It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.4K
- Chat 505
- Study Groups 21
- Petri Nets 9
- Epidemiology 4
- Leaf Modeling 2
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- Baez ACT 2019: Online Course 339
- Baez ACT 2019: Lectures 79
- Baez ACT 2019: Exercises 149
- Baez ACT 2019: Chat 50
- UCR ACT Seminar 4
- General 75
- Azimuth Code Project 111
- Statistical methods 4
- Drafts 10
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 148
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 719

## Comments

That's right :D We have a functor.

The object mapping is \(\langle\mathcal{V}_\bullet\rangle : \mathbb{ME} \to \mathbf{Grp}\)

The morphism mapping is \(\hat{\bullet} : \mathrm{hom}(\mathbb{ME}) \to \mathrm{hom}(\mathbf{Grp}) \).

Let \(\mathcal{G}_G(g,h) = g^{-1}h \in G\) be the enriched category John constructed in #81

Take have a group homomorphims \(\psi : G \to H\) for \(G,H \in \mathbf{Grp}\).

We can turn \(\psi\) into an \(\mathbb{ME}\) morphism with \(\bar{\psi}\) with

\[ \bar{\psi} (g^{-1}h) = \psi(g^{-1})\psi(h) \]

I think \(\langle\mathcal{V}_\bullet \rangle \) is the left inverse of \(\mathcal{G}•\) up to isomorphism.

`> So what you've constructed (I think!) as a morphism \\(\mathcal{X}\to\mathcal{Y}\\) in ME is a function \\(\phi\colon\text{Ob}(\\mathcal{X}) \to \text{Ob}(\\mathcal{Y})\\) and a monoid/group homomorphism \\(\hat\phi\colon \langle \mathcal{V}_X \rangle\to \langle \mathcal{U}_Y \rangle\\), such that \\(\hat\phi(\mathcal{X}(a,b)) = \mathcal{Y}(\phi(a),\phi(b))\\) . That's right :D We have a functor. The object mapping is \\(\langle\mathcal{V}_\bullet\rangle : \mathbb{ME} \to \mathbf{Grp}\\) The morphism mapping is \\(\hat{\bullet} : \mathrm{hom}(\mathbb{ME}) \to \mathrm{hom}(\mathbf{Grp}) \\). ------------------------------------ Let \\(\mathcal{G}_G(g,h) = g^{-1}h \in G\\) be the enriched category John constructed in [#81](https://forum.azimuthproject.org/discussion/comment/18687/#Comment_18687) Take have a group homomorphims \\(\psi : G \to H\\) for \\(G,H \in \mathbf{Grp}\\). We can turn \\(\psi\\) into an \\(\mathbb{ME}\\) morphism with \\(\bar{\psi}\\) with \\[ \bar{\psi} (g^{-1}h) = \psi(g^{-1})\psi(h) \\] I think \\(\langle\mathcal{V}_\bullet \rangle \\) is the left inverse of \\(\mathcal{G}•\\) up to isomorphism.`

Well, my point wasn't that we have a functor, but rather that was this is a more category theoretic way to express what the category \(\mathbb{ME}\) is.

Then, it makes it straight forward to see that we have the functor that you were claiming.

Having said that, \(\mathbb{ME}\) doesn't quite feel like a natural category to me, unless you can give me a place where it crops up naturally.

What feels more natural is the following which I'll call \(\mathbb{ECM}\) (rather than \(\mathbb{ME}2\)).

An object is a pair \((\mathcal{V}, \mathcal{X})\) where \(\mathcal{V}\) is a monoid and \(\mathcal{X}\) is an enriched category over \(\mathcal{V}\).

A morphism \((\mathcal{V}, \mathcal{X}) \to (\mathcal{U}, \mathcal{Y})\) consists of a pair \(\left(\hat\phi\colon \mathcal{V}\to \mathcal{U}, \,\,\phi \colon \text{Ob}(\mathcal{X})\to \text{Ob}(\mathcal{Y})\right)\) where \(\hat\phi\) is a monoid map and we have \(\hat\phi(\mathcal{X}(a,b)) = \mathcal{Y}(\phi(a),\phi(b))\).

Then you still get the functor \(\mathcal{G}\colon \textbf{Grp}\to \mathbb{ECM}\), but I think that I can now construct two natural left inverses.

