The image of the intersection of two lines is the intersection of the image of the lines

Question

Let $P$ be a projective plane and let $\sigma : P \to P$ be a projectivity. Let $r$ and $s$ be to projective lines in $P$ such that $r \neq s$. Why $\sigma (r \cap s) = \sigma(r) \cap \sigma(s)$?

Answer 1

HINT

Normally such proofs are done in 2 steps.

1. Let $x \in \sigma(r) \cap \sigma(s)$. Then $x\in \sigma(r)$ and $x \in \sigma(s)$. So $\exists y_r \in r$ and $y_s \in s$ such that $\sigma(y_r) = x = \sigma(y_s)$. Can you finish to conclude that $x \in \sigma(r \cap s)$? This would show that $$\sigma(r) \cap \sigma(s) \subseteq \sigma(r \cap s).$$

2. Now let $x \in \sigma(r \cap s)$. So $\exists y \in r \cap s$ such that $\sigma(y) = x$. Can you complete the argument and conclude that $x \in \sigma(r) \cap \sigma(s)$? This would show that $$\sigma(r) \cap \sigma(s) \supseteq \sigma(r \cap s).$$

Then, since $\sigma(r) \cap \sigma(s) \subseteq \sigma(r \cap s)$ and $\sigma(r) \cap \sigma(s) \supseteq \sigma(r \cap s)$, you must have $$\sigma(r) \cap \sigma(s) = \sigma(r \cap s),$$ as desired.