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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 
|- ( Homf ` C ) = ( Homf ` ( oppCat ` O ) )

Proof

Step Hyp Ref Expression
1 oppcbas.1 
 |-  O = ( oppCat ` C )
2 eqid 
 |-  ( Homf ` C ) = ( Homf ` C )
3 eqid 
 |-  ( Base ` C ) = ( Base ` C )
4 2 3 homffn 
 |-  ( Homf ` C ) Fn ( ( Base ` C ) X. ( Base ` C ) )
5 fnrel 
 |-  ( ( Homf ` C ) Fn ( ( Base ` C ) X. ( Base ` C ) ) -> Rel ( Homf ` C ) )
6 4 5 ax-mp 
 |-  Rel ( Homf ` C )
7 relxp 
 |-  Rel ( ( Base ` C ) X. ( Base ` C ) )
8 4 fndmi 
 |-  dom ( Homf ` C ) = ( ( Base ` C ) X. ( Base ` C ) )
9 8 releqi 
 |-  ( Rel dom ( Homf ` C ) <-> Rel ( ( Base ` C ) X. ( Base ` C ) ) )
10 7 9 mpbir 
 |-  Rel dom ( Homf ` C )
11 tpostpos2 
 |-  ( ( Rel ( Homf ` C ) /\ Rel dom ( Homf ` C ) ) -> tpos tpos ( Homf ` C ) = ( Homf ` C ) )
12 6 10 11 mp2an 
 |-  tpos tpos ( Homf ` C ) = ( Homf ` C )
13 eqid 
 |-  ( oppCat ` O ) = ( oppCat ` O )
14 1 2 oppchomf 
 |-  tpos ( Homf ` C ) = ( Homf ` O )
15 13 14 oppchomf 
 |-  tpos tpos ( Homf ` C ) = ( Homf ` ( oppCat ` O ) )
16 12 15 eqtr3i 
 |-  ( Homf ` C ) = ( Homf ` ( oppCat ` O ) )