John wrote:

> Matthew wrote:

> > For one, we get a category, which I am going to call \\(\mathbf{Mult}\\)

>

> > - The objects are the morphisms of \\(\mathbf{N}\\)

> > - 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\\)

>

> In a category, each morphism goes from one particular object to one particular object. You're not specifying how this works for \\(\mathbf{Mult}\\).

You are right.

I attempted to clean this up above:

- 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\\)

Since there's only one object, every morphism fixes it.

This category is the same as the monoid \\(\langle \mathbb{N}, \times, 1\rangle\\) Jonathan mentioned in [#1](https://forum.azimuthproject.org/discussion/comment/19163/#Comment_19163).

> But I can guess: morphisms in \\(\mathbf{Mult}\\) are not really functors \\(F : \mathbf{N} \to \mathbf{N}\\); rather, each functor \\(F : \mathbf{N} \to \mathbf{N}\\) gives infinitely many morphisms in \\(\mathbf{Mult}\\), one for each object of \\(\mathbf{Mult}\\). Objects of \\(\mathbf{Mult}\\) can be identified with natural numbers, and if a functor \\(F : \mathbf{N} \to \mathbf{N}\\) maps the object \\(n\\) to the object \\(n'\\), we decree that there's a morphism

>

> \[ (F,n) : n \to n' \]

>

> in \\(\mathbf{Mult}\\). We compose these in the obvious way:

>

> \[ (F',n') \circ (F,n) = (F' \circ F, n) .\]

>

> This is an example of a sort of well-known construction.

This wasn't what I had in mind but I am curious to know the name of this construction.

> Matthew wrote:

> > For one, we get a category, which I am going to call \\(\mathbf{Mult}\\)

>

> > - The objects are the morphisms of \\(\mathbf{N}\\)

> > - 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\\)

>

> In a category, each morphism goes from one particular object to one particular object. You're not specifying how this works for \\(\mathbf{Mult}\\).

You are right.

I attempted to clean this up above:

- 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\\)

Since there's only one object, every morphism fixes it.

This category is the same as the monoid \\(\langle \mathbb{N}, \times, 1\rangle\\) Jonathan mentioned in [#1](https://forum.azimuthproject.org/discussion/comment/19163/#Comment_19163).

> But I can guess: morphisms in \\(\mathbf{Mult}\\) are not really functors \\(F : \mathbf{N} \to \mathbf{N}\\); rather, each functor \\(F : \mathbf{N} \to \mathbf{N}\\) gives infinitely many morphisms in \\(\mathbf{Mult}\\), one for each object of \\(\mathbf{Mult}\\). Objects of \\(\mathbf{Mult}\\) can be identified with natural numbers, and if a functor \\(F : \mathbf{N} \to \mathbf{N}\\) maps the object \\(n\\) to the object \\(n'\\), we decree that there's a morphism

>

> \[ (F,n) : n \to n' \]

>

> in \\(\mathbf{Mult}\\). We compose these in the obvious way:

>

> \[ (F',n') \circ (F,n) = (F' \circ F, n) .\]

>

> This is an example of a sort of well-known construction.

This wasn't what I had in mind but I am curious to know the name of this construction.