Let \(q: X \to Y\) and \(r: Y \to Z\) be covering maps; let \(p = r \circ q\). Show that if \(r^{-1}(z)\) is finite for each \(z \in Z\), then \(p\) is a covering map.
Solution
Fix \(z\in Z\). Since \(r:Y\to Z\) is a covering map, choose an evenly covered neighborhood \(U\) of \(z\): \[ r^{-1}(U)=\bigsqcup_{i=1}^m V_i, \] where \(m=|r^{-1}(z)|<\infty\) and each \(r|_{V_i}:V_i\to U\) is a homeomorphism.
Let \(y_i\in V_i\) be the unique point over \(z\). Since \(q:X\to Y\) is a covering map, for each \(i\) choose an evenly covered neighborhood \(W_i\subset V_i\) of \(y_i\). Because \(r|_{V_i}\) is a homeomorphism, \(r(W_i)\) is an open neighborhood of \(z\) in \(U\). Set \[ U'=\bigcap_{i=1}^m r(W_i). \] This is open because the intersection is finite. Replace each \(W_i\) by \[ W_i'=W_i\cap r^{-1}(U'). \] Then \(q^{-1}(W_i')\) is a disjoint union of open slices, each mapped homeomorphically by \(q\) onto \(W_i'\); composing with \(r|_{W_i'}\) maps each such slice homeomorphically onto \(U'\).
Moreover \[ p^{-1}(U')=q^{-1}(r^{-1}(U'))=\bigsqcup_{i=1}^m q^{-1}(W_i'), \] so these slices partition \(p^{-1}(U')\). Hence \(U'\) is evenly covered by \(p=r\circ q\), and \(p\) is a covering map.