#### Howdy, Stranger!

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

Options

# Lecture 29 - Chapter 2: Enriched Categories

edited February 2020

Now we come to one of the coolest and most powerful ideas in this chapter: enriched categories. But first, beware:

ENRICHED CATEGORIES ARE NOT CATEGORIES!!!

We are just warming up to study categories. When we do, we'll be in a position to understand this strange fact better. For now, you should think of enriched categories as a generalization of preorders.

Given two elements $$x,y$$ of a preorder, $$x\le y$$ is Boolean-valued. That's a fancy way of saying it's either true or false. But secretly it's relying on the Booleans:

$$\mathbf{Bool} = \{ \tt{true}, \tt{false} \} .$$ And if we replace the Booleans by something else, we can get generalizations of the concept of preorder, called "enriched categories", that do very interesting and important things.

To see why this could be a good idea, remember the kind of questions a preorder can answer for us:

1) Is this at least as big as that?

2) Is this at least as expensive than that?

3) Can I get from here to there?

All these are yes-or-no questions. Their answers are Boolean-valued. But many questions in life don't have yes-or-no answers:

4) How much bigger is this than that?

5) How much more expensive is this than that?

6) How long will it take to get from here to there?

7) How many ways are there to get from here to there?

Questions 4)-6) of these have number-valued answers: if we like, we can decree that the answer should be an element of the set of costs:

$$\mathbf{Cost} = [0,\infty] ,$$ that is, nonnegative real numbers together with infinity. We use infinity to handle exceptions like this: if I ask you how long it will take to get from here to there, and it's impossible to get from here to there, you can say "$$\infty$$". If I ask you how much more expensive this painting is than that one, and this painting is not for sale, you can say "$$\infty$$".

Question 7) has a set-valued answer: there will be a set of ways to get from here to there.

We can answer questions 1)-3) with a $$\mathbf{Bool}$$-enriched category. Don't worry: this is just an ordinary preorder! I'm just calling it a $$\mathbf{Bool}$$-enriched category so you can start to understand this new "enriched" stuff. The point is that a $$\mathbf{Bool}$$-enriched category lets you take two things $$x$$ and $$y$$ and get a Boolean: true or false.

We can answer questions 4-6) with a $$\mathbf{Cost}$$-enriched category. This is something new you need to learn about. The point is that a $$\mathbf{Cost}$$-enriched category lets you take two things $$x$$ and $$y$$ and get a cost: a number from $$0$$ to $$\infty$$.

We can question 7) with a $$\mathbf{Set}$$-enriched category. This is just a category!

And this is something we won't talk about until the next chapter. Sorry! I know that hundreds of you are reading this course, refusing to ask any questions or make a any comments until I start talking explicitly about categories. But the book Seven Sketches works its way to categories in a clever way. First it talks about preorders. Then it talks about a special case of enriched categories. Then it talks about enriched categories in general - which include categories as a special case! At this point you'll suddenly shoot forward into modern category theory - the kind of stuff that bigshots like Ross Street and Martin Hyland do. But it should be painless, because you'll already be used to this "enriched" business.

Let's get started. I'll just show you the definition; next time we'll talk about it and start looking at lots of examples. We'll start by picking a monoidal preorder $$(\mathcal{V},\leq,\otimes,I)$$, like $$\mathbf{Bool}$$ or $$\mathbf{Cost}$$. Then:

Definition. A $$\mathcal{V}$$-enriched category $$\mathcal{X}$$ consists of two parts, satisfying two properties. First:

1. one specifies a set $$\mathrm{Ob}(\mathcal{X})$$, elements of which are called objects;

2. for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathcal{V}$$.

Then:

a) for every object $$x\in\text{Ob}(\mathcal{X})$$ we require that

$$I\leq\mathcal{X}(x,x) .$$ b) for every three objects $$x,y,z\in\mathrm{Ob}(\mathcal{X})$$, we require that

$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z).$$ We call $$\mathcal{X}$$ a $$\mathcal{V}$$-category for short. We call $$\mathcal{V}$$ the base of the enrichment for $$\mathcal{X}$$, and we say that $$\mathcal{X}$$ is enriched in $$\mathcal{V}$$.

Puzzle 86. Show that $$\mathbf{Bool}$$ becomes a commutative monoidal poset if we use the usual ordering $$\tt{false} \le \tt{true}$$, take $$\otimes$$ to be $$\wedge$$ - that is, "and" - and take $$I$$ to be $$\tt{true}$$ .

Puzzle 87. Figure out exactly what a $$\mathbf{Bool}$$-enriched category is, starting with the definition above. Here what matters is not your final answer so much as your process of deducing it!

To read other lectures go here.

«13

• Options
1.

I was just thinking, economies based on a Cost-category are basically market-economies, and economies based on a Bool-category is sort of like small-c communism (the currency-less ideal). Just something interesting my brain thought up.

Anyway, I'm going to think on the puzzles.

Comment Source:I was just thinking, economies based on a **Cost**-category are basically market-economies, and economies based on a **Bool**-category is sort of like small-c communism (the currency-less ideal). Just something interesting my brain thought up. Anyway, I'm going to think on the puzzles. 
• Options
2.
edited May 2018

Puzzle 86. Show that $$\mathbf{Bool}$$ becomes a symmetric monoidal poset if we use the usual ordering $$\tt{false} \le \tt{true}$$, take $$\otimes$$ to be $$\wedge$$ - that is, "and" - and take $$I$$ to be $$\tt{true}$$ .

I'm just going to gut out the definition for a monoidal preorder from the book and modify it for a symmetric monoidal poset.

Definition 2.1. A symmetric monoidal structure on a poset (X, ≤) consists of two constituents: (i) an element I ∈ X, called the monoidal unit, and (ii) a function ⊗: X × X → X, called the monoidal product.

Yes, we take $$I$$ to be $$\tt{true}$$ and take $$\otimes$$ to be $$\wedge$$, as was given.

These constituents must satisfy the following properties: (a) for all x1, x2, y1, y2 ∈ X, if x1 ≤ y1 and x2 ≤ y2, then x1 ⊗ x2 ≤ y1 ⊗ y2,

Since $$\otimes$$ is defined to be $$\wedge$$ - that is, "and", this is somewhat trivially true.

$$x_1 \wedge x_2 \leq y_1 \wedge y_2 \Leftrightarrow x_1 \leq y_1 \ \tt{and} \ x_2 \leq y_2$$

(b) for all x ∈ X, the equations I ⊗ x = x and x ⊗ I = x hold,

Since and'ing by the truth does nothing to our object $$x$$, this holds. $$\tt{true}\wedge x = x =x \wedge \tt{true}$$

(c) for all x, y, z ∈ X, the equation (x ⊗ y) ⊗ z = x ⊗ (y ⊗ z) holds,

Follows from the associativity of $$\wedge$$,

$$(x\wedge y) \wedge z = x \wedge (y \wedge z)$$

(d) for all x, y ∈ X, the [equation] x ⊗ y = y ⊗ x holds.

Follows from th symmetry of $$\wedge$$,

$$x \wedge y = y \wedge x.$$ Edit: Whoops almost forgot.

Lastly, for a poset, we need to check anti-symmetry, x ≤ y and y ≤ x.

$$\tt{false} \leq x \ and \ x \leq \tt{false} \Rightarrow x =\tt{false}$$ holds, and likewise so does,

$$x \leq \tt{true} \ \tt{and} \ \tt{true} \leq x \Rightarrow x =\tt{true}$$

Comment Source:>**Puzzle 86.** Show that \$$\mathbf{Bool}\$$ becomes a symmetric monoidal poset if we use the usual ordering \$$\tt{false} \le \tt{true}\$$, take \$$\otimes\$$ to be \$$\wedge\$$ - that is, "and" - and take \$$I\$$ to be \$$\tt{true}\$$ . I'm just going to gut out the definition for a monoidal preorder from the book and modify it for a symmetric monoidal poset. >Definition 2.1. A symmetric monoidal structure on a poset (X, ≤) consists of two constituents: (i) an element I ∈ X, called the monoidal unit, and (ii) a function ⊗: X × X → X, called the monoidal product. Yes, we take \$$I\$$ to be \$$\tt{true}\$$ and take \$$\otimes\$$ to be \$$\wedge\$$, as was given. >These constituents must satisfy the following properties: (a) for all x1, x2, y1, y2 ∈ X, if x1 ≤ y1 and x2 ≤ y2, then x1 ⊗ x2 ≤ y1 ⊗ y2, Since \$$\otimes\$$ is defined to be \$$\wedge\$$ - that is, "and", this is somewhat trivially true. $$x_1 \wedge x_2 \leq y_1 \wedge y_2 \Leftrightarrow x_1 \leq y_1 \ \tt{and} \ x_2 \leq y_2$$ >(b) for all x ∈ X, the equations I ⊗ x = x and x ⊗ I = x hold, Since and'ing by the truth does nothing to our object \$$x\$$, this holds. $$\tt{true}\wedge x = x =x \wedge \tt{true}$$ >(c) for all x, y, z ∈ X, the equation (x ⊗ y) ⊗ z = x ⊗ (y ⊗ z) holds, Follows from the associativity of \$$\wedge\$$, $$(x\wedge y) \wedge z = x \wedge (y \wedge z)$$ >(d) for all x, y ∈ X, the [equation] x ⊗ y = y ⊗ x holds. Follows from th symmetry of \$$\wedge\$$, $$x \wedge y = y \wedge x.$$ Edit: Whoops almost forgot. Lastly, for a poset, we need to check anti-symmetry, x ≤ y and y ≤ x. $$\tt{false} \leq x \ and \ x \leq \tt{false} \Rightarrow x =\tt{false}$$ holds, and likewise so does, $$x \leq \tt{true} \ \tt{and} \ \tt{true} \leq x \Rightarrow x =\tt{true}$$
• Options
3.
edited May 2018

Puzzle 87. Figure out exactly what a $$\mathbf{Bool}$$-enriched category is, starting with the definition above.

There are three posets over Bool - the trivial preorder, the discrete poset, and the lattice from Puzzle 86.

In the case of the trivial preorder and discrete poset there is only one monoid up to isomorphism.

The trivial preorder corresponds to the complete graph, and the monoidal operation can effectively be anything since all objects are equivalent. This is trivially a symmetric preoder.

On the other hand, in the case of the discrete poset we know that $$\mathcal{X}(x,x) = I$$ for all $$x$$ as per (a). Along with (b) we have $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,x) = I$$ . This means that $$\mathcal{X}(x,y) = \mathcal{X}(y,x)$$ - if they were different then the identity law for monoids would be violated. If there are pairs $$a, b, c$$ and $$d$$ such that $$\mathcal{X}(a,b) \neq \mathcal{X}(c,d)$$ then $$\mathtt{true} \otimes \mathtt{true} = \mathtt{false} \otimes \mathtt{false} = I$$. The monoid $$\langle\mathbf{Bool},\otimes, I\rangle$$ must then be isomorphic to the cyclic group $$\mathbb{Z}/(2)$$. Hence all of the monoids for $$\mathbf{Bool}$$-enriched category with discrete posets are symmetric and isomorphic.

