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The following is an exercise taken from "Manifolds and Differenial Geometry" by Jeffrey M. Lee.

Let $\widetilde M$ and M be (connected) $C^r$ manifolds. Let $f: \widetilde M \to M$ be a proper map which is a local diffeomorphism from $\widetilde M$ onto $M$.

Prove: $f$ is a covering map.

Attempt:

Let $x \in M$. $\{x\}$ is compact hence $f^{-1}(x)$ is compact. $f$ is a local diffeomorphism so every $y \in f^{-1}(x)$ is contained in some open neighborhood $U_y$ that is mapped $C^r$-diffeomorphically to $f(U_y)$ that is: $\left.f\right|_{U_y} : U_y \to f(U_y)$ is a $C^r$- diffeomorphism. It follows that $f^{-1}(x) \cup U_y$ = $\{y\}$.

$\{U_y| y \in f^{-1}(x)\}$ is an open cover of $f^{-1}(x)$ which has a finite subcover. Let $\{U_i\}_{ 1\le i \le k}$ be a finite subcover. ($U_i$ denoting the set that contains $y_i)$.

By the preceding arguments $f^{-1}(x)=\{y_1,y_2, \dots,y_k \}$.

This is where i got stuck...

I can't seem to find the appropriate neighborhood of $x$ that will be evenly covered by it's preimage. I know it can't be just $\bigcap_{i=1}^k f(U_i)$...

Any help would be appreciated.

Side note: Only recently did I start to read about manifolds and to find a simple exercise that's so out of my reach is really off putting. Would you say I'm better off putting the book aside and picking up an easier one? (It would be a shame since so far i really like it).

Saal Hardali
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    Look under the header "Related" on the right side of the page, the question with 19 up votes and try to imitate the proof (using properness instead of compactness). – Moishe Kohan Jul 08 '14 at 17:59
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    This is not completely simple. One place you can find a proof is Proposition 4.46 in my Introduction to Smooth Manifolds (2nd ed.). [The other J. M. Lee.] – Jack Lee Jul 08 '14 at 20:01
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    "the question with 19 up votes" is not a time-invariant definition, is better a link :) http://math.stackexchange.com/questions/45990/when-is-a-local-homeomorphism-a-covering-map?rq=1 – user126154 Jul 10 '14 at 08:58

2 Answers2

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In the case of manifolds, you can use the characterization of covering map via the path-lifting property:

$f$ is a covering if for every $x\in M$, for every continuous path $\sigma:[0,1]\to M$ with $\sigma(0)=x$, and for every $\tilde x\in\tilde M$ that prjects to $x$, there is a unique lift $\tilde\sigma:[0,1]\to \tilde M$ that projects to $\sigma$ (i.e. $f(\tilde\sigma(t))=\sigma(t)\ \forall t$).

Now, the properness of $f$ plus the local diffeo porperty ensure the path-lifting property:

Let $x,\tilde,\sigma$ be as above. Let $D\subset[0,1]$ the set of times for which $\tilde\sigma$ exists: $D=\{t\in[0,1]: \tilde\sigma$ is defined on $[0,s]$ for every $s\leq t\}$.

We show that $D=[0,1]$ by showing that it is open and closed and non-empty. Clearly $D\neq\emptyset$ because it contains $0$ by definition of $\tilde x$. The local diffeo property of $f$ implies that $D$ is open.

To see that $D$ is closed we use properness. Let $T=\sup D$ so that $\bar D=[0,T]$ is compact. By properness $f^{-1}([0,1])$ is compact. Thus its connected components are compacts. Let $S$ be the connected component containing $\tilde x$. $S$ is compact hence is closed. Since $T=\sup D$ there is a sequence $t_n\to T$ with $t_n\in D$. Let $s_n=\tilde\sigma(t_n)$ be the corresponding sequence in $S$. As $S$ is compact $s_n$ has an adherence point $s$ in $S$. By continuity $f(s)=\sigma(T)$. The local diffeo property of $f$ now concludes that $T\in D$, whence $\bar D=D$ and $D$ is closed.

user126154
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    Thanks! but i'm afraid this is a bit out of my reach. The book doesn't mention path lifting before this exercise. I'll come back to this answer after being more familiar with the notion. It's always nice to have an additional perspective. – Saal Hardali Jul 11 '14 at 08:37
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Manifolds are compactly generated and a proper map between compactly generated spaces is closed. After establishing that $f$ is closed Sam's answer to this question completes the proof.

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Saal Hardali
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