(Jan ’06) Let \(p:\widetilde X\to X\) be a covering map, and \(f:Y\to X\) be a continuous map. Define \[\widetilde Y=\{(y,\widetilde x)\in Y\times\widetilde X:f(y)=p(\widetilde x)\}\subseteq Y\times\widetilde X\] with the subspace topology inherited from \(Y\times\widetilde X\) and define \(q:\widetilde Y\to Y\) by \(q(y,\widetilde x)=y\). Show that \(q\) is also a covering map.
Solution
Fix \(y \in Y\), let \(U \subseteq X\) be an evenly covered open neighborhood of \(f(y)\) with \(p^{-1}(U) = \bigsqcup_\alpha \widetilde U_\alpha\) and each \(p|_{\widetilde U_\alpha}\colon \widetilde U_\alpha \to U\) a homeomorphism, and put \(V = f^{-1}(U)\), an open neighborhood of \(y\) because \(f\) is continuous. For each \(\alpha\) let \(s_\alpha = (p|_{\widetilde U_\alpha})^{-1}\colon U \to \widetilde U_\alpha\) and \(W_\alpha = \widetilde Y \cap (V \times \widetilde U_\alpha)\).
1. \(q^{-1}(V) = \bigsqcup_\alpha W_\alpha\), a disjoint union of open subsets of \(\widetilde Y\).
Each \(W_\alpha\) is open in \(\widetilde Y\) because \(V \times \widetilde U_\alpha\) is open in \(Y \times \widetilde X\), and the \(W_\alpha\) are pairwise disjoint because the \(\widetilde U_\alpha\) are. If \((y', \widetilde x) \in \widetilde Y\) with \(y' \in V\), then \(p(\widetilde x) = f(y') \in U\), so \(\widetilde x\) lies in exactly one \(\widetilde U_\alpha\), and \((y', \widetilde x) \in W_\alpha\). Conversely every \(W_\alpha\) lies in \(q^{-1}(V)\).
2. For each \(\alpha\), \(q|_{W_\alpha}\colon W_\alpha \to V\) is a homeomorphism.
The map \(\sigma_\alpha\colon V \to W_\alpha\), \(\sigma_\alpha(y') = (y', s_\alpha(f(y')))\), is continuous, lands in \(\widetilde Y\) because \(p(s_\alpha(f(y'))) = f(y')\), and satisfies \(q \circ \sigma_\alpha = \operatorname{id}_V\). If \((y', \widetilde x) \in W_\alpha\), then \(\widetilde x \in \widetilde U_\alpha\) and \(p(\widetilde x) = f(y')\), so \(\widetilde x = s_\alpha(f(y'))\) and \(\sigma_\alpha(q(y', \widetilde x)) = (y', \widetilde x)\). Thus \(\sigma_\alpha\) is a continuous inverse of the continuous map \(q|_{W_\alpha}\).
3. Q.E.D.
By steps 1 and 2, every \(y \in Y\) has an open neighborhood \(V\) evenly covered by \(q\), so \(q\) is a covering map.