Suppose \(X,X_1,X_2\) are path-connected spaces, \(p:X_1\to X\) and \(q:X_2\to X\) are (surjective) covering maps, and \(f:X_1\to X_2\) is a continuous map with \(q\circ f=p\). Show that \(f\) is also a covering map.
Solution
For \(x\in X\), choose a connected open neighborhood \(U\ni x\) evenly covered by both \(p\) and \(q\): take the connected component containing \(x\) of the intersection of evenly covered neighborhoods for \(p\) and \(q\). This component is open when \(X\) is locally connected, and an open subset of an evenly covered set is evenly covered.
1. If \(V\) is a sheet of \(p^{-1}(U)\), then \(f(V)\) lies in a unique sheet \(W\) of \(q^{-1}(U)\), and \(f|_V=(q|_W)^{-1}\circ p|_V\colon V\to W\) is a homeomorphism.
Since \(q\circ f=p\), \(f(V)\subseteq q^{-1}(U)\). The sheets of \(q^{-1}(U)\) are pairwise disjoint open subsets covering \(q^{-1}(U)\), and \(f(V)\) is connected because \(V\cong U\) is connected, so \(f(V)\) lies in one sheet \(W\). Since \(p|_V\colon V\to U\) and \(q|_W\colon W\to U\) are homeomorphisms and \(q\circ f=p\), we get \(f|_V=(q|_W)^{-1}\circ p|_V\).
2. For every sheet \(W\) of \(q^{-1}(U)\), \(f^{-1}(W)\) is empty or a disjoint union of sheets of \(p^{-1}(U)\), each mapped homeomorphically onto \(W\).
By step 1, each sheet of \(p^{-1}(U)\) either maps homeomorphically onto \(W\) or is disjoint from \(f^{-1}(W)\), and \(f^{-1}(W)\subseteq p^{-1}(U)\).
3. \(f(X_1)\) is open and closed in \(X_2\).
If \(y\in X_2\), choose \(U\) as above for \(x=q(y)\) and let \(W\) be the sheet of \(q^{-1}(U)\) containing \(y\). By step 2, either \(W\subseteq f(X_1)\) or \(W\cap f(X_1)=\varnothing\). So both \(f(X_1)\) and its complement are open.
4. Q.E.D.
\(f(X_1)\) is nonempty and, by step 3, clopen in the connected space \(X_2\), so \(f\) is surjective. By step 2, every point of \(X_2\) has an open neighborhood \(W\) evenly covered by \(f\).