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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

1 Answer 1

up vote 23 down vote accepted

There is no such map.

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.

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I'd upvote this if it wouldn't spoil your nice reputation. –  Marc van Leeuwen Aug 31 '12 at 16:20
@MarcvanLeeuwen: I already got a screenshot... :-) –  Asaf Karagila Aug 31 '12 at 16:21
OK there it goes... –  Marc van Leeuwen Aug 31 '12 at 16:23
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
@clark: Well, you can always choose to remove the points which are not at the end. –  Asaf Karagila Aug 31 '12 at 16:40

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