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

I argue this conjecture is *false*

This might be true for functions in traditional ZF set theory.

However, category theorists define morphisms on \\(\mathbf{Set}\\) a little differently. In particular, an *initial object* is considered.

(EDIT: After a little reread, I am wrong about this - everyone is using the same definition after all!)

**Proof.**

To start, I am going to write \\(\mathbf{1}\_\ast\\) as \\(id\_\ast\\). This is because if \\(\circ\\) for \\(\mathbf{N}\\) is like \\(+\\) then \\(\mathbf{1}_\ast\\) feels more like 0...

Let \\(\tilde{2}\\) be some morphism in \\(\mathbf{N}\\) where \\(\tilde{2} \neq id_\ast\\).

For any set \\(Y\\), consider the unique morphism \\(\phi : \varnothing \to Y \\). This unique morphism exists because \\(\varnothing\\) is the *initial object* of \\(\mathbf{Set}\\) (see [nLab](https://ncatlab.org/nlab/show/initial+object#examples)).

Now define the functor \\(F : \mathbf{Set} \to \mathbf{N}\\) such that for all \\(f : X \to Y \\) where \\(X \neq \emptyset\\) then \\( F(f) = id_\ast\\) and \\(F(\phi) = \tilde{2}\\) otherwise.

\\(F\\) obeys the functor laws and doesn't send everything to the identity morphism.

\\(\Box\\)