Let $f: \mathbb R^n \rightarrow \mathbb R^n$, where $n> 1$ be a bijective map such that the image of every line is a line.


Is $f$ continuous?


I think it is, but the proof isn't immediately obvious to me.
Related to [this question on math.SE](https://math.stackexchange.com/q/11232/3638).


Feel free to retag.



Continuity in terms of lines
This is called the fundamental theorem of affine geometry. Let $f : E \to E'$ be a map between affine spaces over a field $K$. Suppose that 


1. $f$ is bijective;
2. $\dim E=\dim E'\ge 2$;
3. If $a, b, c\in E$ are aligned, then so are $f(a), f(b), f(c)$.


Then $f$ is semi-affine: fix some $a\_0\in E$, then there exists a field automorphism $\sigma$ of $K$ such that the map $h: v\mapsto f(a\_0+v)-f(a\_0)$ (which goes from the vector space attached to ${E}$ to that attached to $E'$) is additive and $h(\lambda v)=\sigma(\lambda)h(v)$ for all $v$ and all $\lambda \in K$. I don't have an URL for this theorem, I find it in Jean Fresnel:  *Méthodes Modernes en Géométrie*, Exercise 3.5.7. But I think it is in any standard textbook on affine geometry. 


When $K=\mathbb R$, it is known that $K$ has no non-trivial field automorphism. So your $f$ is an affine function, hence continuous. If $K=\mathbb C$, as pointed out by Kevin in above comments, take any non-trivial automorphism of $\mathbb C$, then you get a semi-affine map $\mathbb C^n \to \mathbb C^n$ which will not be affine, even not continuous (if $\sigma$ is not the conjugation). 


