# Continuous bijection from $\mathbb{R}^n$ to $\mathbb{R}^m$

Is there a continuous bijection from $\mathbb{R}^n$ to $\mathbb{R}^m$, for $n \neq m$?

Such a map would not be an open map, since $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic.

• Ordinary space-filling curves are surjections from a lower-dimensional space to a higher-dimensional space, but I don't think you can make them injective. – Michael Hardy Dec 31 '14 at 19:41
• arxiv.org/abs/1003.1467 says "no" for $m = 2$. – Unit Dec 31 '14 at 19:41
• @MichaelHardy: Correct. An injective space-filling curve would be a continuous bijection from the compact set $[0,1]$ to its Hausdorff image, hence a homeomorphism. – Unit Dec 31 '14 at 19:43
• – user99914 Dec 31 '14 at 19:43
• See en.wikipedia.org/wiki/Invariance_of_domain (there are also numerous discussions of it here on Math.SE) – Grigory M Dec 31 '14 at 19:58

No. There is not such a bijection. Actually, it is possible to prove the following result:

Theorem 1: If there is a continuous bijection $\phi : \mathbb{R}^n\to \mathbb{R}^m$, then $m=n$ and $\phi$ is a homeomorphism.

Proof: Firstly, assume that $m < n$ and $\phi : \mathbb{R} ^n\to \mathbb{R} ^m$ is a bijection. I am going to prove that such a bijection is not continuous. You can take, then, $S ^{m-1}\subset \mathbb{R}^n$. By Jordan theorem, $\mathbb{R} ^m - \phi (S^{m-1})$ has two connected components. While, $\mathbb{R} ^n - S^{m-1}$ is connected. So $\phi$ is not continuous.

Another approach would be using the Borsuk Ulan theorem. In particular, Borsuk Ulam theorem implies that there is no continuous injection $S^m\to \mathbb{R} ^m$. Assuming that $m<n$, there is $S^m\subset\mathbb{R}^n$ and, by Borsuk Ulam theorem, since $\phi |_ {S^m}: S^m\to \mathbb{R}^m$ is an injection, we have that $\phi$ is not continuous.

_

Now, assume that $n<m$ and $\phi$ is a continuous injection (and we are going to prove that it is not a surjection). The key argument is to restrict $\phi: \mathbb{R} ^n\to \mathbb{R}^m$ to each closed ball $B[0, n]\subset \mathbb{R}^n$. This gives us a bijection $\varphi _n$ between $B_n : = B[0,n]$ and $\phi (B_n )$.

We have that $\varphi _n$ is a homeomorphism (because it is a continuous bijection between a compact set and a Hausdorff space). But $\varphi _n (B_n)$ has empty interior, because, if it were nonempty, there would be an oben ball $O\subset B_n$ such that $\varphi_n (O)$ is open (and connected) and (of course) $\varphi_n: O\to \varphi_n(O)$ would be a homeomorphism (this fact contradicts the invariance of domain - but you can see that this would be impossible using the fact that $S^{n-1}$ is not of the same homotopy type of $S^{m-1}$). So, now you know that $\phi (B_n)$ has empty interior for every natural $n$. Now, you have just to apply the Baire theorem -

$\bigcup _ {n\in\mathbb{N}}\phi (B_n)$

has empty interior. Therefore it could not be the whole $\mathbb{R} ^m$.

_

It remains to prove that, if there is a continuous bijection $f : \mathbb{R}^n\to\mathbb{R}^n$, it is necessarily a homeomorphism. It is enough to show that such a continuous bijection is open. Let $B$ be an open ball in $\mathbb{R}^n$. We denote by $S$ its boundary (sphere). By Jordan curve, $\mathbb{R}^n - f(S)$ has two connected components. Since $f(S)$ is compact (and thereby closed in $\mathbb{R} ^n$), $\mathbb{R}^n - f(S)$ is open in $\mathbb{R} ^n$. Thereby, since each connected component of $\mathbb{R}^n - f(S)$ is open in $\mathbb{R}^n - f(S)$ , they are also open in $\mathbb{R}^n$.

Now, observe that $f$ induces a continuous bijection $\mathbb{R}^n - S\to \mathbb{R}^n - f(S)$ between spaces with $2$ connected components. Therefore the image of each connected component is a connected component. In particular, the image of $B$ is open. This completes the proof that $f$ is open.

Obs.: This question is somewhat related to the proposition about the reversibility of $\mathbb{R}^n$. There is an article about it: Reversible Topological Spaces (Rajagopalan and Wilansky).

• It's good to know that dimension is a topological invariant and we aren't all stick people. – GPerez Jan 1 '15 at 1:38