Metamath Proof Explorer


Theorem oppcinv

Description: An inverse 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 φ C Cat
oppcsect.x φ X B
oppcsect.y φ Y B
oppcinv.s I = Inv C
oppcinv.t J = Inv O
Assertion oppcinv φ X J Y = Y I X

Proof

Step Hyp Ref Expression
1 oppcsect.b B = Base C
2 oppcsect.o O = oppCat C
3 oppcsect.c φ C Cat
4 oppcsect.x φ X B
5 oppcsect.y φ Y B
6 oppcinv.s I = Inv C
7 oppcinv.t J = Inv O
8 incom X Sect O Y Y Sect O X -1 = Y Sect O X -1 X Sect O Y
9 eqid Sect C = Sect C
10 eqid Sect O = Sect O
11 1 2 3 5 4 9 10 oppcsect2 φ Y Sect O X = Y Sect C X -1
12 11 cnveqd φ Y Sect O X -1 = Y Sect C X -1 -1
13 eqid Hom C = Hom C
14 eqid comp C = comp C
15 eqid Id C = Id C
16 1 13 14 15 9 3 5 4 sectss φ Y Sect C X Y Hom C X × X Hom C Y
17 relxp Rel Y Hom C X × X Hom C Y
18 relss Y Sect C X Y Hom C X × X Hom C Y Rel Y Hom C X × X Hom C Y Rel Y Sect C X
19 16 17 18 mpisyl φ Rel Y Sect C X
20 dfrel2 Rel Y Sect C X Y Sect C X -1 -1 = Y Sect C X
21 19 20 sylib φ Y Sect C X -1 -1 = Y Sect C X
22 12 21 eqtrd φ Y Sect O X -1 = Y Sect C X
23 1 2 3 4 5 9 10 oppcsect2 φ X Sect O Y = X Sect C Y -1
24 22 23 ineq12d φ Y Sect O X -1 X Sect O Y = Y Sect C X X Sect C Y -1
25 8 24 eqtrid φ X Sect O Y Y Sect O X -1 = Y Sect C X X Sect C Y -1
26 2 1 oppcbas B = Base O
27 2 oppccat C Cat O Cat
28 3 27 syl φ O Cat
29 26 7 28 4 5 10 invfval φ X J Y = X Sect O Y Y Sect O X -1
30 1 6 3 5 4 9 invfval φ Y I X = Y Sect C X X Sect C Y -1
31 25 29 30 3eqtr4d φ X J Y = Y I X