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.

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    $\begingroup$ 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. $\endgroup$ – Michael Hardy Dec 31 '14 at 19:41
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    $\begingroup$ arxiv.org/abs/1003.1467 says "no" for $m = 2$. $\endgroup$ – Unit Dec 31 '14 at 19:41
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    $\begingroup$ @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. $\endgroup$ – Unit Dec 31 '14 at 19:43
  • $\begingroup$ Related: math.stackexchange.com/questions/43096/… $\endgroup$ – user99914 Dec 31 '14 at 19:43
  • $\begingroup$ See en.wikipedia.org/wiki/Invariance_of_domain (there are also numerous discussions of it here on Math.SE) $\endgroup$ – 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).

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

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