The conjecture really is compelling, however. Here's a little modification I was thinking of to get it to work.

Consider \\(\mathbf{Set}^\dagger\\), which is the same as \\(\mathbf{Set}\\) but without \\(\emptyset\\).

**Puzzle MD 1.** Show that any functor \\( F: \mathbf{Set}^\dagger \to \mathbf{N}\\) must send every morphism in \\(\textbf{Set}^\dagger \\) to the identity morphism.

Consider \\(\mathbf{Set}^\dagger\\), which is the same as \\(\mathbf{Set}\\) but without \\(\emptyset\\).

**Puzzle MD 1.** Show that any functor \\( F: \mathbf{Set}^\dagger \to \mathbf{N}\\) must send every morphism in \\(\textbf{Set}^\dagger \\) to the identity morphism.