Let $p : E \to B$ be a covering map with $B$ 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 $b \in B$.
Proof: Suppose there exists $b_0 \in B$ with $p^{-1}(b_0)$ having $k$ elements. Let $A := \{b \in B : p^{-1}(b)$ has $k$ elements $\}$. Observe that
- $A \neq \emptyset$ since $b_0 \in A$.
- $A \cap (B \setminus A) = \emptyset$.
- $A \cup (B \setminus A) = B$.
- Claim 1: $A$ is open.
- Claim 2: $B \setminus A$ is open.
Conclude that $B \setminus A = \emptyset$, otherwise we would have a separation of $B$, a contradiction that $B$ is connected. Since $B \setminus A = \emptyset$, we must have $A = B$ which means every fiber of $p$ (ie, every set $p^{-1}(b)$) has $k$ elements.
Proof of Claim 1: $A$ is open: Let $a \in A$, then since $p$ is a covering map, there exists $U \ni a$ open such that $p^{-1}(U) = \bigsqcup_{\alpha \in \Lambda} V_\alpha$ where each $V_\alpha$ is homeomorphic to $U$ via $p \vert_{V_\alpha}$. Notice that $\Lambda$ has $k$ elements since $p^{-1}(a)$ intersects with each $V_\alpha$ at exactly one point and there are $k$ such points. Hence we can write $p^{-1}(U) = \bigsqcup_{n = 1}^k V_n$. Now let $x \in U$ be arbitrary and notice that $p^{-1}(x) \subseteq p^{-1}(U) = \bigsqcup_{n = 1}^k V_n$ and $p \vert_{V_n}$ is a homeomorphism and hence $p^{-1}(x)$ has $k$ elements as well, so $x \in A$. Conclude that $A$ is open since for each $a \in A$, $a \in U \subseteq A$.
Proof of Claim 2: $B \setminus A$ is open: Same reasoning as above.