Isomorphism

[isomorphism ! definition]

Let \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes then \( \sigma \) and \( \sigma '\) are isomorphic if there exist a bijections
\( f:\mathscr{P} \to \mathscr{P}', F:\mathscr{L} \to \mathscr{L}'\) provided \( (p,L) \in \mathscr{I} \) iff \( (p',L') \in \mathscr{I'} \) .
Isomorphic planes \( \sigma \) and \( \sigma '\) are denoted by \( \sigma \sim \sigma '\).

Note:
Let \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two isomorphic planes, then

• \( f:\mathscr{P} \to \mathscr{P}'\) is bijection
• \( F:\mathscr{L} \to \mathscr{L}'\) is bijection
• \( f \) preserves the collinearity

Theorem: If \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes and \( f:\mathscr{P} \to \mathscr{P}'\) is bijection such that points \( a_1,a_2,a_3, \cdots \in P \) are collinear if and only if \( f(a_1), f(a_2), f(a_3),\cdots \in P'\) are collinear then \( \sigma \sim \sigma '\).

[isomorphism ! theorem]

Proof
Given, \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes with a bijection \( f:\mathscr{P} \to \mathscr{P}'\) and preserves collinearity.
Now, we show \( F:\mathscr{L} \to \mathscr{L}'\) is a bijection.

• F is well defined and one-one
  Let \(L_1 \in \mathscr{L}\), and \( L_2 \in \mathscr{L'}\), then then \( x_1,y_1 \in L_1 \) and \( x_2,y_2 \in L_2\).
  Suppose, \(L_1=L_2 \), then, \(x_1,y_1, x_2,y_2 \) are collinear. This implies that
  \(f(x_1),f(y_1),f(x_2),f(y_2)\) are collinear, so \(F(L_1)=F(L_2)\).
  Hence, \( F:\mathscr{L} \to \mathscr{L}'\) is well defined and one-one.
• F is onto
  For each \( L' \in \mathscr{L'} \) there exists \( x',y' \in L' \) such that
  \(f(x)=x', f(y)=y'\)
  Since, \(f(x),f(y)\) are collinear, therefore \(x,y\) are collinear
  Thus, there exists a line \( L \in \mathscr{L}\) such that \(F(L)=L'\), where \(L=xy\)
  Hence, \(F\) is onto

This completes the proof.

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