Let \(f:U\to V\) be a proper local homeomorphism between open sets \(U,V\subset\mathbb C\). Prove that \(f\) is a covering map. Here proper means that \(f^{-1}(K)\) is compact whenever \(K\subset V\) is compact.
Solution
1. Under the convention that a covering map is surjective, the statement as written needs an additional hypothesis, for example that \(V\) is connected and \(U\) is nonempty, or more directly that \(f\) is surjective.
Let \[ V=D(0,1)\sqcup D(3,1), \qquad U=D(0,1), \] and let \(f:U\hookrightarrow V\) be the inclusion. It is a local homeomorphism. It is also proper: \(U\) is both open and closed in \(V\), so for every compact \(K\subset V\) the set \[ f^{-1}(K)=K\cap U \] is closed in the compact set \(K\), hence compact. But \(f\) is not surjective onto \(V\), so it is not a covering map under the stated convention.
2. For any proper local homeomorphism \(f:U\to V\), the image \(f(U)\) is both open and closed in \(V\).
It is open because every local homeomorphism is an open map.
To prove that it is closed, let \(y_n=f(x_n)\) be a sequence in \(f(U)\) converging to \(y\in V\). Choose \(r>0\) such that the closed disk \[ K=\overline{D(y,r)} \] lies in \(V\). For all sufficiently large \(n\), we have \(y_n\in K\), hence \(x_n\in f^{-1}(K)\). Properness makes \(f^{-1}(K)\) compact, so a subsequence \(x_{n_j}\) converges to some \(x\in f^{-1}(K)\). Continuity gives \[ f(x)=\lim_j f(x_{n_j})=y. \] Thus \(y\in f(U)\), proving that \(f(U)\) is closed.
Consequently \(f(U)\) is a union of connected components of \(V\). In particular, if \(V\) is connected and \(U\ne\varnothing\), then \(f(U)=V\).
3. Every fiber of \(f\) is finite.
Fix \(y\in f(U)\). Since \(\{y\}\) is compact and \(f\) is proper, the fiber \[ F=f^{-1}(y) \] is compact. Local homeomorphy makes \(F\) discrete: for each \(x\in F\) there is a neighborhood on which \(f\) is injective, and that neighborhood meets \(F\) only at \(x\). A compact discrete space is finite, so write \[ F=\{x_1,\ldots,x_m\}. \]
4. Every \(y\in f(U)\) has an evenly covered neighborhood.
For each point \(x_i\) of the finite fiber in 3, choose pairwise disjoint open neighborhoods \(W_i\subset U\) such that \[ f|_{W_i}:W_i\longrightarrow O_i \] is a homeomorphism onto an open neighborhood \(O_i\) of \(y\).
Choose \(r>0\) with \(\overline{D(y,r)}\subset V\) and put \[ K=\overline{D(y,r)}, \qquad C=f^{-1}(K)\setminus\bigcup_{i=1}^m W_i. \] The set \(f^{-1}(K)\) is compact by properness, and \(C\) is closed in it, so \(C\) is compact. No point of \(C\) maps to \(y\), because the whole fiber over \(y\) is contained in the union of the \(W_i\). Hence \(f(C)\) is compact, therefore closed in \(V\), and \(y\notin f(C)\).
Now set \[ N=D(y,r)\cap\bigcap_{i=1}^m O_i\setminus f(C). \] This is an open neighborhood of \(y\). If \(x\in f^{-1}(N)\), then \(f(x)\in K\), while \(f(x)\notin f(C)\); hence \(x\notin C\) and therefore \(x\in W_i\) for some \(i\). Thus \[ f^{-1}(N) =\bigsqcup_{i=1}^m\bigl(W_i\cap f^{-1}(N)\bigr). \] Because \(N\subset O_i\), each restriction \[ f:W_i\cap f^{-1}(N)\longrightarrow N \] is a homeomorphism. Therefore \(N\) is evenly covered.
5. Hence a proper local homeomorphism is a covering map onto its image, and the source claim follows whenever \(V\) is connected and \(U\) is nonempty.
By 2, the image is a union of connected components of \(V\), and by 4 every point of the image has an evenly covered neighborhood. Thus \[ f:U\longrightarrow f(U) \] is a covering map. If \(V\) is connected and \(U\ne\varnothing\), then 2 gives \(f(U)=V\), so \(f:U\to V\) is a covering map.