Metamath Proof Explorer


Theorem homfeq

Description: Condition for two categories with the same base to have the same hom-sets. (Contributed by Mario Carneiro, 6-Jan-2017)

Ref Expression
Hypotheses homfeq.h H=HomC
homfeq.j J=HomD
homfeq.1 φB=BaseC
homfeq.2 φB=BaseD
Assertion homfeq φHom𝑓C=Hom𝑓DxByBxHy=xJy

Proof

Step Hyp Ref Expression
1 homfeq.h H=HomC
2 homfeq.j J=HomD
3 homfeq.1 φB=BaseC
4 homfeq.2 φB=BaseD
5 eqid Hom𝑓C=Hom𝑓C
6 eqid BaseC=BaseC
7 5 6 1 homffval Hom𝑓C=xBaseC,yBaseCxHy
8 eqidd φxHy=xHy
9 3 3 8 mpoeq123dv φxB,yBxHy=xBaseC,yBaseCxHy
10 7 9 eqtr4id φHom𝑓C=xB,yBxHy
11 eqid Hom𝑓D=Hom𝑓D
12 eqid BaseD=BaseD
13 11 12 2 homffval Hom𝑓D=xBaseD,yBaseDxJy
14 eqidd φxJy=xJy
15 4 4 14 mpoeq123dv φxB,yBxJy=xBaseD,yBaseDxJy
16 13 15 eqtr4id φHom𝑓D=xB,yBxJy
17 10 16 eqeq12d φHom𝑓C=Hom𝑓DxB,yBxHy=xB,yBxJy
18 ovex xHyV
19 18 rgen2w xByBxHyV
20 mpo2eqb xByBxHyVxB,yBxHy=xB,yBxJyxByBxHy=xJy
21 19 20 ax-mp xB,yBxHy=xB,yBxJyxByBxHy=xJy
22 17 21 bitrdi φHom𝑓C=Hom𝑓DxByBxHy=xJy