show there is no C^1 homeomorphism from R^3 to R^2 Show that there is no C^1 homeomorphism from $R^3$ to $R^2$.
I am fully aware that in general $R^m$ and $R^n$ are not homeomorphism by homology theory. I wonder if we add the condition $C^1$ we can have a proof using differential calculus.
Here is what i've tried just in case someone ask: 
I figured that by rank theorem the differential of such map can't have rank 2. But i don't see any contradictions if the rank is 1 or 0. I don't think i go through the right way. 

a) If the linear map $df_A:\mathbb R^3\to \mathbb R$ has rank $2$ at a single point $A\in \mathbb R^3$, then locally near $A$   the rank theorem says that up to a change of coordinates $f(x,y,z)=(x,y)$, so that $f$ is not injective and thus even less a homeomorphism.      
b) If $f$ is nowhere of rank $2$, then  all points of $\mathbb R^3$ are critical .
But then $f(\mathbb R^3)$ has zero Lebesgue mesure by Sard's theorem and thus $f$ is hilariously far from being surjective anf even farther from being a homeomorphism.
[The use of Sard requires the slightly stronger hypothesis  that $f$ be at least  $C^2$] 