Comment Source:> **Puzzle 87.** Figure out exactly what a \$$\mathbf{Bool}\$$-enriched category is, starting with the definition above. (My partial answer) There are three posets over **Bool** - the trivial preorder, the discrete poset, and the lattice from **Puzzle 86**. In the case of the trivial preorder and discrete poset there is only one monoid up to isomorphism. The trivial preorder corresponds to the complete graph, and the monoidal operation can effectively be anything since all objects are equivalent. This is trivially a symmetric preoder. On the other hand, in the case of the discrete poset we know that \$$\mathcal{X}(x,x) = I\$$ for all \$$x\$$ as per (a). Along with (b) we have \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,x) = I\$$ . This means that \$$\mathcal{X}(x,y) = \mathcal{X}(y,x)\$$ - if they were different then the identity law for monoids would be violated. If there are pairs \$$a, b, c\$$ and \$$d\$$ such that \$$\mathcal{X}(a,b) \neq \mathcal{X}(c,d) \$$ then \$$\mathtt{true} \otimes \mathtt{true} = \mathtt{false} \otimes \mathtt{false} = I\$$. The monoid \$$\langle\mathbf{Bool},\otimes, I\rangle\$$ must then be isomorphic to the [cyclic group \$$\mathbb{Z}/(2)\$$](https://en.wikipedia.org/wiki/Cyclic_group#Integer_and_modular_addition). Hence all of the monoids for \$$\mathbf{Bool}\$$-enriched category with discrete posets are symmetric and isomorphic. 
• Options
4.

Matthew Doty - in Puzzle 87 I was implicitly making Bool into a symmetric monoidal poset as in Puzzle 86. With this choice, it's not true that Bool-enriched categoreis are symmetric monoidal preorders.

What are they really?

Comment Source:Matthew Doty - in Puzzle 87 I was implicitly making **Bool** into a symmetric monoidal poset as in Puzzle 86. With this choice, it's not true that **Bool**-enriched categoreis are symmetric monoidal preorders. What are they really?
• Options
5.

Keith - yes, that's a great solution to Puzzle 86!

Comment Source:Keith - yes, that's a great solution to Puzzle 86!
• Options
6.

Keith wrote:

I was just thinking, economies based on a Cost-category are basically market-economies, and economies based on a Bool-category is sort of like small-c communism (the currency-less ideal). Just something interesting my brain thought up.

That's a fun analogy. In a Cost-economy it takes a certain amount of money to do something; in a Bool-economy you can either do it or you can't: that's all.

Comment Source:Keith wrote: > I was just thinking, economies based on a **Cost**-category are basically market-economies, and economies based on a **Bool**-category is sort of like small-c communism (the currency-less ideal). Just something interesting my brain thought up. That's a fun analogy. In a **Cost**-economy it takes a certain amount of money to do something; in a **Bool**-economy you can either do it or you can't: that's all.
• Options
7.
edited May 2018

John Baez wrote:

That's a fun analogy.

It's fun until you realize that people have actually injured, killed others over a disagreement over which enrichment to use.

Maybe that's another reason to learn category theory and the ideas that follow from it: to build bridges and to avoid needless suffering.

Comment Source:John Baez wrote: >That's a fun analogy. It's fun until you realize that people have actually injured, killed others over a disagreement over which enrichment to use. Maybe that's another reason to learn category theory and the ideas that follow from it: to build bridges and to avoid needless suffering. 
• Options
8.
edited May 2018

I am a bit confused on the notation.

1. one specifies a set $$\mathrm{Ob}(\mathcal{X})$$, elements of which are called objects;

2. for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathcal{V}$$.

According to the definition above, $$x,y \in \mathrm{Ob}(\mathcal{X})$$. But in order for $$\mathcal{X}(x,y)$$ to be an element of $$\mathcal{V}$$, $$x,y \in V$$ must be true also? Then what is the difference between $$\mathrm{Ob}(\mathcal{X})$$ and $$V$$?

Comment Source:I am a bit confused on the notation. >1. one specifies a set \$$\mathrm{Ob}(\mathcal{X})\$$, elements of which are called **objects**; >2. for every two objects \$$x,y\$$, one specifies an element \$$\mathcal{X}(x,y)\$$ of \$$\mathcal{V}\$$. According to the definition above, \$$x,y \in \mathrm{Ob}(\mathcal{X})\$$. But in order for \$$\mathcal{X}(x,y)\$$ to be an element of \$$\mathcal{V}\$$, \$$x,y \in V\$$ must be true also? Then what is the difference between \$$\mathrm{Ob}(\mathcal{X})\$$ and \$$V\$$?
• Options
9.
edited May 2018

@Michael: Let's take as example the divisibility poset on positive integers. We can construct this as a Bool-enriched category as follows.

Define $$\mathrm{Ob}(\mathcal{X}) = \mathbb{Z}^+$$. This is the set of objects; in other words, every positive integer is an object.

Define $$\mathcal{X}(x, y) = \mathrm{true}$$ iff $$x \le y$$, and define $$\mathcal{X}(x, y) = \mathrm{false}$$ iff $$x \not\le y$$. This associates to every pair of objects -- natural numbers -- a value in Bool, corresponding to whether or not these objects are related by $$\le$$. As an example, $$\mathcal{X}(4, 8) = \mathrm{true}$$, since $$4$$ divides $$8$$.

Only thing is, I think we need to use $$\lor$$ and $$\mathrm{false}$$ for the monoidal operator and identity. By the chosen order, $$\mathrm{true}$$ is the greatest element of the preorder, so we're tacitly requiring that $$\mathcal{X}(x, y) = \mathrm{true}$$ for all $$x, y$$, since $$I \le \mathcal{X}(x, y)$$ needs to hold. Is this a typo, John, or am I misreading things?

If we use $$\lor$$ and $$\mathrm{false}$$, then we can check the other conditions. First, $$\mathrm{false}$$ is the least element of Bool, so the first property is trivially satisfied. And we have that $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ because divisibility is transitive.

EDIT: But this actually fails the transitivity property once we switch to $$\lor$$. I think we might actually want to use the order $$\mathrm{true} \le \mathrm{false}$$, along with $$\land$$ and $$\mathrm{true}$$. Editing! (EDIT 2: nope, didn't work, see my next post below)

Comment Source:@[Michael](https://forum.azimuthproject.org/discussion/comment/18388/#Comment_18388): Let's take as example the divisibility poset on positive integers. We can construct this as a **Bool**-enriched category as follows. Define \$$\mathrm{Ob}(\mathcal{X}) = \mathbb{Z}^+\$$. This is the set of objects; in other words, every positive integer is an object. Define \$$\mathcal{X}(x, y) = \mathrm{true}\$$ iff \$$x \le y\$$, and define \$$\mathcal{X}(x, y) = \mathrm{false}\$$ iff \$$x \not\le y\$$. This associates to every pair of objects -- natural numbers -- a value in **Bool**, corresponding to whether or not these objects are related by \$$\le\$$. As an example, \$$\mathcal{X}(4, 8) = \mathrm{true}\$$, since \$$4\$$ divides \$$8\$$. Only thing is, I think we need to use \$$\lor\$$ and \$$\mathrm{false}\$$ for the monoidal operator and identity. By the chosen order, \$$\mathrm{true}\$$ is the _greatest_ element of the preorder, so we're tacitly requiring that \$$\mathcal{X}(x, y) = \mathrm{true}\$$ for all \$$x, y\$$, since \$$I \le \mathcal{X}(x, y)\$$ needs to hold. Is this a typo, John, or am I misreading things? If we use \$$\lor\$$ and \$$\mathrm{false}\$$, then we can check the other conditions. First, \$$\mathrm{false}\$$ is the least element of **Bool**, so the first property is trivially satisfied. And we have that \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)\$$ because divisibility is transitive. **EDIT:** But this actually fails the transitivity property once we switch to \$$\lor\$$. I think we might actually want to use the order \$$\mathrm{true} \le \mathrm{false}\$$, along with \$$\land\$$ and \$$\mathrm{true}\$$. Editing! (**EDIT 2:** nope, didn't work, see my next post below)
• Options
10.
edited May 2018

I was curious how the choice of symmetric monoidal poset affects the resulting $$\mathbf{Bool}$$-enriched category.

From any poset over $$\mathbf{Bool}$$, we can make it symmetric monoidal by picking an $$I$$ and a $$\otimes$$. For $$\leq$$, I will just pick the usual $$\tt{false} \le \tt{true}$$. Although as Matthew notes, there is also the trivial poset where $$\forall x y, x \le y$$ also and the discrete poset where $$\forall x, x \le x$$.

The choice for $$I$$ over $$\mathbf{Bool}$$ must be either $$\tt{true}$$ or $$\tt{false}$$. I'll interpret the effect of each using property a) of $$\mathcal{V}$$-enriched categories:

$$\text{for all objects } x, I\leq\mathcal{X}(x,x) .$$ If $$I = \tt{false}$$, we've added no restrictions, since $$\tt{false} \leq \tt{anything}$$. However, if $$I = \tt{true}$$, then $$\mathcal{X}(x,x)$$ must be $$\tt{true}$$ also. So setting $$I = \tt{true}$$ effectively forces every identity relation to be $$\tt{true}$$. In other words:

$$\begin{array}{c|c} I & \text{effect of a) on \mathcal{X}(x,y)} \\ \hline \tt{false} & \text{No restrictions} \\ \tt{true} & \mathcal{X}(x,x) = \tt{true} \end{array}$$ There are 16 possible binary relations $$\mathbf{Bool} \times \mathbf{Bool} \to \mathbf{Bool}$$, but only 8 symmetric binary relations. We can actually name these 8 relations:

$$\{ C_F (\tt{constant~false}), \downarrow (\tt{nor}), \oplus (\tt{xor}), \uparrow (\tt{nand}), \wedge (\tt{and}), \leftrightarrow (\tt{xnor}), \vee (\tt{or}), C_T (\tt{constant~true}) \}$$ $$\{ C_F, \downarrow, \uparrow, C_T \}$$ fail to satisfy the monoidal unit laws, so we can't use them for our monoidal product $$\otimes$$. So we are left with $${ \oplus, \wedge, \leftrightarrow, \vee }$$. There is one valid choice of identity for each of them, giving us four symmetric monoidal structures over our $$\mathbf{Bool}$$ poset:

$${ (\oplus, \tt{false}), (\wedge, \tt{true}), (\leftrightarrow, \tt{true}), (\vee, \tt{false}) }$$ With these monoids in hand, let's explore property b) of $$\mathcal{V}$$-enriched categories.

$$\text{for all objects } x,y,z, \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z).$$ This sets restrictions on $$\mathcal{X}(x,y)$$ in our category, based on how $$\otimes$$ acts.

For example, if $$\otimes = \vee$$, then with $$\mathcal{X}(x,y) = \tt{true}$$, we can derive $$\forall z, \mathcal{X}(y,z) = \tt{true}$$ and $$\forall w, \mathcal{X}(w,x) = \tt{true}$$. So if any $$\mathcal{X}(x,y) = \tt{true}$$, then all $$\mathcal{X}(x,y) = \tt{true}$$!

If $$\otimes = \wedge$$, then $$\mathcal{X}(x,z) = \tt{true}$$ whenever $$\mathcal{X}(x,y) = \tt{true}$$ and $$\mathcal{X}(y,z) = \tt{true}$$. This kind of looks like the choice of $$\otimes$$ gives $$\mathcal{X}$$ closure over composition.

