$f:\mathbb{R}^n\to\mathbb{R}^n$ is diffeomorphism iff $f$ is local diffeomorphism and proper map

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?

Thanks.

• 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. Sep 23, 2019 at 14:46

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 $$0. Because $$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.