Metamath Proof Explorer


Theorem fthpropd

Description: If two categories have the same set of objects, morphisms, and compositions, then they have the same faithful functors. (Contributed by Mario Carneiro, 27-Jan-2017)

Ref Expression
Hypotheses fullpropd.1 φ Hom 𝑓 A = Hom 𝑓 B
fullpropd.2 φ comp 𝑓 A = comp 𝑓 B
fullpropd.3 φ Hom 𝑓 C = Hom 𝑓 D
fullpropd.4 φ comp 𝑓 C = comp 𝑓 D
fullpropd.a φ A V
fullpropd.b φ B V
fullpropd.c φ C V
fullpropd.d φ D V
Assertion fthpropd φ A Faith C = B Faith D

Proof

Step Hyp Ref Expression
1 fullpropd.1 φ Hom 𝑓 A = Hom 𝑓 B
2 fullpropd.2 φ comp 𝑓 A = comp 𝑓 B
3 fullpropd.3 φ Hom 𝑓 C = Hom 𝑓 D
4 fullpropd.4 φ comp 𝑓 C = comp 𝑓 D
5 fullpropd.a φ A V
6 fullpropd.b φ B V
7 fullpropd.c φ C V
8 fullpropd.d φ D V
9 relfth Rel A Faith C
10 relfth Rel B Faith D
11 1 2 3 4 5 6 7 8 funcpropd φ A Func C = B Func D
12 11 breqd φ f A Func C g f B Func D g
13 1 homfeqbas φ Base A = Base B
14 13 raleqdv φ y Base A Fun x g y -1 y Base B Fun x g y -1
15 13 14 raleqbidv φ x Base A y Base A Fun x g y -1 x Base B y Base B Fun x g y -1
16 12 15 anbi12d φ f A Func C g x Base A y Base A Fun x g y -1 f B Func D g x Base B y Base B Fun x g y -1
17 eqid Base A = Base A
18 17 isfth f A Faith C g f A Func C g x Base A y Base A Fun x g y -1
19 eqid Base B = Base B
20 19 isfth f B Faith D g f B Func D g x Base B y Base B Fun x g y -1
21 16 18 20 3bitr4g φ f A Faith C g f B Faith D g
22 9 10 21 eqbrrdiv φ A Faith C = B Faith D