I've discussed the easy two, but I haven't found a good interpretation of the properties of the other two ($$(\oplus, \tt{false})$$ and $$(\leftrightarrow, \tt{true})$$).

$$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both.} \\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false}, or \mathcal{X}(x,y) = \tt{true}, for all objects x,y.} \\ \end{array}$$ Maybe we can get even more interesting $$\mathbf{Bool}$$-categories by considering the other two posets.

Comment Source:I was curious how the choice of symmetric monoidal poset affects the resulting \$$\mathbf{Bool}\$$-enriched category. From any poset over \$$\mathbf{Bool}\$$, we can make it symmetric monoidal by picking an \$$I\$$ and a \$$\otimes\$$. For \$$\leq\$$, I will just pick the usual \$$\tt{false} \le \tt{true}\$$. Although as Matthew notes, there is also the trivial poset where \$$\forall x y, x \le y \$$ also and the discrete poset where \$$\forall x, x \le x \$$. The choice for \$$I\$$ over \$$\mathbf{Bool}\$$ must be either \$$\tt{true}\$$ or \$$\tt{false}\$$. I'll interpret the effect of each using property a) of \$$\mathcal{V}\$$-enriched categories: $\text{for all objects } x, I\leq\mathcal{X}(x,x) .$ If \$$I = \tt{false}\$$, we've added no restrictions, since \$$\tt{false} \leq \tt{anything}\$$. However, if \$$I = \tt{true}\$$, then \$$\mathcal{X}(x,x)\$$ must be \$$\tt{true}\$$ also. So setting \$$I = \tt{true}\$$ effectively forces every identity relation to be \$$\tt{true}\$$. In other words: $$\begin{array}{c|c} I & \text{effect of a) on \mathcal{X}(x,y)} \\\\ \hline \tt{false} & \text{No restrictions} \\\\ \tt{true} & \mathcal{X}(x,x) = \tt{true} \end{array}$$ There are 16 possible binary relations \$$\mathbf{Bool} \times \mathbf{Bool} \to \mathbf{Bool}\$$, but only 8 symmetric binary relations. We can actually name these 8 relations: $\\{ C_F (\tt{constant~false}), \downarrow (\tt{nor}), \oplus (\tt{xor}), \uparrow (\tt{nand}), \wedge (\tt{and}), \leftrightarrow (\tt{xnor}), \vee (\tt{or}), C_T (\tt{constant~true}) \\}$ \$$\\{ C_F, \downarrow, \uparrow, C_T \\} \$$ fail to satisfy the monoidal unit laws, so we can't use them for our monoidal product \$$\otimes\$$. So we are left with \$$\{ \oplus, \wedge, \leftrightarrow, \vee \} \$$. There is one valid choice of identity for each of them, giving us four symmetric monoidal structures over our \$$\mathbf{Bool}\$$ poset: $\{ (\oplus, \tt{false}), (\wedge, \tt{true}), (\leftrightarrow, \tt{true}), (\vee, \tt{false}) \}$ With these monoids in hand, let's explore property b) of \$$\mathcal{V}\$$-enriched categories. $\text{for all objects } x,y,z, \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z).$ This sets restrictions on \$$\mathcal{X}(x,y)\$$ in our category, based on how \$$\otimes\$$ acts. For example, if \$$\otimes = \vee\$$, then with \$$\mathcal{X}(x,y) = \tt{true}\$$, we can derive \$$\forall z, \mathcal{X}(y,z) = \tt{true}\$$ and \$$\forall w, \mathcal{X}(w,x) = \tt{true}\$$. So if any \$$\mathcal{X}(x,y) = \tt{true}\$$, then all \$$\mathcal{X}(x,y) = \tt{true}\$$! If \$$\otimes = \wedge\$$, then \$$\mathcal{X}(x,z) = \tt{true}\$$ whenever \$$\mathcal{X}(x,y) = \tt{true}\$$ and \$$\mathcal{X}(y,z) = \tt{true}\$$. This kind of looks like the choice of \$$\otimes\$$ gives \$$\mathcal{X}\$$ closure over composition. I've discussed the easy two, but I haven't found a good interpretation of the properties of the other two (\$$(\oplus, \tt{false})\$$ and \$$(\leftrightarrow, \tt{true})\$$). $$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\\\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both.} \\\\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\\\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z).} \\\\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false}, or \mathcal{X}(x,y) = \tt{true}, for all objects x,y.} \\\\ \end{array}$$ Maybe we can get even more interesting \$$\mathbf{Bool}\$$-categories by considering the other two posets.
• Options
11.
edited May 2018

Okay, I tried recovering the attempt in my previous post and failed to get anywhere. I think there's some kind of fundamental issue between the two properties we're supposed to obey for an enriched category. Transitivity of divisibility gives us that "if x divides y and y divides z, then x divides z"; we want this to translate into the second property, so $$\mathcal{X}(x, y)$$ should assign true iff x divides y, and $$\otimes$$ should be "and", $$\land$$. The identity of $$\land$$ is $$\mathrm{true}$$, so everything we assign should be bounded below by true. If we pick the order $$\mathrm{false} \le \mathrm{true}$$, we have to assign everything $$\mathrm{true}$$, which is not what we want. And if we pick the order $$\mathrm{true} \le \mathrm{false}$$, we violate the translation of the second property -- it would say that if x divides y and y divides z, it's okay if x doesn't divide z.

So that leaves us with two options: the discrete preorder and the codiscrete preorder. The discrete preorder won't do either, since it's okay if x doesn't divide y or y doesn't divide z for x to still divide z. But the codiscrete preorder isn't strong enough to guarantee transitivity (or much of anything, really). I guess it just trivially satisfies both properties. But this feels really unsatisfying, and I can't imagine this is what was intended.

So what do we do? I feel like this argument has exhausted all options for representing the poset of divisibility on the positive integers as a Bool-enriched category. Am I just way off track?

Comment Source:Okay, I tried recovering the attempt in my previous post and failed to get anywhere. I think there's some kind of fundamental issue between the two properties we're supposed to obey for an enriched category. Transitivity of divisibility gives us that "if x divides y and y divides z, then x divides z"; we want this to translate into the second property, so \$$\mathcal{X}(x, y)\$$ should assign true iff x divides y, and \$$\otimes\$$ should be "and", \$$\land\$$. The identity of \$$\land\$$ is \$$\mathrm{true}\$$, so everything we assign should be bounded below by true. If we pick the order \$$\mathrm{false} \le \mathrm{true}\$$, we have to assign everything \$$\mathrm{true}\$$, which is _not_ what we want. And if we pick the order \$$\mathrm{true} \le \mathrm{false}\$$, we violate the translation of the second property -- it would say that if x divides y and y divides z, it's okay if x doesn't divide z. So that leaves us with two options: the discrete preorder and the codiscrete preorder. The discrete preorder won't do either, since it's okay if x doesn't divide y or y doesn't divide z for x to still divide z. But the codiscrete preorder isn't strong enough to guarantee transitivity (or much of anything, really). I guess it just trivially satisfies both properties. But this feels _really_ unsatisfying, and I can't imagine this is what was intended. So what do we do? I feel like this argument has exhausted all options for representing the poset of divisibility on the positive integers as a **Bool**-enriched category. Am I just way off track?
• Options
12.
edited May 2018

@Jonathan – re "everything we assign should be bounded below by true" – I don't think this is the case. Surely $$\mathcal{X}(x, y)$$ should be $$\tt{true}$$ if $$x \mid y$$ and $$\tt{false}$$ otherwise.

Comment Source:@Jonathan – re "everything we assign should be bounded below by true" – I don't think this is the case. Surely \$$\mathcal{X}(x, y)\$$ should be \$$\tt{true}\$$ if \$$x \mid y\$$ and \$$\tt{false}\$$ otherwise.
• Options
13.

I'm wondering why we insist that $$\mathcal{V}$$ be symmetric in the definition of a $$\mathcal{V}$$-enriched category. The symmetry doesn't seem to play any explicit role in the definition.

Comment Source:I'm wondering why we insist that \$$\mathcal{V}\$$ be symmetric in the definition of a \$$\mathcal{V}\$$-enriched category. The symmetry doesn't seem to play any explicit role in the definition.
• Options
14.

Surely $$\mathcal{X}(x, y)$$ should be $$\tt{true}$$ if $$x \mid y$$ and $$\tt{false}$$ otherwise.

Aha, I misread Property 1! It says that $$I \leq \mathcal{X}(x,x)$$, but I swapped one $$x$$ out for a $$y$$, which would mean that everything has to be less than or equal to $$\mathrm{true}$$ if we take that as the identity. But as you've made me aware, that's not the case -- only $$\mathcal{X}(x, x)$$ is so constrained, and this corresponds to how categories require the presence of all identity arrows.

Let me see if I can recover this once more!

Comment Source:[Anindya wrote](https://forum.azimuthproject.org/discussion/comment/18393/#Comment_18393): > Surely \$$\mathcal{X}(x, y)\$$ should be \$$\tt{true}\$$ if \$$x \mid y\$$ and \$$\tt{false}\$$ otherwise. Aha, I misread Property 1! It says that \$$I \leq \mathcal{X}(x,x)\$$, but I swapped one \$$x\$$ out for a \$$y\$$, which would mean that everything has to be less than or equal to \$$\mathrm{true}\$$ if we take that as the identity. But as you've made me aware, that's not the case -- only \$$\mathcal{X}(x, x)\$$ is so constrained, and this corresponds to how categories require the presence of all identity arrows. Let me see if I can recover this once more!
• Options
15.

Once again, let's consider the divisibility poset on positive integers. We can construct this as a Bool-enriched category as follows.

First, let's consider Bool to be $$\langle \{\mathrm{true}, \mathrm{false}\}, \le, \land, \mathrm{true} \rangle$$, where the order is given by $$false \le true$$. This is clearly a poset; it's monoidal because $$\land$$ is a meet under this order; and it's symmetric because $$\land$$ is commutative.

Now let's render our divisiblity poset as an enriched category. Define $$\mathrm{Ob}(\mathcal{X}) = \mathbb{Z}^+$$. This is the set of objects in our Bool-enriched category; in other words, every positive integer is an object.

Define $$\mathcal{X}(x, y) = \mathrm{true}$$ iff $$x \le y$$, and define $$\mathcal{X}(x, y) = \mathrm{false}$$ iff $$x \not\le y$$. This associates to every pair of objects -- natural numbers -- a value in Bool, corresponding to whether or not these objects are related by $$\le$$. As an example, $$\mathcal{X}(4, 8) = \mathrm{true}$$, since $$4$$ divides $$8$$.

Now we can check the two conditions. First, since $$x \le x$$ is always true in our preorder, we trivially have that $$I = \mathcal{X}(x, x)$$, satisfying the first condition. And we have that $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$ because divisibility is transitive: this property translates to "if x divides y and y divides z, then x divides z" under our definitions. (If one or the other does not hold, then the left side is $$\mathrm{false}$$, leaving the right side entirely unconstrained.)

So $$\langle \mathbb{Z}^+, \mid \rangle$$ is just a Bool-enriched category!

