Let \(p: E \to B\) be a covering map; let \(B\) be 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 \(b \in B\). In such a case, \(E\) is called a \(k\)-fold covering of \(B\).
Solution
1. Local constancy of fiber cardinality:For every \(b \in B\), there exists an open neighborhood \(U\) of \(b\) such that \(|p^{-1}(y)| = |p^{-1}(b)|\) for all \(y \in U\).Proof:
1.1. Since \(p: E \to B\) is a covering map, \(b\) has an evenly covered open neighborhood \(U \subseteq B\).
1.2. By definition of evenly covered, \(p^{-1}(U) = \bigsqcup_{\alpha \in J} V_\alpha\) is a disjoint union of open sets \(V_\alpha \subseteq E\), where each restriction \(p|_{V_\alpha}: V_\alpha \to U\) is a homeomorphism.
1.3. For any \(y \in U\) and each \(\alpha \in J\), the restriction \(p|_{V_\alpha}\) is a bijection onto \(U\), so the fiber \(p^{-1}(y)\) contains exactly one element in each slice \(V_\alpha\):\[|p^{-1}(y) \cap V_\alpha| = 1 \quad \text{for all } \alpha \in J.\]
1.4. Since the slices \(\{V_\alpha\}_{\alpha \in J}\) are pairwise disjoint and cover \(p^{-1}(U)\):\[|p^{-1}(y)| = \sum_{\alpha \in J} |p^{-1}(y) \cap V_\alpha| = |J|.\]
1.5. Setting \(y = b\) gives \(|p^{-1}(b)| = |J|\).
1.6. Thus \(|p^{-1}(y)| = |p^{-1}(b)| = |J|\) for all \(y \in U\).
2. The set of \(b\) with \(|p^{-1}(b)| = |p^{-1}(b_0)|\) is all of \(B\):Proof:
2.1. Let \(k = |p^{-1}(b_0)|\), and define the level set:\[S_k = \ts{b \in B \mid |p^{-1}(b)| = k}.\]
2.2. \(S_k\) is non-empty because \(b_0 \in S_k\).
2.3. \(S_k\) is open in \(B\): for any \(b \in S_k\), 1 provides an evenly covered open neighborhood \(U\) of \(b\) with \(|p^{-1}(y)| = |p^{-1}(b)| = k\) for all \(y \in U\), so \(U \subseteq S_k\).
2.4. The complement \(B \setminus S_k = \bigcup_{m \neq k} S_m\) is a union of open sets, hence open in \(B\). Thus \(S_k\) is closed.
2.5. Since \(B\) is connected and \(S_k\) is a non-empty clopen subset of \(B\), we must have \(S_k = B\).
3. Q.E.D.By step 2, \(|p^{-1}(b)| = k\) for every \(b \in B\).