Anindya wrote:

> $(\check{F})^\text{op} = \widehat{F^\text{op}}$

Wow, that's cool! As you clearly realize, this reduces the study of conjoints to the study of companions together with the study of 'op' for both functors and profunctors!

I guess we also have

$(\widehat{F})^{\text{op}} = \check{F^\text{op}} .$

This should follow from \$$\text{op} \circ \text{op} = 1 \$$.