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=oppCatC
oppchomf.h H=Hom𝑓C
Assertion oppchomf tposH=Hom𝑓O

Proof

Step Hyp Ref Expression
1 oppcbas.1 O=oppCatC
2 oppchomf.h H=Hom𝑓C
3 eqid HomC=HomC
4 3 1 oppchom yHomOx=xHomCy
5 4 a1i yBaseCxBaseCyHomOx=xHomCy
6 5 mpoeq3ia yBaseC,xBaseCyHomOx=yBaseC,xBaseCxHomCy
7 eqid Hom𝑓O=Hom𝑓O
8 eqid BaseC=BaseC
9 1 8 oppcbas BaseC=BaseO
10 eqid HomO=HomO
11 7 9 10 homffval Hom𝑓O=yBaseC,xBaseCyHomOx
12 2 8 3 homffval H=xBaseC,yBaseCxHomCy
13 12 tposmpo tposH=yBaseC,xBaseCxHomCy
14 6 11 13 3eqtr4ri tposH=Hom𝑓O