I have to prove that a proper local diffeomorphism in $\mathbb{R}^n$ is a diffeomorphism. I'm trying to show that it is injective but I just have that the preimage of every $y\in\mathbb{R}^n$ is a finite set.

Could anybody help me?


  • 1
    $\begingroup$ Try to prove that it is a covering map. If this is the case, then since $\mathbf R^n$ is simply connected, it is in fact a diffeomorphism. $\endgroup$ Sep 23, 2019 at 14:46

1 Answer 1


First remember that a proper map is closed. Because $f$ is a local diffomrphism its open hence $f(\mathbb{R}^n)$ is a closed and open set so $f(\mathbb{R}^n)=\mathbb{R}^n$. Now we will want to show that $f$ is injective.

Let's show that if $f$ is local diffeomorphism and proper map we can lift any path uniquely once we choose a base point. Let $\alpha:[0,l] \to \mathbb{R}^n$ be a path. Once we choose a base point $p$ such that $f(p)=\alpha(0)$, we can use the fact that $f$ is a local diffeomorphism to lift $\alpha$ to $\tilde{\alpha}:[0,a) \to \mathbb{R}^n$ for some $0<a \le l$. Because $f$ is a local diffeomorphism the set $A$ of $a \in [0,l]$ such that $\tilde{\alpha}:[0,a) \to \mathbb{R}^n$ is open in $[0,l]$. if we can show that its closed we have shown that $A=[0,l]$ because $[0,l]$ is connected.

Assume to the contrary then exist $t_0 \in [0,l]$ such that $A=[0,t_0)$. Let $\{t_i\} \subset [0,t_0)$ be such that $\lim_{i \to \infty }t_i=t_0$. Since $f$ is proper and $\alpha([0,l])$ is compact we can deduce that $\{\tilde{\alpha}(t_i)\} \subset K$ where $K$ is compact. Let $r$ be an accumulation point of $\{\tilde{\alpha}(t_i)\} $ and let $V$ be a neighborhood of $r$ small enough such that $f_{\restriction V}$ is a diffeomorphism. Then $\tilde{\alpha}(t_0 ) \in f(V)$ and, by continuity, there exists an interval $I \subset [0, l]$, $t_0 \in I$, such that $\alpha(I) \subset f (V)$. Choose an index $n$ such that $\tilde{\alpha} \in V$ and consider the lifting $g$ of $\alpha$ on $I$ passing through $r$. The liftings $g$ and $\tilde{\alpha}$ coincide on $[O, t_n) \cap I$,because $f_{\restriction V}$ is bijective. Therefore, $g$ is an extension of $\tilde{\alpha}$ to $I$, hence $\tilde{\alpha}$ is defined at $t_0$ and $t_0 \in A$. This shows that $A=[0,l]$.

Now I get lazy and state a theorem form do Carmo curves and surfaces 387 (Its a similar construction to what I have done above) Let $B$ be arcwise connected and let $\pi: \tilde{B} \to B$ be a local homeomorphism with the property of lifting arcs. Let $\alpha_0, \alpha_1: [0, l] \to B$ be two arcs of $B$ joining the points $p$ and $q$, let $H: [0, l]×[0, 1] \to B$ be a homotopy between $\alpha_0$ and $\alpha_1$, and let $\tilde{p} \in \tilde{B}$ be a point of $\tilde{B}$ such that $\pi(\tilde{p}) = p$. Then there exists a unique lifting $\tilde{H}$ of $H$ with origin at $\tilde{p}$.

To complete the proof: Take $p_1 ,p_2$ such that $f(p_1)=f(p_2)=b$. Let $c$ be an arc from $p_1$ to $p_2$ so $f\circ c$ is a loop based in $b$. $\mathbb{R}^n$ is simply connected so $f\circ c$ is homotopy equivalent to the constant loop $\alpha(t)=b$ for all $t \in [0,1]$. But this shows that $c$ is homotopy equivalent to a constant path and this can only happen if $p_1=p_2$. This shows that $f$ is injective. As differentiability is a local property we can deduce that $f^{-1}$ is a differentiable function hence $f$ is a diffeomorphism.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .