Metamath Proof Explorer


Theorem oppgoppcid

Description: The converted opposite monoid has the same identity morphism as that of the opposite category. Example 3.6(2) of Adamek p. 25. (Contributed by Zhi Wang, 22-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
Assertion oppgoppcid ⊢ φ → Id ⁡ D ⁡ X = Id ⁡ O ⁡ 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 eqid ⊢ opp 𝑔 ⁡ M = opp 𝑔 ⁡ M
8 eqid ⊢ 0 M = 0 M
9 7 8 oppgid ⊢ 0 M = 0 opp 𝑔 ⁡ M
10 9 a1i ⊢ φ → 0 M = 0 opp 𝑔 ⁡ M
11 eqid ⊢ Base C = Base C
12 4 11 oppcbas ⊢ Base C = Base O
13 12 eqcomi ⊢ Base O = Base C
14 13 a1i ⊢ φ → Base O = Base C
15 1 2 mndtccat ⊢ φ → C ∈ Cat
16 eqid ⊢ Id ⁡ C = Id ⁡ C
17 4 16 oppcid ⊢ C ∈ Cat → Id ⁡ O = Id ⁡ C
18 15 17 syl ⊢ φ → Id ⁡ O = Id ⁡ C
19 1 2 14 6 18 mndtcid ⊢ φ → Id ⁡ O ⁡ Y = 0 M
20 7 oppgmnd ⊢ M ∈ Mnd → opp 𝑔 ⁡ M ∈ Mnd
21 2 20 syl ⊢ φ → opp 𝑔 ⁡ M ∈ Mnd
22 eqidd ⊢ φ → Base D = Base D
23 eqidd ⊢ φ → Id ⁡ D = Id ⁡ D
24 3 21 22 5 23 mndtcid ⊢ φ → Id ⁡ D ⁡ X = 0 opp 𝑔 ⁡ M
25 10 19 24 3eqtr4rd ⊢ φ → Id ⁡ D ⁡ X = Id ⁡ O ⁡ Y