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Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be bijective map with following properties:

1) $f|_{\mathbb{Q}^2}=id$;

2) Image of any line under map $f$ is again a line.

Is it right that $f=id$?

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Alex: maybe you should ask over on MathOverflow (or, if you want, I can ask for you). The solution to me is not obvious, and if a counterexample were to be found, it is possible that it won't be constructive. Very interesting question, this is. – Willie Wong Nov 21 '10 at 19:53
My conjecture is that (2) implies that $f$ is continuous, and that would imply that $f=id$. I think I'll ask about this more general conjecture on MO. – trutheality Nov 21 '10 at 19:57
My question is equivalent to following one. Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be bijective map such that image of any line under map $f$ is again a line. Is it right that $f$ is affine transformation of $\mathbb{R}^2$? Yes, if you will ask my question on MathOverflow I will be grateful. – Alex Nelson Nov 21 '10 at 20:22
Looks like the answer to my question answers yours:… – trutheality Nov 21 '10 at 21:12
up vote 5 down vote accepted

(Posting this as CW so Alex can accept the answer.)

Trutheality re-asked the question on MathOverflow, and it turns out the answer is given by what is known as the "Fundamental Theorem of Affine Geometry". See

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To minimise clickthrough: the answer turns out to be “Yes! (2) implies continuity, which together with (1) implies $f = \mathit{id}$. Getting continuity from (2) is quite non-trivial; the Fund. Thm. of Aff. Geom. says that the only possible obstruction to continuity is an automorphism of the ambient field, but in fact $\mathbb{R}$ has no field automorphisms so for $\mathbb{R}$ (2) implies continuity.” (I would edit this in but I don’t have enough rep.) – Peter LeFanu Lumsdaine Nov 21 '10 at 22:04
@Peter: you were just 1 rep short! I think now you can. (I don't know the proof of the Fund. Thm myself, so I didn't feel comfortable elaborating). Cheers – Willie Wong Nov 21 '10 at 22:22

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