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
|- ( ph -> C = ( MndToCat ` M ) )
mndtccat.m
|- ( ph -> M e. Mnd )
oppgoppchom.d
|- ( ph -> D = ( MndToCat ` ( oppG ` M ) ) )
oppgoppchom.o
|- O = ( oppCat ` C )
oppgoppchom.x
|- ( ph -> X e. ( Base ` D ) )
oppgoppchom.y
|- ( ph -> Y e. ( Base ` O ) )
Assertion oppgoppcid
|- ( ph -> ( ( Id ` D ) ` X ) = ( ( Id ` O ) ` Y ) )

Proof

Step Hyp Ref Expression
1 mndtccat.c
 |-  ( ph -> C = ( MndToCat ` M ) )
2 mndtccat.m
 |-  ( ph -> M e. Mnd )
3 oppgoppchom.d
 |-  ( ph -> D = ( MndToCat ` ( oppG ` M ) ) )
4 oppgoppchom.o
 |-  O = ( oppCat ` C )
5 oppgoppchom.x
 |-  ( ph -> X e. ( Base ` D ) )
6 oppgoppchom.y
 |-  ( ph -> Y e. ( Base ` O ) )
7 eqid
 |-  ( oppG ` M ) = ( oppG ` M )
8 eqid
 |-  ( 0g ` M ) = ( 0g ` M )
9 7 8 oppgid
 |-  ( 0g ` M ) = ( 0g ` ( oppG ` M ) )
10 9 a1i
 |-  ( ph -> ( 0g ` M ) = ( 0g ` ( oppG ` 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
 |-  ( ph -> ( Base ` O ) = ( Base ` C ) )
15 1 2 mndtccat
 |-  ( ph -> C e. Cat )
16 eqid
 |-  ( Id ` C ) = ( Id ` C )
17 4 16 oppcid
 |-  ( C e. Cat -> ( Id ` O ) = ( Id ` C ) )
18 15 17 syl
 |-  ( ph -> ( Id ` O ) = ( Id ` C ) )
19 1 2 14 6 18 mndtcid
 |-  ( ph -> ( ( Id ` O ) ` Y ) = ( 0g ` M ) )
20 7 oppgmnd
 |-  ( M e. Mnd -> ( oppG ` M ) e. Mnd )
21 2 20 syl
 |-  ( ph -> ( oppG ` M ) e. Mnd )
22 eqidd
 |-  ( ph -> ( Base ` D ) = ( Base ` D ) )
23 eqidd
 |-  ( ph -> ( Id ` D ) = ( Id ` D ) )
24 3 21 22 5 23 mndtcid
 |-  ( ph -> ( ( Id ` D ) ` X ) = ( 0g ` ( oppG ` M ) ) )
25 10 19 24 3eqtr4rd
 |-  ( ph -> ( ( Id ` D ) ` X ) = ( ( Id ` O ) ` Y ) )