Metamath Proof Explorer


Theorem oppchomf

Description: Hom-sets of the opposite category. (Contributed by Mario Carneiro, 17-Jan-2017)

Ref Expression
Hypotheses oppcbas.1 O = oppCat C
oppchomf.h H = Hom 𝑓 C
Assertion oppchomf tpos H = Hom 𝑓 O

Proof

Step Hyp Ref Expression
1 oppcbas.1 O = oppCat C
2 oppchomf.h H = Hom 𝑓 C
3 eqid Hom C = Hom C
4 3 1 oppchom y Hom O x = x Hom C y
5 4 a1i y Base C x Base C y Hom O x = x Hom C y
6 5 mpoeq3ia y Base C , x Base C y Hom O x = y Base C , x Base C x Hom C y
7 eqid Hom 𝑓 O = Hom 𝑓 O
8 eqid Base C = Base C
9 1 8 oppcbas Base C = Base O
10 eqid Hom O = Hom O
11 7 9 10 homffval Hom 𝑓 O = y Base C , x Base C y Hom O x
12 2 8 3 homffval H = x Base C , y Base C x Hom C y
13 12 tposmpo tpos H = y Base C , x Base C x Hom C y
14 6 11 13 3eqtr4ri tpos H = Hom 𝑓 O