Metamath Proof Explorer


Theorem oppcsect2

Description: A section in the opposite category. (Contributed by Mario Carneiro, 3-Jan-2017)

Ref Expression
Hypotheses oppcsect.b
|- B = ( Base ` C )
oppcsect.o
|- O = ( oppCat ` C )
oppcsect.c
|- ( ph -> C e. Cat )
oppcsect.x
|- ( ph -> X e. B )
oppcsect.y
|- ( ph -> Y e. B )
oppcsect.s
|- S = ( Sect ` C )
oppcsect.t
|- T = ( Sect ` O )
Assertion oppcsect2
|- ( ph -> ( X T Y ) = `' ( X S Y ) )

Proof

Step Hyp Ref Expression
1 oppcsect.b
 |-  B = ( Base ` C )
2 oppcsect.o
 |-  O = ( oppCat ` C )
3 oppcsect.c
 |-  ( ph -> C e. Cat )
4 oppcsect.x
 |-  ( ph -> X e. B )
5 oppcsect.y
 |-  ( ph -> Y e. B )
6 oppcsect.s
 |-  S = ( Sect ` C )
7 oppcsect.t
 |-  T = ( Sect ` O )
8 2 1 oppcbas
 |-  B = ( Base ` O )
9 eqid
 |-  ( Hom ` O ) = ( Hom ` O )
10 eqid
 |-  ( comp ` O ) = ( comp ` O )
11 eqid
 |-  ( Id ` O ) = ( Id ` O )
12 2 oppccat
 |-  ( C e. Cat -> O e. Cat )
13 3 12 syl
 |-  ( ph -> O e. Cat )
14 8 9 10 11 7 13 4 5 sectss
 |-  ( ph -> ( X T Y ) C_ ( ( X ( Hom ` O ) Y ) X. ( Y ( Hom ` O ) X ) ) )
15 relxp
 |-  Rel ( ( X ( Hom ` O ) Y ) X. ( Y ( Hom ` O ) X ) )
16 relss
 |-  ( ( X T Y ) C_ ( ( X ( Hom ` O ) Y ) X. ( Y ( Hom ` O ) X ) ) -> ( Rel ( ( X ( Hom ` O ) Y ) X. ( Y ( Hom ` O ) X ) ) -> Rel ( X T Y ) ) )
17 14 15 16 mpisyl
 |-  ( ph -> Rel ( X T Y ) )
18 relcnv
 |-  Rel `' ( X S Y )
19 18 a1i
 |-  ( ph -> Rel `' ( X S Y ) )
20 1 2 3 4 5 6 7 oppcsect
 |-  ( ph -> ( f ( X T Y ) g <-> g ( X S Y ) f ) )
21 vex
 |-  f e. _V
22 vex
 |-  g e. _V
23 21 22 brcnv
 |-  ( f `' ( X S Y ) g <-> g ( X S Y ) f )
24 20 23 bitr4di
 |-  ( ph -> ( f ( X T Y ) g <-> f `' ( X S Y ) g ) )
25 17 19 24 eqbrrdv
 |-  ( ph -> ( X T Y ) = `' ( X S Y ) )