In a book I am reading on transformational Euclidean geometry, the author defines a collineation as a bijection of the plane which takes lines to lines -- that is, for a collineation $F$, if $L$ is a line, then $F[L]$ is a line. A line is understood as a set of points satisfying a linear equation.
The author remarks that for a collineation $F$, $F[L]$ is a line and $F(p)$ lies on $F[L]$ if and only if point $p$ lies on line $L$.
My question is: is it possible to establish the "only if" here given that we are not assuming an "only if" in the definition of collineation?
I believe this remark is equivalent to each of the following:
- If F is a collineation, then the inverse of F is also a collineation
- If F is a collineation, then F is (naturally induces) a bijection on the set of all lines
I am trying to determine if this definition is incorrect or if I am overlooking something trivial.
This was just me being silly. At the location of the remark, I was not sure if we could safely assume certain things (like that there is a unique line through any two distinct points), but I see now the author obviously intends these to just be derived from the analytic definitions presented later.
Given a collineation $F$ and a line $L$, we can fix distinct points $p$ and $q$ on $L$, and since $F$ is a bijection, $p'=F^{-1}(p)$ and $q'=F^{-1}(q)$ are distinct. Let $L'$ be the line through $p'$ and $q'$. Then $F[L']$ is a line through $p$ and $q$, so it must be $L$. This shows that $F$ is surjective on lines. It is clear that $F$ is also injective on lines, so it is bijective, which is equivalent to the remark as noted.