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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.

  • 0
    Linearly dependent functions/equations when Jacobian zero – 2017-01-20
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    Since any idea is welcome here is a naive approach: the volume of image of any region is an integral of determinant of Jacobian and so it is zero. So the volume of the image of the whole space is zero and the image can not be $\mathbb{R}^n$ - it's volume is infinite. But this is not rigorous at all. – 2017-01-20

1 Answers 1

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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.