Owen wrote:

> Puzzle 124: I don't think there's just one functor from \\(\mathcal{D}\\) to \\(\mathcal{C}\\). There is just one graph homomorphism, though.

You're right!

I'll fix the puzzle by indicating the functor I'm actually interested in: the one that maps the morphisms \\(\textrm{FriendOf}\\) and \\(\textrm{FriendOf}^\prime\\) in \\(\mathcal{D}\\) to the morphism \\(\textrm{FriendOf}\\) in \\(\mathcal{C}\\).

> (Feel free to delete this paragraph after reading it.)

I'll leave it in! I don't want to convey the false impression that I'm infallible. I do, however, want to fix up my "lectures" so that students don't get confused by mistakes.

Anindya wrote:

> Could we not send one or both FriendOf in \\(\mathcal{D}\\) to the identity map in \\(\mathcal{C}\\)?

Yes, that's one of infinitely many other functors from \\(\mathcal{D}\\) to \\(\mathcal{C}\\) I hadn't noticed when first stating Puzzle 124.