Functions – Theorems
Statement: If f: A → B, g: B → C are two one one onto functions then gof: A → C is also one one onto.
i) If f: A → B, g: B → C are two one one functions then gof: A → C is also one one.
Proof: Let x₁, x₂ Є A; (gof) (x₁) = (gof) (x₂)
x₁, x₂ Є A, f: A → B → f(x₁), f(x₂) Є B
(gof) (x₁) = (gof)(x₂) → g[f(x₁)] = g[f(x₂)]
f(x₁), f(x₂) Є B, g[f(x₁)] = g[f(x₂)], g : B → C is one one → f(x₁) = f(x₂)
x₁, x₂ Є A, (gof) (x₁) = (gof) (x₂) → x₁ = x₂.
ii) If f: A → B, g: B → C are two functions and gof: A → C is one one then f: A → B is one one.
Proof: Let x₁, x₂ Є A and f(x₁) = f(x₂)
x₁, x₂ Є A, f: A → B → f(x₁), f(x₂) Є B
f(x₁), f(x₂) Є B, g: B → C, f(x₁) = f(x₂) → g[f(x₁)] = g[f(x₂)] → (gof) (x₁) = (gof)(x₂)
x₁, x₂ Є A, (gof)(x₁) = (gof) (x₂), gof: A → C is one one → x₁ = x₂.
∴ x₁, x₂ Є A, f(x₁) = f(x₂) → x₁ = x₂.
∴ f: A → B is one one.
iii) If f: A → B, g: B → C are two onto functions, then gof: A → C is also onto.
Proof: Let z Є C
z Є C, g: B → C is onto → ∃ y Є B ϶ g(y) = z
y Є B, f: A → B is onto → ∃ x Є A ϶ f(x) = y
now (gof)(x) = g[f(x)] = g(y) = z
∴ z Є C → ∃ x Є A ϶ (gof) (x) = z
∴ gof: A → C is onto.