# Existence of an injective $C^1$ map between $\mathbb R^2$ and $\mathbb R$

I am getting bored waiting for the train so I'm thinking whether there can exist a $C^1$ injective map between $\mathbb{R}^2$ and $\mathbb{R}$. It seems to me that the answer is no but I can't find a proof or a counterexample... Can you help me?

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You should make your title more precise: the existence of a map is not a very interesting point :-) –  Mariano Suárez-Alvarez Aug 31 '12 at 16:35
Here is a related question. I think all answers apply to your question. –  David Mitra Aug 31 '12 at 17:19

If $f\colon\mathbb R^2\to\mathbb R$ is continuous then its image is connected, that is an interval in $\mathbb R$. Note that this is a non-degenerate interval since the function is injective.
However if you remove any point from $\mathbb R^2$ it remains connected, however if we remove a point whose image is in the interior of the interval then the image cannot be still connected if the function is injective.
why if you remove one point the image cannot be connected? Could not be like this $f( \mathbb{R^2} / {x_0}) =(a,b)$ and $f(x_0)=b$ then $f(\mathbb{R^2}) = (a,b]$. I can see that the same argument works if you remove three points from $\mathbb{R^2}$ I have some problems with one. –  clark Aug 31 '12 at 16:34