Metamath Proof Explorer


Theorem fulloppc

Description: The opposite functor of a full functor is also full. Proposition 3.43(d) in Adamek p. 39. (Contributed by Mario Carneiro, 27-Jan-2017)

Ref Expression
Hypotheses fulloppc.o O=oppCatC
fulloppc.p P=oppCatD
fulloppc.f φFCFullDG
Assertion fulloppc φFOFullPtposG

Proof

Step Hyp Ref Expression
1 fulloppc.o O=oppCatC
2 fulloppc.p P=oppCatD
3 fulloppc.f φFCFullDG
4 fullfunc CFullDCFuncD
5 4 ssbri FCFullDGFCFuncDG
6 3 5 syl φFCFuncDG
7 1 2 6 funcoppc φFOFuncPtposG
8 eqid BaseC=BaseC
9 eqid HomD=HomD
10 eqid HomC=HomC
11 3 adantr φxBaseCyBaseCFCFullDG
12 simprr φxBaseCyBaseCyBaseC
13 simprl φxBaseCyBaseCxBaseC
14 8 9 10 11 12 13 fullfo φxBaseCyBaseCyGx:yHomCxontoFyHomDFx
15 forn yGx:yHomCxontoFyHomDFxranyGx=FyHomDFx
16 14 15 syl φxBaseCyBaseCranyGx=FyHomDFx
17 ovtpos xtposGy=yGx
18 17 rneqi ranxtposGy=ranyGx
19 9 2 oppchom FxHomPFy=FyHomDFx
20 16 18 19 3eqtr4g φxBaseCyBaseCranxtposGy=FxHomPFy
21 20 ralrimivva φxBaseCyBaseCranxtposGy=FxHomPFy
22 1 8 oppcbas BaseC=BaseO
23 eqid HomP=HomP
24 22 23 isfull FOFullPtposGFOFuncPtposGxBaseCyBaseCranxtposGy=FxHomPFy
25 7 21 24 sylanbrc φFOFullPtposG