Metamath Proof Explorer


Theorem fcof1od

Description: A function is bijective if a "retraction" and a "section" exist, see comments for fcof1 and fcofo . Formerly part of proof of fcof1o . (Contributed by Mario Carneiro, 21-Mar-2015) (Revised by AV, 15-Dec-2019)

Ref Expression
Hypotheses fcof1od.f ⊢ φ → F : A ⟶ B
fcof1od.g ⊢ φ → G : B ⟶ A
fcof1od.a ⊢ φ → G ∘ F = I ↾ A
fcof1od.b ⊢ φ → F ∘ G = I ↾ B
Assertion fcof1od ⊢ φ → F : A ⟶ 1-1 onto B

Proof

Step Hyp Ref Expression
1 fcof1od.f ⊢ φ → F : A ⟶ B
2 fcof1od.g ⊢ φ → G : B ⟶ A
3 fcof1od.a ⊢ φ → G ∘ F = I ↾ A
4 fcof1od.b ⊢ φ → F ∘ G = I ↾ B
5 fcof1 ⊢ F : A ⟶ B ∧ G ∘ F = I ↾ A → F : A ⟶ 1-1 B
6 1 3 5 syl2anc ⊢ φ → F : A ⟶ 1-1 B
7 fcofo ⊢ F : A ⟶ B ∧ G : B ⟶ A ∧ F ∘ G = I ↾ B → F : A ⟶ onto B
8 1 2 4 7 syl3anc ⊢ φ → F : A ⟶ onto B
9 df-f1o ⊢ F : A ⟶ 1-1 onto B ↔ F : A ⟶ 1-1 B ∧ F : A ⟶ onto B
10 6 8 9 sylanbrc ⊢ φ → F : A ⟶ 1-1 onto B