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