Metamath Proof Explorer


Theorem cofid2

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
cofid1.f ⊢ φ → F D Func E G
cofid1.k ⊢ φ → K E Func D L
cofid1.o ⊢ φ → K L ∘ func F G = I
cofid2.y ⊢ φ → Y ∈ B
cofid2.h ⊢ H = Hom ⁡ D
cofid2.r ⊢ φ → R ∈ X H Y
Assertion cofid2 ⊢ φ → F ⁡ X L F ⁡ Y ⁡ X G 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 cofid1.f ⊢ φ → F D Func E G
5 cofid1.k ⊢ φ → K E Func D L
6 cofid1.o ⊢ φ → K L ∘ func F G = I
7 cofid2.y ⊢ φ → Y ∈ B
8 cofid2.h ⊢ H = Hom ⁡ D
9 cofid2.r ⊢ φ → R ∈ X H Y
10 5 func2nd ⊢ φ → 2 nd ⁡ K L = L
11 4 func1st ⊢ φ → 1 st ⁡ F G = F
12 11 fveq1d ⊢ φ → 1 st ⁡ F G ⁡ X = F ⁡ X
13 11 fveq1d ⊢ φ → 1 st ⁡ F G ⁡ Y = F ⁡ Y
14 10 12 13 oveq123d ⊢ φ → 1 st ⁡ F G ⁡ X 2 nd ⁡ K L 1 st ⁡ F G ⁡ Y = F ⁡ X L F ⁡ Y
15 4 func2nd ⊢ φ → 2 nd ⁡ F G = G
16 15 oveqd ⊢ φ → X 2 nd ⁡ F G Y = X G Y
17 16 fveq1d ⊢ φ → X 2 nd ⁡ F G Y ⁡ R = X G Y ⁡ R
18 14 17 fveq12d ⊢ φ → 1 st ⁡ F G ⁡ X 2 nd ⁡ K L 1 st ⁡ F G ⁡ Y ⁡ X 2 nd ⁡ F G Y ⁡ R = F ⁡ X L F ⁡ Y ⁡ X G Y ⁡ R
19 df-br ⊢ F D Func E G ↔ F G ∈ D Func E
20 4 19 sylib ⊢ φ → F G ∈ D Func E
21 df-br ⊢ K E Func D L ↔ K L ∈ E Func D
22 5 21 sylib ⊢ φ → K L ∈ E Func D
23 1 2 3 20 22 6 7 8 9 cofid2a ⊢ φ → 1 st ⁡ F G ⁡ X 2 nd ⁡ K L 1 st ⁡ F G ⁡ Y ⁡ X 2 nd ⁡ F G Y ⁡ R = R
24 18 23 eqtr3d ⊢ φ → F ⁡ X L F ⁡ Y ⁡ X G Y ⁡ R = R