# Does There exist a bijection between $\mathbb{R}^2$ and $\mathbb{R}$ such that it is differentiable

I was thinking recently about Louville's Theorem and the fact that there exist a bijection between $\mathbb{R}^2$ and $\mathbb{R}$.

Since $\mathbb{C}$ is nothing but $\mathbb{R}^2$ via the construction of $\mathbb{C}$ we can use those terms interchangeably. I thought of the following tho transformations. First $$f_1:\mathbb{R}^2 \to \mathbb{R}$$

which we know exists

and then $$f_2:\mathbb{R} \to S^1$$ witch is a well known bijection and is differentiable (as far as I know).

the composition of the functions $f_1$ and $f_2$ (let's call it $f_c$) should hence be a bijection between $\mathbb{R}^2$ and $\mathbb{R}$.

Louville's theorem states that any function such that $|f(z)|<M$ and $f\in H(\mathbb{C})$ must be equal to a constant function. aka $f(z)=c, c\in\mathbb{C}$ Our function $f_c$ is not a constant function because it ascribes a unique number in $S^1$ for each number in $\mathbb{C}$ and also $f_c$ is bounded $(f_c(z)<2$ for every $z\in \mathbb{C})$. The only thing that could make Louville's theorem not work in this case would be the fact that $f_1$ is not a holomorphic function. Which would mean that there is no holomorphic bijection between $\mathbb{R}^2$ and $\mathbb{R}$.

Is this argument true? If it is, does anyone know any restraints when it comes to differentiability of bijections between $\mathbb{R}^2$ and $\mathbb{R}$?

• I'm confused on the ordering of your maps; you seem to end in $S^1$, but claim you have a differentiable map $R^2$ to $R$. – TomGrubb Mar 12 '18 at 19:30
• There is no continuous bijection $\mathbb R\to S^1$. – Christoph Mar 12 '18 at 19:31
• The pre-image of any point $x\in\Bbb R$ of a map $\Bbb R^2\to\Bbb R$ is either empty or "at least one-dimensional". As such there exists no injective differentiable map from $\Bbb R^2$ into $\Bbb R$. – s.harp Mar 12 '18 at 19:40
• @ThomasGrubb I claim the opposite that there is no differentiable map from $R^2$ to $R$ – Alexandar Solženjicin Mar 12 '18 at 19:43
• @Christoph I would like to see a proof(not that I don't believe you) if you know a link, book reference.. – Alexandar Solženjicin Mar 12 '18 at 19:46

There are no continuous injective maps from $\mathbb R^2$ to $\mathbb R$ (which implies there are no differentiable maps with this property).

Proof: Suppose there is such a map $f.$ Then for each $y\in \mathbb R,$ $f$ is continuous and injective on the line $\mathbb R\times \{y\}.$ Since each of these lines is connected, so is each $f(\mathbb R\times \{y\}).$ Thus each $f(\mathbb R\times \{y\})$ is an interval with nonempty interior, and therefore contains a rational. But the collection of all of these intervals is uncountable and pairwise disjoint. This implies there are uncountably many distinct rational numbers, contradiction.

Holomorphic functions are open. So if a holomorphic function is a bijection, it automatically is a homeomorphism.
But there is no homeomorphism from $\mathbb R^2$ to $\mathbb R$.

You cannot even do this for a continuous mapping. You may be thinking of the existence of space filling curves: continuous onto mappings from $[0,1]$ to $[0,1] \times [0,1]$, but these are not 1-1 mappings.

A theorem relevant to your question is Brouwer's Invariance of DomainTheorem. If $f:\mathbb{R}^2 \to \mathbb{R}$ is a continuous 1-1 map, then so is the composite with an inclusion $i:\mathbb{R} \to \mathbb{R}^2$ given by $i(x) = (x,0)$. Then $i \circ f:\mathbb{R}^2 \to \mathbb{R}^2$ is a continuous 1-1 map. By Invariance of Domain, the image is an open set. But this is not the case since the image is contained in the image of $i$, namely the $x$-axis.

It may be possible to give a much simpler proof if you assume the map is differentiable. (The Invariance of Domain Theorem, while unsurprising, is very difficult to prove without knowing some geometric or algebraic topology.)

• I edited my previous answer. It now gives a simple proof for the continuous case. – zhw. Aug 24 '18 at 19:16