Keith wrote:

> Why not simply take \\(\mathbf{N}\\) as the free category on one object and one (non-identity) morphism?

Okay, that's one good way. The category with one object \\(\star\\), one morphism for each natural number, and addition as composition!

Now, you were talking about functors \\( F: \mathbf{Set} \to \mathbf{N}\\).

**Puzzle.** What's a functor \\( F: \mathbf{Set} \to \mathbf{N}\\) if we make \\(\mathbf{N}\\) into a category as you've suggested?

This is actually a fun question I've never thought about. Let me start. \\(\mathbf{N}\\) has just one object \\(\star\\) so \\(F\\) must send every set to that object.

(So, if you were trying to use such a functor to count the number of elements in sets, it's not going to work.)

\\(\mathbf{N}\\) has just one invertible morphism, namely the identity \\(1_\star\\), so \\(F\\) must send every invertible morphism in \\(\mathbf{Set}\\) to this identity morphism. You see, functors map invertible morphisms to invertible morphisms. And the invertible morphisms in \\(\mathbf{Set}\\) are the bijections. So, \\(F\\) must send every bijection to the identity!

Based on this tiny bit of information, and my intuition, I'll make a wild guess:

**Conjecture.** Any functor \\( F: \mathbf{Set} \to \mathbf{N}\\) must send every morphism in \\(\textbf{Set}\\) to the identity morphism.

Can someone prove it or find a counterexample?