Non-existence of $C^1$ injective mapping $\mathbb{R}^3 \to \mathbb{R}^2$. A friend of mine did a test yesterday where it asked to prove that there does not exist a $C^1$ injective mapping $\mathbb{R}^3 \to \mathbb{R}^2$.
This is an immediate result from invariance of domain, but since this is a real analysis test (where people are being introduced to derivation in $\mathbb{R}^n$), I tried to come up with an elementary solution. However, none came to mind.
I thought about using the local form of submersions (which, by the way, I wouldn't expect in this point in the course my friend is taking anyway), but we would need to have a regular value which is on the image, and this is not given by the hypotheses, neither by Sard's theorem.
Since this was on the test, I have the feeling I may be letting something slip. My question therefore is to prove the given statement with only tools of differentiation in $\mathbb{R}^n$ (inverse function theorem, chain rule etc).
 A: Here's the most elementary argument I can find.  Suppose $f:\mathbb{R}^3\to\mathbb{R}^2$ is $C^1$.  By lower semicontinuity of rank, we can find a nonempty open subset $U\subset\mathbb{R}^3$ on which $df$ has constant rank.  If you know the constant rank theorem, you're done: since the constant rank of $df$ on $U$ is less than $3$, $f$ will not be injective on $U$.  (Incidentally, if you are assuming that students were expected to know the inverse function theorem, it seems reasonable to me that they might also know the constant rank theorem.)
Without using the constant rank theorem, you can finish the proof as follows using the Peano existence theorem for ODEs (you can use the more standard Picard existence theorem if you know that $f$ is $C^2$).  I will assume the constant rank of $df$ on $U$ is $2$; the other cases are similar.  Fix a point $p\in U$ and let $u,v,w\in\mathbb{R}^3$ be vectors such that $df_p(u)=0$ and $df_p(v)$ and $df_p(w)$ are linearly independent.  Shrinking $U$, we may assume that in fact $df_q(v)$ and $df_q(w)$ are linearly independent for all $q\in U$.  Define $h:U\to\mathbb{R}^3$ by $h(q)=u+av+bw$, where $a$ and $b$ are the unique scalars such that $df_q(u+av+bw)=0$.  Continuity of $df$ (and continuity of matrix inversion) implies that $h$ is continuous.
By the Peano existence theorem, there exists $g:(-\epsilon,\epsilon)\to U$ such that $g(0)=p$ and $g'(t)=h(g(t))$.  Since $df_q(h(q))=0$ for all $q\in U$, we find that the derivative of $f\circ g$ vanishes identically.  So $f$ is constant on the image of $g$.  Since $h$ never vanishes, $g'$ never vanishes, so this image has more than one point.  Thus $f$ is not injective.
