Anindya wrote:

> [I]t 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'm not sure if this relevant for the current discussion, but I wanted to point out that if a map is functorial in each argument it's not a sufficient condition to conclude that it's a functor.
We need an additional coherence condition, known as the "interchange law"; for example, see the bifunctor lemma stated in [chapter 7]( from Awodey's book.