Give the definition of a covering space \(\widehat{X}\) (and covering map \(p:\widehat{X}\to X\)) for a topological space \(X\). Prove that if \(X\) is simply connected, and \((\widehat{X},p)\) is a covering space for \(X\), then \(p\) is a homeomorphism. Make your proof as elementary as possible.
Solution
1. A covering map is a continuous surjection\[ p:\widehat X\to X \]such that every \(x\in X\) has an open neighborhood \(U\) for which\[ p^{-1}(U)=\coprod_{\alpha\in A}V_\alpha \]is a disjoint union of open subsets of \(\widehat X\), and every restriction\[ p|_{V_\alpha}:V_\alpha\to U \]is a homeomorphism.The pair \((\widehat X,p)\) is then called a covering space of \(X\), and \(U\) is said to be evenly covered.
This is the standard definition. In particular, a covering map is a local homeomorphism.
2. With this definition, the conclusion printed in the problem is false unless an additional hypothesis is imposed on \(\widehat X\).
Take \[ X=\mathbb R \] and \[ \widehat X=\mathbb R\amalg\mathbb R. \] Let \(p\) restrict to the identity map on each copy of \(\mathbb R\). Then \(p\) is a two-sheeted covering map of the simply connected space \(\mathbb R\), but it is not injective and hence is not a homeomorphism.
Thus the printed statement is false for arbitrary covering spaces. Its conclusion holds when \(\widehat X\) is path-connected.
3. Assume henceforth that \(X\) is simply connected and that \(\widehat X\) is path-connected.Then every fiber of \(p\) contains at most one point.
Fix \(x\in X\) and suppose \[ \widehat x_0,\widehat x_1\in p^{-1}(x). \] Since \(\widehat X\) is path-connected, choose a path \[ \alpha:[0,1]\to\widehat X \] from \(\widehat x_0\) to \(\widehat x_1\). Its projection \[ \beta=p\circ\alpha \] is a loop in \(X\) based at \(x\).
Because \(X\) is simply connected, \(\beta\) is homotopic relative to its endpoints to the constant loop \(c_x\). We use the homotopy-lifting property for covering maps: if two paths in the base are homotopic relative to their endpoints, then their lifts with the same initial point have the same terminal point.
The path \(\alpha\) is the lift of \(\beta\) beginning at \(\widehat x_0\). The constant path \(c_{\widehat x_0}\) is the lift of \(c_x\) beginning at the same point. Therefore their terminal points agree: \[ \widehat x_1 = \alpha(1) = c_{\widehat x_0}(1) = \widehat x_0. \] Hence \(p^{-1}(x)\) has at most one point.
4. Under the path-connectedness hypothesis, \(p\) is a homeomorphism.
By the definition in 1, \(p\) is surjective. By 3, every fiber has at most one point, so \(p\) is injective. Thus \(p\) is bijective.
A covering map is a local homeomorphism, hence an open map: if \(O\subseteq\widehat X\) is open and \(\widehat x\in O\), choose an evenly covered neighborhood \(U\) of \(p(\widehat x)\) and the sheet \(V\) over \(U\) containing \(\widehat x\). Then \[ p(O\cap V) \] is open in \(U\) and therefore open in \(X\); these sets cover \(p(O)\) as \(\widehat x\) ranges over \(O\). So \(p(O)\) is open.
Consequently a bijective covering map is a continuous open bijection, and its inverse is continuous. Therefore \[ \boxed{p:\widehat X\xrightarrow{\cong}X} \] is a homeomorphism.
5. Hence a covering map onto a simply connected space is a homeomorphism when its total space \(\widehat X\) is path-connected; without that hypothesis, 2 is a counterexample.
This is exactly 2 together with 4.