> "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 \$$