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What conditions are sufficient to prove a transformation in $\mathbb{R}^2$ is affine?

I currently have shown that the transformation is bijective, points and lines are preserved and also the following:

If $M$, $N$, $X$, $Y$ are points such that $MN \parallel XY$ and $MN$ is parallel to one of the coordinate axes, then the images $M'$, $N'$, $X'$, $Y'$ also satisfy $M'N' \parallel X'Y'$ and $M'N'$ is parallel to one of the coordinate axes.

Do I still need to show parallel-ness in general is preserved or not?

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  • $\begingroup$ I think the question is OK. When we have just proved, with Chubarev, the theorem referred to in my answer below, we did not have any idea about the existence of the fundamental theorem of geometry -- we found it in the literature only after we proved our result. $\endgroup$ Commented Mar 15, 2018 at 2:29
  • $\begingroup$ Possible duplicate of Line-preserving bijection of ${\mathbb{R}}^n$ onto itself $\endgroup$ Commented Mar 15, 2018 at 2:42
  • $\begingroup$ @FrancoisZiegler : As explained on page 2737 in the paper cited in my answer below, the proof in the book by Berger you cited at mathoverflow.net/questions/139978/… contains gaps/errors. Also, as noted in my answer, one only needs surjectivity, rather than the bijectivity condition assumed in Berger's book. $\endgroup$ Commented Mar 15, 2018 at 2:50
  • $\begingroup$ In my first comment here, I wrote "When we have just proved". Of course, I meant "When we had just proved". $\endgroup$ Commented Mar 15, 2018 at 3:35
  • $\begingroup$ @IosifPinelis Yes; I still wonder if everything here, including your interesting answer and comment, might not find a better home at the other question. Cf. also the references at MR 3471078. $\endgroup$ Commented Mar 15, 2018 at 5:17

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By the fundamental theorem of geometry, if a transformation of $\mathbb R^d$ with $d\ge2$ is bijective and maps lines onto lines, then it is affine. The main result in AMS Proc. 1999 implies that, more generally, if a transformation of $\mathbb R^d$ with $d\ge2$ is surjective and maps lines into lines, then it is affine.

So, you don't need the condition "If $M$, $N$, $X$, $Y$ are points such that $MN \parallel XY$ ..." at all. Also, you can relax the conditions of bijectivity and preservation of lines to the conditions of surjectivity and preservation of collinearity -- and you don't need any other conditions to get the affinity.

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