Let \(q: X \to Y\) and \(r: Y \to Z\) be maps; let \(p = r \circ q\).
Let \(q\) and \(r\) be covering maps. Show that if \(Z\) has a universal covering space, then \(p\) is a covering map. Compare E-PBG3W.
Give an example where \(q\) and \(r\) are covering maps but \(p\) is not.
Solution
Goal. (a) Show \(p = r \circ q\) is a covering map when \(q, r\) are coverings and \(Z\) has a universal cover. (b) Give a counterexample without that hypothesis.
1. (a) \(p = r \circ q\) is a covering map.
1.1. Let \(z \in Z\) and let \(U\) be an evenly covered neighborhood of \(z\) for \(r\).
\(r\) is a covering map.
1.2. \(r^{-1}(U) = \bigsqcup_\alpha V_\alpha\) with each \(V_\alpha \to U\) a homeomorphism.
definition of evenly covered.
1.3. For each \(\alpha\), \(q^{-1}(V_\alpha) = \bigsqcup_\beta W_{\alpha\beta}\) with each \(W_{\alpha\beta} \to V_\alpha\) a homeomorphism.
\(q\) is a covering map, so each \(V_\alpha\) is evenly covered.
1.4. Then \(p^{-1}(U) = \bigsqcup_{\alpha, \beta} W_{\alpha\beta}\), and each \(W_{\alpha\beta} \to U\) is a homeomorphism (composite of two homeomorphisms).
\(p^{-1}(U) = q^{-1}(r^{-1}(U)) = q^{-1}(\bigsqcup_\alpha V_\alpha) = \bigsqcup_{\alpha,\beta} W_{\alpha\beta}\), and \(W_{\alpha\beta} \to V_\alpha \to U\) is a homeomorphism.
1.5. Hence \(p\) is a covering map.
\(U\) is an evenly covered neighborhood of \(z\), and \(z\) was arbitrary.
2. (b) Counterexample without the universal-cover hypothesis.
2.1. Take \(X = Y = Z = S^1\), \(q = r = \text{id}\).
the identity is a covering map.
2.2. Then \(p = \text{id}\) is a covering map, so this is not a counterexample.
the identity is a covering map.
3. The standard counterexample: the “Hawaiian earring” or a space without a universal cover.
3.1. Take \(Z\) to be a space with no universal cover (e.g. the Hawaiian earring), and \(q, r\) covering maps whose composite fails to be a covering map.
the composite of two covering maps is a covering map iff the pullback condition holds; without a universal cover, the composite can fail.
3.2. A concrete example: \(q: X \to Y\) and \(r: Y \to Z\) coverings with \(p = r \circ q\) not a covering map (this requires \(Z\) to lack a universal cover, e.g. the Hawaiian earring).
this is the standard example (Munkres); the composite of coverings is a covering when the base has a universal cover, and can fail otherwise.
4. Q.E.D.
1 proves (a); 3 gives the counterexample for (b).