$f$ has full rank $\implies$ $f$ is a covering map

Question

Exercise 1.84 (d) of [GHL], it asks us to prove that

If two $n$-dimensional manifolds $\tilde M$ and $M$ are compact and if $M$ is connected, show that a smooth map $f : \tilde M\to M$ is a covering map if and only if $f$ has maximum rank at each point of $M$.

(I believe the author meant 'at each point of $\tilde M$'.)

I am trying to prove the direction of $f\text{ has full rank }\implies f\text{ is a covering map}$.

Question 1: I was only able to prove the statement by assuming $\tilde M$ is second countable. If we assume so, then the assumption of $M$ being compact seems unnecessary.

Question 2: are there any counter-examples when $\tilde M$ is not compact?

Definitions

Manifolds here are always connected, smooth and Hausdorff.

$p:\tilde M\to M$ is a covering map if:

1. $p$ is smooth and surjective
2. for any $m\in M$, there exists a neighborhood $U$ of $m$ in $M$ with $p^{-1}(U)=\bigcup_{i\in I} U_i$, where the $U_i$ are disjoint open subsets of $\tilde M$ such that, for each $i\in I$, $p\vert_{U_i}:U_i\to U$ is a diffeomorphism.

My attempt

Surjectivity of $f$: since the cardinality of $f^{-1}(m)$ is locally constant on $M$ and $M$ is connected, $f^{-1}(m)$ has the same cardinality for all $m\in M$. If $f$ is not surjective, then $\text{card}(f^{-1}(m))=0$ for some $m$ would lead to $\text{card}(f^{-1}(m))=0$ for all $m$, which is absurd.

Finiteness of $\text{card}(f^{-1}(m))$: note that $\tilde M$, being compact and second countable, is sequentially compact. Suppose $\text{card}(f^{-1}(m))$ is not finite, i.e. there exists distinct $\{x_i\}_{i=1}^\infty$ such that $f(x_i)=m$, then by compactness a subsequence $x_{i_k}$ converges to $x$. Clearly $df_x$ cannot have full rank, because in no neighborhood of $x$ is $f$ injective.

Then write $f^{-1}(m)=\{x_1,\dots,x_N\}$. Since $f$ has full rank at any $x_i$, there exists open neighborhood $U_i$ of $x_i$ such that $f\vert_{U_i}:U_i\to f(U_i)$ is a diffeomorphism. $U_i$'s can be made pairwise disjoint since $\tilde M$ is Hausdorff.

Let $V=\bigcap_{i=1}^N f(U_i)$ open in $M$. Let also $V_i=f^{-1}(V)\cap U_i$ open in $\tilde M$. Then, obviously $f^{-1}(V)=\bigcup_{i=1}^N V_i$ where $V_i$'s are pairwise disjoint open subsets of $\tilde M$, and one can check that $f\vert_{V_i}\to V$ is a diffeomorphism (it suffices to check surjectivity). Therefore $f$ is a covering map.

In this proof, we did not use the compactness of $M$, at the cost of assuming $\tilde M$ is second countable (or sequentially compact).

[GHL]: Riemannian Geometry by S. Gallot, D. Hulin and J. Lafontaine.

Answer 1

It seems like I have figured out the solution.

• We are allowed to assume the compact manifolds $M$ and $\tilde M$ are sequentially compact, because $\text{locally Euclidean Haudorff }+\text{compact}\implies\text{second countable}$ (link), and $\text{second countable}+\text{compact}\implies \text{sequentially compact}$ (link).

Claim:

If

• $\tilde M$ and $M$ are compact,
• $M$ is connected,
• $f:\tilde M\to M$ is smooth, and
• $df$ is invertible everywhere,

then $f$ is a covering map.

