Let \(p: E \to X\) be a covering map such that \(\pi_1(X, x_0)\) is countable. Show that if \(X\) is regular with a countable basis, so is \(E\). [Hint: Let \(\mathcal{B}\) be a countable basis for \(X\) consisting of path-connected sets. Let \(\mathcal{C}\) be the collection of path components of \(p^{-1}(B)\), for \(B \in \mathcal{B}\). Compare Exercise 6 of §53.]
Solution
Choose a countable basis \(\mathcal B\) for \(X\) consisting of path-connected evenly covered open sets; this is possible by local path connectedness and refinement of a countable basis.
For \(B\in\mathcal B\), the components of \(p^{-1}(B)\) are precisely the sheets over \(B\). Fix \(b\in B\). Each sheet contains exactly one point of the fiber \(p^{-1}(b)\). Since \(E\) is connected, points of this fiber correspond to cosets of \[ p_*\pi_1(E,e_0) \] in \(\pi_1(X,x_0)\), so the fiber is at most countable because \(\pi_1(X,x_0)\) is countable. Hence every \(p^{-1}(B)\) has at most countably many sheets.
Let \(\mathcal C\) be the collection of all these sheets as \(B\) ranges over \(\mathcal B\). It is a countable union of countable families, hence countable. Since \(p\) restricts to a homeomorphism from each sheet onto its basis element \(B\), the family \(\mathcal C\) is a basis for \(E\). Thus \(E\) is second countable.
It remains to prove regularity. Let \(e\in E\) and let \(O\) be an open neighborhood of \(e\). Choose an evenly covered open set \(U\subset X\) such that the sheet \(V\) containing \(e\) satisfies \[ e\in V\subset O. \] Because \(X\) is regular, choose an open neighborhood \(W\) of \(p(e)\) with \[ \overline W\subset U. \] Let \[ V_W=(p|_V)^{-1}(W). \] Then \(V_W\) is open, contains \(e\), and \[ \overline{V_W}\subset (p|_V)^{-1}(\overline W)\subset V\subset O. \] Therefore \(E\) is regular. Hence \(E\) is regular with a countable basis.