Metamath Proof Explorer


Theorem 2oppchomf

Description: The double opposite category has the same morphisms as the original category. Intended for use with property lemmas such as monpropd . (Contributed by Mario Carneiro, 3-Jan-2017)

Ref Expression
Hypothesis oppcbas.1 O=oppCatC
Assertion 2oppchomf Hom𝑓C=Hom𝑓oppCatO

Proof

Step Hyp Ref Expression
1 oppcbas.1 O=oppCatC
2 eqid Hom𝑓C=Hom𝑓C
3 eqid BaseC=BaseC
4 2 3 homffn Hom𝑓CFnBaseC×BaseC
5 fnrel Hom𝑓CFnBaseC×BaseCRelHom𝑓C
6 4 5 ax-mp RelHom𝑓C
7 relxp RelBaseC×BaseC
8 4 fndmi domHom𝑓C=BaseC×BaseC
9 8 releqi ReldomHom𝑓CRelBaseC×BaseC
10 7 9 mpbir ReldomHom𝑓C
11 tpostpos2 RelHom𝑓CReldomHom𝑓CtpostposHom𝑓C=Hom𝑓C
12 6 10 11 mp2an tpostposHom𝑓C=Hom𝑓C
13 eqid oppCatO=oppCatO
14 1 2 oppchomf tposHom𝑓C=Hom𝑓O
15 13 14 oppchomf tpostposHom𝑓C=Hom𝑓oppCatO
16 12 15 eqtr3i Hom𝑓C=Hom𝑓oppCatO