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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 :-)2012-08-31
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    [Here](http://math.stackexchange.com/questions/116350/continuous-injective-map-f-mathbbr3-to-mathbbr?rq=1) is a related question. I think all answers apply to your question.2012-08-31

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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.2012-08-31
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    @MarcvanLeeuwen: I already got a screenshot... :-)2012-08-31
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    OK there it goes...2012-08-31
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    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.2012-08-31
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    @clark: Well, you can always choose to remove the points which are *not* at the end.2012-08-31
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    yes that works:) {+1}2012-08-31
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    And if we drop the continuity hypothesis?2012-09-01
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    @Glen: Of course, the two sets are equipotent and therefore there exists a bijection between them. However the question was about a *differentiable* function, which by definition has to be continuous.2012-09-01
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    I guess I was too subtle in my suggestion that the addition of a note to the effect of "The existence of a bijection is clear, it is the additional continuity hypothesis that is at work here." could improve the answer.2012-09-03
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    @Glen: Well, I figured that if you ask about a *continuous* injection you probably know by that point that there is *an* injection to begin with.2012-09-03
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For what its worth, there isn't even any continuous injection from $\mathbb{R}^m$ into $\mathbb{R}$ for $m > 1$

The proof follows the exact same argument as Asaf's does.

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    Even from $R^m$ into $R^n$ for $n2018-09-26