TITLE: The fiber of a covering space over a connected space has constant cardinality
QUESTION [8 upvotes]: Let $p: E\to B$ be a covering map; let $B$ connected. Show that if $p^{-1}(b_0)$ has $k$ elements for some $b_0 \in B$, then $p^{-1}(b)$ has $k$ elements for every $b \in B$.

I know that $E$ has a unique slice because $B$ is connected, but I don't know what to do next.

For the sake of providing some context, this is Section 53, Exercise 3 of Munkres' Topology.

REPLY [11 votes]: Let $m\in \mathbb N \cup \{\infty\} $ and $B_m\subset B$ be the set of points $b\in B$ such that the cardinality of $p^{-1}(b)$ is $m$.
From the definition of "covering map" it is clear that each $B_m$ is open in $B$ and from the definition of "set-theoretic map" it follows that $B$ is the  disjoint union $B=\sqcup_{m=0}^\infty B_m$.
The definition of connected applied to $B$ forces every $B_m$  save one, necessarily  $B_k$,  to be empty.
We have then proved what we wanted: that $B=B_k$, i.e. that all fibers of $p$ have $k$ elements.