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