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