Try to prove that is not proper using as little special properties of as possible, the general idea should follow from there.
Could anyone help me solve the following problem ?
Show that a covering map is proper if and only if it is finite-sheeted.
This is the problem 11-10 from "Introduction to Topological Manifolds" by John M. Lee. The definition of a covering map in this book is the following.
A covering map is a continuous surjective map such that is connected and locally path-connected, and every point of has an evenly covered
My first attempt was to show that the image of any sequence in that diverges to infinity diverges to infinity in . But, I noticed that for this to be a sufficient condition for properness of , should be a second countable Hausdorff space, and that no additional condition is assumed for other than connectedness and local path-connectedness.
I don't have a clue any more.
Thank you for your help.
Your mapping is not proper since the preimage of a point in is an infinite discrete set in . I understand that to prove a continuous map being proper, it is necessary to show that it is a closed map in addition to showing that it has compact fibers. The former is harder for me.
The definition of proper maps in the second edition is the same.
I referred to Proposition 4.93 (c).
Proposition 4.93 (Sufficient Conditions for Properness)
Suppose and are topological spaces, and is a continuous map.
(c) If is a closed map with compact fibers, then is proper.
You don't need that proposition: The idea is that you associate with a covering of the preimage of a compact set in X a finite refinement (here's where you use the finite-sheeted-ness) using that the set in X is compact and the special open sets one can define associated with points in X.
Edit: I'm assuming you have trouble with finite-sheeted implies proper.
But, I'm not sure about the converse too.
What troubles me is the following.
Given a compact set in and an open covering of ,
since is an open covering of , one can choose from it a finite covering of . But, is not necessarily a covering of .
Okay, here's a bigger hint: Let compact and a covering of the inverse image. For every there is an such that where the are open and disjoint and is an homeomorphism (this is just the definition of a covering map). Try to shrink the so that they are inside one but in a way that they are still preimages of a set and have analogous properties to . Now use the compacteness of to finish the argument.
Thanks to your guidance, I managed to come up with something resembling a proof of the first half of the problem.
Let compact and a covering of . One can construct a covering of satisfying the condition that
(i) is evenly covered by for each ,
(ii) for each there exists such that ,
For the condition (i), one can use the fact that every connected open
subset of an evenly covered open subset is itself evenly covered.
It follows from the compactness of that there exists a finite set such that . Since are evenly covered by , we have where are disjoint open sets such that is a homeomorphism. Let be a subset of defined by
Finally, the index set can be constructed by choosing one element arbitrarily from that is not empty for . Then is a finite covering of .
I'm looking forward to your comment.
[I corrected a mistake in the definition of the index set .]
I noticed a serious error in my "proof".
By the definition of , only one necessary open set in is guaranteed to be chosen for a finite cover of . Other open sets necessary in the finite cover on the other sheets can be lost, if they intersect with but don't contain it.
I revised my proof.
Let be compact and an open cover of . For each there exists an open set evenly covered by . Let be the preimages of by . For each there exists such that .
Define by . Then, is a cover of . Since is compact, there exists a finite subcover of it. Let it denoted by . Since is evenly covered by , we have where are disjoint open sets such that is a homeomorphism. Let be a subset of defined by
. Finally, the index set can be constructed by choosing one element arbitrarily from for each . Then is a finite cover of .