Metamath Proof Explorer


Theorem yonpropd

Description: If two categories have the same set of objects, morphisms, and compositions, then they have the same Yoneda functor. (Contributed by Mario Carneiro, 26-Jan-2017)

Ref Expression
Hypotheses hofpropd.1 φ Hom 𝑓 C = Hom 𝑓 D
hofpropd.2 φ comp 𝑓 C = comp 𝑓 D
hofpropd.c φ C Cat
hofpropd.d φ D Cat
Assertion yonpropd φ Yon C = Yon D

Proof

Step Hyp Ref Expression
1 hofpropd.1 φ Hom 𝑓 C = Hom 𝑓 D
2 hofpropd.2 φ comp 𝑓 C = comp 𝑓 D
3 hofpropd.c φ C Cat
4 hofpropd.d φ D Cat
5 1 oppchomfpropd φ Hom 𝑓 oppCat C = Hom 𝑓 oppCat D
6 1 2 oppccomfpropd φ comp 𝑓 oppCat C = comp 𝑓 oppCat D
7 eqid oppCat C = oppCat C
8 7 oppccat C Cat oppCat C Cat
9 3 8 syl φ oppCat C Cat
10 eqid oppCat D = oppCat D
11 10 oppccat D Cat oppCat D Cat
12 4 11 syl φ oppCat D Cat
13 eqid Hom F oppCat C = Hom F oppCat C
14 eqid SetCat ran Hom 𝑓 C = SetCat ran Hom 𝑓 C
15 fvex Hom 𝑓 C V
16 15 rnex ran Hom 𝑓 C V
17 16 a1i φ ran Hom 𝑓 C V
18 ssidd φ ran Hom 𝑓 C ran Hom 𝑓 C
19 7 13 14 3 17 18 oppchofcl φ Hom F oppCat C C × c oppCat C Func SetCat ran Hom 𝑓 C
20 1 2 5 6 3 4 9 12 19 curfpropd φ C oppCat C curry F Hom F oppCat C = D oppCat D curry F Hom F oppCat C
21 5 6 9 12 hofpropd φ Hom F oppCat C = Hom F oppCat D
22 21 oveq2d φ D oppCat D curry F Hom F oppCat C = D oppCat D curry F Hom F oppCat D
23 20 22 eqtrd φ C oppCat C curry F Hom F oppCat C = D oppCat D curry F Hom F oppCat D
24 eqid Yon C = Yon C
25 24 3 7 13 yonval φ Yon C = C oppCat C curry F Hom F oppCat C
26 eqid Yon D = Yon D
27 eqid Hom F oppCat D = Hom F oppCat D
28 26 4 10 27 yonval φ Yon D = D oppCat D curry F Hom F oppCat D
29 23 25 28 3eqtr4d φ Yon C = Yon D