# 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?

## Answers and Replies

Mentor
2021 Award
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.