Metamath Proof Explorer


Theorem 2fvidf1od

Description: A function is bijective if it has an inverse function. (Contributed by AV, 15-Dec-2019)

Ref Expression
Hypotheses 2fvcoidd.f ⊢ φ → F : A ⟶ B
2fvcoidd.g ⊢ φ → G : B ⟶ A
2fvcoidd.i ⊢ φ → ∀ a ∈ A G ⁡ F ⁡ a = a
2fvidf1od.i ⊢ φ → ∀ b ∈ B F ⁡ G ⁡ b = b
Assertion 2fvidf1od ⊢ φ → F : A ⟶ 1-1 onto B

Proof

Step Hyp Ref Expression
1 2fvcoidd.f ⊢ φ → F : A ⟶ B
2 2fvcoidd.g ⊢ φ → G : B ⟶ A
3 2fvcoidd.i ⊢ φ → ∀ a ∈ A G ⁡ F ⁡ a = a
4 2fvidf1od.i ⊢ φ → ∀ b ∈ B F ⁡ G ⁡ b = b
5 1 2 3 2fvcoidd ⊢ φ → G ∘ F = I ↾ A
6 2 1 4 2fvcoidd ⊢ φ → F ∘ G = I ↾ B
7 1 2 5 6 fcof1od ⊢ φ → F : A ⟶ 1-1 onto B