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?
Nonexistence of an injective $C^1$ map between $\mathbb R^2$ and $\mathbb R$
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$\begingroup$
real-analysis
general-topology
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0[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
2 Answers
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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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0@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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0Even from $R^m$ into $R^n$ for n
– 2018-09-26