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

Question

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?

You should make your title more precise: the existence of a map is not a very interesting point :-) — Mariano Suárez-Álvarez, 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

Asaf Karagila · Accepted Answer · 2012-08-31T16:40:56.233

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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.

edited Aug 31 '12 at 16:40

answered Aug 31 '12 at 16:16

Asaf Karagila

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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
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@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
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@clark: Well, you can always choose to remove the points which are not at the end. – Asaf Karagila Aug 31 '12 at 16:40
yes that works:) {+1} – clark Aug 31 '12 at 16:41
And if we drop the continuity hypothesis? – Glen Wheeler Sep 01 '12 at 20:23
@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. – Asaf Karagila Sep 01 '12 at 20:26
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. – Glen Wheeler Sep 03 '12 at 13:17
@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. – Asaf Karagila Sep 03 '12 at 14:02
@AsafKaragila, Warner's Differentiable Manifold asks if there is a smooth map from $\mathbb{R}^2 \to \mathbb{R}$ which is injective. I think continuity is sufficent from your answer. – permutation_matrix Dec 23 '22 at 10:39
@permutation_matrix Yes, that is correct. – Asaf Karagila Dec 23 '22 at 10:51

score 1 · Answer 2 · answered Sep 22 '17 at 11:13

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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.

answered Sep 22 '17 at 11:13

Perturbative

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Even from $R^m$ into $R^n$ for $n<m$ – mathemather Sep 26 '18 at 02:03

score 0 · Answer 3 · 2025-07-01T07:27:14.373

Another simple proof is this: as $f(0,1)\not = f(0,0)$, according to the mean value theorem, there is a value $t\in (0,1)$ such that $\partial_y f(0,t)\not = 0$.

By the implicit function theorem, there is a neighborhood $V$ of $0$ in $\mathbb R$ a neighborhood $W$ of $t$ in $\mathbb R$ and a function $g:V\to W$ such that for $(x, y)\in V\times W$, the equation $f(x, y) = f(0, t)$ is equivalent to $y = g(x)$.

This implies that the equation $f(x, y) = f(0, t)$ has infinitely many solutions $(x, g(x))$, contradicting injectivity.

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

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