Show that if \(p:X\to\CP^n\) is a covering space map, then \(X\) is homeomorphic to \(\CP^n\).
Solution
1. As printed, the statement is false if covering spaces are allowed to be disconnected.
Let \[ X=\CP^n\sqcup\CP^n \] and define \[ p:X\longrightarrow\CP^n \] to be the identity map on each component. Every sufficiently small open set \(U\subseteq\CP^n\) has inverse image \[ p^{-1}(U)=U\sqcup U, \] with each copy mapped homeomorphically onto \(U\). Thus \(p\) is a two-sheeted covering map.
But \(X\) is disconnected whereas \(\CP^n\) is connected, so \[ X\not\cong\CP^n. \] Hence the conclusion \(X\cong\CP^n\) requires \(X\) to be connected.
2. The space \(\CP^n\) is simply connected for every \(n\ge0\).
For \(n=0\), \(\CP^0\) is a point.
For \(n\ge1\), complex projective space has the standard CW decomposition \[ \CP^n=e^0\cup e^2\cup e^4\cup\cdots\cup e^{2n}. \] In particular its \(1\)-skeleton is a point. By cellular approximation, every loop \[ S^1\longrightarrow\CP^n \] is homotopic to a cellular map and therefore to a map into the \(1\)-skeleton. Such a map is constant. Hence \[ \pi_1(\CP^n)=0. \]
3. Assume now that \(X\) is connected and that\[ p:X\longrightarrow\CP^n \]is a covering map.Then every fiber of \(p\) contains exactly one point.
The base \(\CP^n\) is locally path-connected, so the covering space \(X\) is locally path-connected as well. Since \(X\) is connected, it is therefore path-connected.
Fix \(y\in\CP^n\) and suppose \[ x_0,x_1\in p^{-1}(y). \] Choose a path \[ \widetilde\gamma:[0,1]\longrightarrow X \] from \(x_0\) to \(x_1\). Then \[ \gamma=p\circ\widetilde\gamma \] is a loop at \(y\). By 2, \(\gamma\) is homotopic relative endpoints to the constant loop at \(y\).
Lift such a homotopy through the covering, starting at \(x_0\). Homotopy lifting implies that the endpoints of the lifts of the two endpoint-homotopic paths agree. The lift of the constant loop beginning at \(x_0\) is constant and ends at \(x_0\), while \(\widetilde\gamma\) ends at \(x_1\). Therefore \[ x_1=x_0. \] Thus every fiber has at most one point; by surjectivity of a covering map, every fiber has exactly one point.
4. If \(X\) is connected, the covering map \(p\) is a homeomorphism.
By 3, \(p\) is bijective. A covering map is a local homeomorphism. A bijective local homeomorphism is an open continuous bijection, hence a homeomorphism. Therefore \[ \boxed{X\cong\CP^n.} \]
5. Consequently, a covering map \(X\to\CP^n\) is a homeomorphism exactly when \(X\) is connected.
Step 4 proves that a covering map with connected total space is a homeomorphism, and a space homeomorphic to \(\CP^n\) is connected; step 1 gives a disconnected covering that is not a homeomorphism.