Equivalence of definitions of collineations. I've seen two definitions of collineations:

*

*A collineation on $\mathbb R^n$ is a bijection $\mathbb R^n \to \mathbb R^n$ that takes lines to lines.


*A collineation on $\mathbb R^n$ is a bijection $\mathbb R^n \to \mathbb R^n$ that takes triples of collinear points to triples of collinear points.
(1) obviously implies (2), and there's a neat proof that (2) implies (1) in the two-dimensional case here. So the question is: does (2) imply (1) in higher dimensions, too, or is there a counterexample?
 A: Let's use induction to prove this. The collineation is called $f$ in the following content.
Suppose a $1$D linear manifold $a^1$ is mapped into $1$D linear manifold $b^1$ but not onto, then $\exists P_1 \notin a^1 \, , \, f(P_1) \in b^1$. According to the referenced proof, the $2$D linear manifold $a^2$, containing $a^1$ and $P_1$, is mapped into $b^1$.
Arbitrarily choose a $2$D linear manifold $b^2$ containing $b^1$, and due to the fact that $a^2$ is mapped into $b^2$ but not onto, it is also the truth that $\exists P_2 \notin a^2 \, , \, f(P_2) \in b^2$.
Consider the $3$D linear manifold $a^3$ containing $a^2$ and $P_2$. Similar to the referenced proof, for any $1$D linear manifold $l$ through $P_2$, either $l$ intersect $a^2$: Then $2$ different points on $l$ (namely $P_2$ and the intersection) are mapped into $b^2$, causing $l$ mapped into $b^2$; or $l$ is parallel to $a^2$: Then choose $Q \in a^3$ such that $PQ$ intersects $a^2$, meaning $f(Q) \in b^2$. Simple analysis on $1$D linear manifolds through $Q$ and points on $l$ (which apparently intersect $a^2$) will imply that $l$ is also mapped into $b^2$.
Then we get a $3$D linear manifold $a^3$ mapped into a $2$D linear manifold $b^2$, marking the success of the induction.
Note that the whole process doesn't rely on the base field.
Addition: It's quite interesting when it comes to vector space of infinite dimension. I'm not familiar with transfinite induction so I don't know whether the same method helps in this case.
