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 φHom𝑓A=Hom𝑓B
fullpropd.2 φcomp𝑓A=comp𝑓B
fullpropd.3 φHom𝑓C=Hom𝑓D
fullpropd.4 φcomp𝑓C=comp𝑓D
fullpropd.a φAV
fullpropd.b φBV
fullpropd.c φCV
fullpropd.d φDV
Assertion fthpropd φAFaithC=BFaithD

Proof

Step Hyp Ref Expression
1 fullpropd.1 φHom𝑓A=Hom𝑓B
2 fullpropd.2 φcomp𝑓A=comp𝑓B
3 fullpropd.3 φHom𝑓C=Hom𝑓D
4 fullpropd.4 φcomp𝑓C=comp𝑓D
5 fullpropd.a φAV
6 fullpropd.b φBV
7 fullpropd.c φCV
8 fullpropd.d φDV
9 relfth RelAFaithC
10 relfth RelBFaithD
11 1 2 3 4 5 6 7 8 funcpropd φAFuncC=BFuncD
12 11 breqd φfAFuncCgfBFuncDg
13 1 homfeqbas φBaseA=BaseB
14 13 raleqdv φyBaseAFunxgy-1yBaseBFunxgy-1
15 13 14 raleqbidv φxBaseAyBaseAFunxgy-1xBaseByBaseBFunxgy-1
16 12 15 anbi12d φfAFuncCgxBaseAyBaseAFunxgy-1fBFuncDgxBaseByBaseBFunxgy-1
17 eqid BaseA=BaseA
18 17 isfth fAFaithCgfAFuncCgxBaseAyBaseAFunxgy-1
19 eqid BaseB=BaseB
20 19 isfth fBFaithDgfBFuncDgxBaseByBaseBFunxgy-1
21 16 18 20 3bitr4g φfAFaithCgfBFaithDg
22 9 10 21 eqbrrdiv φAFaithC=BFaithD