Let \(p: X \to B\) be a covering map (not necessarily regular); let \(G\) be its group of covering transformations.
Show that the action of \(G\) on \(X\) is properly discontinuous.
Let \(\pi: X \to X/G\) be the projection map. Show there is a covering map \(k: X/G \to B\) such that \(k \circ \pi = p\).
Solution
Assume \(X\) is connected, as in the classification of covering spaces, and let \(G\) be the group of covering transformations of \(p:X\to B\).
Fix \(x\in X\), and choose an evenly covered neighborhood \(U\) of \(p(x)\). Let \(V\) be the sheet over \(U\) containing \(x\). If \(g\in G\) and \[ g(V)\cap V\ne\varnothing, \] then \(g(V)\) and \(V\) are sheets over the same \(U\), hence they are equal. Since \(p|_V\) is injective and \(pg=p\), this forces \(g|_V=1\). A covering transformation that agrees with the identity at one point (indeed on an open set) is the identity on connected \(X\). Thus \[ g(V)\cap V=\varnothing\qquad(g\ne1). \] So the action is properly discontinuous.
Define \[ k:X/G\to B,\qquad k([x])=p(x). \] This is well defined since every \(g\in G\) satisfies \(pg=p\). If \(\pi:X\to X/G\) is the orbit projection, then \[ k\pi=p. \] Because \(\pi\) is a quotient map and \(p\) is continuous, \(k\) is continuous.
Let \(U\subset B\) be evenly covered by \(p\), say \[ p^{-1}(U)=\coprod_{i\in I}V_i. \] The group \(G\) permutes the sheets \(V_i\). Choose one representative \(V_j\) from each \(G\)-orbit of sheets. By part (a), no nonidentity element stabilizes a sheet. Therefore \(\pi|_{V_j}\) is injective, and \[ k:\pi(V_j)\longrightarrow U \] is a homeomorphism with inverse \(\pi\circ(p|_{V_j})^{-1}\).
Moreover, \[ k^{-1}(U)=\coprod_{[j]}\pi(V_j), \] where \([j]\) ranges over the \(G\)-orbits of sheets. These sets are open and pairwise disjoint: their inverse images under the quotient map are unions of sheets. Thus \(U\) is evenly covered by \(k\). Hence \[ \boxed{k:X/G\to B\text{ is a covering map and }k\pi=p.} \]