Metamath Proof Explorer


Theorem oppgoppchom

Description: The converted opposite monoid has the same hom-set as that of the opposite category. Example 3.6(2) of Adamek p. 25. (Contributed by Zhi Wang, 21-Sep-2025)

Ref Expression
Hypotheses mndtccat.c φ C = MndToCat M
mndtccat.m φ M Mnd
oppgoppchom.d φ D = MndToCat opp 𝑔 M
oppgoppchom.o O = oppCat C
oppgoppchom.x φ X Base D
oppgoppchom.y φ Y Base O
oppgoppchom.h φ H = Hom D
oppgoppchom.j φ J = Hom O
Assertion oppgoppchom φ X H X = Y J Y

Proof

Step Hyp Ref Expression
1 mndtccat.c φ C = MndToCat M
2 mndtccat.m φ M Mnd
3 oppgoppchom.d φ D = MndToCat opp 𝑔 M
4 oppgoppchom.o O = oppCat C
5 oppgoppchom.x φ X Base D
6 oppgoppchom.y φ Y Base O
7 oppgoppchom.h φ H = Hom D
8 oppgoppchom.j φ J = Hom O
9 eqid opp 𝑔 M = opp 𝑔 M
10 eqid Base M = Base M
11 9 10 oppgbas Base M = Base opp 𝑔 M
12 11 a1i φ Base M = Base opp 𝑔 M
13 eqid Base C = Base C
14 4 13 oppcbas Base C = Base O
15 14 eqcomi Base O = Base C
16 15 a1i φ Base O = Base C
17 eqidd φ Hom C = Hom C
18 1 2 16 6 6 17 mndtchom φ Y Hom C Y = Base M
19 9 oppgmnd M Mnd opp 𝑔 M Mnd
20 2 19 syl φ opp 𝑔 M Mnd
21 eqidd φ Base D = Base D
22 3 20 21 5 5 7 mndtchom φ X H X = Base opp 𝑔 M
23 12 18 22 3eqtr4rd φ X H X = Y Hom C Y
24 eqid Hom C = Hom C
25 24 4 oppchom Y Hom O Y = Y Hom C Y
26 23 25 eqtr4di φ X H X = Y Hom O Y
27 8 oveqd φ Y J Y = Y Hom O Y
28 26 27 eqtr4d φ X H X = Y J Y