> "Mild problem" - like it completely doesn't work?
>
> Simon is much nicer than I am.

In the scale of mathematical disasters this construction really just suffered a minor glitch IMO.

If you want to see total disaster, one could look at David Hilbert's "proof" of the continuum hypothesis in [*On the Infinite* (1926)](https://math.dartmouth.edu/~matc/Readers/HowManyAngels/Philosophy/Philosophy.html).

Hilbert is so revered, and that proof is such a failure, the mathematicians I know are reluctant to talk about it.

I am no Hilbert. I know I make mistakes, even grave ones.

I came here to learn.

For this I am thankful for the patience and guidance of the people on this forum as I muddle about.

> **Revised Conjecture.** Every functor \\(F: \mathbf{Set} \to \mathbb{N}\\) is of this form: \\(F\\) sends every object to \\(\star\\), it sends every morphism \\(f: \emptyset \to Y\\) to the same morphism \\(n : \star \to \star\\), and it sends every morphism \\(f: X \to Y\\) with \\(X \ne \emptyset\\) to the identity morphism \\(1\_\star : \star \to \star\\).

I am reasonably confident this revised conjecture is true.

The second half is essentially the puzzle I wrote:

> Let \\(\mathbf{Set}^\dagger\\) be 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.

Here's my attempt at an answer.

**Proof.**

Take any function \\(f : A \to B\\) in \\(\mathbf{Set}^\dagger\\).

We have to prove \\(F(f) = id_{\star}\\).

Since \\(A\\) is not empty, there is some \\(a \in A\\).

Let \\(g : B \to A\\) be the constant function where \\(g(b) := a\\) for all \\(b \in B\\).

We have

\[(g \circ f) \circ (g \circ f) = (g \circ f)\]

Applying the functor \\(F\\) to either side gives:

\[ F((g \circ f) \circ (g \circ f)) = F(g \circ f) \]

Since \\(F\\) is a functor, it distributes across morphism composition:

\[ F(g \circ f) \circ F(g \circ f) = F(g \circ f) \]

There's only one [idempotent](https://en.wikipedia.org/wiki/Idempotence) element of \\(\mathbf{N}\\) and it is \\(id_{\star}\\). Hence

\[ F(g \circ f) = id_{\star} \]

Distributing gives

\[ F(g) \circ F(f) = id_{\star} \]

In the monoid \\(\langle \mathbb{N}, 0, + \rangle\\), if \\(x + y = 0\\) then \\(x = y = 0\\). We also have the monoid \\(\langle \mathbf{N}, id_{\star}, \circ \rangle \\) is isomorphic to \\(\langle \mathbb{N}, 0, + \rangle\\). Hence:

\[ F(g) = F(f) = id_{\star} \]

Which gives \\(F(f) = id_{\star}\\) as desired.

\\(\Box \\)