Let me apologize in advance as this is possibly an extremely stupid question: can one prove or disprove the existence of a bijection from the plane to itself, such that the image of any circle becomes a square? Or, more generally, are there any shapes other than a square such that a bijection does exist? (obviously, a linear map sends a circle to an ellipse of fixed dimensions and orientation)

28$\begingroup$ You can have two squares that intersect at 8 points (for example  take two congruent squares on top of each other and rotate by $45^\circ$ around their common center. Under the inverse of such a bijection, your squares would have to go to two distinct circles with 8 common points, which is impossible. $\endgroup$– amakelovCommented Apr 10, 2017 at 19:27

2$\begingroup$ I guess any two circles have 1, 2 or infinitely many common points, whereas squares can intersect in 4 points. $\endgroup$– Denis TCommented Apr 10, 2017 at 19:27

$\begingroup$ Maybe some interesting related questions are: For which $\mathcal{S}$ does this proposition hold, where $\mathcal{S}$ is the set of boundaries of some class of convex bodies? Is number of intersection points the only obstruction? $\endgroup$– NealCommented Apr 10, 2017 at 19:33

1$\begingroup$ As you tagged this as AG, I guess you may be interested in some works of Vladlen Timorin, who wrote a few papers on classification of $\mathbb {RP}^n$ selfmaps taking lines into plane curves of some degree. See arxiv.org/abs/math/0212098 and later. $\endgroup$– Denis TCommented Apr 10, 2017 at 19:34

$\begingroup$ I'd like to use the sudden interest in this question to generate some overflow interest into a different but similar question on MSE: math.stackexchange.com/q/2221362/101420 from a few days ago. It is not mine, but I am still curious to see an answer. $\endgroup$– VincentCommented Apr 12, 2017 at 11:46
2 Answers
There is no such bijection.
To see this, imagine four circles all tangent to some line at some point $p$, but all of different radii, so that any two of them intersect only at the point $p$. (E.g., any four circles from this picture.) Under your hypothetical bijection, these four circles would map to four squares, any two of which have exactly one point in common, the same point for any two of them. You can easily convince yourself that no collection of four squares has this property.


$\begingroup$ I wonder what a three or higher dimensional version would be like. Perhaps mapping nested spheres to various hypothetical shapes with square cross sections? Gerhard "Four Doesn't Suffice In 3D" Paseman, 2017.04.10. $\endgroup$ Commented Apr 10, 2017 at 20:54

3$\begingroup$ And this extends easily to prove that there is no bijection that maps every circle to some regular polygon. $\endgroup$ Commented Apr 11, 2017 at 6:34

$\begingroup$ Wouldn't said bijection only map one of those circles to a square though? Or does a bijection from the plan to itself that maps one of the circles to a square automatically map all the others to a square as well? Would this depend on the nature of the bijection? $\endgroup$– D. W.Commented Apr 14, 2017 at 2:38

$\begingroup$ @D.W. The problem in the original question asks about the existence or nonexistence of a bijection which maps ANY circle to a square. $\endgroup$ Commented Apr 15, 2017 at 5:31
The problem is that two distinct squares can have more than two common points (easy to make an example), and under such a bijection these squares would have to go to two distinct circles with more than two common points  an impossibility.

23$\begingroup$ The question was whether there is a bijection sending circles to squares. As far as I can tell, your argument only shows that there is no bijection sending squares to circles. (It is not obvious to me that if a bijection makes every circle a square, then its inverse must make every square a circle.) Am I missing something? $\endgroup$ Commented Apr 10, 2017 at 19:32

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$\begingroup$ @WillBrian: Doesn't a bijection have to be invertible by definition though? If A maps into B one way, B has to map back to A the other. So showing that a mapping doesn't exist one way should be enough to prove that there can be no such bijection. $\endgroup$ Commented Apr 11, 2017 at 14:20

2$\begingroup$ @MattiVirkkunen : A bijection has to be invertible, but just because a bijection maps every circle to a square doesn't mean that its inverse maps every square to a circle. "Bijection" just means that points are put into onetoone correspondence, and it doesn't automatically mean that the set of circles is put into bijection with the set of squares. The bijection could inject the set of circles into the set of squares. $\endgroup$ Commented Apr 11, 2017 at 14:38

5$\begingroup$ further, for example, we have a bijection $\mathbb{R}\to\mathbb{R}$ given by $x\mapsto 4x$ for which it is true that the image of any integer is an even integer, but yet it is not true that the inverse image of any even integer is an integer $\endgroup$– amakelovCommented Apr 11, 2017 at 14:53