Discrete-math 7

\begin{matrix} B \subseteq A \\ f : P (A) \to P (A) \\ f (C) = C △ B \\ f (f (C)) = (C △ B) △ B = C △ (B △ B) = C △ \emptyset = C \\ ⟹ f = f^{- 1} \end{matrix}

\begin{matrix} f, g : A \to A \\ \exists (f \circ (g \circ f))^{- 1} \\ (f \circ (g \circ f)) is surjective ⟹ f is surjective \\ ((f \circ g) \circ f) is injective ⟹ f is injective \\ f is bijective ⟹ \exists f^{- 1} \\ (f^{- 1} \circ ((f \circ (g \circ f)) \circ f^{- 1})) = ((f^{- 1} \circ f) \circ g \circ (f \circ f^{- 1}))) = g \\ g is a composition of invertible functions ⟹ \exists g^{- 1} \end{matrix}

Set of functions

\begin{matrix} B^{A} = {f \subseteq A \times B | f is a function} \\ | B^{A} | = | B |^{| A |} \end{matrix}

\begin{matrix} Let A \neq \emptyset \\ F : A^{A} \to A \\ G : A \to A^{A} \\ Find F which is surjective \\ Find G which is injective \\ Solution: \\ \exists a \in A ⟹ F (f) = f (a) \\ Let x \in A \\ Let f_{x} : A \to A, \forall b \in A : f_{x} (b) = x \\ F (f_{x}) = f_{x} (a) = x \\ \forall a \in A : G (a) = f_{a} \end{matrix}