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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.

Thanks in advance

T. Eskin
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Seth
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  • prolly should be merged with : http://math.stackexchange.com/questions/179290/there-is-no-c1-function-f-mapping-an-open-interval-in-mathbbr-onto-ope – still_learning Apr 11 '13 at 21:44

1 Answers1

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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 $f : M \rightarrow N$.

Willie Wong
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achille hui
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  • Thank you for the answer. I would also like to know is it possible to do the question without resorting to Morse-Sard Theorem. Suppose I did not know that f(R) has measure zero in R^2 and I all i know are both the implicit and inverse function theorem. – Seth Jan 25 '13 at 15:55
  • @WillieWong

    Thank you for editing my post to be more readable. I am not very good at typing math on online forums.

     – Seth Jan 25 '13 at 16:13
  • @Seth: you should probably thank Thomas, not me. Cheers. – Willie Wong Jan 25 '13 at 17:01
  • @ThomasE

    Thanks for editing. When i saw two people that had edited the post, i thought Willie Wong had did the editing first. Anyways, thank you both.

     – Seth Jan 25 '13 at 17:11
  • @Seth: Sorry. I don't know other proof that is not essentially using Morse-Sard theorem in disguise. – achille hui Jan 25 '13 at 17:45
  • @achillehui

    That is okay if you don't know. I am surprised that such a simple statement requires such powerful theorem to prove it.

     – Seth Jan 25 '13 at 18:07
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    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 02:56
  • @achillehui

    Yes, i have been taught Baire. I am not sure how it would help in this situation. When i was looking at this question, all i thought about were the implicit and inverse function theorems for the case where the dimension of domain and codomain are not equal.

     – Seth Jan 27 '13 at 01:48
  • @Seth: stumble across this Why isn't R2 a countable union of ranges of curves? which answer this question using "Baire category theorem". – achille hui Jan 29 '13 at 23:39