TITLE: Proving differentiable mapping is onto
QUESTION [10 upvotes]: Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a differentiable mapping such that there is a $C > 0$ satisfying $C|x-y| \leq |f(x) - f(y)|$ for all $x,y \in \mathbb{R}^n$.  Prove that $\det f'(x) \neq 0$ for all $x \in \mathbb{R}^n$ and then deduce that $f(\mathbb{R}^n) = \mathbb{R}^n$.
I am really stuck on this, but it's a problem I've been working on for a while.  I have used the first inequality to show that $f$ is one-to-one.  Am I on the right track, or is there something additional I can determine from this inequality?

REPLY [3 votes]: 1) If $\det f'(x) = 0$, then there is a vector $h$ such that $f'(x)h = 0$. What happens now, from the definition of the derivative?
2) The easiest way I can see to do this is use invariance of domain. Since you've shown $f$ is 1-1, you know that $f(\mathbb{R}^n)$ is an open set. (I suppose you secretly don't need invariance of domain, only the IFT.) Now, I claim you can also show it's closed, using your condition. (Recall some facts about Cauchy sequences for this.) Hence it must be all of $\mathbb{R}^n$.