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Problem: Suppose that $f:\;\mathbb{R}^2\to\mathbb{R}^2$ is an injective mapping from the 2-dimensional Euclidean plane into itself which maps lines into (instead of onto) lines and whose range contains three non-collinear points. Can we say that f is an affine transformation?

Note that neither of two "into"s in assumptions means "onto". If either of two "into"s replaced by "onto", then the anwer is yes which can be deduced from the result in Li Baokui and Wang Yuefei's paper, and Chubarev and Pinelis' paper respectively.

On the other hand, if removing the injectivity, we can construct certain counterexample.

Counterexample: $f:\mathbb{R}^2→\mathbb{R}^2.$ Let f fix every point in some line $L$ and map the complement to one point outside $L$. $f$ is not an affine transformation.

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You actually do not need injectivity, only the assumption that the image is not contained in a single line. Then, Von Staudt'a original proof of fundamental theorem of projective geometry will go through and yield the required conclusion. –  Misha Dec 28 '12 at 23:13
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Can you provide more details on the source of Von Staudt'a original proof, say, the title of his article, the journal name or the book name, volume #, pages? Thank you in advance. –  woodbass Dec 29 '12 at 5:09
    
I read the proof in Hartshorne, Foundations of projective geometry. It probably could be found in other textbooks on projective geometry. Von Staudt's proof is based on encoding algebraic operations into projective configurations. Then, after normalizing the map so it preserves "standard quadrangle", he shows that the induced map of one of the projective lines is a field endomorphism of real numbers, hence, the identity. –  Misha Dec 29 '12 at 13:15
    
@woobbass: You should replace "geometric topology" tag with "geometry" or/and "projective-geometry", since geometric topology deals with very different issues. (Think, say, of the Poincare conjecture or geometric structures on manifolds.) –  Misha Dec 29 '12 at 14:54
    
Sorry, I realized that von Staudt's argument does not work because of non-surjectivity: Algebraic operations defined geometrically will not be everywhere defined. –  Misha Dec 29 '12 at 18:28
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2 Answers

Some very close statements are contained in the paper

MR2558789 Li, Baokui; Wang, Yuefei A new characterization for isometries by triangles, and in the reference list of this paper. For example, your conjecture is true if the map is surjective:

MR1657778 Chubarev, Alexander; Pinelis, Iosif Fundamental theorem of geometry without the $1$-to-$1$ assumption. Proc. Amer. Math. Soc. 127 (1999), no. 9, 2735–2744.

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In fact, much more is true, see Theorem 3.1 here. In particular, any map $RP^n\to RP^n$ preserving collinearity, whose image contains at least $n+1$ points which do not belong to a common projective hyperplane, is an element of $PGL(n,R)$. Note that injectivity is not required.

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There is an essential gap in his proof. I have written to him to point out the gap and he acknowledged his mistake. A simple counterexample to attack his "Theorem 3.1" is as follows (for simplicity let $n=2$). $f:RP^2\to RP^2$, let $f$ fix every point in some line $L$ and map the complement to one point outside $L$. –  woodbass Jan 1 '13 at 6:57
    
In his proof, he actually made use of certain injectivity in some hyperplane because he thought that $f$ preserves parallelism which is not assumed in advance. Even if his "Theorem 3.1$ is true, we cannot deduce any definite answer to my problem. –  woodbass Jan 1 '13 at 7:07
    
I thank woodbass for pointing out the error in Theorem 3.1 of my thesis. I have been pondering these issues since he sent me that message, and I believe I have a modified statement which is true: namely, a map $RP^{n}\rightarrow RP^{n}$ preserving collineaity, whose image contains at least $n+2$ points any $n+1$ of which are in general position, is an element of $PGL(n,R)$. I hope to write up the proof of this soon. I also believe that I can answer the question originally posed in this thread in the affirmative. I will communicate with woodbass about this shortly. –  Rupert Jan 11 '13 at 12:48
    
@Rupert:Hi,Rupert. I have noticed you by email that I have proved your above statement but I haven't heard from you so far. I am drafting the proof. –  woodbass Jan 12 '13 at 19:03
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