# Two skew lines and two lines in \mathbb{P}^3

Given two lines $r, r' \in \mathbb{P}^3$ such that $r \cap r' = \varnothing$ and two other lines $s, s'$ such that $A=r \cap s, B=r \cap s', C = r' \cap s, D = r' \cap s', A≠B≠C≠D$.

Show that $A, B, C, D$ are linearly independent, $s \cap s' = \varnothing$ and if a projectivity $f: \mathbb{P}^3 \to \mathbb{P}^3$ leaves invariant all points of the lines $r, r', s, s'$ it is necessarily the identity map.

My solution

If $A, B, C, D$ are linearly dependent $\Rightarrow C, D \in AB = r$ $\Rightarrow r \cap r' ≠ \varnothing \ \unicode{x21af}$ $\Rightarrow A,B,C,D$ are linearly independent.

If $\exists$ point $P \in s \cap s'$ $\Rightarrow s, s'$ are coplanar $\Rightarrow r, r', s, s'$ are coplanar $\Rightarrow r \cap r' ≠ \varnothing \ \unicode{x21af}$ $\Rightarrow s \cap s' = \varnothing$.

Taking points $R \in r, R' \in r', S' \in s'$ such that $A≠R≠B, C≠R'≠D, B≠S'≠D$ $\Rightarrow A, C, R, R', S'$ are in general position. $\Rightarrow \exists! f: \mathbb{P}^3 \to \mathbb{P}^3, A \mapsto A, C \mapsto C, R \mapsto R, R' \mapsto R', S' \mapsto S'$. And this projectivity is necessarily the identity map.

Did i miss something?