Comment Source:Once again, let's consider the divisibility poset on positive integers. We can construct this as a **Bool**-enriched category as follows. First, let's consider **Bool** to be \$$\langle \\{\mathrm{true}, \mathrm{false}\\}, \le, \land, \mathrm{true} \rangle\$$, where the order is given by \$$false \le true\$$. This is clearly a poset; it's monoidal because \$$\land\$$ is a meet under this order; and it's symmetric because \$$\land\$$ is commutative. Now let's render our divisiblity poset as an enriched category. Define \$$\mathrm{Ob}(\mathcal{X}) = \mathbb{Z}^+\$$. This is the set of objects in our **Bool**-enriched category; in other words, every positive integer is an object. Define \$$\mathcal{X}(x, y) = \mathrm{true}\$$ iff \$$x \le y\$$, and define \$$\mathcal{X}(x, y) = \mathrm{false}\$$ iff \$$x \not\le y\$$. This associates to every pair of objects -- natural numbers -- a value in **Bool**, corresponding to whether or not these objects are related by \$$\le\$$. As an example, \$$\mathcal{X}(4, 8) = \mathrm{true}\$$, since \$$4\$$ divides \$$8\$$. Now we can check the two conditions. First, since \$$x \le x\$$ is always true in our preorder, we trivially have that \$$I = \mathcal{X}(x, x)\$$, satisfying the first condition. And we have that \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)\$$ because divisibility is transitive: this property translates to "if x divides y and y divides z, then x divides z" under our definitions. (If one or the other does not hold, then the left side is \$$\mathrm{false}\$$, leaving the right side entirely unconstrained.) So \$$\langle \mathbb{Z}^+, \mid \rangle\$$ is just a **Bool**-enriched category!
• Options
16.

Jonathan

So my question was actually

According to the definition above, $$x,y \in \mathrm{Ob}(\mathcal{X})$$. But in order for $$\mathcal{X}(x,y)$$ to be an element of $$\mathcal{V}$$, $$x,y \in V$$ must be true also? Then what is the difference between $$\mathrm{Ob}(\mathcal{X})$$ and $$V$$?

And you showed me that the answer is no: the set $$V$$ is the set which you enrich the arrows of the category. So there are two layers to this configuration, object and arrows of the category and the monoidal preorder which decorates the arrows.

I think I got confused because I thought both sets were part of a preorder but one was actually the objects of a category which brings me to another question.

Can we enrich a preorder? To me its seems the category itself is also a preorder or can be made into a preorder since we are trying to answer the quantified version of the same question. Why do we have to define it as category instead of a preorder?

Comment Source:Jonathan So my question was actually >According to the definition above, \$$x,y \in \mathrm{Ob}(\mathcal{X})\$$. But in order for \$$\mathcal{X}(x,y)\$$ to be an element of \$$\mathcal{V}\$$, \$$x,y \in V\$$ must be true also? Then what is the difference between \$$\mathrm{Ob}(\mathcal{X})\$$ and \$$V\$$? And you showed me that the answer is no: the set \$$V\$$ is the set which you enrich the arrows of the category. So there are two layers to this configuration, object and arrows of the category and the monoidal preorder which decorates the arrows. I think I got confused because I thought both sets were part of a preorder but one was actually the objects of a category which brings me to another question. Can we enrich a preorder? To me its seems the category itself is also a preorder or can be made into a preorder since we are trying to answer the quantified version of the same question. Why do we have to define it as category instead of a preorder? 
• Options
17.
edited May 2018

This is a minor point, but since Michael Hong brought into discussion the notation, I think the definition:

for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathcal{V}$$.

should say "an element $$\mathcal{X}(x,y)$$ of $$V$$" rather than "[...] of $$\mathcal{V}$$" (which denotes a tuple, not a set).

Comment Source:This is a minor point, but since Michael Hong brought [into discussion the notation](https://forum.azimuthproject.org/discussion/comment/18388/#Comment_18388), I think the definition: > for every two objects \$$x,y\$$, one specifies an element \$$\mathcal{X}(x,y)\$$ of \$$\mathcal{V}\$$. should say "an element \$$\mathcal{X}(x,y)\$$ of \$$V\$$" rather than "[...] of \$$\mathcal{V}\$$" (which denotes a tuple, not a set). 
• Options
18.
edited May 2018

Puzzle TF3. Let $$(X,\leq,\otimes,I)$$ be a symmetric monoidal poset, and $$\in X$$ an arbitrary element. (I'm only naming the element suggestively, no further properties should be required.) Now for given $$x,y\in X$$, define $$\mathcal{X}(x,y)$$ to be the smallest natural number $$n$$ such that $$x \leq y \otimes ^{\otimes n}$$ if such an $$n$$ exists, and $$\infty$$ otherwise. Prove that this makes $$X$$ into a Cost-enriched category! What is its resource-theoretic interpretation?

Comment Source:**Puzzle TF3.** Let \$$(X,\leq,\otimes,I)\$$ be a symmetric monoidal poset, and \$$\\in X\$$ an arbitrary element. (I'm only naming the element suggestively, no further properties should be required.) Now for given \$$x,y\in X\$$, define \$$\mathcal{X}(x,y)\$$ to be the smallest natural number \$$n\$$ such that \$$x \leq y \otimes \^{\otimes n}\$$ if such an \$$n\$$ exists, and \$$\infty\$$ otherwise. Prove that this makes \$$X\$$ into a **Cost**-enriched category! What is its resource-theoretic interpretation?
• Options
19.
edited May 2018

@Dan: Mathematicians often pun the algebraic structure and the underlying set when context allows. We like to think of $$\mathcal{V}$$ as primarily a set, but carrying extra structure, so we still say that $$\mathcal{V}$$ has elements, even though formally we might define it as a tuple of a set $$V$$ and some operators and relations.

For instance, when talking about $$\mathbb{R}$$ and $$\mathbb{C}$$, it's understood that we're referring to the "unique Dedekind-complete ordered field" and the unique algebraic extension of the former by $$\sqrt{-1}$$, respectively, even though we still write $$x \in \mathbb{R}$$. And the standard topology for the cartesian product $$X \times Y$$ of two topological spaces is the component-wise topology, even though the lexicographic topology is also completely valid -- so we still just say $$(x, y) \in X \times Y$$, leaving the structure implicit.

It starts to matter more when talking about homomorphisms between structures sharing the same underlying set. This is when you start to see things like $$\ln : \langle \mathbb{R}, \times \rangle \to \langle \mathbb{R}, + \rangle$$, where we want to be very explicit about how the morphism interacts with the structure.

Comment Source:@[Dan](https://forum.azimuthproject.org/discussion/comment/18398/#Comment_18398): Mathematicians often pun the algebraic structure and the underlying set when context allows. We like to think of \$$\mathcal{V}\$$ as primarily a set, but carrying extra structure, so we still say that \$$\mathcal{V}\$$ has elements, even though formally we might define it as a tuple of a set \$$V\$$ and some operators and relations. For instance, when talking about \$$\mathbb{R}\$$ and \$$\mathbb{C}\$$, it's understood that we're referring to the "unique Dedekind-complete ordered field" and the unique algebraic extension of the former by \$$\sqrt{-1}\$$, respectively, even though we still write \$$x \in \mathbb{R}\$$. And the standard topology for the cartesian product \$$X \times Y\$$ of two topological spaces is the component-wise topology, even though the lexicographic topology is also completely valid -- so we still just say \$$(x, y) \in X \times Y\$$, leaving the structure implicit. It starts to matter more when talking about homomorphisms between structures sharing the same underlying set. This is when you start to see things like \$$\ln : \langle \mathbb{R}, \times \rangle \to \langle \mathbb{R}, + \rangle\$$, where we want to be very explicit about how the morphism interacts with the structure.
• Options
20.

Thanks a lot, Jonathan, for the detailed explanation! It does make sense and I see why this notation is convenient; I guess that as a beginner I tend to be over-explicit. :)

Comment Source:Thanks a lot, Jonathan, for [the detailed explanation](https://forum.azimuthproject.org/discussion/comment/18400/#Comment_18400)! It does make sense and I see why this notation is convenient; I guess that as a beginner I tend to be over-explicit. :)
• Options
21.
edited May 2018

As an exceptionally silly case, what does a Unit-enriched category look like? Unit is the single-element set $$\{\cdot\}$$ with the trivial preorder -- the only one it could have, after all -- and the natural monoidal structure -- again, the only one it could possibly have.

For a given set $$S$$, we can define the Unit-enriched category $$\mathcal{X}$$ by $$\mathrm{Ob}(\mathcal{X}) = S$$ and $$\mathcal{X}(x, y) = \cdot$$ for all $$x, y \in S$$. This satisfies both conditions trivially.

I'm tempted to identify this construction with the codiscrete preorder on $$S$$.

Comment Source:As an exceptionally silly case, what does a **Unit**-enriched category look like? **Unit** is the single-element set \$$\\{\cdot\\}\$$ with the trivial preorder -- the only one it could have, after all -- and the natural monoidal structure -- again, the only one it could possibly have. For a given set \$$S\$$, we can define the **Unit**-enriched category \$$\mathcal{X}\$$ by \$$\mathrm{Ob}(\mathcal{X}) = S\$$ and \$$\mathcal{X}(x, y) = \cdot\$$ for all \$$x, y \in S\$$. This satisfies both conditions trivially. I'm tempted to identify this construction with the codiscrete preorder on \$$S\$$.
• Options
22.

I’m tempted to identify this construction with the codiscrete preorder on $$S$$

It’s similar to the codiscrete order I mention above for Bool-Catgeories. I would second that they are the same structure.

Comment Source:> I’m tempted to identify this construction with the codiscrete preorder on \$$S\$$ It’s similar to the codiscrete order I mention above for Bool-Catgeories. I would second that they are the same structure.
• Options
23.
edited May 2018

These "edge cases" are often useful, I find, no matter how silly they may seem. I'd say in this case that the enrichment tells us absolutely nothing, so we're just looking at the plain set $$S$$. Which tells us that plain sets are a special case of enriched categories.

Comment Source:These "edge cases" are often useful, I find, no matter how silly they may seem. I'd say in this case that the enrichment tells us absolutely nothing, so we're just looking at the plain set \$$S\$$. Which tells us that plain sets are a special case of enriched categories.
• Options
24.
edited May 2018

@John Baez wrote:

in Puzzle 87 I was implicitly making Bool into a symmetric monoidal poset as in Puzzle 86. With this choice, it's not true that Bool-enriched categoreis are symmetric monoidal preorders.

You are right.

I believe the structure in Puzzle 86 is the only interesting underlying monoid.

Comment Source:@John Baez wrote: > in Puzzle 87 I was implicitly making **Bool** into a symmetric monoidal poset as in Puzzle 86. With this choice, it's not true that **Bool**-enriched categoreis are symmetric monoidal preorders. You are right. I believe the structure in **Puzzle 86** is the only *interesting* underlying monoid.
• Options
25.
edited May 2018

I'm hitting a blocker with @Tobias's Puzzle TF3.

Suppose for simplicity that $$= I$$, so every $$\mathcal{X}(x,y)$$ is either $$0$$ (if $$x \leq y$$) or $$\infty$$ (if $$x \nleq y$$).

Suppose further that we have $$x, y, z \in X$$ with $$x \leq z$$, $$y \leq z$$, but $$x \nleq y$$.

