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$D:\tilde{M}\rightarrow \Omega$ is a local isometry (developing map) onto a connected manifold $\Omega$. $\tilde{M}$ is simply connected, has constant sectional curvature and there exists a deck group $\pi(M)$ on $\tilde{M}$.

$\tilde{M}$ is the universal cover of a compact manifold $M$. The inverse set $D^{-1}(x)$ can be shown to be finite.

I want to show that $D$ is in fact a covering map, and to do so I would have to show that every inverse set $D^{-1}(x)$ has the same cardinality, i.e. the same number of finite points.

Is there a counterexample to this, or how could this be done?

Vertex
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  • is $M=\Omega$? If not, what is $\Omega$ good for? – Thomas Jan 24 '16 at 15:48
  • $M\not=\Omega$, as $\Omega \subset X$. The problem is that $D$ might not be surjective onto $X$.

    Perhaps I should have formulated $D:\tilde{M}\rightarrow D(\tilde{M})=\Omega\subset X$!?

     – Vertex Jan 24 '16 at 15:53

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I disagree with the claim that $D^{-1}(x)$ can be shown to be finite. Why not take the covering map $\mathbb{R}^2 \to S^1 \times S^1$? (And put the flat metric on the torus so it this map is a local isometry, and satisfies your curvature conditions.)

What is your condition about the deck group? That is not clear.

If you assume that $\tilde{M}$ is compact this result is easier. In that case, the hint would be: find a neighborhood for each point in the fiber which is a local isometry. Try to refine these neighborhoods to produce local trivializations. (A proper local diffeomorphism between connected manifolds is a covering map. Really this is a topological fact: When is a local homeomorphism a covering map?)

Maybe you should look in Do Carmo, in the section on space forms. He has the following theorem, which is related in spirit to your question:

(Thm. 4.1 in chapter 8) Let $M^n$ be a complete Riemannian manifold with constant sectional curvature $K$. Then the universal covering, with the coverin gmetric, is isoemetric to hyperbolic space, Euclidean space, or $S^n$, with their natural constant curvature metrics, and curvatures $-1$, $0$ and $1$ respectively.

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Elle Najt
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  • So as I perhaps should have mentioned, $M$ is not a Riemannian manifold, but a Lorentzian one. M is also compact. The argument for finite inverse set goes as follows: If a sequence ${\tilde{x}_n}$ exists in $\tilde{M}$, we can find a convergent sequence $k_n\tilde{x}_n$ in $\tilde{M}$. Since $D^{-1}(x)$ cannot have a convergent subsequence, it follows that this will be a contradiction. – Vertex Jan 24 '16 at 15:45
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    I read the statement "$D^{-1}(x)$ can shown to be finite" as a given fact. If that's true there is no point to disagree with that. – Thomas Jan 24 '16 at 15:46
  • Yes Thomas, in the given setting this is true. The setting is the (G,X)-structure, or Ehresmann structure, which is this:

    $D:\tilde{M}\rightarrow X$ is a local isometry to $X$, a homogeneous space with isometry group $G$, and

    $h:\pi(M)\simeq D\rightarrow h(D)=\Gamma \subset G$ is a group homomorphism, such that $D$ is equivariant:

    $D(k(p))=h(k)D(p)$, where $k\in D$.

     – Vertex Jan 24 '16 at 15:50
  • @Vertex I see. Well, in that case I don't know. Compactness is used in the proof of the result I linked, not just finite fibers. Good luck! (You will get better answers from more knowledgeable users if you are explicit about your hypothesis in the question.) – Elle Najt Jan 24 '16 at 15:53
  • @Thomas I disagree with your disapproval. Disagreement is always an opportunity to learn something. – Elle Najt Jan 24 '16 at 15:54