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Prove that a local homeomorphism between compact, connected, Hausdorff spaces is a covering map of finite degree.

Attempt at solution: Let $f:M\rightarrow N$ be the local homeomorphism. Since $N$ is Hausdorff, the singleton ${n}$ is closed, hence $f^{-1}(n)$ is a closed subset of a compact space hence compact. Each point in the inverse image has a neighborhood $U$ such that $f|_U :U \rightarrow f(U)$ is a homeomorphism.

Hence in particular, the points in the inverse image must all have disjoint open neighborhoods (since, say if every neighborhood of the point $x$ also contained the point $y$ and they were mapped to the same point $n$ then it would not be a homeomorphism). Taking intersections with these neighborhoods, we get disjoint homeomorphically mapped neighborhoods ${U_i}$.

Further, since the primage is compact and the points in the preimage all have disjoint open neighborhoods, there must be finitely many points in the preimage. Now, let $V=\bigcap_{i=1}^n f(U_i)$ which is an open subset of $N$ as the finite intersection of open sets (since each $f(U_i)$ is open by definition of local homeomorphism). Then this is our trivially covered neighborhood.

One question I have is, is it possible to relax the Hausdorff condition? It seems not, if we consider the space two intersecting disks, with a topology defined by the two disks and their intersection being open, and a map sending each disk to an open disk with the indiscrete topology. The two spaces are compact and connected, and the map f a local homeomorphism but not a covering map. However, this was a quals question where the Hausdorff condition was not given so I would like to make sure.

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  • $\begingroup$ If I understand your construction the two space are Hausdorff (any clarification for the two spaces and the map is welcome). $\endgroup$ – Hamou Aug 15 '14 at 20:49
  • $\begingroup$ So imagine the disks as unit two disks, but not with the usual topology. So the topology on the bottom space is indiscrete, so there is only one open set, hence no two points have disjoint open neighborhoods. On the top you have two intersecting disks, and the open sets are the disks themselves and their intersection where the intersection is nonempty (so imagine two 2 disks intersecting like a Venn diagram). This satisfies the axioms for a topology, but it is not Hausdorff, since two points not in the intersection and in one disk do not have disjoint neighborhoods. Then project down. $\endgroup$ – TheManWhoNeverSleeps Aug 15 '14 at 20:56

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