# Prove that exist bijection between inverse image of covering space [duplicate]

Let $B$ be path-connected and $p:E\to B$ covering map (with $E$ as covering space). Prove that $\forall a,b\in B$ exist 1-1 injection correspondence between $p^{-1}(a)$ and $p^{-1}(b)$

I thought somehow taking the path between $a$ to $b$ and take on that the inverse means $p^{-1}(\gamma(t))$ but that’s not well defined. How can I prove that?

Declare an equivalence relation on $B$ by saying $a \sim b$ if there exists a bijection between $p^{-1}(a)$ and $p^{-1}(b)$. Try to prove that the equivalence class $[x]$ of $x \in B$ is both open and closed. Since $B$ is path-connected (and hence connected), this implies that $[x] = B$, which completes the proof. To prove that $[x]$ is open, use the definition of a covering map. To prove it’s closed, use the fact that the equivalence relation creates a partition of the set.
The set $p^{-1}(x)$ is often called the fiber above $x$, and this result is often stated by saying that all the fibers have the same cardinality.