Okay, so continuing on... let \\(\mathcal{D}\\) be the free category on this graph:

and let \\(\mathcal{C}\\) be the free category on this graph:

We got an obvious 'inclusion' functor \\(G : \mathcal{D} \to \mathcal{C}\\). Now, let's try to think of this as a 'forgetful' functor. There's no technical definition of a forgetful functor, so this is just a matter of _attitude_. We need to imagine the objects of \\(\mathcal{C}\\) and \\(\mathcal{D}\\) as 'mathematical objects' of some kind. Their names fight against this way of thinking! That's why this problem is tough. But try.

The functor \\(G\\\) is full, faithful, but not essentially surjective. This basically says that \\(\mathcal{C}\\) has more objects than \\(\mathcal{D}\\), but if you have two objects of \\(\mathcal{C}\\) that came from objects in \\(\mathcal{D}\\), the morphisms between them in \\(\mathcal{D}\\) are the same as the morphisms between them in \\(\mathcal{C}\\)

A great mathematical example of this sort of situation is where\\( \mathcal{D} \\) is the category of abelian groups and \\(\mathcal{C}\\) is the category of groups. Here the forgetful functor \\(G : \mathcal{D} \to \mathcal{C}\\) forgets the property of being abelian.

In _our_ example, therefore, \\(G\\) is forgetting the property of being "not \\(\textbf{Department}\\)". That's a property that all objects of \\(\mathcal{D}\\) have, but not all objects of \\(\mathcal{C}\\).

I know this sounds weird. But it's just a psychological thing. If we regard the objects of \\(\mathcal{D}\\) as mathematical gadgets of some sort, and the objects of \\(\mathcal{C}\\) as mathematical gadgets with some extra property, then it's not so bad.