Let $P$ be a projective plane, and $p_{1},p_{2}$ different points of $P$. Consider now a projective line $L\subset P$ not passing for the aforementioned points. Let $F(p_{i})\,,\, i \in\{1,2\}$ be the pencil of lines in $P$ passing through $p_i$ and define \begin{equation} f:F(p_1)\rightarrow F(p_2) \end{equation} such that, for every $R\in F(p_1)$, $f(R)$ is the line passing trough both $p_2$ and $L\cap R$ . Show that $f$ is a projective transformation. It is suggested to do so using the properties of dual projective spaces, but I couldn't manage to do so .
2026-03-26 10:38:47.1774521527
Projective applications and projective duality
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Let's just dualize the assumptions (point $p$ becomes line $p'$, line $L$ becomes point $L'$, etc).
Let $P$ be a projective plane, and $p'_{1},p'_{2}$ different lines of $P$. Consider now a projective point $L'\subset P$ not incident with the aforementioned lines. Let $F'(p_{i})\,,\, i \in\{1,2\}$ be the range of points in $P$ lying on $p'_i$ and define \begin{equation} f':F'(p'_1)\rightarrow F'(p'_2) \end{equation} such that, for every $R'\in F'(p'_1)$, $f'(R')$ is the intersection of $p'_2$ and $L'\vee R'$.
Then $f'$ is a projective transformation (it is a perspectivity with center $L'$). By duality $f$ is projective.