Metamath Proof Explorer


Theorem uobeq

Description: If a full functor (in fact, a full embedding) is a section of a functor (surjective on objects), then the sets of universal objects are equal. (Contributed by Zhi Wang, 17-Nov-2025)

Ref Expression
Hypotheses uobffth.b B = Base D
uobffth.x φ X B
uobffth.f φ F C Func D
uobffth.g φ K func F = G
uobffth.y φ 1 st K X = Y
uobeq.i I = id func D
uobeq.k φ K D Full E
uobeq.n φ L func K = I
uobeq.l φ L E Func D
Assertion uobeq Could not format assertion : No typesetting found for |- ( ph -> dom ( F ( C UP D ) X ) = dom ( G ( C UP E ) Y ) ) with typecode |-

Proof

Step Hyp Ref Expression
1 uobffth.b B = Base D
2 uobffth.x φ X B
3 uobffth.f φ F C Func D
4 uobffth.g φ K func F = G
5 uobffth.y φ 1 st K X = Y
6 uobeq.i I = id func D
7 uobeq.k φ K D Full E
8 uobeq.n φ L func K = I
9 uobeq.l φ L E Func D
10 relfunc Rel D Func E
11 fullfunc D Full E D Func E
12 11 7 sselid φ K D Func E
13 1st2nd Rel D Func E K D Func E K = 1 st K 2 nd K
14 10 12 13 sylancr φ K = 1 st K 2 nd K
15 12 func1st2nd φ 1 st K D Func E 2 nd K
16 9 func1st2nd φ 1 st L E Func D 2 nd L
17 12 9 cofu1st2nd φ L func K = 1 st L 2 nd L func 1 st K 2 nd K
18 17 8 eqtr3d φ 1 st L 2 nd L func 1 st K 2 nd K = I
19 6 15 16 18 cofidfth φ 1 st K D Faith E 2 nd K
20 df-br 1 st K D Faith E 2 nd K 1 st K 2 nd K D Faith E
21 19 20 sylib φ 1 st K 2 nd K D Faith E
22 14 21 eqeltrd φ K D Faith E
23 7 22 elind φ K D Full E D Faith E
24 1 2 3 4 5 23 uobffth Could not format ( ph -> dom ( F ( C UP D ) X ) = dom ( G ( C UP E ) Y ) ) : No typesetting found for |- ( ph -> dom ( F ( C UP D ) X ) = dom ( G ( C UP E ) Y ) ) with typecode |-