I am trying to answer the following question:

Does there exist a one-to-one $C^1$ function $F: \mathbb{R}^m \to \mathbb{R}^n$ with $m > n$?

I believe the answer to be no, but am having difficulties proving it. This is how I have tried thus far:

Let $k = m - n$, and give $F$ the coordinates $(x,y)\in \mathbb{R}^{k+n}$. Suppose there exists a point $(a,b) \in \mathbb{R}^{k+n}$ such that $J_{F,y}=[(\partial F_i/\partial y_j)(a,b)]$ is invertible.

Define $G(x,y) = F(x,y) - F(a,b)$. Then $G$ is $C^1$, $J_{G,y} = J_{F,y}$, and $G(a,b) = 0$.

By the implicit mapping theorem, there exists a neighborhood $U \subset \mathbb{R}^k$ of the point $a$ and a $C^1$ function $h: U \to \mathbb{R}^n$ such that $$G(x, h(x)) = 0 \ \ \implies \ \ F(x, h(x)) = F(a,b)$$ for all $x \in U$. Hence $F$ is not one-to-one.

The problem can be similarly addressed if there exists a point where $J_{F,x}=[(\partial F_i/\partial x_j)(a,b)]$ is invertible.

However, I am confused about the case where $J_{F,x}$ and $J_{F,y}$ are singular everywhere. I must either show that the function is not one-to-one, or not $C^1$, and am leaning towards the latter currently.

Is it possible to have a $C^1$ function where these matrices are everywhere singular?

  • $\begingroup$ What about $C^0$ already? Would you be happy with a proof that there can't be such a $C^0$ injective map? $\endgroup$ – Arnaud Mortier Mar 9 '18 at 18:27
  • $\begingroup$ @ArnaudMortier: I was hoping to utilize my current methodology, but that would be nice as well since every $C^1$ function is $C^0$. $\endgroup$ – infinitylord Mar 9 '18 at 18:32
  • $\begingroup$ See here for an answer math.stackexchange.com/questions/2524150/… $\endgroup$ – qbert Mar 9 '18 at 19:46
  • $\begingroup$ but implicit function theorem is not the play here I believe, since you are looking for a sort of converse (global singularity implies not injective) which the theorem does not provide $\endgroup$ – qbert Mar 9 '18 at 19:47
  • $\begingroup$ @qbert: Thank you for that reference. I didn't understand some of it (I have never studied differential geometry, so have no idea what a canonical smooth immersion is), but it did lead me to figure out a solution using the theorem they provided. $\endgroup$ – infinitylord Mar 9 '18 at 21:21

Here is a proof that there is no injective $C^0$ mapping $\Bbb R^m\to \Bbb R^n$ when $m>n$.

Assume that there is one - call it $\phi$.

Consider the canonical injection $\iota :\Bbb R^n\hookrightarrow \Bbb R^m$.

Consider the map $\iota\circ\phi|_{S^{m-1}}$, and denote its range by $K$.

Fact 1. $$\iota\circ\phi|_{S^{m-1}} : S^{m-1} \to \Bbb R^m$$ is injective continuous, therefore by the Jordan-Brouwer theorem $\Bbb R^m\setminus K$ has two connected components.

Fact 2. $K$ is a compact subset of the strict subspace $\iota(\Bbb R^n)\varsubsetneq \Bbb R^m$, therefore $\Bbb R^m\setminus K$ is path-connected.

This is a contradiction.


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