$C^1$ surjective map from $\Bbb{R}\to\Bbb{R}^2$? I now there is a continuous surjective map from $\Bbb{R}\to\Bbb{R}^2$ thanks to Peano curve.
My question is simple: does there exist a $C^1$ surjective map from $\Bbb{R}\to\Bbb{R}^2$ ? I think that the answer is no and I have seen this long time ago but I was too young to understand the proof. Unfortunately I cannot think of a proof now.
Any idea ?
 A: The image of a $C^1$ map from $\Bbb R$ to $\Bbb R^2$ has Lebesgue measure zero, so it cannot be surjective, nor have non-empty interior.
A: Notation: $f''S=\{f(x):x\in S\}$ when $S$ is a subset of the domain of $f.$
Let $f:\Bbb R\to \Bbb R^2$ be a continuous surjection.  For $n\in \Bbb Z,$ each $[n,n+1]$ is compact so $f''[n,n+1]$ is compact and therefore closed in $\Bbb R^2.$ Since $\Bbb R^2=\cup_{n\in \Bbb Z}f''[n,n+1],$ the Baire category theorem implies that some $f''[n,n+1]$ has non-empty interior. So for some $n$ we have $f''[n,n+1]\supset [a,a+b]\times [a',a'+b']$  with positive $b,b'$. 
To simplify the notation we will assume WLOG (by a change of scale and a change of variable)  that $f''[0,1]\supset [0,1]^2.$
For $1<n\in \Bbb N,$ consider $[0,1]^2$ as an  $n\times n$ checkerboard of closed sub-squares  with sides of length $1/n.$  Let $C_n$ be the set of centers of these sub-squares. (The point is that $C_n$ has $n^2$ members and if $(u,v),(u',v')$ are distinct members of $C_n$ then $|u-u'|\geq 1/n$ or $|v-v'|\geq 1/n$...or both).
Let $B_n \subset [0,1]$ where $B_n$ has $n^2$ members and $f''B_n=C_n.$  Choose distinct $x_n,x'_n\in B_n$ such that $|x_n-x'_n|\leq (n^2-1)^{-1}.$  Let $f(x_n)=(u_n,v_n)$ and $f(x'_n)=(u'_n,v'_n).$ We have $|u_n-u'_n|\geq 1/n$ or $|v_n-v'_n|\geq 1/n$ (or both). 
Let $A$ be an infinite subset of $\Bbb N$ \ $\{1\}$ such that $\forall n\in A\;(|u_n-u'_n|\geq 1/n)$ or $\forall n\in A\; (|v_n-v'_n|\geq 1/n\} .$ 
WLOG assume $\forall n\in A\; (|u_n-u'_n|\geq 1/n).$
Suppose $f$ were continuously differentiable. Then with $f(x)=(f_1(x),f_2(x)),$ the function $f'_1(x)$ is continuous. Now  $\frac {f_1(x_n)-f_1(x'_n)}{x_n-x'_n}= f'_1(y_n)$ for some $y_n$ between $x_n$ and $x'_n$ by the MVT.  
For all $n\in A$ we have $|f'_1(y_n)|=$ $\frac {|u_n-u'_n|}{|x_n-x'_n|}\geq$ $ \frac {1/n}{(n^2-1)^{-1}}=n-n^{-1}.$ 
But $A$ is infinite; therefore  $\{|f'_1(x)|: x\in [0,1]\}$ is unbounded  above, which is impossible if $f'_1$ is continuous.
