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 φF:AB
fcof1od.g φG:BA
fcof1od.a φGF=IA
fcof1od.b φFG=IB
Assertion fcof1od φF:A1-1 ontoB

Proof

Step Hyp Ref Expression
1 fcof1od.f φF:AB
2 fcof1od.g φG:BA
3 fcof1od.a φGF=IA
4 fcof1od.b φFG=IB
5 fcof1 F:ABGF=IAF:A1-1B
6 1 3 5 syl2anc φF:A1-1B
7 fcofo F:ABG:BAFG=IBF:AontoB
8 1 2 4 7 syl3anc φF:AontoB
9 df-f1o F:A1-1 ontoBF:A1-1BF:AontoB
10 6 8 9 sylanbrc φF:A1-1 ontoB