Consider the Euclidean plane $\mathbb{R}^2$. A collineation on $\mathbb{R}^2$ is a bijective function $f$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ such that the image of every line under $f$ is also a line. Does there exist a collineation which does not preserve betweenness? That is, there exists at least one triple of distinct points $(A,B,C)$ such that $B$ is between $A$ and $C$, but $f(B)$ is not between $f(A)$ and $f(C)$.
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No nondegenerate cases. Nope. (Degenerate: map a line to a point.) – David G. Stork Apr 18 '22 at 05:24
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@DavidG.Stork Since points aren't lines, that's not a collineation. – Noah Schweber Apr 18 '22 at 05:31
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The degenerate case is to map a line to another line of length $0$. But if that isn't allowed, then the answer is a clear NO. – David G. Stork Apr 18 '22 at 05:57
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Let consider $\mathbb{R}\to\mathbb{R}$ subcase. Let $f:\mathbb{R}\to\mathbb{R}$ is bijection and $f(x)$ maps $(-\infty;0]$ to $(0;1]$, $(0;1]$ to $(-\infty;0]$ and $(1;\infty)$ to $(1;\infty)$. Then for $A\in(-\infty;0]$, $B\in(0;1]$, $C\in(1;\infty)$: $B$ is between $A$ and $C$, but $f(B)$ is not between $f(A)$ and $f(C)$. But maybe this subcase cannot be extended to $\mathbb{R}^2$. – Ivan Kaznacheyeu Apr 18 '22 at 07:18