>**Puzzle.** If \\(\mathbf{N}\\) is the free category on the graph

>what is the endofunctor category \\(\mathbf{N}^\mathbf{N}\\)?

**Summary**: Objects in \\(\mathbf{N}^\mathbf{N}\\) correspond to scalar multipliers \\(k \times \cdot : \mathbb{N} \to \mathbb{N}\\), while morphisms act like scalar addition \\(c+ \cdot : \mathbb{N} \to \mathbb{N} \\). There is no morphism from one functor to another in \\(\mathbf{N}^\mathbf{N}\\).


We have seen that functors \\(F: \mathbf{N} \to \mathbf{N}\\) are just scalar multipliers.

From Wikipedia's definition of a [*Functor Category*](https://en.wikipedia.org/wiki/Functor_category), we have:

> In category theory, a branch of mathematics, the functors between two given categories form a category, where the objects are the *functors* and the morphisms are *natural transformations* between the functors.

Wikipedia gives the following definition of *natural transformations*:

> A natural transformation \\(\eta\\) from functors \\(F: \mathbf{C} \to \mathbf{D}\\) to \\(G: \mathbf{C} \to \mathbf{D}\\) is a family of morphisms that satisfies two requirements.

> 1. The natural transformation must associate to every object \\(X\\) in \\(\mathbf{C}\\) a morphism \\(\eta_X : F(X) → G(X)\\) between objects of \\(\mathbf{D}\\). The morphism \\(\eta_X\\) is called the component of \\(\eta\\) at \\(X\\).
> 2. Components must be such that for every morphism \\(f : X \to Y\\) in \\(\mathbf{C}\\) we have:
> $$\eta_{Y}\circ_{\mathbf{D}} F(f) =G(f)\circ_{\mathbf{D}} \eta _{X}$$

There is only one object in \\(\mathbf{N}\\) and it is \\(\star\\). The definition of natural transformation simplifies to:

\[ \eta _{\star} \circ\_{\mathbf{N}} F(f) =G(f) \circ\_{\mathbf{N}} \eta _{\star} \tag{✴} \]

Since both \\(\mathbf{C}\\) and \\(\mathbf{D}\\) are \\(\mathbf{N}\\) for functors in \\(\mathbf{N} \to \mathbf{N}\\), we have \\(\circ\_{\mathbf{D}} = \circ\_{\mathbf{N}}\\).

Note that \\(\circ\\) in \\(\mathbf{N}\\) reflects addition and \\(\eta _{\star}\\) corresponds to some \\(c \in \mathbb{N}\\).

Also note that \\(F \in \mathbf{N}^\mathbf{N} \\) reflects \\(k \times \cdot: \mathbb{N} \to \mathbb{N}\\) for some \\(k \in \mathbb{N}\\) and \\(G \in \mathbf{N}^\mathbf{N} \\) reflects \\(l \times \cdot: \mathbb{N} \to \mathbb{N}\\) for some \\(l \in \mathbb{N}\\).

This means (✴) is corresponds to, for all \\(n \in \mathbb{N}\\):

\[ c + k \times n = l \times n + c\]

However, we know that addition is commutative and all scalar additions \\(c + \cdot : \mathbb{N} \to \mathbb{N}\\) are injective. So we may cancel \\(c\\) on either side of (✴). Thus we have for all morphisms \\(f \in \mathbf{Mor}(\mathbf{N}) \\):

\[ k \times n =l \times n \]

Hence \\(k = l\\), and thus \\(F = G\\).

This means that every natural transformation \\(\eta : F \to F\\) is an endomorphism in \\(\mathbf{N}^\mathbf{N}\\).

We also have that every \\(c \in \mathbf{Mor}(\mathbf{N})\\) gives rise to some endomorphism for a functor \\(F : \mathbf{N} \to \mathbf{N}\\).

To see this, note that for every functor there is constant \\(k \in \mathbb{N} \\) such that \\(F(f) \in \mathbf{N} \\) reflects \\(k \times f \in \mathbb{N}\\). Since addition commutes in arithmetic, for all \\(c \in \mathbb{N} \\) we have:

\[ c + k \times f = k \times f + c \]

which corresponds to:

\[ \eta \circ_{\mathbf{N}} F(f) = F(f) \circ_{\mathbf{N}} \eta \]

For all \\(\eta \in \mathbf{Mor}(\mathbf{N}) \\).