21
$\begingroup$

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?

  • 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 2

50

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.

  • 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
0

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.

  • 0
    Even from $R^m$ into $R^n$ for n2018-09-26