(Purdue Jan ’09) Let \(p:E\to B\) be a covering map with \(B\) connected. Suppose that \(p^{-1}(b_0)\) is finite for some \(b_0\in B\). Prove that, for every \(b\in B\), \(p^{-1}(b)\) has the same number of elements as \(p^{-1}(b_0)\).
Solution
1. Every \(x \in B\) has an open neighborhood \(U\) on which \(y \mapsto |p^{-1}(y)|\) is constant.
Let \(U\) be an evenly covered open neighborhood of \(x\), with \(p^{-1}(U) = \bigsqcup_{\alpha \in A} V_\alpha\) and each \(p|_{V_\alpha}\colon V_\alpha \to U\) a homeomorphism. For \(y \in U\), each \(V_\alpha\) contains exactly one point of \(p^{-1}(y)\), namely \((p|_{V_\alpha})^{-1}(y)\), and \(p^{-1}(y) \subseteq p^{-1}(U)\). Hence \(|p^{-1}(y)| = |A|\) for all \(y \in U\).
2. With \(k = |p^{-1}(b_0)|\), the set \(S = \{b \in B : |p^{-1}(b)| = k\}\) is nonempty, open, and closed.
\(b_0 \in S\). If \(b \in S\), the neighborhood \(U\) of \(b\) from step 1 lies in \(S\), so \(S\) is open. If \(b \notin S\), the neighborhood \(U\) of \(b\) from step 1 lies in \(B \setminus S\), so \(B \setminus S\) is open.
3. Q.E.D.
\(B\) is connected and \(S\) is a nonempty clopen subset by step 2, so \(S = B\): every fiber has \(k = |p^{-1}(b_0)|\) elements.