Metamath Proof Explorer


Theorem cofu2

Description: Value of the morphism part of the functor composition. (Contributed by Mario Carneiro, 28-Jan-2017)

Ref Expression
Hypotheses cofuval.b B = Base C
cofuval.f φ F C Func D
cofuval.g φ G D Func E
cofu2nd.x φ X B
cofu2nd.y φ Y B
cofu2.h H = Hom C
cofu2.y φ R X H Y
Assertion cofu2 φ X 2 nd G func F Y R = 1 st F X 2 nd G 1 st F Y X 2 nd F Y R

Proof

Step Hyp Ref Expression
1 cofuval.b B = Base C
2 cofuval.f φ F C Func D
3 cofuval.g φ G D Func E
4 cofu2nd.x φ X B
5 cofu2nd.y φ Y B
6 cofu2.h H = Hom C
7 cofu2.y φ R X H Y
8 1 2 3 4 5 cofu2nd φ X 2 nd G func F Y = 1 st F X 2 nd G 1 st F Y X 2 nd F Y
9 8 fveq1d φ X 2 nd G func F Y R = 1 st F X 2 nd G 1 st F Y X 2 nd F Y R
10 eqid Hom D = Hom D
11 relfunc Rel C Func D
12 1st2ndbr Rel C Func D F C Func D 1 st F C Func D 2 nd F
13 11 2 12 sylancr φ 1 st F C Func D 2 nd F
14 1 6 10 13 4 5 funcf2 φ X 2 nd F Y : X H Y 1 st F X Hom D 1 st F Y
15 fvco3 X 2 nd F Y : X H Y 1 st F X Hom D 1 st F Y R X H Y 1 st F X 2 nd G 1 st F Y X 2 nd F Y R = 1 st F X 2 nd G 1 st F Y X 2 nd F Y R
16 14 7 15 syl2anc φ 1 st F X 2 nd G 1 st F Y X 2 nd F Y R = 1 st F X 2 nd G 1 st F Y X 2 nd F Y R
17 9 16 eqtrd φ X 2 nd G func F Y R = 1 st F X 2 nd G 1 st F Y X 2 nd F Y R