Metamath Proof Explorer


Theorem cofull

Description: The composition of two full functors is full. Proposition 3.30(d) in Adamek p. 35. (Contributed by Mario Carneiro, 28-Jan-2017)

Ref Expression
Hypotheses cofull.f φ F C Full D
cofull.g φ G D Full E
Assertion cofull φ G func F C Full E

Proof

Step Hyp Ref Expression
1 cofull.f φ F C Full D
2 cofull.g φ G D Full E
3 relfunc Rel C Func E
4 fullfunc C Full D C Func D
5 4 1 sselid φ F C Func D
6 fullfunc D Full E D Func E
7 6 2 sselid φ G D Func E
8 5 7 cofucl φ G func F C Func E
9 1st2nd Rel C Func E G func F C Func E G func F = 1 st G func F 2 nd G func F
10 3 8 9 sylancr φ G func F = 1 st G func F 2 nd G func F
11 1st2ndbr Rel C Func E G func F C Func E 1 st G func F C Func E 2 nd G func F
12 3 8 11 sylancr φ 1 st G func F C Func E 2 nd G func F
13 eqid Base D = Base D
14 eqid Hom E = Hom E
15 eqid Hom D = Hom D
16 relfull Rel D Full E
17 2 adantr φ x Base C y Base C G D Full E
18 1st2ndbr Rel D Full E G D Full E 1 st G D Full E 2 nd G
19 16 17 18 sylancr φ x Base C y Base C 1 st G D Full E 2 nd G
20 eqid Base C = Base C
21 relfunc Rel C Func D
22 5 adantr φ x Base C y Base C F C Func D
23 1st2ndbr Rel C Func D F C Func D 1 st F C Func D 2 nd F
24 21 22 23 sylancr φ x Base C y Base C 1 st F C Func D 2 nd F
25 20 13 24 funcf1 φ x Base C y Base C 1 st F : Base C Base D
26 simprl φ x Base C y Base C x Base C
27 25 26 ffvelrnd φ x Base C y Base C 1 st F x Base D
28 simprr φ x Base C y Base C y Base C
29 25 28 ffvelrnd φ x Base C y Base C 1 st F y Base D
30 13 14 15 19 27 29 fullfo φ x Base C y Base C 1 st F x 2 nd G 1 st F y : 1 st F x Hom D 1 st F y onto 1 st G 1 st F x Hom E 1 st G 1 st F y
31 eqid Hom C = Hom C
32 relfull Rel C Full D
33 1 adantr φ x Base C y Base C F C Full D
34 1st2ndbr Rel C Full D F C Full D 1 st F C Full D 2 nd F
35 32 33 34 sylancr φ x Base C y Base C 1 st F C Full D 2 nd F
36 20 15 31 35 26 28 fullfo φ x Base C y Base C x 2 nd F y : x Hom C y onto 1 st F x Hom D 1 st F y
37 foco 1 st F x 2 nd G 1 st F y : 1 st F x Hom D 1 st F y onto 1 st G 1 st F x Hom E 1 st G 1 st F y x 2 nd F y : x Hom C y onto 1 st F x Hom D 1 st F y 1 st F x 2 nd G 1 st F y x 2 nd F y : x Hom C y onto 1 st G 1 st F x Hom E 1 st G 1 st F y
38 30 36 37 syl2anc φ x Base C y Base C 1 st F x 2 nd G 1 st F y x 2 nd F y : x Hom C y onto 1 st G 1 st F x Hom E 1 st G 1 st F y
39 7 adantr φ x Base C y Base C G D Func E
40 20 22 39 26 28 cofu2nd φ x Base C y Base C x 2 nd G func F y = 1 st F x 2 nd G 1 st F y x 2 nd F y
41 eqidd φ x Base C y Base C x Hom C y = x Hom C y
42 20 22 39 26 cofu1 φ x Base C y Base C 1 st G func F x = 1 st G 1 st F x
43 20 22 39 28 cofu1 φ x Base C y Base C 1 st G func F y = 1 st G 1 st F y
44 42 43 oveq12d φ x Base C y Base C 1 st G func F x Hom E 1 st G func F y = 1 st G 1 st F x Hom E 1 st G 1 st F y
45 40 41 44 foeq123d φ x Base C y Base C x 2 nd G func F y : x Hom C y onto 1 st G func F x Hom E 1 st G func F y 1 st F x 2 nd G 1 st F y x 2 nd F y : x Hom C y onto 1 st G 1 st F x Hom E 1 st G 1 st F y
46 38 45 mpbird φ x Base C y Base C x 2 nd G func F y : x Hom C y onto 1 st G func F x Hom E 1 st G func F y
47 46 ralrimivva φ x Base C y Base C x 2 nd G func F y : x Hom C y onto 1 st G func F x Hom E 1 st G func F y
48 20 14 31 isfull2 1 st G func F C Full E 2 nd G func F 1 st G func F C Func E 2 nd G func F x Base C y Base C x 2 nd G func F y : x Hom C y onto 1 st G func F x Hom E 1 st G func F y
49 12 47 48 sylanbrc φ 1 st G func F C Full E 2 nd G func F
50 df-br 1 st G func F C Full E 2 nd G func F 1 st G func F 2 nd G func F C Full E
51 49 50 sylib φ 1 st G func F 2 nd G func F C Full E
52 10 51 eqeltrd φ G func F C Full E