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.
 A: 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.
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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 $.
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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).
