Existence of the bijective map from set of points to set of lines Let $P$ be a set of all points in a plane and $L$ set of all lines in that plane.
Is there a bijective map $f:P\to L$ such that for all colinear points $A,B,C$ the lines$f(A),f(B),f(C)$ are parallel or concurrent?

If a plane is projective then a polar map (with respect to any circle) does the job. But what if the plane is euclidean? I have no idea how to approach.
That's not possible. 

A way to prove this is to form a 2-bijection from $P$ to the antipodally punctured unit sphere (APUS) with axis perpendicular to $P$ that has contradictory properties.

We form this by selecting a point $a$ outside $P$. We take a point in $P$ maps it to a line, then we take the plane through $a$ containing that line and finally the two unit normals of that plane (being two values, but they are each others negation or opposites).

This mapping has some properties that aren't really compatible:


It's bijective (meaning that it never maps to the same pair of opposite vectors twice)
It maps lines to great circles (allowing any intersection of the APUS and a plane through the center of the APUS including those that actually aren't circles - that is they're punctured)


These are contradictory since two lines that doesn't intersect must map to great circles that doesn't intersect which is only possible if they are punctured. On the other side two lines that do intersect must map to great circles that do intersect.

This is the contradiction since any line must map to a great circle that is both punctured  and not punctured.

As for your example with the projective plane I guess that the mapping constructed above would work out since it would map to the non-punctured unit sphere instead, lines in $P$ would always intersect and great circles would always intersect.



Most of the proof is quite straight forward and there's nothing really difficult, but there are some points that's probably not that obvious.

The fact that lines would map to great circles follows from three bits put together:


If three lines are parallel in $P$ they won't intersect and therefore the intersections of the planes through $a$ containing the lines would be a lines and that lines has to be parallel to $P$. Or put in another way the cross product of the normals of the planes must be parallel to $P$. Or put in another way two of the normals of those planes and the normal of $P$ are linearly dependent which makes the normals of those planes linearly dependent which in turn puts them all in the same plane.
If three lines are concurrent in $P$ the planes through $a$ containing these line will both contain $a$ and the intersection point of the lines. This means that there is a line common to all these planes and their normal must be perpendicular to this line. This puts the normals all in the same plane.
The image of a line is therefore contained in some great circle. If the image was not the entire great circle there would be a point in the great circle whose preimage wouldn't be on the line. Now pick any point in $P$ then that point, the preimage mentioned and a point on the line would be colinear and must map to the great circle with the exception that the mentioned preimage and the picked point is on the parallel line or that the point on the line is antipodal (to the image of the mentioned preimage). This exception is not enough to make the mapping surjective.
Having a bijective map implies that it is invertible. If you were given parallel lines $f(A),f(B),f(C)$, then $f^{-1} : L \rightarrow P$ should give you colinear points $A,B,C$. However there is not one obvious choice for which colinear points should be chosen (there are uncountably many). This should tell you that a map from the set of points to the set of lines and vice versa would not be bijective.