> For a category \\(\mathcal{C}\\) an endofunctor \\(F: \mathbf{Hom}(\mathcal{C}) \to \mathbf{Hom}(\mathcal{C})\\) maps morphisms to morphisms.
> If I have three endofunctors \\(A,B,C\\) and a morphism \\(f\\) then we have associativity of functorial composition:

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

I think the difference between our constructions amounts to the difference between [Puzzle 99 and Puzzle 100](https://forum.azimuthproject.org/discussion/2198/lecture-34-chapter-3-categories/p1). In Puzzle 100, we can _label_ arrows from one object to another with particular endomorphisms (group elements), but _every_ object has some out-arrow labeled with this element. There isn't a one-to-one correspondence between endomorphisms and arrows, because we're applying \\(F\\) specially to each object and relating the single input to the single output.

Put differently, if we take each \\(s^{k}\\) as an object in our new category, and we induce all arrows \\(s^k \to s^{nk}\\) for each endofunctor \\(n\\), we end up with a partial order (except for zero). Indeed, this is the divisibility lattice on the positive natural numbers, augmented with a lonely \\(0\\) object that behaves like the single-object category I described, with one endomorphism for each \\(n\\).

(**EDIT:** Some confusion might arise from the fact that we often construct the naturals by \\(z, s(z), s(s(z)), \cdots\\), and we have a morphism \\(s\\) and an object \\(z\\) in evidence here. But \\(s(z)\\) is a nonsensical construction in \\(\textbf{N}\\)! Morphisms are not applied to objects. Objects represent the domain and codomain of morphisms. If \\(z\\) were a set, then we could take \\(x \in z\\) and have \\(s(x) \in z\\).)

> I don't think this is my construction. Saunder's 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.

> I don't think it makes a big difference for the puzzle I invented...

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.

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.

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.