Collineations of an affine geometry

[samkolb]

An affine geometry is a nonempty set of points A, together with a set of lines, where each line connects one or more of the points in A.

A collineation of A is a bijection f: A --> A that carries lines to lines. That is, if P,Q are points in A lying on the same line, then f(P), f(Q) are points in A lying on the same line.

My question:

Does this definition imply that each pair of points not lying on the same line is carried to a pair of points not on the same line? That is, if P, Q do not lie on the same line, then do f(P) and f(Q) also not lie on the same line?

 

[fresh_42]

No, this does not follow. The condition says nothing about points which are not collinear. But, and this is somehow missing in the definition of a collineation, what about $f^{-1}$. You would expect from a collineation as a bijection, that the inverse function is also a collineation. This should have been stated.

So if we assume, that $f^{-1}$ is a collineation, too, then the situation is a different one:
If $f(P)$ and $f(Q)$ are collinear, then $f^{-1}(f(P))=P$ and $f^{-1}(f(Q))=Q$ are collinear because $f^{-1}$ is a collineation. This is equivalent to: If $P$ and $Q$ are not collinear, then $f(P)$ and $f(Q)$ can't be collinear.

Hence we need that $f^{-1}$ is a collineation. This should either be part of the definition, or a theorem proven by other means, will say other axioms.