Metamath Proof Explorer


Theorem 2ndfpropd

Description: If two categories have the same set of objects, morphisms, and compositions, then they have same second projection functors. (Contributed by Zhi Wang, 20-Nov-2025)

Ref Expression
Hypotheses 1stfpropd.1 ⊢ φ → Hom 𝑓 ⁡ A = Hom 𝑓 ⁡ B
1stfpropd.2 ⊢ φ → comp 𝑓 ⁡ A = comp 𝑓 ⁡ B
1stfpropd.3 ⊢ φ → Hom 𝑓 ⁡ C = Hom 𝑓 ⁡ D
1stfpropd.4 ⊢ φ → comp 𝑓 ⁡ C = comp 𝑓 ⁡ D
1stfpropd.a ⊢ φ → A ∈ Cat
1stfpropd.b ⊢ φ → B ∈ Cat
1stfpropd.c ⊢ φ → C ∈ Cat
1stfpropd.d ⊢ φ → D ∈ Cat
Assertion 2ndfpropd ⊢ φ → A 2 nd F C = B 2 nd F D

Proof

Step Hyp Ref Expression
1 1stfpropd.1 ⊢ φ → Hom 𝑓 ⁡ A = Hom 𝑓 ⁡ B
2 1stfpropd.2 ⊢ φ → comp 𝑓 ⁡ A = comp 𝑓 ⁡ B
3 1stfpropd.3 ⊢ φ → Hom 𝑓 ⁡ C = Hom 𝑓 ⁡ D
4 1stfpropd.4 ⊢ φ → comp 𝑓 ⁡ C = comp 𝑓 ⁡ D
5 1stfpropd.a ⊢ φ → A ∈ Cat
6 1stfpropd.b ⊢ φ → B ∈ Cat
7 1stfpropd.c ⊢ φ → C ∈ Cat
8 1stfpropd.d ⊢ φ → D ∈ Cat
9 1 2 3 4 5 6 7 8 xpcpropd ⊢ φ → A × c C = B × c D
10 9 fveq2d ⊢ φ → Base A × c C = Base B × c D
11 10 reseq2d ⊢ φ → 2 nd ↾ Base A × c C = 2 nd ↾ Base B × c D
12 9 fveq2d ⊢ φ → Hom ⁡ A × c C = Hom ⁡ B × c D
13 12 oveqd ⊢ φ → x Hom ⁡ A × c C y = x Hom ⁡ B × c D y
14 13 reseq2d ⊢ φ → 2 nd ↾ x Hom ⁡ A × c C y = 2 nd ↾ x Hom ⁡ B × c D y
15 10 10 14 mpoeq123dv ⊢ φ → x ∈ Base A × c C , y ∈ Base A × c C ⟼ 2 nd ↾ x Hom ⁡ A × c C y = x ∈ Base B × c D , y ∈ Base B × c D ⟼ 2 nd ↾ x Hom ⁡ B × c D y
16 11 15 opeq12d ⊢ φ → 2 nd ↾ Base A × c C x ∈ Base A × c C , y ∈ Base A × c C ⟼ 2 nd ↾ x Hom ⁡ A × c C y = 2 nd ↾ Base B × c D x ∈ Base B × c D , y ∈ Base B × c D ⟼ 2 nd ↾ x Hom ⁡ B × c D y
17 eqid ⊢ A × c C = A × c C
18 eqid ⊢ Base A × c C = Base A × c C
19 eqid ⊢ Hom ⁡ A × c C = Hom ⁡ A × c C
20 eqid ⊢ A 2 nd F C = A 2 nd F C
21 17 18 19 5 7 20 2ndfval ⊢ φ → A 2 nd F C = 2 nd ↾ Base A × c C x ∈ Base A × c C , y ∈ Base A × c C ⟼ 2 nd ↾ x Hom ⁡ A × c C y
22 eqid ⊢ B × c D = B × c D
23 eqid ⊢ Base B × c D = Base B × c D
24 eqid ⊢ Hom ⁡ B × c D = Hom ⁡ B × c D
25 eqid ⊢ B 2 nd F D = B 2 nd F D
26 22 23 24 6 8 25 2ndfval ⊢ φ → B 2 nd F D = 2 nd ↾ Base B × c D x ∈ Base B × c D , y ∈ Base B × c D ⟼ 2 nd ↾ x Hom ⁡ B × c D y
27 16 21 26 3eqtr4d ⊢ φ → A 2 nd F C = B 2 nd F D