Metamath Proof Explorer


Theorem fulloppc

Description: The opposite functor of a full functor is also full. Proposition 3.43(d) in Adamek p. 39. (Contributed by Mario Carneiro, 27-Jan-2017)

Ref Expression
Hypotheses fulloppc.o O = oppCat C
fulloppc.p P = oppCat D
fulloppc.f φ F C Full D G
Assertion fulloppc φ F O Full P tpos G

Proof

Step Hyp Ref Expression
1 fulloppc.o O = oppCat C
2 fulloppc.p P = oppCat D
3 fulloppc.f φ F C Full D G
4 fullfunc C Full D C Func D
5 4 ssbri F C Full D G F C Func D G
6 3 5 syl φ F C Func D G
7 1 2 6 funcoppc φ F O Func P tpos G
8 eqid Base C = Base C
9 eqid Hom D = Hom D
10 eqid Hom C = Hom C
11 3 adantr φ x Base C y Base C F C Full D G
12 simprr φ x Base C y Base C y Base C
13 simprl φ x Base C y Base C x Base C
14 8 9 10 11 12 13 fullfo φ x Base C y Base C y G x : y Hom C x onto F y Hom D F x
15 forn y G x : y Hom C x onto F y Hom D F x ran y G x = F y Hom D F x
16 14 15 syl φ x Base C y Base C ran y G x = F y Hom D F x
17 ovtpos x tpos G y = y G x
18 17 rneqi ran x tpos G y = ran y G x
19 9 2 oppchom F x Hom P F y = F y Hom D F x
20 16 18 19 3eqtr4g φ x Base C y Base C ran x tpos G y = F x Hom P F y
21 20 ralrimivva φ x Base C y Base C ran x tpos G y = F x Hom P F y
22 1 8 oppcbas Base C = Base O
23 eqid Hom P = Hom P
24 22 23 isfull F O Full P tpos G F O Func P tpos G x Base C y Base C ran x tpos G y = F x Hom P F y
25 7 21 24 sylanbrc φ F O Full P tpos G