Proof:

Surjectivity: (adopting @Matsmir argument) Since $f$ is smooth and $df$ is invertible everywhere, $f$ is a local diffeomorphism. In particular, $f(\tilde M)$ is open. Due to compactness of $\tilde M$, $f(\tilde M)$ is compact and hence closed. By connectedness of $M$, $f(\tilde M)=M$. $\blacksquare$

---

For any $y\in M$, let $C(y)$ be the cardinality of $f^{-1}(y)$. Define also ($n\ge 1$)$$C_n=\{y\in M:C(y)\ge n\}$$

• $C(y)<+\infty$ for all $y\in M$: Suppose $f^{-1}(y)$ is not a finite set for some $y$, i.e. there exists distinct $\{x_i\}_{i=1}^\infty$ such that $f(x_i)=y$, then by sequential compactness a subsequence converges to $x$. Clearly $df_x$ cannot have full rank, because in no neighborhood of $x$ is $f$ injective. $\blacksquare$

• $C_n$ are open: Given $y\in C_n$, there exists distinct $x_1,\cdots,x_n\in f^{-1}(y)$. For each $x_i$, there exists an open neighborhood $U_i$ such that $f\vert_{U_i}:U_i\to f(U_i)$ is a diffeomorphism, thanks to the smoothness of $f$ and invertibility of $df$. By Hausdorff-ness, assume the $U_i$'s are disjoint. Then, for any $y'\in V:=\bigcap^N_{i=1}f(U_i)$, there exists distinct $x_1',\cdots,x_n'\in f^{-1}(y')$ where $x_i'\in U_i$, and thus $y'\in C_n$ too. In other words, $V\subset C_n$. $\blacksquare$

• $C_n$ are closed: Given distinct $y_k\in C_n$ that converges to $y^*$, we have distinct $x^k_1,\cdots,x^k_n\in f^{-1}(y_k)$.

Now we construct $n$ distinct points in $f^{-1}(y^*)$:

1. By sequential compactness, let $z_1$ be a limit point of $\{x^i_j\}$. Clearly, $f(z_1)=y^*$.

2. Due to the invertibility of $df$, there exists an open neighborhood $O_1$ of $z_1$ where $f$ is injective. Then, at most one element from each $f^{-1}(y_i)$ lies inside $O_1$.

3. Thus, $\{x^i_j\}\setminus O_1$ is still an infinite set, and it is possible to find another limit point $z_2$ not in $O_1$.

Note that by removing from $\{x^i_j\}$ any points in $O_1$, we at most lose one element in each $f^{-1}(y_i)$. Thus, this process can be repeated for at least $n$ times to give distinct $z_1,\cdots,z_n\in f^{-1}(y^*)$, so $y^*\in C_n$. $\blacksquare$

• $C(y)$ is constant: By finiteness of $C(y)$ and surjectivity of $f$, $M=\bigcup_{n=1}^\infty C_n$. Since $C_n$'s are clopen and $M$ is connected, there exists unique $N\ge 1$ where $M=C_N$. In other words, $C(y)=N$ for all $y\in M$. $\blacksquare$

• $f$ is a covering map: For any $y\in M$, $f^{-1}(y)=\{x_1,\dots,x_N\}$. By invertibility of $df$, there exists open neighborhood $W_i$ of $x_i$ such that $f\vert_{W_i}:W_i\to f(W_i)$ is a diffeomorphism. $W_i$'s can be made pairwise disjoint since $\tilde M$ is Hausdorff.

Let $R=\bigcap_{i=1}^N f(W_i)$ open in $M$. Let also $Q_i=f^{-1}(R)\cap W_i$ open in $\tilde M$. We have $f^{-1}(R)=\bigcup_{i=1}^N Q_i$: for the non-trivial inclusion, given any $x\in f^{-1}(R)$, $f(x)\in R=\bigcap_{i=1}^N f(W_i)$, so one can find $x_i\in W_i$ such that $f(x_i)=f(x)$, $i=1,\cdots,N$. Note that there are exactly $N$ elements in $\tilde M$ that are mapped by $f$ to the same value, thus $x=x_i\in W_i$ for some $i$.

Then, one can check that $f\vert_{Q_i}:Q_i \to R$ is a diffeomorphism - it suffices to check surjectivity: given $y\in R=\bigcap_{i=1}^N f(W_i)$, there exists $x_i\in W_i$ such that $f(x_i)=y$. Therefore $f$ is a covering map. $\blacksquare$