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?