!)
We now claim that $p^{-1}(b_0) = \{e_0, \dots, e_n\}$. So, let $e \in p^{-1}(b_0)$ be arbitrary. Then $[p \circ \gamma_e] \in X$, so there is some $n \in \mathbb{N}$ such that $[p \circ \gamma_e] = [p \circ \gamma_{e_n}]$.
Claim (For you to fill in!) $\gamma_e$ is homotopic to $\gamma_{e_n}$ (rel. endpoints), and thus $e = e_n$.
(Hint: use the homotopy lifting property)
This will complete the proof!
