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.
Feel free to retag.
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
$f$ is bijective;
$\dim E=\dim E'\ge 2$;
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).