Then we have $$\mathcal{X}(x, y) = \infty, \mathcal{X}(y, z) = 0$$ so $$\mathcal{X}(x, y) + \mathcal{X}(y, z) = \infty \nleq 0 = \mathcal{X}(x, z)$$

... which contravenes rule (b) for enriched categories.

EDIT to add: in fact it seems in general we have $$\mathcal{X}(x, y) + \mathcal{X}(y, z) \geq \mathcal{X}(x, z)$$ rather than $$\leq$$

EDIT to add: aaaargh just checked the Fong/Spivak book and of course the ordering is $$\geq$$ not $$\leq$$ on Cost. d'oh!

Comment Source:I'm hitting a blocker with @Tobias's **Puzzle TF3**. Suppose for simplicity that \$$\ = I\$$, so every \$$\mathcal{X}(x,y)\$$ is either \$$0\$$ (if \$$x \leq y\$$) or \$$\infty\$$ (if \$$x \nleq y\$$). Suppose further that we have \$$x, y, z \in X\$$ with \$$x \leq z\$$, \$$y \leq z\$$, but \$$x \nleq y\$$. Then we have \$$\mathcal{X}(x, y) = \infty, \mathcal{X}(y, z) = 0\$$ so \$$\mathcal{X}(x, y) + \mathcal{X}(y, z) = \infty \nleq 0 = \mathcal{X}(x, z)\$$ ... which contravenes rule (b) for enriched categories. **EDIT** to add: in fact it seems in general we have \$$\mathcal{X}(x, y) + \mathcal{X}(y, z) \geq \\mathcal{X}(x, z)\$$ rather than \$$\leq\$$ **EDIT** to add: aaaargh just checked the Fong/Spivak book and of course the ordering *is* \$$\geq\$$ not \$$\leq\$$ on **Cost**. d'oh!
• Options
26.
edited May 2018

@Anindya: whoops, you're right! What I should have added is that you should consider Cost $$= [0,\infty]$$ to be a symmetric monoidal preorder where the order is given by the opposite of the usual one. I suspect that this is also secretly what John has in mind as the definition of Cost. But perhaps I'm running a little ahead of the course and already giving away things that will appear in Lecture 30.

Comment Source:@Anindya: whoops, you're right! What I should have added is that you should consider **Cost** \$$= [0,\infty]\$$ to be a symmetric monoidal preorder where the order is given by the *opposite* of the usual one. I suspect that this is also secretly what John has in mind as the definition of **Cost**. But perhaps I'm running a little ahead of the course and already giving away things that will appear in Lecture 30.
• Options
27.
edited May 2018

Regarding comment #4:

Matthew Doty - in Puzzle 87 I was implicitly making Bool into a symmetric monoidal poset as in Puzzle 86. With this choice, it's not true that Bool-enriched categories are symmetric monoidal preorders.

What are they really?

My take is that $$\mathbf{Bool}$$-enriched categories are preorders, but they are not monoidal.

$$\mathbf{Bool}$$-enriched categories are preorders, because we can construct a preorder from a $$\mathbf{Bool}$$-enriched category, and viceversa; and the two constructions are inverses of each other (we have to prove this in exercise 2.31). For example, we can obtain a preorder $$(X, \le_X)$$ from a $$\mathbf{Bool}$$-enriched category $$\mathcal{X}$$ as follows:

• $$X = \mathrm{Ob}(\mathcal{X})$$
• $$x \le_X y$$ iff. $$\mathcal{X}(x, y) = \tt{true}$$.

However, I don't think $$\mathbf{Bool}$$-enriched categories are necessarily monoidal: we have seen in puzzle 66 from lecture 21 that there exist preorders that cannot be given monoidal structure (for example, Matthew's and Anindya's examples).

Comment Source:Regarding [comment #4](https://forum.azimuthproject.org/discussion/comment/18377/#Comment_18377): > Matthew Doty - in Puzzle 87 I was implicitly making **Bool** into a symmetric monoidal poset as in Puzzle 86. With this choice, it's not true that **Bool**-enriched categories are symmetric monoidal preorders. > > What are they really? My take is that \$$\mathbf{Bool}\$$-enriched categories are preorders, but they are not monoidal. \$$\mathbf{Bool}\$$-enriched categories are preorders, because we can construct a preorder from a \$$\mathbf{Bool}\$$-enriched category, and viceversa; and the two constructions are inverses of each other (we have to prove this in [exercise 2.31](https://forum.azimuthproject.org/discussion/1988)). For example, we can obtain a preorder \$$(X, \le_X)\$$ from a \$$\mathbf{Bool}\$$-enriched category \$$\mathcal{X}\$$ as follows: - \$$X = \mathrm{Ob}(\mathcal{X}) \$$ - \$$x \le_X y \$$ iff. \$$\mathcal{X}(x, y) = \tt{true} \$$. However, I don't think \$$\mathbf{Bool}\$$-enriched categories are necessarily monoidal: we have seen in puzzle 66 from [lecture 21](https://forum.azimuthproject.org/discussion/2082/lecture-21-chapter-2-monoidal-preorders/p1) that there exist preorders that cannot be given monoidal structure (for example, [Matthew's](https://forum.azimuthproject.org/discussion/comment/17926/#Comment_17926) and [Anindya's](https://forum.azimuthproject.org/discussion/comment/17927/#Comment_17927) examples).
• Options
28.
edited May 2018

I wrote:

1. for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathcal{V}$$.

Michael wrote:

According to the definition above, $$x,y \in \mathrm{Ob}(\mathcal{X})$$.

Right.

But in order for $$\mathcal{X}(x,y)$$ to be an element of $$\mathcal{V}$$, $$x,y \in V$$ must be true also?

No. I didn't say that.

Comment Source:I wrote: > 2. for every two objects \$$x,y\$$, one specifies an element \$$\mathcal{X}(x,y)\$$ of \$$\mathcal{V}\$$. Michael wrote: > According to the definition above, \$$x,y \in \mathrm{Ob}(\mathcal{X})\$$. Right. > But in order for \$$\mathcal{X}(x,y)\$$ to be an element of \$$\mathcal{V}\$$, \$$x,y \in V\$$ must be true also? No. I didn't say that. 
• Options
29.

Dan wrote:

My take is that $$\mathbf{Bool}$$-enriched categories are preorders, but they are not monoidal.

Right. There's nothing in the definition of enriched category that gives a way to multiply objects, so there's nothing "monoidal" about them. In the next lecture I'll show a $$\mathbf{Bool}$$-enriched category is a preorder, nothing more and nothing less.

Later we will meet monoidal enriched categories, but we're not there yet!

Comment Source:Dan wrote: > My take is that \$$\mathbf{Bool}\$$-enriched categories are preorders, but they are not monoidal. Right. There's nothing in the definition of enriched category that gives a way to multiply objects, so there's nothing "monoidal" about them. In the next lecture I'll show a \$$\mathbf{Bool}\$$-enriched category is a preorder, nothing more and nothing less. Later we will meet _monoidal_ enriched categories, but we're not there yet!
• Options
30.

Dan wrote:

This is a minor point, but since Michael Hong brought into discussion the notation, I think the definition:

for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathcal{V}$$.

should say "an element $$\mathcal{X}(x,y)$$ of $$V$$" rather than "[...] of $$\mathcal{V}$$" (which denotes a tuple, not a set).

That's true, given Fong and Spivak's unfortunate decision to use different names for the monoidal preorder $$\mathcal{V}$$ and its set of elements, $$V$$.

This is mainly unfortunate because they never did it before when discussing monoidal preorders! Bringing it in now seems more confusing than helpful to me.

Second, I'm never in favor of building such distinctions into the notation. For example, I don't say "a group $$\mathcal{G} = (G, \cdot, 1)$$", distinguishing the group $$\mathcal{G}$$ and its underlying set $$G$$. It's a waste of brain cells. There are times when this distinction is very important, but there are other ways to convey it.

Comment Source:Dan wrote: > This is a minor point, but since Michael Hong brought [into discussion the notation](https://forum.azimuthproject.org/discussion/comment/18388/#Comment_18388), I think the definition: > > for every two objects \$$x,y\$$, one specifies an element \$$\mathcal{X}(x,y)\$$ of \$$\mathcal{V}\$$. > should say "an element \$$\mathcal{X}(x,y)\$$ of \$$V\$$" rather than "[...] of \$$\mathcal{V}\$$" (which denotes a tuple, not a set). That's true, given Fong and Spivak's unfortunate decision to use different names for the monoidal preorder \$$\mathcal{V}\$$ and its set of elements, \$$V\$$. This is mainly unfortunate because they never did it before when discussing monoidal preorders! Bringing it in now seems more confusing than helpful to me. Second, I'm never in favor of building such distinctions into the notation. For example, I don't say "a group \$$\mathcal{G} = (G, \cdot, 1) \$$", distinguishing the group \$$\mathcal{G}\$$ and its underlying set \$$G\$$. It's a waste of brain cells. There are times when this distinction is very important, but there are other ways to convey it. 
• Options
31.
edited May 2018

In fact, for now I'm going to eliminate the use of $$V$$ to stand for the underlying set of the monoidal preorder $$\mathcal{V}$$. I've changed my lecture as follows:

We'll start by picking a symmetric monoidal poset $$( \mathcal{V},\leq,\otimes,I)$$, like $$\mathbf{Bool}$$ or $$\mathbf{Cost}$$.

I may regret this, but this is more like what we've been saying up to now, except for the funky new calligraphic font.

Comment Source:In fact, for now I'm going to eliminate the use of \$$V\$$ to stand for the underlying set of the monoidal preorder \$$\mathcal{V}\$$. I've changed my lecture as follows: > We'll start by picking a symmetric monoidal poset \$$( \mathcal{V},\leq,\otimes,I)\$$, like \$$\mathbf{Bool}\$$ or \$$\mathbf{Cost}\$$. I may regret this, but this is more like what we've been saying up to now, except for the funky new calligraphic font.
• Options
32.
edited May 2018

Matthew, observe that $$\mathcal{X}(\cdot, \cdot)$$ effectively gives a function $$\mathcal{X} : \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{X}) \to \mathcal{V}$$. Therefore, the two properties on $$\mathcal{V}$$-enriched categories are statements about how objects in $$\mathcal{V}$$ interact, not how objects in $$\mathrm{Ob}(\mathcal{X})$$ interact. We do not have any obvious way of producing a monoidal structure on $$\mathcal{X}$$.

(I would argue that, at least at this stage, a $$\mathcal{V}$$-enriched category is little more than a certain kind of function $$\mathcal{X} : S \times S \to \mathcal{V}$$ from a product of sets to a monoidal preorder, since we can always recover $$\mathrm{Ob}(\mathcal{X})$$ from the diagonal elements $$(x, x) \in \mathrm{Dom}(\mathcal{X})$$.)

