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Here are the definitions:

  • An $\textbf{affine plane}$ is a geometric model where there is a unique line between each pair of distinct points; through every point there is a unique parallel to a given line; and there are three points that are pairwise not collinear.
  • A $\textbf{collineation}$ is a bijection from the set of points of an affine plane to itself, so that collinear points are mapped to collinear points. Equivalently, it maps lines $\textit{into}$ lines.
  • A bijection from the set of points of an affine plane to itself $\textbf{preserves lines}$ if the image of each line is a line. Equivalently, it maps lines $\textit{onto}$ lines.

In the script of a course there is the statement (without proof) that each collineation preserves lines, which does not look trivial to me. Maybe it could even be false, although I cannot find a counterexample.

Can anyone help me either by confirming that the statement is true or by disproving it?

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Let $r$ be a line in the plane and $r'$ the line containing the image of $r$ under a collineation. Suppose that the mapping $r\to r'$ in not onto, that is there exists a point $P'$ on $r'$ which is not the image of any point of $r$. As the collineation is bijective, $P'$ must then be the image of some point $P$ in the plane not belonging to line $r$.

Consider now any line $a$ passing through $P$. If $a$ meets $r$ at some point $A$, then $a$ contains two points ($A$ and $P$) whose image is in $r'$ and the whole of $a$ is mapped to $r'$. If $a$ is parallel to $r$, take a point $B$ on it and a point $C$ not belonging to $a$ or $r$: line $BC$ meets $r$ at some point $D$ whose image is on $r'$, but the image of $C$ is also on $r'$ by the above reasoning (because line $PC$ meets $r$), so the image of $B$ must be on $r'$ too.

We come then to the conclusion that if the mapping $r\to r'$ is not onto, then all the plane is mapped to a single line $r'$. This is not possible, because the collineation is a bijective mapping of the plane onto itself. It follows that a collineation preserves lines.

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  • $\begingroup$ Thanks a lot! This actually shows that a collineation does preserve straight lines, since the third axiom implies that there is a point not lying on $r'$, and the collineation must be bijective. $\endgroup$ – 57Jimmy Aug 28 '16 at 7:52
  • $\begingroup$ You are right: I've just edited my answer to incorporate that observation. $\endgroup$ – Aretino Aug 28 '16 at 11:15
  • $\begingroup$ Just a detail: if $r'$ is the image of $r$ then by definition all points in $r'$ are an image of some point in $r$, the mapping $r\to r'$ is then onto. What is (probably) actually meant is that $r'$ is the line that contains the image of $r$. $\endgroup$ – skyking Sep 1 '17 at 12:49
  • $\begingroup$ @skyking You are right: I'll edit my answer accordingly. $\endgroup$ – Aretino Sep 1 '17 at 14:51

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