My point here is that \(\mathbb{ECM}\) feels more like an enriched category theory construction, and easily generalises to enriching over arbitary monoidal categories. However, that's not to say you won't have a naturally occurring mathematical situation in which you want to deal with \(\mathbb{ME}\).

`Well, my point wasn't that we have a functor, but rather that was this is a more category theoretic way to express what the category \\(\mathbb{ME}\\) is. Then, it makes it straight forward to see that we have the functor that you were claiming. Having said that, \\(\mathbb{ME}\\) doesn't quite feel like a natural category to me, unless you can give me a place where it crops up naturally. What feels more natural is the following which I'll call \\(\mathbb{ECM}\\) (rather than \\(\mathbb{ME}2\\)). - An object is a pair \\((\mathcal{V}, \mathcal{X})\\) where \\(\mathcal{V}\\) is a monoid and \\(\mathcal{X}\\) is an enriched category over \\(\mathcal{V}\\). - A morphism \\((\mathcal{V}, \mathcal{X}) \to (\mathcal{U}, \mathcal{Y})\\) consists of a pair \\(\left(\hat\phi\colon \mathcal{V}\to \mathcal{U}, \,\,\phi \colon \text{Ob}(\mathcal{X})\to \text{Ob}(\mathcal{Y})\right)\\) where \\(\hat\phi\\) is a monoid map and we have \\(\hat\phi(\mathcal{X}(a,b)) = \mathcal{Y}(\phi(a),\phi(b))\\). Then you still get the functor \\(\mathcal{G}\colon \textbf{Grp}\to \mathbb{ECM}\\), but I think that I can now construct two natural left inverses. My point here is that \\(\mathbb{ECM}\\) feels more like an enriched category theory construction, and easily generalises to enriching over arbitary monoidal categories. However, that's not to say you won't have a naturally occurring mathematical situation in which you want to deal with \\(\mathbb{ME}\\).`

Anyway, you came up with a cool construction.

`Anyway, you came up with a cool construction.`

Isn't \(\mathbb{ECM}\) isomorphic to the category of groups?

`> What feels more natural is the following which I'll call \\(\mathbb{ECM}\\) (rather than \\(\mathbb{ME}2\\)). > > - An object is a pair \\((\mathcal{V}, \mathcal{X})\\) where \\(\mathcal{V}\\) is a monoid and \\(\mathcal{X}\\) is an enriched category over \\(\mathcal{V}\\). > > - A morphism \\((\mathcal{V}, \mathcal{X}) \to (\mathcal{U}, \mathcal{Y})\\) consists of a pair \\(\left(\hat\phi\colon \mathcal{V}\to \mathcal{U}, \,\,\phi \colon \text{Ob}(\mathcal{X})\to \text{Ob}(\mathcal{Y})\right)\\) where \\(\hat\phi\\) is a monoid map and we have \\(\hat\phi(\mathcal{X}(a,b)) = \mathcal{Y}(\phi(a),\phi(b))\\). Isn't \\(\mathbb{ECM}\\) isomorphic to the category of groups?`

What's the isomorphism?

`What's the isomorphism?`

Never mind, now I see you just wanted a clean up my formulation of \(\mathbb{ME}\) now that I reread this.

I think we both see now how to embed \(\mathbf{Grp}\) into \(\mathbb{ECM}\).

It looks like \(\mathbb{ECM}\) is a subcategory of \(\mathbb{ECP}\):

Now if we restrict ourselves now to \(\hat{\phi} = id\) and \(\mathcal{V} = \mathbf{Cost}\) then we recover just the \(\mathbf{Cost}\)-categories and functors.

I see why you say \(\mathbb{ECM}\) feels more like an enriched category theory construction.

