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 = oppCat C
Assertion 2oppchomf Hom 𝑓 C = Hom 𝑓 oppCat O

Proof

Step Hyp Ref Expression
1 oppcbas.1 O = oppCat C
2 eqid Hom 𝑓 C = Hom 𝑓 C
3 eqid Base C = Base C
4 2 3 homffn Hom 𝑓 C Fn Base C × Base C
5 fnrel Hom 𝑓 C Fn Base C × Base C Rel Hom 𝑓 C
6 4 5 ax-mp Rel Hom 𝑓 C
7 relxp Rel Base C × Base C
8 4 fndmi dom Hom 𝑓 C = Base C × Base C
9 8 releqi Rel dom Hom 𝑓 C Rel Base C × Base C
10 7 9 mpbir Rel dom Hom 𝑓 C
11 tpostpos2 Rel Hom 𝑓 C Rel dom Hom 𝑓 C tpos tpos Hom 𝑓 C = Hom 𝑓 C
12 6 10 11 mp2an tpos tpos Hom 𝑓 C = Hom 𝑓 C
13 eqid oppCat O = oppCat O
14 1 2 oppchomf tpos Hom 𝑓 C = Hom 𝑓 O
15 13 14 oppchomf tpos tpos Hom 𝑓 C = Hom 𝑓 oppCat O
16 12 15 eqtr3i Hom 𝑓 C = Hom 𝑓 oppCat O