> So we have \\(A, B, C : \mathbf{Hom}(\mathcal{C}) \to \mathbf{Hom}(\mathcal{C})\\). Are there endofunctors with any other signature?

No, there are not. All endofunctors have that signature.

Now that I've slept on this, I believe you are right. I made a level slip.

The revised definition of the category \\(\mathbf{Mult}\\) should be:

> - The objects of this category are defined to be \\(\mathrm{Obj}(\mathbf{Mult}) = \lbrace\mathbf{Mor}(\mathbf{N})\rbrace\\). In other words there is a single object.
> - The identity is the identity functor \\(\mathbf{1}_{\bullet}\\), corresponding to the scaling \\(1 \times \cdot\\)
> - Morphisms in this category are functors \\(F : \mathbf{N} \to \mathbf{N}\\)
> - Morphism composition is functor composition \\(\bullet\\)

------------------------------------

> > I don't think this is my construction. Saunders MacLane defines exactly the same thing on page 128 of [*Categories for the Working Mathematician* (1968)](https://books.google.com/books?id=MXboNPdTv7QC&pg=PA138&lpg=PA138&dq=%22monoid+in+the+category+of+endofunctors%22+mac+lane&source=bl&ots=feQWTkH2Uw&sig=tv-1JwaMOygKGmFE2vM2FhJVS9o&hl=en&ei=5iWsTJCkBIPSsAPQwJ36Aw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q&f=false).
>
> He seems to be doing something with a single fixed endofunctor \\(T\\) and some natural transformations \\(\eta, \mu\\), the latter of which I have no familiarity with. We have a whole family of endofunctors and no obvious maps between any of them.

First, I was wrong. Saunders MacLane has a different construction, based on the [Category of Functors](https://en.wikipedia.org/wiki/Functor_category). He is considering a whole family of endofunctors, however.

Moreover, I think you *may* have seen the \\(\eta\\) and \\(\mu\\) MacLane is talking about using different names. In Haskell, we write \\(\eta\\) as [`return`](https://hackage.haskell.org/package/base-4.11.1.0/docs/Prelude.html#v:return) (from the `Prelude`) and \\(\mu\\) as [`Control.Monad.join`](https://hackage.haskell.org/package/base-4.11.1.0/docs/Control-Monad.html#v:join).

> If \\(\textbf{Mult}\\) is the divisibility lattice augmented with zero, then we have essentially two classes of functor \\(F : \mathbf{N} \to \mathbf{Mult}\\). One functor maps \\(z\\) onto a non-zero element, forcing all morphisms onto the identity morphism. It doesn't much matter which number we map onto; the image of the functor will be a single object with its identity morphism. The other functors are covered by the case below.

I don't think we can map to any other constant other than \\(1_{\mathbf{Mult}}\\), sadly.

> If \\(\textbf{Mult}\\) is the single-object category with functors \\(\mathbf{N} \to \mathbf{N}\\) as morphisms, then functors \\(\mathbf{N} \to \mathbf{Mult}\\) are monoid homomorphisms between \\(\langle \mathbb{N}, + \rangle\\) and \\(\langle \mathbb{N}, \cdot \rangle\\). It suffices to consider where \\(1\\) is mapped to. If \\(1\\) maps to \\(n\\) for a choice of \\(n\\), then \\(k\\) maps to \\(n^k\\). In other words, we get exponentiation.

Exactly!

> If we do this again, what do we get? There's a [name](https://en.wikipedia.org/wiki/Hyperoperation#Examples) for what's happening here.

I am not sure how to do this again.

Both \\(\mathbf{Nat}\\) and \\(\mathbf{Mult}\\) are monoids.

However, the exponential functors \\(E : \mathbf{Nat} \to \mathbf{Mult}\\) don't compose, so they don't form a monoid.

In particular, if I have \\(3^k\\) and \\(5^k\\), there is no natural number \\(n\\) where \\(n^k = 3^{5^k}\\) for all \\(k\\).

But I know what you are getting at, so I want to try to puzzle it out.

One strategy might be to make the functors \\(E : \mathbf{Nat} \to \mathbf{Mult}\\) into a category with natural transformations as morphisms, following the [Category of Functors](https://en.wikipedia.org/wiki/Functor_category) construction I mentioned. Then look at endofunctors on this and see what those are. And then we get *hyped* up about it if this works! But I am not sure if that will work, it's just an idea.