Metamath Proof Explorer


Theorem cofid2a

Description: Express the morphism part of ( G o.func F ) = I explicitly. (Contributed by Zhi Wang, 15-Nov-2025)

Ref Expression
Hypotheses cofid1a.i I = id func D
cofid1a.b B = Base D
cofid1a.x φ X B
cofid1a.f φ F D Func E
cofid1a.g φ G E Func D
cofid1a.o φ G func F = I
cofid2a.y φ Y B
cofid2a.h H = Hom D
cofid2a.r φ R X H Y
Assertion cofid2a φ 1 st F X 2 nd G 1 st F Y X 2 nd F Y R = R

Proof

Step Hyp Ref Expression
1 cofid1a.i I = id func D
2 cofid1a.b B = Base D
3 cofid1a.x φ X B
4 cofid1a.f φ F D Func E
5 cofid1a.g φ G E Func D
6 cofid1a.o φ G func F = I
7 cofid2a.y φ Y B
8 cofid2a.h H = Hom D
9 cofid2a.r φ R X H Y
10 6 fveq2d φ 2 nd G func F = 2 nd I
11 10 oveqd φ X 2 nd G func F Y = X 2 nd I Y
12 11 fveq1d φ X 2 nd G func F Y R = X 2 nd I Y R
13 2 4 5 3 7 8 9 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
14 4 func1st2nd φ 1 st F D Func E 2 nd F
15 14 funcrcl2 φ D Cat
16 1 2 15 8 3 7 9 idfu2 φ X 2 nd I Y R = R
17 12 13 16 3eqtr3d φ 1 st F X 2 nd G 1 st F Y X 2 nd F Y R = R