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>