Comment Source:[Matthew](https://forum.azimuthproject.org/discussion/comment/18420/#Comment_18420), observe that \$$\mathcal{X}(\cdot, \cdot)\$$ effectively gives a function \$$\mathcal{X} : \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{X}) \to \mathcal{V}\$$. Therefore, the two properties on \$$\mathcal{V}\$$-enriched categories are statements about how objects in \$$\mathcal{V}\$$ interact, not how objects in \$$\mathrm{Ob}(\mathcal{X})\$$ interact. We do not have any obvious way of producing a monoidal structure on \$$\mathcal{X}\$$. (I would argue that, at least at this stage, a \$$\mathcal{V}\$$-enriched category is little more than a certain kind of function \$$\mathcal{X} : S \times S \to \mathcal{V}\$$ from a product of sets to a monoidal preorder, since we can always recover \$$\mathrm{Ob}(\mathcal{X})\$$ from the diagonal elements \$$(x, x) \in \mathrm{Dom}(\mathcal{X})\$$.)
• Options
33.
edited May 2018

Daniel Wang wrote:

$$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\\\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both} \\\\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false}, or \mathcal{X}(x,y) = \tt{true}, for all objects x,y.} \\\\ \end{array}$$

Excellent, and good use of LaTeX too!

In Lecture 30 I consider the second row in detail: with this particular monoidal structure on $$\mathbf{Bool}$$, a $$\mathbf{Bool}$$-enriched category is just a preorder, since

If $$\mathcal{X}(x,y) = \mathcal{X}(y,z)$$, then $$\mathcal{X}(x,z)$$

is the good old transitive law. Let me think a bit about the other three: I haven't really studied them, but they must be good for something!

Comment Source:Daniel Wang wrote: > $$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\\\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both} \\\\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false}, or \mathcal{X}(x,y) = \tt{true}, for all objects x,y.} \\\\ \end{array}$$ Excellent, and good use of LaTeX too! In [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories) I consider the second row in detail: with this particular monoidal structure on \$$\mathbf{Bool}\$$, a \$$\mathbf{Bool}\$$-enriched category is just a preorder, since <center> If \$$\mathcal{X}(x,y) = \mathcal{X}(y,z)\$$, then \$$\mathcal{X}(x,z)\$$ </center> is the good old transitive law. Let me think a bit about the other three: I haven't really studied them, but they must be good for something!
• Options
34.
edited May 2018

Michael wrote:

Can we enrich a preorder?

Yes, but not today.

To me its seems the category itself is also a preorder or can be made into a preorder since we are trying to answer the quantified version of the same question. Why do we have to define it as category instead of a preorder?

An enriched category has an element of $$\mathcal{V}$$ for each pair of objects $$x,y$$. For a preorder we have a truth value for each pair $$x,y$$: either $$x \le y$$ is true or $$x \le y$$ is false. (See Lecture 30 for a detailed explanation of the second sentence here.)

If you have a way to turn elements of $$\mathcal{V}$$ into truth values, and this way obeys some nice properties, then you can turn your enriched category into a preorder. Sometimes this is an interesting thing to do! But not always.

Remember, $$\mathcal{V}$$ could be any symmetric monoidal preorder whatsoever. Think of all the dozens of examples you've memorized, now that you've resolved to acquire a good "zoo" of examples of every concept. For example, take the real numbers with addition and the usual notion of $$\le$$. How are you going to turn real numbers into truth values? The only way that has nice properties that are required to get a preorder is the completely boring way: send every real number to "true".

Comment Source:Michael wrote: > Can we enrich a preorder? Yes, but not today. > To me its seems the category itself is also a preorder or can be made into a preorder since we are trying to answer the quantified version of the same question. Why do we have to define it as category instead of a preorder? An enriched category has an _element of \$$\mathcal{V}\$$_ for each pair of objects \$$x,y\$$. For a preorder we have a _truth value_ for each pair \$$x,y\$$: either \$$x \le y\$$ is true or \$$x \le y\$$ is false. (See [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories) for a detailed explanation of the second sentence here.) If you have a way to turn elements of \$$\mathcal{V}\$$ into truth values, and this way obeys some nice properties, then you can turn your enriched category into a preorder. Sometimes this is an interesting thing to do! But not always. Remember, \$$\mathcal{V}\$$ could be any symmetric monoidal preorder whatsoever. Think of all the dozens of examples you've memorized, now that you've resolved to acquire a good "zoo" of examples of every concept. For example, take the real numbers with addition and the usual notion of \$$\le\$$. How are you going to turn real numbers into truth values? The only way that has nice properties that are required to get a preorder is the completely boring way: send every real number to "true".
• Options
35.

Anindya wrote:

I'm wondering why we insist that $$\mathcal{V}$$ be symmetric in the definition of a $$\mathcal{V}$$-enriched category. The symmetry doesn't seem to play any explicit role in the definition.

Excellent question! There is no reason to insist that $$\mathcal{V}$$ be symmetric.. Some more advanced theorems require it, but the concept of $$\mathcal{V}$$-enriched category makes perfect sense whenever $$\mathcal{V}$$ is a monoidal poset.

In fact, it works just fine whenever $$\mathcal{V}$$ is a monoidal preorder! The only reason I said "symmetric monoidal poset" is that Fong and Spivak restrict their attention to this case, so I'm assuming they will use these hypotheses at some point, and I don't want to have to worry about exactly when they need them.

Comment Source:Anindya wrote: > I'm wondering why we insist that \$$\mathcal{V}\$$ be symmetric in the definition of a \$$\mathcal{V}\$$-enriched category. The symmetry doesn't seem to play any explicit role in the definition. Excellent question! There is no reason to insist that \$$\mathcal{V}\$$ be symmetric.. Some more advanced theorems require it, but the concept of \$$\mathcal{V}\$$-enriched category makes perfect sense whenever \$$\mathcal{V}\$$ is a monoidal poset. In fact, it works just fine whenever \$$\mathcal{V}\$$ is a monoidal preorder! The only reason I said "symmetric monoidal poset" is that Fong and Spivak restrict their attention to this case, so I'm assuming they will use these hypotheses at some point, and I don't want to have to worry about exactly when they need them. 
• Options
36.
edited May 2018

Matthew, observe that $$\mathcal{X}(\cdot, \cdot)$$ effectively gives a function $$\mathcal{X} : \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{X}) \to \mathcal{V}$$. Therefore, the two properties on $$\mathcal{V}$$-enriched categories are statements about how objects in $$\mathcal{V}$$ interact, not how objects in $$\mathrm{Ob}(\mathcal{X})$$ interact. We do not have any obvious way of producing a monoidal structure on $$\mathcal{X}$$.

Got it, I have updated my posts now that I've been set straight. Thank you Dan!

Comment Source:> [Matthew](https://forum.azimuthproject.org/discussion/comment/18420/#Comment_18420), observe that \$$\mathcal{X}(\cdot, \cdot)\$$ effectively gives a function \$$\mathcal{X} : \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{X}) \to \mathcal{V}\$$. Therefore, the two properties on \$$\mathcal{V}\$$-enriched categories are statements about how objects in \$$\mathcal{V}\$$ interact, not how objects in \$$\mathrm{Ob}(\mathcal{X})\$$ interact. We do not have any obvious way of producing a monoidal structure on \$$\mathcal{X}\$$. Got it, I have updated my posts now that I've been set straight. Thank you Dan!
• Options
37.

@John

$$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\\\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both} \\\\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false}, or \mathcal{X}(x,y) = \tt{true}, for all objects x,y.} \\\\ \end{array}$$ In Lecture 30 I consider the second row in detail: with this particular monoidal structure on $$\mathbf{Bool}$$, a $$\mathbf{Bool}$$-enriched category is just a preorder, since

If $$\mathcal{X}(x,y) = \mathcal{X}(y,z)$$, then $$\mathcal{X}(x,z)$$

is the good old transitive law. Let me think a bit about the other three: I haven't really studied them, but they must be good for something!

Hopefully I can recover a tiny bit from my screw-ups here.

Theorem. We can always define a preorder over $$\mathrm{Obj}(\mathcal{X})$$ from any Bool-enriched category using the following rule:

$$x \preceq_{\mathcal{X}} y \iff \mathcal{X}(x,y) = \mathcal{X}(x,x)$$ This generic construction works for the case considered by you in Lecture 30, as well as the other cases Dan and I have been considering.

Proof.

The proof goes by considering each preorder over Bool. Reflexivity is trivial to see by the rule, so the only thing we need to show is transitivity.

In the case of the trivial preorder, since $$\mathcal{X}(a,b) = \mathcal{X}(c,d)$$ for all $$a,b,c,d \in \mathrm{Obj}(\mathcal{X})$$. So the preorder generated by the rule is trivial preorder over $$\mathrm{Obj}(\mathcal{X})$$.

In the case of the discrete poset, we know that $$\mathcal{X}(x,x) = I$$ for all $$x$$ (see my post # 3 in this thread). Suppose that $$a \preceq_{\mathcal{X}} b$$ and $$b \preceq_{\mathcal{X}} c$$. Then it must be that $$\mathcal{X}(a,b) = \mathcal{X}(b,c) = I$$. By the rule (b) of Bool-enriched categories we have:

$$\mathcal{X}(a,b) \otimes \mathcal{X}(b,c) \leq \mathcal{X}(a,c)$$ Since $$\mathcal{X}(a,b) \otimes \mathcal{X}(b,c) = I$$, it must be that $$I \leq \mathcal{X}(a,c)$$. Because we are consider the discrete poset, that means that $$\mathcal{X}(a,c) = I = \mathcal{X}(a,a)$$. Since $$\mathcal{X}(a,b) = \mathcal{X}(b,a)$$, this is an equivalence relation.

The next case is for the lattice $$\mathtt{false} \leq \mathtt{true}$$. Either $$I = \mathtt{true}$$ or $$I = \mathtt{false}$$.

Assume that $$I = \mathtt{true}$$. It must be that $$\otimes = \wedge$$. We can figure this out by just going through the truth table for $$\wedge$$. We know that $$\mathtt{true} \otimes \mathtt{true} = \mathtt{true}$$, and $$\mathtt{true} \otimes \mathtt{false} = \mathtt{false} \otimes \mathtt{true} = \mathtt{false}$$. Finally, since $$\mathtt{false} \leq \mathtt{true}$$ it must be $$\mathtt{false} \otimes \mathtt{false} \leq \mathtt{false} \otimes \mathtt{true}$$, hence $$\mathtt{false} \otimes \mathtt{false} = \mathtt{false}$$. The rule follows from the proof John writes in Lecture 30 along with the fact that $$\mathcal{X}(a,a) = I.\forall a$$.

Finally, consider the case where $$I = \mathtt{false}$$. Following logic similar to the above argument, it must be that $$\otimes = \vee$$. Next, we have a kind of collapse argument: $$\forall a,b,c,d. \mathcal{X}(a,b) = \mathcal{X}(c,d)$$. For suppose that some pair $$\mathcal{X}(a,b) = \mathtt{true}$$, then since $$\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)$$, it must be $$\mathcal{X}(a,c) = \mathtt{true}$$ since $$\otimes = \vee$$. However, we also have $$\mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)$$, hence $$\mathcal{X}(c,d) = \mathtt{true}$$. Since every element is comparable to every other element, this is just a trivial preorder.

$$\Box$$

As an aside, the case considered in Lecture 30 is the only interesting case in the above proof.

