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# Lecture 29 - Chapter 2: Enriched Categories

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101.
edited June 2018

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

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.

Comment Source: > 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.
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102.
edited June 2018

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 $$\mathbb{ME}$$.

Comment Source: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}\$$.
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103.
edited June 2018

Anyway, you came up with a cool construction.

Comment Source:Anyway, you came up with a cool construction.
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104.

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?

Comment Source:> 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?
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105.

What's the isomorphism?

Comment Source:What's the isomorphism?
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106.

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.

Comment Source: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.
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107.

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

Comment Source:Thank you so much Simon for taking the time to work through this thing with me.
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108.
edited June 2018

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.

• 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.

Comment Source: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.
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109.
edited June 2018

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:

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,

not $$I\leq\mathcal{X}(x,x)$$

does not imply

$$I\gt \mathcal{X}(x,x)$$ .

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 $$X$$,

not $$S \subseteq T$$

does not imply

$$S \supset T$$ .
Comment Source: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>
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110.
edited June 2018

In comment #88 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, 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 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 $$\mathrm{Hom}(X,Y)$$.

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.

Comment Source: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 &mdash; and others in the book &mdash; 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.