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Mathematics
General Math
Collineations of an affine geometry
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[QUOTE="fresh_42, post: 6178263, member: 572553"] 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. [/QUOTE]
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Mathematics
General Math
Collineations of an affine geometry
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