Comment Source:@John > $$\begin{array}{c|c} \text{monoid} & \text{effect of b) on \mathcal{X}(x,y)} \\\\ \hline (\oplus, \tt{false}) & \text{If \mathcal{X}(x,y), then for all objects z, either \mathcal{X}(x,z) or \mathcal{X}(z,x) or both} \\\\ (\wedge, \tt{true}) & \text{If \mathcal{X}(x,y) and \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\leftrightarrow, \tt{true}) & \text{If \mathcal{X}(x,y) = \mathcal{X}(y,z), then \mathcal{X}(x,z)} \\\\ (\vee, \tt{false}) & \text{Either \mathcal{X}(x,y) = \tt{false}, or \mathcal{X}(x,y) = \tt{true}, for all objects x,y.} \\\\ \end{array}$$ > > In [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories) I consider the second row in detail: with this particular monoidal structure on \$$\mathbf{Bool}\$$, a \$$\mathbf{Bool}\$$-enriched category is just a preorder, since > > If \$$\mathcal{X}(x,y) = \mathcal{X}(y,z)\$$, then \$$\mathcal{X}(x,z)\$$ > > is the good old transitive law. Let me think a bit about the other three: I haven't really studied them, but they must be good for something! Hopefully I can recover a tiny bit from my screw-ups here. **Theorem**. We can always define a preorder over \$$\mathrm{Obj}(\mathcal{X})\$$ from any **Bool**-enriched category using the following rule: $$x \preceq_{\mathcal{X}} y \iff \mathcal{X}(x,y) = \mathcal{X}(x,x)$$ This generic construction works for the case considered by you in [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories), as well as the other cases Dan and I have been considering. **Proof**. The proof goes by considering each preorder over **Bool**. Reflexivity is trivial to see by the rule, so the only thing we need to show is transitivity. In the case of the trivial preorder, since \$$\mathcal{X}(a,b) = \mathcal{X}(c,d)\$$ for all \$$a,b,c,d \in \mathrm{Obj}(\mathcal{X})\$$. So the preorder generated by the rule is trivial preorder over \$$\mathrm{Obj}(\mathcal{X})\$$. In the case of the discrete poset, we know that \$$\mathcal{X}(x,x) = I\$$ for all \$$x\$$ (see my post [# 3](https://forum.azimuthproject.org/discussion/comment/18372/#Comment_18372) in this thread). Suppose that \$$a \preceq_{\mathcal{X}} b\$$ and \$$b \preceq_{\mathcal{X}} c\$$. Then it must be that \$$\mathcal{X}(a,b) = \mathcal{X}(b,c) = I\$$. By the rule (b) of **Bool**-enriched categories we have: $$\mathcal{X}(a,b) \otimes \mathcal{X}(b,c) \leq \mathcal{X}(a,c)$$ Since \$$\mathcal{X}(a,b) \otimes \mathcal{X}(b,c) = I\$$, it must be that \$$I \leq \mathcal{X}(a,c)\$$. Because we are consider the discrete poset, that means that \$$\mathcal{X}(a,c) = I = \mathcal{X}(a,a)\$$. Since \$$\mathcal{X}(a,b) = \mathcal{X}(b,a)\$$, this is an equivalence relation. The next case is for the lattice \$$\mathtt{false} \leq \mathtt{true}\$$. Either \$$I = \mathtt{true}\$$ or \$$I = \mathtt{false}\$$. Assume that \$$I = \mathtt{true}\$$. It must be that \$$\otimes = \wedge\$$. We can figure this out by just going through the truth table for \$$\wedge\$$. We know that \$$\mathtt{true} \otimes \mathtt{true} = \mathtt{true}\$$, and \$$\mathtt{true} \otimes \mathtt{false} = \mathtt{false} \otimes \mathtt{true} = \mathtt{false}\$$. Finally, since \$$\mathtt{false} \leq \mathtt{true}\$$ it must be \$$\mathtt{false} \otimes \mathtt{false} \leq \mathtt{false} \otimes \mathtt{true}\$$, hence \$$\mathtt{false} \otimes \mathtt{false} = \mathtt{false}\$$. The rule follows from the proof John writes in [Lecture 30](https://forum.azimuthproject.org/discussion/2124/lecture-30-chapter-1-preorders-as-enriched-categories) along with the fact that \$$\mathcal{X}(a,a) = I.\forall a\$$. Finally, consider the case where \$$I = \mathtt{false}\$$. Following logic similar to the above argument, it must be that \$$\otimes = \vee\$$. Next, we have a kind of collapse argument: \$$\forall a,b,c,d. \mathcal{X}(a,b) = \mathcal{X}(c,d)\$$. For suppose that some pair \$$\mathcal{X}(a,b) = \mathtt{true}\$$, then since \$$\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)\$$, it must be \$$\mathcal{X}(a,c) = \mathtt{true}\$$ since \$$\otimes = \vee\$$. However, we also have \$$\mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)\$$, hence \$$\mathcal{X}(c,d) = \mathtt{true}\$$. Since every element is comparable to every other element, this is just a trivial preorder. \$$\Box\$$ As an aside, the case considered in Lecture 30 is the only *interesting* case in the above proof.
• Options
38.
edited May 2018

I was not understanding what was being 'enriched' and what was doing the 'enriching'. Based on comments I think I am not the only one. Here is my informal definition of $$\mathcal{X}$$ a $$\mathcal{V}$$-enriched category or $$\mathcal{V}$$-category. [Which may not turn out to be a category at all :P .]

You start with a container $$\mathcal{X}$$ which contains, in part, of a set of objects $$\text{Ob}(\mathcal{X})$$. You want to say something about the structure of objects in this container. You decide that structure has something to do with the relationship between pairs of objects $$\mathcal{X}(x, y) | x, y \in Ob(\mathcal{X})$$ .

Stop for a moment and notice that we have not mentioned the enriching $$\mathcal{V}$$ yet.

We introduce the enriching $$\mathcal{V} := (V, \le, I, \otimes )$$. Now we map each of the pairs $$\mathcal{X}(x, y) \rightarrow V$$ and in doing we say that $$\mathcal{X}$$ is enriched in $$\mathcal{V}$$. [Why don't we say 'enriched by'?] Well almost enriched, the mapping is constrained by two properties.

a) We want the enrichment to reinforce the identity of each object. For the identity pair $$\mathcal{X}(x, x) | x \in Ob( \mathcal{X} )$$ we have $$I \le \mathcal{X}(x, x)$$ where the $$I$$ and $$\le$$ come from $$\mathcal{V}$$.

b) We also want to have some notion of composition. When we have three objects $$x, y, z \in \text{Ob}( \mathcal{X} )$$, with pairs $$\mathcal{X}(x, y) \text{ and } \mathcal{X}(y, z)$$. Using the order relation and monoidal operation $$\le , \otimes$$ from $$\mathcal{V}$$ we require that they compose in the following way $$\mathcal{X}(x, y) \otimes \mathcal{X}(y, z) \le \mathcal{X}(x, z)$$.

Puzzle 87. A preorder is isomorphic with Bool-category. What makes this odd it that the number of pairs in the Bool-category appears to be greater than the number of arrows in the corresponding preorder. This discrepancy is explained by the fact that 'false' pairs are not expressed in the preorder.

Comment Source:I was not understanding what was being 'enriched' and what was doing the 'enriching'. Based on comments I think I am not the only one. Here is my informal definition of \$$\mathcal{X} \$$ a \$$\mathcal{V}\$$-enriched category or \$$\mathcal{V}\$$-category. [Which may not turn out to be a category at all :P .] You start with a container \$$\mathcal{X} \$$ which contains, in part, of a set of objects \$$\text{Ob}(\mathcal{X}) \$$. You want to say something about the structure of objects in this container. You decide that structure has something to do with the relationship between pairs of objects \$$\mathcal{X}(x, y) | x, y \in Ob(\mathcal{X}) \$$ . Stop for a moment and notice that we have not mentioned the enriching \$$\mathcal{V} \$$ yet. We introduce the enriching \$$\mathcal{V} := (V, \le, I, \otimes ) \$$. Now we map each of the pairs \$$\mathcal{X}(x, y) \rightarrow V \$$ and in doing we say that \$$\mathcal{X} \$$ is enriched in \$$\mathcal{V} \$$. [Why don't we say 'enriched by'?] Well almost enriched, the mapping is constrained by two properties. a) We want the enrichment to reinforce the identity of each object. For the identity pair \$$\mathcal{X}(x, x) | x \in Ob( \mathcal{X} ) \$$ we have \$$I \le \mathcal{X}(x, x) \$$ where the \$$I\$$ and \$$\le \$$ come from \$$\mathcal{V} \$$. b) We also want to have some notion of composition. When we have three objects \$$x, y, z \in \text{Ob}( \mathcal{X} ) \$$, with pairs \$$\mathcal{X}(x, y) \text{ and } \mathcal{X}(y, z) \$$. Using the order relation and monoidal operation \$$\le , \otimes \$$ from \$$\mathcal{V} \$$ we require that they compose in the following way \$$\mathcal{X}(x, y) \otimes \mathcal{X}(y, z) \le \mathcal{X}(x, z) \$$. **Puzzle 87.** A preorder is isomorphic with **Bool**-category. What makes this odd it that the number of pairs in the **Bool**-category appears to be greater than the number of arrows in the corresponding preorder. This discrepancy is explained by the fact that 'false' pairs are not expressed in the preorder. 
• Options
39.

re Puzzle 87, @John mentioned that the "process of deducing" the answer is important... so I thought it might be instructive to simply replace $$\mathcal{V}$$ with $$\mathbf{Bool}$$ in his definition to see where it gets us:

A $$\mathbf{Bool}$$-enriched category $$\mathcal{X}$$ consists of two parts, satisfying two properties. First:

1. one specifies a set $$\mathrm{Ob}(\mathcal{X})$$, elements of which are called objects;

2. for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathbf{Bool}$$.

Then:

a) for every object $$x\in\text{Ob}(\mathcal{X})$$ we require that

$${\tt{true}} \leq \mathcal{X}(x, x)$$ b) for every three objects $$x,y,z\in\mathrm{Ob}(\mathcal{X})$$, we require that $$\mathcal{X}(x,y)\wedge \mathcal{X}(y,z)\leq\mathcal{X}(x,z)$$

It's pretty clear that (a) is equivalent to $\mathcal{X}(x, x) = \tt{true}$ and (b) is equivalent to $\mathcal{X}(x,y) = {\tt{true}} {\textrm{ and }} \mathcal{X}(y,z) = {\tt{true}} {\textrm{ implies }} \mathcal{X}(x,z) = {\tt{true}}$ ie, the reflexive and transitive laws.

Comment Source:re **Puzzle 87**, @John mentioned that the "process of deducing" the answer is important... so I thought it might be instructive to simply replace \$$\mathcal{V}\$$ with \$$\mathbf{Bool}\$$ in his definition to see where it gets us: > A **\$$\mathbf{Bool}\$$-enriched category** \$$\mathcal{X}\$$ consists of two parts, satisfying two properties. First: > > 1. one specifies a set \$$\mathrm{Ob}(\mathcal{X})\$$, elements of which are called **objects**; > > 2. for every two objects \$$x,y\$$, one specifies an element \$$\mathcal{X}(x,y)\$$ of \$$\mathbf{Bool}\$$. > > Then: > > a) for every object \$$x\in\text{Ob}(\mathcal{X})\$$ we require that > > ${\tt{true}} \leq \mathcal{X}(x, x)$ > > b) for every three objects \$$x,y,z\in\mathrm{Ob}(\mathcal{X})\$$, we require that > > $\mathcal{X}(x,y)\wedge \mathcal{X}(y,z)\leq\mathcal{X}(x,z)$ It's pretty clear that (a) is equivalent to \$\mathcal{X}(x, x) = \tt{true}\$ and (b) is equivalent to \$\mathcal{X}(x,y) = {\\tt{true}} {\textrm{ and }} \mathcal{X}(y,z) = {\\tt{true}} {\textrm{ implies }} \mathcal{X}(x,z) = {\\tt{true}}\$ ie, the reflexive and transitive laws.
• Options
40.
edited May 2018

You don't enrich by, but rather enrich in.

