I have been confused by the 'op' thing for quite a while. On reading [Anindya's answer](https://forum.azimuthproject.org/discussion/comment/19538/#Comment_19538):

>If you think about how this might works on pairs of morphisms \\((f, g)\\) you'll notice it has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from \\(\mathcal{C}^{op} \times \mathcal{C}\\) to \\(\textbf{Set}\\).

I reckon this is a good way to approach the issue, but my brain fails to think out anything. :-/ @Anindya Could you perhaps illustrate a bit more details of the thinking process (even just some informal hints)?

On that note, I notice many textbooks use the 'op' notation without really explicitly explaining why, perhaps because it is too obvious for mathematically better-equipped readers. I see @John might still explain this in his latest lectures, so this course is really making a difference for beginner-beginners. =D>

>If you think about how this might works on pairs of morphisms \\((f, g)\\) you'll notice it has to be contravariant in the first variable and covariant in the second one. Hence it's a functor from \\(\mathcal{C}^{op} \times \mathcal{C}\\) to \\(\textbf{Set}\\).

I reckon this is a good way to approach the issue, but my brain fails to think out anything. :-/ @Anindya Could you perhaps illustrate a bit more details of the thinking process (even just some informal hints)?

On that note, I notice many textbooks use the 'op' notation without really explicitly explaining why, perhaps because it is too obvious for mathematically better-equipped readers. I see @John might still explain this in his latest lectures, so this course is really making a difference for beginner-beginners. =D>