Given maps \(X \to Y \to Z\) such that both \(Y \to Z\) and the composition \(X \to Z\) are covering spaces, show that \(X \to Y\) is a covering space if \(Z\) is locally path-connected, and show that this covering space is normal if \(X \to Z\) is a normal covering space.
Solution
Write \[ p:X\to Y, \qquad q:Y\to Z, \qquad r=q\circ p:X\to Z. \]
1. For every \(z\in Z\) there is a path-connected open neighborhood \(U\) of \(z\) that is evenly covered by both \(q\) and \(r\).
Choose evenly covered neighborhoods \(U_q\) and \(U_r\) of \(z\) for the two covering maps. Their intersection is an open neighborhood of \(z\). Since \(Z\) is locally path connected, choose a path-connected open neighborhood \[ z\in U\subseteq U_q\cap U_r. \] Restricting an evenly covered neighborhood to a connected open subset preserves the sheet decomposition, so \(U\) is evenly covered by both maps.
2. Let \(W\) be a sheet of \(r^{-1}(U)\).Then \(p(W)\) lies in a single sheet \(V\) of \(q^{-1}(U)\).
The restriction \[ r|_W:W\to U \] is a homeomorphism, and \(U\) is path connected, so \(W\) is path connected. The image \(p(W)\) is therefore connected.
Since \[ q(p(W))=r(W)=U, \] one has \[ p(W)\subseteq q^{-1}(U), \] which is a disjoint union of open sheets. A connected subset of this disjoint union lies in one sheet, say \(V\).
3. The restriction\[ p|_W:W\to V \]is a homeomorphism.
On \(W\) one has \[ q|_V\circ p|_W=r|_W. \] Both \[ q|_V:V\to U \quad\text{and}\quad r|_W:W\to U \] are homeomorphisms. Hence \[ p|_W=(q|_V)^{-1}\circ r|_W, \] which is a homeomorphism from \(W\) onto \(V\).
4. For each sheet \(V\) of \(q^{-1}(U)\),\[ p^{-1}(V) \]is a disjoint union of sheets \(W\) of \(r^{-1}(U)\), each mapped homeomorphically onto \(V\).
Every point of \(p^{-1}(V)\) lies in a unique \(r\)-sheet \(W\) over \(U\). By 2, the entire connected sheet \(W\) maps into one \(q\)-sheet; since it contains a point mapping into \(V\), that sheet must be \(V\). Conversely every \(r\)-sheet that maps into \(V\) lies in \(p^{-1}(V)\). The homeomorphism statement is 3.
5. Therefore \(p:X\to Y\) is a covering map.
The sheets \(V\) arising from all path-connected neighborhoods \(U\) in 1 form an open cover of \(Y\). By 4, every such \(V\) is evenly covered by \(p\).
6. Now assume \(r:X\to Z\) is a normal covering.Let \(x_1,x_2\in X\) satisfy\[ p(x_1)=p(x_2)=y. \]Then there is a deck transformation \(h\) of \(r\) with\[ h(x_1)=x_2. \]
The two points lie in the same fiber of \(r\) because \[ r(x_i)=q(p(x_i))=q(y). \] For a connected normal covering, the deck transformation group acts transitively on each fiber.
7. Every such deck transformation \(h\) is a deck transformation of \(p\) whenever it sends one point of a \(p\)-fiber to another point of the same \(p\)-fiber.
The maps \[ p\circ h, \qquad p:X\to Y \] are both lifts through \(q\) of the same map \(r:X\to Z\), since \[ qph=rh=r=qp. \] At \(x_1\) they agree: \[ (ph)(x_1)=p(x_2)=y=p(x_1). \] Since \(X\) is path connected for a normal covering, uniqueness of lifts gives \[ p\circ h=p. \] Thus \(h\) is a deck transformation of \(p\).
8. The deck group of \(p\) acts transitively on every fiber, so \(p\) is normal.
For arbitrary \(x_1,x_2\) in one \(p\)-fiber, 6 supplies a deck transformation \(h\) of \(r\) carrying \(x_1\) to \(x_2\), and 7 shows that the same \(h\) is a deck transformation of \(p\). This is the transitivity criterion for normal coverings.