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 \(y_0\in Y\), choose an evenly covered open neighborhood \(U\) of \(f(y_0)\) with \(p^{-1}(U)=\coprod_{\alpha\in A}U_\alpha\) and each \(p_\alpha=p|_{U_\alpha}\colon U_\alpha\to U\) a homeomorphism, and put \(V=f^{-1}(U)\), an open neighborhood of \(y_0\). For \(\alpha\in A\) let \(W_\alpha=\widetilde Y\cap(V\times U_\alpha)\).
1. \(q\) is continuous and surjective.
\(q\) is the restriction of the projection \(Y\times\widetilde X\to Y\). For \(y\in Y\), surjectivity of \(p\) gives \(\widetilde x\) with \(p(\widetilde x)=f(y)\), and \(q(y,\widetilde x)=y\).
2. The \(W_\alpha\) are pairwise disjoint open subsets of \(\widetilde Y\) with \(q^{-1}(V)=\coprod_{\alpha\in A}W_\alpha\).
\(V\times U_\alpha\) is open in \(Y\times\widetilde X\), and the \(U_\alpha\) are pairwise disjoint. If \((y,\widetilde x)\in q^{-1}(V)\), then \(p(\widetilde x)=f(y)\in U\), so \(\widetilde x\) lies in exactly one \(U_\alpha\).
3. Each \(q|_{W_\alpha}\colon W_\alpha\to V\) is a homeomorphism.
Define \(r_\alpha\colon V\to W_\alpha\) by \(r_\alpha(y)=(y,p_\alpha^{-1}(f(y)))\); it lands in \(\widetilde Y\) because \(p(p_\alpha^{-1}(f(y)))=f(y)\), and it is continuous because its coordinates are. \(q(r_\alpha(y))=y\), and for \((y,\widetilde x)\in W_\alpha\), \(p_\alpha(\widetilde x)=f(y)\) gives \(r_\alpha(y)=(y,\widetilde x)\). So \(r_\alpha\) is a continuous inverse of \(q|_{W_\alpha}\).
4. Q.E.D.
By steps 2 and 3, \(V\) is evenly covered by \(q\). Since \(y_0\) was arbitrary and \(q\) is continuous and surjective by step 1, \(q\) is a covering map.