Metamath Proof Explorer


Theorem diagpropd

Description: If two categories have the same set of objects, morphisms, and compositions, then they have same diagonal 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 diagpropd ⊢ φ → A Δ func C = B Δ func 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 eqid ⊢ A × c C = A × c C
10 eqid ⊢ A 1 st F C = A 1 st F C
11 9 5 7 10 1stfcl ⊢ φ → A 1 st F C ∈ A × c C Func A
12 1 2 3 4 5 6 7 8 11 curfpropd ⊢ φ → A C curry F A 1 st F C = B D curry F A 1 st F C
13 eqid ⊢ A Δ func C = A Δ func C
14 13 5 7 diagval ⊢ φ → A Δ func C = A C curry F A 1 st F C
15 eqid ⊢ B Δ func D = B Δ func D
16 15 6 8 diagval ⊢ φ → B Δ func D = B D curry F B 1 st F D
17 1 2 3 4 5 6 7 8 1stfpropd ⊢ φ → A 1 st F C = B 1 st F D
18 17 oveq2d ⊢ φ → B D curry F A 1 st F C = B D curry F B 1 st F D
19 16 18 eqtr4d ⊢ φ → B Δ func D = B D curry F A 1 st F C
20 12 14 19 3eqtr4d ⊢ φ → A Δ func C = B Δ func D