(This question relates to my incomplete answer at https://math.stackexchange.com/a/892212/168832.)
Is the following true (for all n)?
"If $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable and satisfies $\det(f'(x)) = 0$ for all $x$, then $f$ is not injective."
If so, what's the most elementary proof you can think of?
It is clearly true for $n=1$. In lieu of a proof for the general case, I'll accept answers for other small $n$.
I have a simple intuitive argument: pick a path in $\mathbb R^n$ such that at each point of the path, it points along some vector in the kernel of $Df$ at that point. (Remember, $\det(f'(x)) = 0$ for all $x$.) Now take the integral of the directional derivative (along the curve) of $f$ over the curve. It describes a difference between two values in the range of $f$, and it should come out to zero (QED). I have a problem showing that such a path exists and is suitable for the purpose described.
Note: "elementary" means stuff that comes before chapter 3 in Spivak's "Calculus on Manifolds". However, note also that Spivak seems to assume that integrals and elementary facts about functions in one variable are available to the reader. Here's a list of things that were not covered in Spivak at the point where the problem came up: constant rank theorem, implicit function theorem. The inverse function theorem was introduced in the same chapter as the problem was given, so that would be ok to use.
Note: This is not quite the initial problem from Spivak, and does not necessarily need to be proved to solve the original problem (see the link).
Best Answer
Very **non**elementary proof:
If $f$ were injective, it would be an open map by invariance of domain.
But by Sards theorem, the set of critical values (in this case, this is equal to the range of $f$) is a null-set, contradiction.
EDIT: Note that Sard's theorem is indeed applicable, because $f : \Bbb{R}^n \to \Bbb{R}^n$, see http://en.m.wikipedia.org/wiki/Sard's_theorem