Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be continuous and let there exist $\alpha > 0$ such that $||f(\mathbf{x}) - f(\mathbf{y})|| \geq \alpha || \mathbf{x} - \mathbf{y}||$ for all $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$. Prove that $f$ is one-one, onto and that $f^{-1}$ is continuous.
One-one is trivial. It is onto-ness that I can't show.
Write $S = f(\mathbb{R}^n)$. Using sequential continuity, it is possible to show that $S$ is closed. If I could show $S$ is open, I would be done, but I can't.
Also, writing $g(\mathbf{x}) := \dfrac{f(\mathbf{x})}{\alpha}$, the condition can be converted to that of proper expansive map, $||g(\mathbf{x}) - g(\mathbf{y})|| \geq || \mathbf{x} - \mathbf{y}||$. But since $\mathbb{R}^n$ is not compact, I cannot use the result here.
Any help is appreciated!
EDIT: As commented below, the Invariance of Domain theorem seems to work in this case, but that result does not use the expansive-type condition provided here (except for showing the injectivity), and so it appears that an easier proof would be possible.
Actually the Invariance of Domain theorem solves your problem, but not the way you think : indeed, if you apply it with $U = \mathbb{R}^n$, you only get that $f$ is a homeomorphism on its image, and you don't get that its image is $\mathbb{R}^n$.
Here is the way you can use it correctly, and your expansion condition is really necessary :
For $r > 0$, denote by $B(a,r)$ the open ball of center $a \in \mathbb{R}^n$ and radius $r$. For all $r > 0$, the image $f(B(0,r))$ is open, by the Invariance of Domain theorem. So $f(B(0,r)) \cap B(f(0), \alpha r)$ is open in $B(f(0), \alpha r)$.
But $f(\overline{B(0,r)}) \cap B(f(0), \alpha r) = f(B(0,r)) \cap B(f(0), \alpha r)$ : indeed, if $||x|| = r$, then $||f(x)-f(0)|| \geq \alpha r$, so $f(x) \notin B(f(0), \alpha r)$. So you get that $f(B(0,r)) \cap B(f(0), \alpha r)$ is also closed in $B(f(0), \alpha r)$.
By connectivity, $f(B(0,r)) \cap B(f(0), \alpha r)$ is open and closed in $B(f(0), \alpha r)$, and not empty (because it contains $f(0)$), so it is equal to $B(f(0), \alpha r)$. In other words you have, for all $r > 0$, $$ B(f(0), \alpha r) \subset f(B(0,r)) $$
Obviously this implies that $f$ is surjective.