This is from advanced calculus.
Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable. The Jacobian of $f$ has zero determinant over $\mathbb{R}^n$. For example in 2-dimensional case, $f(x,y)=(y,y)$ is such a function. Let's suppose $n \ge 2$ since $n=1$ case gives a constant function.
My question is :is it true that $f$ cannot be 1-1?
My best attempt is that at each $x \in \mathbb{R}^n$ some directional derivative is zero, and perhaps an integral over some path will work, but I am not sure how to do it.
Any idea is welcome. Thank you very much.
Best Answer
There are (at least) two ways to prove this:
The first is "easy", but uses theorems which are more difficult: Assume that $f$ is 1-1. Then, by invariance of domain, it follows that $f$ is an open map, i.e., if $U \subset \Bbb{R}^n$ is open, then so is $f(U)$. In particular, $f(\Bbb{R}^n)$ is nonempty and open and thus has positive Lebesgue measure. But Sard's theorem shows that $f(\Bbb{R}^n)$ is a set of measure zero, contradiction.
For a more elementary proof, one might argue as follows: Let $k := \max_{x \in \Bbb{R}^n} \text{rank}(Df(x))$ be the maximal rank of $f$. It is then not too hard to see that $U := \{x \,:\, \text{rank}(Df(x)) = k\}$ is open and nonempty.
Now, the constant rank theorem (see e.g. here) shows for arbitrary $x_0 \in U$ that there is a neighorhood $V$ of $x$ and a neighborhood $W$ of $f(x)$ such that $f(V) \subset W$ and there are $C^1$ diffeomorphisms $\Phi : W \to W'$ and $\Psi : V \to V'$ such that $$ (\Phi \circ f \circ \Psi^{-1})(x_1, \dots, x_n) = (x_1, \dots, x_k, 0,...0) $$ for all $(x_1, \dots, x_n) \in V'$. Hence, $\Phi \circ f \circ \Psi^{-1}$ is not 1-1, which easily shows that $f$ is not 1-1 either.