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If $f:{\mathbb{R}}^n\to{\mathbb{R}}^n$ $(n\ge2)$ is a bijection such that the image of every line is a line (continuity of $f$ not assumed), must $f$ be an affinity?

Assuming continuity would certainly suffice, even assuming that $f$ is order-preserving on each line. Is there a counterexample if we drop the assumption that $f$ is a bijection? Any references?

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  • $\begingroup$ The answer is yes and it is very standard. I saw the proof in a couple books, but I can remember where. $\endgroup$
    – Anton Petrunin
    Commented Aug 21, 2013 at 0:35
  • $\begingroup$ @AntonPetrunin Can you give an idea of the proof? At least in two dimensions? $\endgroup$
    – Wlodek Kuperberg
    Commented Aug 21, 2013 at 0:43
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    $\begingroup$ Why does this have the open-problem tag? $\endgroup$
    – Nate Eldredge
    Commented Aug 21, 2013 at 3:12

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Yes $f$ must be an affinity – this is called the fundamental theorem of affine geometry and is found e.g. on page 52 of M. Berger's Geometry. (For other treatments and history, see this question.)

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    $\begingroup$ That link is excellent, but for the lazy, here's the gist of the argument: if you fix a point in your affine space to make it a vector space, vector addition can be defined in terms of parallel lines by the "parallelogram law". If you fix a line through the origin and a point on it to call 1, scalar multiplication by points of that line can also be defined using parallel lines (by similar triangles). This multiplication now gives a binary operation on that line, and any line-preserving bijection gives a ring automorphism. The only ring automorphism of $\mathbb{R}$ is the identity. $\endgroup$
    – Eric Wofsey
    Commented Aug 21, 2013 at 1:08
  • $\begingroup$ @EricWofsey Fixing a point is fine, that is, you can assume that the origin is fixed under $f$, but after that it seems to me you are assuming here something more about $f$ than just what the question allows. $\endgroup$
    – Wlodek Kuperberg
    Commented Aug 21, 2013 at 1:38
  • $\begingroup$ @EricWofsey Vector addition "defined in terms of parallel lines" works only if vectors added are non-parallel. Is that not a problem? $\endgroup$
    – Wlodek Kuperberg
    Commented Aug 21, 2013 at 2:16
  • $\begingroup$ As I said, I was only giving the gist. When I say "fix a line and a point on it to call 1", I don't mean to assume $f$ fixes every point of the line, only that it fixes the line as a set and it fixes the point 1. This is equivalent to just fixing the two points 0 and 1. To add parallel vectors $v$ and $v'$ you can first add a non-parallel vector $w$ to $v$, then add $v'$ to $v+w$, then subtract $w$. $\endgroup$
    – Eric Wofsey
    Commented Aug 21, 2013 at 2:32
  • $\begingroup$ @EricWofsey OK, I see. $\endgroup$
    – Wlodek Kuperberg
    Commented Aug 21, 2013 at 3:11
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As Francois Ziegler pointed out, the result that you need is the so-called fundamental theorem of geometry (FTG), which states the following: 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. As shown in Remark 11 in that paper, the surjectivity condition cannot be dropped. As the example given in Lev Borisov's answer shows, the surjectivity condition cannot be dropped even if the transformation is assumed to be continuous.

As far as the rendition of the FTG on page 52 in the book by Berger Geometry (cited in Francois Ziegler's answer) is concerned, it is explained on page 2737 in the mentioned paper AMS Proc. 1999 that the proof in Berger's book contains gaps/errors. Also, as noted above, one only needs the surjectivity, rather than the bijectivity condition assumed in Berger's book.

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    $\begingroup$ I am puzzled by the surjectivity condition cannot be dropped even if the transformation is assumed to be bijective. By definition, any bijective transformation is surjective, isn’t it? $\endgroup$
    – Emil Jeřábek
    Commented Mar 15, 2018 at 16:04
  • $\begingroup$ @EmilJeřábek : Fixed that. I need to get some sleep. :-) $\endgroup$
    – Iosif Pinelis
    Commented Mar 15, 2018 at 16:19
  • $\begingroup$ Interesting! Following MR’s “cited by” further leads to Li-Wang (2016) (about not assuming surjectivity), and from there to Frank (1992) and Lenz (1958). (The problem in Berger is only over fields other than R, right?) $\endgroup$
    – Francois Ziegler
    Commented Mar 15, 2018 at 19:53
  • $\begingroup$ @FrancoisZiegler : Thank you for the references. I'll try to look at them, even though my current interests are far from those matters. As for Berger, looking at what is written on page 2737 in the mentioned AMS Proc. 1999 paper, there appears to be a gap in the proof in Berger's book even when the space is over $\mathbb R$: the proof that every 2-plane is mapped into a 2-plane appears to be missing there. $\endgroup$
    – Iosif Pinelis
    Commented Mar 15, 2018 at 20:43
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Do you assume that $f$ is surjective? Else, $f(x,y)=(x^3+y,0)$ would send any line to the $x$-axis.

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  • $\begingroup$ Yes, I assume $f$ is bijective, that is, both injective and surjective. But you gave a simple counterexample for my additional question about dropping this assumption. $\endgroup$
    – Wlodek Kuperberg
    Commented Aug 21, 2013 at 1:51
  • $\begingroup$ This is not really an answer to the question. I think it should have been a comment. $\endgroup$
    – David White
    Commented Jan 27 at 1:38

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