Matthew wrote:

> I argue this conjecture is _false_.

Hey, I think you're right! If I get the gist of your argument, it goes like this. Pick a nonempty set \\(S\\) and define a functor \\(F: \mathbf{Set} \to \mathbf{N}\\) that sends the unique function \\(f: \emptyset \to S\\) to some non-identity morphism in \\(\mathbf{N}\\) and all other morphisms to the identity. Check that this is a functor.

(We need \\(S\\) to be nonempty so that \\(f\\) isn't an identity morphism, so that \\(F(f)\\) can be chosen to be a non-identity morphism.)

Very nice!

> I argue this conjecture is _false_.

Hey, I think you're right! If I get the gist of your argument, it goes like this. Pick a nonempty set \\(S\\) and define a functor \\(F: \mathbf{Set} \to \mathbf{N}\\) that sends the unique function \\(f: \emptyset \to S\\) to some non-identity morphism in \\(\mathbf{N}\\) and all other morphisms to the identity. Check that this is a functor.

(We need \\(S\\) to be nonempty so that \\(f\\) isn't an identity morphism, so that \\(F(f)\\) can be chosen to be a non-identity morphism.)

Very nice!