# Non surjectivity of $C^1$ function. [duplicate]

I am trying to figure out how to do the following question, but i seems to not have any success.

If $f:\mathbb{R}^1\to \mathbb{R}^2$ is of class $C^1$ mapping, show that $f$ does not carry $\mathbb{R}^1$ onto $\mathbb{R}^2$.

I know that if suppose that $f$ is onto, then let $g$ be a right inverse of $f$ where $f(g(x,y))=x$. If I differentiate $f(g(x,y))=x$, using the chain rule, i get some expression, but there is a problem because i am not sure if $g(x,y)$ is itself differentiable to begin with.

Alternatively, if i let the mapping of $f$ be written as $f(t)=(\phi_1(t), \phi_2(t))$, then the Jacobian matrix $Df(t)$ is a $2$ by $1$ matrix. So the column rank of $Df(t)$ is $1$. But how can i use this information to conclude that $f$ is not onto, since if $f$ were onto in the first place, won't the column rank of its Jacobian be equal to the dimension of the function's codomain.

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## marked as duplicate by vonbrand, Micah, Amzoti, rschwieb, Asaf KaragilaApr 11 '13 at 23:20

prolly should be merged with : math.stackexchange.com/questions/179290/… – MSEoris Apr 11 '13 at 21:44

The easiest way to prove the non-surjectivity is to show $f(\mathbb{R})$ has measure zero in $\mathbb{R}^2$.
If $|f'(x)| < M$ for $x \in [a,b]$, one can cover $f([a,b])$ by a square of length $M|b-a|$ (hint: mean value theorem). If one subdivide $[n,n+1]$ by $N$ small intervals, one get: $$\mu(f([n,n+1])) \le \sup( |f'(x)| : x \in [n,n+1] )^2 / N$$ Taking $N \rightarrow +\infty$, we get $\mu(f([n,n+1])) = 0$. Being a countable union of sets of measure zero, $f(\mathbb{R})$ itself has measure zero and hence cannot equal to $\mathbb{R}^2$.
FYI: The general form of these sort of results is known as Morse-Sard theorem, it is very helpful in middle of a proof where you need to pick a point outside the image of some $C^k$ map $f : M \rightarrow N$.
Have you been taught "Baire category theorem"? In particular, the statement that $\mathbb{R}^2$ cannot be covered by countable union of nowhere dense set. At least for the case $f'(a) \ne 0$, it is relatively easy to see $f([a-\epsilon,a+\epsilon])$ is nowhere dense for sufficiently small $\epsilon$. Of course, "Baire Category theorem" is another very powerful tool. – achille hui Jan 26 '13 at 2:56