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Proper local homeomorphisms and covering maps

I had never really thought about what more it takes beyond being a local homeomorphism for a map to be a covering map, except of course, thinking it shouldn’t be much, but I learned one possible answer today1 in McMullen’s course on Dynamics and Moduli Spaces: mainly properness2! I wrote this little note to work out what additional conditions are needed.

First of all, let’s give examples of local homeomorphisms that are not covering maps: take any space \(X\) and any proper open subset \(U\), the inclusion \(U \to X\) is an example. If it feels like cheating because the map is not surjective, consider an open cover \(\{U_i\}\) of \(X\) and the map \(\coprod_i U_i \to X\) from the disjoint union of the \(U_i\) which is the inclusion on each \(U_i\). If that still feels like cheating because the domain of the map is not connected, just take a chunk of a covering space that covers \(X\) more than once, but less that twice, say, like \((0,7) \to S^1\), \(t \mapsto e^{i t}\). For a crazier example, take \(Y\) to be the real line with the origin doubled3, \(X = \mathbb{R}\) and \(Y \to X\) the map that just identifies the two origins.

Now the result:

If \(f : Y \to X\) is a proper local homeomorphism between locally compact Hausdorff spaces then \(f\) is a (finite) covering map.

Added [2017-07-09 Sun]: Looking for a published reference for this result? Thanks to Jeremy Brazas for pointing out Lemma 2 in this paper:

Chung-Wu Ho, A note on proper maps in Proc. AMS., Volume 51, Number 1, August 1975.

Notice that we don’t really need to assume that \(Y\) is locally compact, since it follows from \(X\) being locally compact Hasudorff and there being a local homeomorphism. However, the example of the line with the doubled origin shows that we do need to assume \(Y\) is Hausdorff. I don’t know whether the assumptions on \(X\) can be relaxed. In the proof I’ll carefully mention where the hypothesis are used, because I got a little confused when thinking about this.

Let \(x \in X\) be any point and consider its preimage \(f^{-1}(x)\). This set is compact because \(f\) is proper. And since \(f\) is a local homeomorphism, it follows the set \(f^{-1}(x)\) is also discrete. Indeed, given any \(y \in f^{-1}(x)\), there is a neighborhood \(y \in V\), such that \(f|_V : V \to f(V)\) is a homeomorphism and in particular, \(y\) is the only preimage of \(x\) inside \(V\). Combining these facts, we get that the preimage of \(x\) is finite, say, \(f^{-1}(x) = \{y_1, \ldots, y_n\}\).

Now, since \(Y\) is Hausdorff, there are pairwise disjoint open neighborhoods \(V_i\) of the \(y_i\), and by shrinking them if necessary we can assume that for each \(i\), \(f|_{V_i} : V_i \to f(V_i)\) is a homeomorphism to an open subset of \(X\) containing \(x\). Now we’d like to do something like take \(\bigcap_i f(V_i)\) to get an evenly covered neighborhood of \(x\), but we have to worry about \(f^{-1}(\bigcap_i f(V_i))\) not being wholly contained in \(\bigcup_i V_i\). So instead, take an open neighborhood \(W\) of \(x\) in \(X\) such that \(\overline{W}\) is compact and contained in \(\bigcap_i f(V_i)\) (this exists because \(X\) is locally compact and Hausdorff), and let \(U = W \setminus f(f^{-1}(\overline{W}) \setminus \bigcup_i V_i)\). Then \(U\) is the required evenly covered open neighborhood of \(x\).

Indeed, \(U\) is open because \(f(f^{-1}(\overline{W}) \setminus \bigcup_i V_i)\) is compact (by properness of \(f\)) and thus closed (since \(X\) is Hausdorrf). And \(x\in U\) because \(\bigcup_i V_i\) contains all preimages of \(x\). Finally, by construction, \(f^{-1}(U)\) is the disjoint union of the \(f^{-1}(U) \cap V_i\), each of which is homeomorphically mapped onto \(U\) by \(f\).


1

September 11, 2012

2

A map is proper if the inverse images of compact sets are compact.

3

\(Y = \mathbb{R} \cup \{0'\}\) where a basis of neighborhoods of \(0'\) is given by the sets of the form \(((-\epsilon,\epsilon) \cup \{0'\}) \setminus \{0\}\).

Omar Antolín Camarena