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