`Never mind, now I see you just wanted a clean up my formulation of \\(\mathbb{ME}\\) now that I reread this. I think we both see now how to embed \\(\mathbf{Grp}\\) into \\(\mathbb{ECM}\\). It looks like \\(\mathbb{ECM}\\) is a subcategory of \\(\mathbb{ECP}\\): > - An object is a pair \\((\mathcal{V}, \mathcal{X})\\) where \\(\mathcal{V}\\) is a monoidal preorder and \\(\mathcal{X}\\) is an enriched category over \\(\mathcal{V}\\). > > - A morphism \\((\mathcal{V}, \mathcal{X}) \to (\mathcal{U}, \mathcal{Y})\\) consists of a pair \\(\left(\hat\phi\colon \mathcal{V}\to \mathcal{U}, \,\,\phi \colon \text{Ob}(\mathcal{X})\to \text{Ob}(\mathcal{Y})\right)\\) where \\(\hat\phi\\) is a monoidal preorder map and we have \\(\hat\phi(\mathcal{X}(a,b)) \leq \mathcal{Y}(\phi(a),\phi(b))\\). Now if we restrict ourselves now to \\(\hat{\phi} = id\\) and \\(\mathcal{V} = \mathbf{Cost}\\) then we recover just the \\(\mathbf{Cost}\\)-categories and functors. I see why you say \\(\mathbb{ECM}\\) feels more like an enriched category theory construction.`

Thank you so much Simon for taking the time to work through this thing with me.

`Thank you so much Simon for taking the time to work through this thing with me.`

Exactly.

To generalize beyond monoidal preorders, we can write \(\mathbb{ECP}\) in terms of 'change of base' that John talked about around Puzzle 94 in Lecture 33.

`Exactly. To generalize beyond monoidal preorders, we can write \\(\mathbb{ECP}\\) in terms of 'change of base' that John talked about around Puzzle 94 in [Lecture 33](https://forum.azimuthproject.org/discussion/2192/lecture-33-chapter-2-tying-up-loose-ends). > - An object is a pair \\((\mathcal{V}, \mathcal{X})\\) where \\(\mathcal{V}\\) is a monoidal preorder and \\(\mathcal{X}\\) is an enriched category over \\(\mathcal{V}\\). > > - A morphism \\((\mathcal{V}, \mathcal{X}) \to (\mathcal{U}, \mathcal{Y})\\) consists of a pair \\(\left( f\colon \mathcal{V}\to \mathcal{U}, \,\, F\colon \mathcal{X}_f \to \mathcal{Y}\right)\\) where \\(f\\) is a monoidal preorder map and \\(F\\) is a \\(\mathcal{U}\\)-functor.`

Julio - maybe someone already said what I'm about to say, or maybe you already know it, and it's probably not your main concern, but anyway.. in comment #88 you wrote:

This makes me feel nervous. Why? Because these are not the only two alternatives! In a poset,

does not imply

Only in a poset that is totally ordered does the former imply the latter. In fact this is the definition of a totally ordered set!

We're interested in lots of posets that aren't totally ordered, like the poset \(P(X)\) consisting of all subsets of \(X\) . If \(S\) and \(T\) are subsets of \(X\),

does not imply

`Julio - maybe someone already said what I'm about to say, or maybe you already know it, and it's probably not your main concern, but anyway.. in [comment #88](https://forum.azimuthproject.org/discussion/comment/18708/#Comment_18708) you wrote: > I managed to understand the adoption of "≤" in \\( \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \\) via this angle, but am still not sure I fully grasp its necessity in \\(I\leq\mathcal{X}(x,x)\\) - I (think I) see that \\(\mathcal{X}(x,x)\\) need not be exactly \\(I\\), but why must it be _at least_ that much? What would happen if \\(I>\mathcal{X}(x,x)\\)? This makes me feel nervous. Why? Because these are not the only two alternatives! In a poset, <center> not \\(I\leq\mathcal{X}(x,x)\\) </center> does not imply <center> \\(I\gt \mathcal{X}(x,x)\\) .</center> Only in a poset that is [totally ordered](https://en.wikipedia.org/wiki/Total_order) does the former imply the latter. In fact this is the definition of a totally ordered set! We're interested in lots of posets that aren't totally ordered, like the poset \\(P(X)\\) consisting of all subsets of \\(X\\) . If \\(S\\) and \\(T\\) are subsets of \\(X\\), <center> not \\(S \subseteq T\\) </center> does not imply <center> \\( S \supset T \\) .</center>`

In comment #88 Julio wrote:

As Simon pointed out, there's no "necessity" in math: we make up axioms that handle examples we're interested in, and give nice theorems. If we want to study examples that don't obey those axioms, we make up other axioms, and hope those also give nice theorems.

