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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.

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    By the same argument, $df$ cannot have rank 2 on an open nonempty subset. Therefore, $df$ has rank $\le 1$ everywhere. Now argue that $df$ cannot have rank 1 on an open nonempty subset... – 2017-01-06
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    An other kind of proof is the following: You just take away a line in $\mathbb{R}^2$ and a line in $\mathbb{R}^3$. But then $\mathbb{R}^2\setminus \mbox { line}$ is not path-connected, whereas $\mathbb{R}^3\setminus \mbox { line}$ is. – 2017-01-06

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Let me expand on my comment: I thought it was clear, Georges comment clearly shows that my comment was clearly unclear.

Claim. If $f: R^n\to R^{n-1}$ is $C^1$-smooth, then $f$ cannot be injective.

Proof. Let $k\le n-1$ denote the maximum of all $i$'s such that there exists a point $p\in R^n$ such that $rank(df_p)=i$. Then, by continuity of the determinant function (applied to $k\times k$ submatrices of $df$), there exists an open subset $U\subset R^n$ such that $df$ has rank $k$ on $U$. Now, apply the constant rank theorem to the restriction $f|U$ and conclude that $f: U\to R^{n-1}$ cannot be injective. qed

Note. No need to appeal to Sard's theorem in this argument.

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    Dear @Moishe: this is indeed quite correct and pleasantly elementary:+1. (I have deleted my previous comment) – 2017-01-06
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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 $C^2$]