Bijective functions from $\mathbb{R}^2$ to $\mathbb{R}$

Question

Consider a differentiable function from $\mathbb{R}^2$ to $\mathbb{R}$. Is it correct to say that a function of this type can never be bijective?

I have attempted to prove this by contradiction in the following way, but I'm not sure:

Consider the generic level set $I_z = \{ (x,y)\in \mathbb{R}^2: f(x,y) = z, z \in \mathbb{R} \}$. If $f$ is supposed bijective, then for each $z \in \mathbb{R}$ the cardinality $|I_z|$ = 1. The auxiliary function $g_z(x,y):= f(x,y) - z$ has only one zero, so its sign never changes. Whatever $g_z$'s sign is, the point $(x_z,y_z)\in \mathbb{R}^2$ such that $g_z(x_z,y_z) = 0$ has to be a maximum or a minimum of $g_z$, so $(x_z,y_z)$ has to be a critical point of $g_z$ and $f$. But repeating this for each $z\in \mathbb{R}$ makes the function $f$ constant, so $f$ is not bijective.

I've also found out that usually the implicit function theorem can be used to prove such statements, but I do not know how to use it in this case.

@anomaly i think op just wants a smooth map not a diffeomorphism — Ajin Shaji Jose, Aug 09 '24 at 12:15
@AjinShajiJose: Right, but a continuous bijection $f:\mathbb{R^2} \to \mathbb{R}$ would give a continuous injection $if:\mathbb{R}^2 \to \mathbb{R^2}$ for some convenient embedding $i:\mathbb{R} \to \mathbb{R}^2$, forcing $if(\mathbb{R}^2) = i(\mathbb{R})$ to be open in $\mathbb{R}^2$. That works without assuming $f^{-1}$ is smooth, or even continuous. — anomaly, Aug 09 '24 at 13:24
What do you think about this: https://www.youtube.com/watch?v=XcMZsF4vDbo ? — Ataulfo, Aug 09 '24 at 14:39
@Piquito That is a nice example. But I'm not sure that it is differentiable — Tom Avery, Aug 09 '24 at 15:10
@AjinShajiJose: It forces $i(\mathbb{R})$ to be open in $\mathbb{R}^2$, which it is not. — anomaly, Aug 09 '24 at 17:13
@anomaly i get what ur saying but how is $i(\mathbb{R})$ open in $\mathbb{R}^2$? — Ajin Shaji Jose, Aug 09 '24 at 17:57
@AjinShajiJose: Invariance of domain guarantees that $if(\mathbb{R}^2) = i(\mathbb{R})$ is open, being the image of the open set $\mathbb{R}^2$ under the embedding $if:\mathbb{R}^2 \to \mathbb{R}^2$. — anomaly, Aug 09 '24 at 18:00
@Tom Avery.- I believe (without be entirely sur) that it is not possible with even continuous. — Ataulfo, Aug 09 '24 at 21:54
This results hold in just the continuous category, but you can also use the constant rank theorem here (and note that the rank is either $0$ or $1$ at each point). — anomaly, Aug 10 '24 at 04:48

score 2 · Answer 1 · answered Aug 09 '24 at 16:04

2

A continuous image of a connected set is connected, Proof of "the continuous image of a connected set is connected"

If your function is bijective and you remove one point from $R^2$ then the image is not connected. So the answer is no.

answered Aug 09 '24 at 16:04

Steen82

1,258

Could you clarify? I do not understand how can this theorem be used to answer my question – Tom Avery Aug 09 '24 at 16:44
If the function is differentiable it is continuous, $R^2\setminus{p}$ is connected and since $f$ is bijective $R\setminus{f(p)}$ is not connected. Contradicting that the image of a connected set is a connected set. – Steen82 Aug 09 '24 at 18:28

Bijective functions from $\mathbb{R}^2$ to $\mathbb{R}$

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