We get lots of nice examples of enriched categories, and lots of nice theorems about enriched categories, if we assume

$$ \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) $$ and

$$ I\leq\mathcal{X}(x,x) .$$ You should look at all the examples and theorems I've stated — and others in the book — and see why these axioms are true in the examples, and how we use them in the theorems. You'll see these axioms work very well!

If we turn around _both) inequalities, and assume

$$ \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\geq\mathcal{X}(x,z) $$ and

$$ I\geq\mathcal{X}(x,x) $$ everything works just as well! The reason is that every preorder has an "opposite", where we define inequality the opposite way. It's just an arbitrary convention whether we write \(\le\) or \(\ge\). It's like deciding to drive on the left or on the right of the road: neither one is better, but you have collisions if people don't agree on a choice!

If you want to explore some new territory, see what happens if you turn around just

oneinequality. For example, assume$$ \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) $$ and

$$ I\geq\mathcal{X}(x,x) .$$ I believe this territory will be an ugly swamp. But I haven't explored it, so I could be wrong!

Ultimately the answer to your question may be "wait a bit, and learn more!" For example, Matthew gave you a great answer. However, his answer only makes sense if you know about categories enriched in

monoidal categories, not just in monoidal preorders. He wrote approximately this:If this doesn't make sense now, it should later. You can hold on to this as a future goal.

`In [comment #88](https://forum.azimuthproject.org/discussion/comment/18708/#Comment_18708) Julio wrote: > I managed to understand the adoption of "≤" in \\( \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \\) via this angle, but am still not sure I fully grasp its necessity in \\(I\leq\mathcal{X}(x,x)\\). As [Simon pointed out](https://forum.azimuthproject.org/discussion/comment/18711/#Comment_18711), there's no "necessity" in math: we make up axioms that handle examples we're interested in, and give nice theorems. If we want to study examples that don't obey those axioms, we make up other axioms, and hope those also give nice theorems. We get lots of nice examples of enriched categories, and lots of nice theorems about enriched categories, if we assume \[ \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \] and \[ I\leq\mathcal{X}(x,x) .\] You should look at all the examples and theorems I've stated — and others in the book — and see why these axioms are true in the examples, and how we use them in the theorems. You'll see these axioms work very well! If we turn around _both) inequalities, and assume \[ \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\geq\mathcal{X}(x,z) \] and \[ I\geq\mathcal{X}(x,x) \] everything works just as well! The reason is that every preorder has an "opposite", where we define inequality the opposite way. It's just an arbitrary convention whether we write \\(\le\\) or \\(\ge\\). It's like deciding to drive on the left or on the right of the road: neither one is better, but you have collisions if people don't agree on a choice! If you want to explore some new territory, see what happens if you turn around just _one_ inequality. For example, assume \[ \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \] and \[ I\geq\mathcal{X}(x,x) .\] I believe this territory will be an ugly swamp. But I haven't explored it, so I could be wrong! Ultimately the answer to your question may be "wait a bit, and learn more!" For example, [Matthew](https://forum.azimuthproject.org/discussion/comment/18702/#Comment_18702) gave you a great answer. However, his answer only makes sense if you know about categories enriched in _monoidal categories_, not just in monoidal preorders. He wrote approximately this: > In category theory, when we have two objects \\(X\\) and \\(Y\\) in a category we often care about the set of morphisms between them. This is the [*hom-set*](https://en.wikipedia.org/wiki/Morphism#Hom-set) \\(\mathrm{Hom}(X,Y)\\). > While this is a powerful generalization, it can be more general. Why do we demand the morphisms between \\(X\\) and \\(Y\\) be a set? Why not \\(\\{\mathtt{false}, \mathtt{true}\\}\\), or a number, or a member of a group? > _Enriched categories substitute the sets in hom-sets with objects in an arbitrary monoidal category_. Ordinary categories can be seen as **Set**-enriched categories, where **Set** is the category of sets. > In order to make this happen, you need two things: composition of arrows and identity morphisms. > \\( \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \\) ensures morphisms are closed under composition. > \\( I\leq\mathcal{X}(x,x) \\) effectively ensures the presence of identity morphisms. If this doesn't make sense now, it should later. You can hold on to this as a future goal.`