That is to say, we're starting with $$\mathcal{V} := (V, \le, I, \otimes )$$ first, and then we build a new kind of $$\mathcal{V}$$-category, called $$\mathcal{X}$$, out of the appropriate $$\mathcal{V}$$-building blocks following certain constraints, if that makes sense.

Comment Source:You don't enrich *by*, but rather enrich *in*. That is to say, we're starting with \$$\mathcal{V} := (V, \le, I, \otimes ) \$$ first, and then we build a new kind of \$$\mathcal{V}\$$-category, called \$$\mathcal{X} \$$, out of the appropriate \$$\mathcal{V}\$$-building blocks following certain constraints, if that makes sense.
• Options
41.
edited May 2018

Keith is right.

Fredrick wrote:

I was not understanding what was being 'enriched' by what. Based on comments I think I am not the only one. Here is my informal definition of $$\mathcal{X}$$ a $$\mathcal{V}$$-enriched category. You start with a category \mathcal{X}\).

No! An enriched category is not, in general, a category.

Don't be fooled by the terminology, or your eagerness to start talking about categories!

A category has a set for any pair of objects, but our enriched categories have a mere element of the set $$\mathcal{V}$$ for any pair of objects!

Later we will look at more general enriched categories, which include both categories and the enriched categories we're studying now as special cases. However, even when we get to that situation, you must be careful. An enriched category will not, in general, be a category.

It might been wiser to call the things we're studying so far "enriched preorders". A plain old preorder has a truth value for any pair of objects - that is, an element of $$\textbf{Bool}$$. We are "enriching" this idea - in the sense of making it more exciting - by replacing $$\textbf{Bool}$$ by an arbitrary symmetric monoidal preorder $$\mathcal{V}$$.

However, even the term "enriched preorder" could confuse you into thinking that every enriched preorder is a preorder. It's not! After all, there's no reason that having an element of $$\mathcal{V}$$ gives you an element of $$\textbf{Bool}$$ in any interesting way.

I suddenly remember being confused by this issue myself, once upon a time. I was confused because there's a certain bunch of examples of enriched categories that are categories... which we haven't talked about in this course yet. That's the reason for this terminology. But don't let the terminology fool you! [-X

Comment Source:Keith is right. Fredrick wrote: > I was not understanding what was being 'enriched' by what. Based on comments I think I am not the only one. Here is my informal definition of \$$\mathcal{X}\$$ a \$$\mathcal{V}\$$-enriched category. You start with a category \mathcal{X}\\). No! _<font color = "red"><b>An enriched category is not, in general, a category</b></font>_. Don't be fooled by the terminology, or your eagerness to start talking about categories! A category has a _set_ for any pair of objects, but our enriched categories have a mere _element of the set \$$\mathcal{V}\$$_ for any pair of objects! Later we will look at more general enriched categories, which include both categories and the enriched categories we're studying now as special cases. However, even when we get to that situation, you must be careful. _<font color = "red"><b>An enriched category will not, in general, be a category</b></font>_. It might been wiser to call the things we're studying so far "enriched preorders". A plain old preorder has a _truth value_ for any pair of objects - that is, an element of \$$\textbf{Bool}\$$. We are "enriching" this idea - in the sense of making it more exciting - by replacing \$$\textbf{Bool}\$$ by an arbitrary symmetric monoidal preorder \$$\mathcal{V}\$$. However, even the term "enriched preorder" could confuse you into thinking that every enriched preorder is a preorder. _It's not!_ After all, there's no reason that having an element of \$$\mathcal{V}\$$ gives you an element of \$$\textbf{Bool}\$$ in any interesting way. I suddenly remember being confused by this issue myself, once upon a time. I was confused because there's a certain bunch of examples of enriched categories that _are_ categories... which we haven't talked about in this course yet. That's the reason for this terminology. But don't let the terminology fool you! [-X 
• Options
42.
edited May 2018

Ok, I will make an edit, I figured that if it was called a category that must be what it was. :-P Do we know that $$\mathcal{X}$$ will be a preorder? Doesn't that depend on the monoidal preorder $$\mathcal{V}$$ that we select. Are you saying that any monoidal preorder $$\mathcal{V}$$ will cause $$\mathcal{X}$$ to be a preorder? That seems likely. How to prove that...

Comment Source:Ok, I will make an edit, I figured that if it was called a category that must be what it was. :-P Do we know that \$$\mathcal{X} \$$ will be a preorder? Doesn't that depend on the monoidal preorder \$$\mathcal{V} \$$ that we select. Are you saying that any monoidal preorder \$$\mathcal{V} \$$ will cause \$$\mathcal{X} \$$ to be a preorder? That seems likely. How to prove that...
• Options
43.
edited May 2018

When I start talking about categories in this course, it'll be with a huge thunderclap, and a peal of trumpets. I have added a warning to the start of this lecture, so nobody else falls into the trap that Fredrick did.

Comment Source:When I start talking about categories in this course, it'll be with a huge thunderclap, and a peal of trumpets. I have added a warning to the start of this lecture, so nobody else falls into the trap that Fredrick did.
• Options
44.
edited May 2018

Puzzle KEP: What kind of structure do we get when we enrich the point ( $${ \bullet }$$ )? That is to say, what is a $${ \bullet }$$-enriched category?

Note: The point ( $${ \bullet }$$ ) can be viewed as the discrete preorder on 1 object.

Comment Source:**Puzzle KEP:** What kind of structure do we get when we enrich the point ( \$$\{ \bullet \}\$$ )? That is to say, what is a \$$\{ \bullet \}\$$-enriched category? Note: The point ( \$$\{ \bullet \}\$$ ) can be viewed as the discrete preorder on 1 object. 
• Options
45.

Puzzle KEP: I addressed this in my earlier comment, which Anindya then followed up on.

Comment Source:**Puzzle KEP:** I addressed this in [my earlier comment](https://forum.azimuthproject.org/discussion/comment/18403/#Comment_18403), which Anindya then [followed up on](https://forum.azimuthproject.org/discussion/comment/18408/#Comment_18408).
• Options
46.
edited May 2018

You guys are no fun. lol

Actually, I don't think that the answer that was given is right.

Comment Source:You guys are no fun. lol Actually, I don't think that the answer that was given is right. 
• Options
47.
edited May 2018

You guys are no fun. lol

Au contraire, you're simply late to the party. ;D

Ok, let's try something else.

Puzzle JMC8 (unsolved): What do we get if we enrich over the lattice given by $$\{\bot \le \mathrm{false} \le \top, \bot \le \mathrm{true} \le \top\}$$, with identity $$\mathrm{\bot}$$ and monoidal operator given by join ($$\vee$$)? Some questions I've been looking at myself work over similar structures, so it might be fun to see if we get anything interesting when we enrich by them.

(In particular, the preorder here is an elementary domain (plus a top element); there are many other domains, and we generally like to look at directed-complete partial orders.)

Comment Source:> You guys are no fun. lol _Au contraire_, you're simply late to the party. ;D Ok, let's try something else. **Puzzle JMC8** (unsolved): What do we get if we enrich over the lattice given by \$$\\{\bot \le \mathrm{false} \le \top, \bot \le \mathrm{true} \le \top\\}\$$, with identity \$$\mathrm{\bot}\$$ and monoidal operator given by join (\$$\vee\$$)? Some questions I've been looking at myself work over similar structures, so it might be fun to see if we get anything interesting when we enrich by them. (In particular, the preorder here is an [elementary domain](https://en.wikipedia.org/wiki/Domain_theory#Special_types_of_domains) (plus a top element); there are many other domains, and we generally like to look at [directed-complete partial orders](https://en.wikipedia.org/wiki/Complete_partial_order).)
• Options
48.
edited May 2018

Actually, I don't think that the answer that was given is right.

The enrichment doesn't give us any information about how objects are related, so we can only really get a set, as Anindya points out -- and there's a trivial information-free preorder, the codiscrete preorder, which we can endow on any set. Can you explain where this logic falls through?

EDIT: changed "discrete" to "codiscrete" -- all arrows need to be the same, but the discrete preorder has "existing" identity arrows and "nonexistent" other arrows. The codiscrete preorder has all arrows.

Comment Source:> Actually, I don't think that the answer that was given is right. The enrichment doesn't give us any information about how objects are related, so we can only really get a set, as Anindya points out -- and there's a trivial information-free preorder, the codiscrete preorder, which we can endow on any set. Can you explain where this logic falls through? **EDIT:** changed "discrete" to "codiscrete" -- all arrows need to be the same, but the discrete preorder has "existing" identity arrows and "nonexistent" other arrows. The codiscrete preorder has all arrows.
• Options
49.
edited May 2018

Fredrick wrote:

Do we know that $$\mathcal{X}$$ will be a preorder?

No. In in comment #41 wrote:

It might been wiser to call the things we're studying so far "enriched preorders". A plain old preorder has a truth value for any pair of objects - that is, an element of $$\textbf{Bool}$$. We are "enriching" this idea - in the sense of making it more exciting - by replacing $$\textbf{Bool}$$ by an arbitrary symmetric monoidal preorder $$\mathcal{V}$$.

However, even the term "enriched preorder" could confuse you into thinking that every enriched preorder is a preorder. It's not! After all, there's no reason that having an element of $$\mathcal{V}$$ gives you an element of $$\textbf{Bool}$$ in any interesting way.

Comment Source:Fredrick wrote: > Do we know that \$$\mathcal{X} \$$ will be a preorder? No. In in comment #41 wrote: > It might been wiser to call the things we're studying so far "enriched preorders". A plain old preorder has a _truth value_ for any pair of objects - that is, an element of \$$\textbf{Bool}\$$. We are "enriching" this idea - in the sense of making it more exciting - by replacing \$$\textbf{Bool}\$$ by an arbitrary symmetric monoidal preorder \$$\mathcal{V}\$$. > However, even the term "enriched preorder" could confuse you into thinking that every enriched preorder is a preorder. _It's not!_ After all, there's no reason that having an element of \$$\mathcal{V}\$$ gives you an element of \$$\textbf{Bool}\$$ in any interesting way. 
• Options
50.

The enrichment doesn't give us any information about how objects are related, so we can only really get a set, as Anindya points out -- and there's a trivial information-free preorder, the codiscrete preorder, which we can endow on any set. Can you explain where this logic falls through?

Try using set union ($$\cup )$$ as your monoidal product, and then apply the definition John gave in the first post. The enriched structure that comes out should be an old friend.

Comment Source:>The enrichment doesn't give us any information about how objects are related, so we can only really get a set, as Anindya points out -- and there's a trivial information-free preorder, the codiscrete preorder, which we can endow on any set. Can you explain where this logic falls through? Try using set union (\$$\cup )\$$ as your monoidal product, and then apply the definition John gave in the first post. The enriched structure that comes out should be an old friend.