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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.

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What do you mean by "every line is mapped to a line"? Do you mean that the restriction of $f$ to any line is linear and injective, or do you just mean that the image of every line is a line? If you mean the latter, then it is easy to construct counterexamples even for $n=1$. – Andy Putman Nov 21 '10 at 20:15
Take the second meaning, and restrict to $n > 1$, please. The original question at math.SE is about $\mathbb{R}^2$. – Willie Wong Nov 21 '10 at 20:17
Ah, I see. That is an interesting question! – Andy Putman Nov 21 '10 at 20:19
Thank you Willie Wong for editing accordingly. – trutheality Nov 21 '10 at 20:23
I'm sure it's standard---but what is a "line"? Is [0,1] or (0,1) a line in $\mathbf{R}$, or does a line go off to infinity in both directions? – Kevin Buzzard Nov 21 '10 at 20:23
up vote 22 down vote accepted

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).

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Another textbook reference: M.Berger "Geometry", vol.1, section 2.6 "The fundamental theorem of affine geometry". – Sergei Ivanov Nov 21 '10 at 21:05
Yes, <a href="… is a link.</a> – Qing Liu Nov 21 '10 at 21:08

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