Let $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ be an injection for $n>m$. Can $f$ be continuous? Why?


I got this question in mind when I was trying to find a continuous map from $\mathbb{R}^{2}$ to $\mathbb{R}$.



Injective maps $\mathbb{R}^{n} \to \mathbb{R}^{m}$
If $f$ is injective and continuous from $\mathbb{R}^n$ to $\mathbb{R}^m$
where $n>m$ then $f$ restricts to a continuous bijection from $S^{n-1}$,
the unit sphere in $R^n$, to a compact subset $K$ of $\mathbb{R}^m$.
Thus you can embed $S^{n-1}$, and a foriori $S^m$ in $\mathbb{R}^m$.


But there are homological obstructions to embedding $S^m$ in $\mathbb{R}^m$.
Using the arguments of this excellent paper


Albrecht Dold,
A simple proof of the Jordan-Alexander complement theorem,
*Amer. Math. Monthly* **100** (1993), 856-85.


(essentially a cunning use of the Mayer-Vietoris theorem)
it would entail the homology of the space $\mathbb{R}^m-K$ being nonzero
in negative dimension, which is absurd.


**Added** As I replied in haste I forgot the sledgehammer
that cracks this little nut, namely [Alexander duality](http://en.wikipedia.org/wiki/Alexander_duality).


**Added later** In fact this result also follows from Brouwer's
Invariance of Domain. 
This is Theorem 2B.3 on page 172 of [Hatcher's book](http://www.math.cornell.edu/~hatcher/AT/ATpage.html).
This implies that if one has an embedding from
$\mathbb{R}^n$ to itself, then its image is open. One gets such an
embedding by composing your putative embedding with the natural embedding of
$\mathbb{R}^m$ in $\mathbb{R}^n$. Adapting the proof, gives a swift proof
that the answer of your original question is no.


If you have a continuous injection from $\mathbb{R}^n$ to $\mathbb{R}^m$,
with $n>m$ then you have an embedding of $S^{n-1}$ into a nontrivial
hyperplane in $\mathbb{R}^n$. By the Jordan-Brouwer separation theorem
the image $K$ of this embedding separates $\mathbb{R}^n$ but it's easy to
see that since the image is in a hyperplane any two points of
the complement of $K$ can be connected by a path (exercise for reader :-)).


