> **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.

**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\$$