Howdy, Stranger!

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

Options

Exercise 70 - Chapter 2

edited June 2018

Just like we can talk of opposite preorders, we can talk of the opposite of any V-category. The opposite of a $$\mathcal{V}$$-category $$\mathcal{X}$$ is denoted $$\mathcal{X}^{op}$$ and is defined by

(i.) $$Ob(\mathcal{X}^{op}) := Ob(\mathcal{X})$$, and

(ii.) for all $$x, y \in \mathcal{X}$$, we have $$\mathcal{X}^{op}(x, y) := \mathcal{X}(y, x)$$.

Similarly, we can talk of daggers. A dagger structure on a $$\mathcal{V}$$-category $$\mathcal{X}$$ is a functor $$\dagger : \mathcal{X} \rightarrow \mathcal{X}^{op}$$ that is identity on objects and such that $$\dagger . \dagger = id_\mathcal{X}$$ .

Recall that an ordinary metric space $$(X, d)$$ is a Lawvere metric space with some extra properties; see Definition 2.32. One of these properties is symmetry: $$d(x, y) = d(y, x)$$ for every $$x, y \in X$$. What if we have a Lawvere metric space $$(X, d)$$ such that the identity function $$id_X : X \rightarrow X$$ is a Cost-functor $$(X, d) \rightarrow (X, d)^{op}$$ . Is this exactly the same as the symmetry property?

1. Show that a skeletal dagger Cost-category is a metric space.

2. Make sense of the following analogy: “preorders are to sets as Lawvere metric spaces are to metric spaces.”

• Options
1.
edited June 2018

I'm having trouble understanding what a dagger Cost-category is.

From the definition above I understand what a dagger structure on a category is: it's a special type of a functor. But I'm not sure how to interpret it as a part of my metric space $$X$$. Is $$id_X$$ the $$\mathcal{X}(x, x)$$? But it can't be since it's a functor to the opposite category? I feel like I'm missing something obvious!

One more remark: It's a bit confusing that a metric space is a Lawvere metric space with some extra properties; it doesn't make sense linguistically. Addition of prefix to a word usually signifies the existence of some extra structure to it. Here, it signifies removal of it. In other words, not every Lawvere metric space is a metric space and this bugs me more than it should :)

And one more remark: Can we define a new type of preorder, where objects are any arbitrary algebraic structures and the $$\leq$$ operation is incluson? For example, every preorder is a category, but not vice versa so $$\textrm{Category} \leq \textrm{Preorder}$$. Also, $$\textrm{Preorder} \leq \textrm{Monoidal Preorder} \leq \textrm{Symmetric monoidal preorder} \leq \textrm{Quantale}$$. This reminds me of the level shifting section from the end of chapter one.

We're defining a preorder relation on mathematical structures, one of which is the preorder itself!

What would a Hasse diagram of that look?

Comment Source:I'm having trouble understanding what a dagger **Cost**-category is. From the definition above I understand what a dagger structure on a category is: it's a special type of a functor. But I'm not sure how to interpret it as a part of my metric space \$$X \$$. Is \$$id_X \$$ the \$$\mathcal{X}(x, x) \$$? But it can't be since it's a functor to the opposite category? I feel like I'm missing something obvious! --- One more remark: It's a bit confusing that a metric space is a Lawvere metric space with some extra properties; it doesn't make sense linguistically. Addition of prefix to a word usually signifies the existence of some extra structure to it. Here, it signifies removal of it. In other words, not every Lawvere metric space is a metric space and this bugs me more than it should :) --- And one more remark: Can we define a new type of preorder, where objects are any arbitrary algebraic structures and the \$$\leq \$$ operation is incluson? For example, every preorder is a category, but not vice versa so \$$\textrm{Category} \leq \textrm{Preorder} \$$. Also, \$$\textrm{Preorder} \leq \textrm{Monoidal Preorder} \leq \textrm{Symmetric monoidal preorder} \leq \textrm{Quantale} \$$. This reminds me of the level shifting section from the end of chapter one. _We're defining a preorder relation on mathematical structures, one of which is the preorder itself!_ What would a Hasse diagram of that look? 
• Options
2.
edited June 2018

I will make a guess here and claim that the dagger functor is actually a pair of functors: $$\dagger : \mathcal{X} \rightarrow \mathcal{X}^{op} \text{ and } \dagger : \mathcal{X}^{op} \rightarrow \mathcal{X}$$ Were it not so, the statement $$\dagger . \dagger = id_\mathcal{X}$$ would not make sense.

There will be other objects with the same problem. The $$\mathcal{V}$$-category is similarly less restrictive than a standard category. As it turns out the standard category is the $$\mathbf{Set}$$-category.

I do not know of an alternate name for the standard metric-space.

I think the Hasse diagram would be very similar to the diagram of $$\mathbb{N}$$ where each number would be represented as the set of its factors. But rather than factors it would be the axioms that apply.

Comment Source:I will make a guess here and claim that the dagger functor is actually a pair of functors: $\dagger : \mathcal{X} \rightarrow \mathcal{X}^{op} \text{ and } \dagger : \mathcal{X}^{op} \rightarrow \mathcal{X}$ Were it not so, the statement $\dagger . \dagger = id_\mathcal{X}$ would not make sense. --- There will be other objects with the same problem. The \$$\mathcal{V}\$$-category is similarly less restrictive than a standard category. As it turns out the standard category is the \$$\mathbf{Set}\$$-category. I do not know of an alternate name for the standard metric-space. --- I think the Hasse diagram would be very similar to the diagram of \$$\mathbb{N}\$$ where each number would be represented as the set of its factors. But rather than factors it would be the axioms that apply.
• Options
3.
edited June 2018

Take $$Ob(\mathcal{X}) = \lbrace x , y \rbrace$$, and $$d(x,y)=0=d(y,x)$$. Is there any way to prove that $$x=y$$? Using that the category is skeletal, we would need that $$x\le y$$ and $$y\le x$$. But I don't see how to get those. There is no order on $$\mathcal{X}$$, the order is in the poset $$[0,\infty]$$.

Comment Source:Take \$$Ob(\mathcal{X}) = \lbrace x , y \rbrace \$$, and \$$d(x,y)=0=d(y,x)\$$. Is there any way to prove that \$$x=y\$$? Using that the category is skeletal, we would need that \$$x\le y\$$ and \$$y\le x\$$. But I don't see how to get those. There is no order on \$$\mathcal{X}\$$, the order is in the poset \$$[0,\infty]\$$.
• Options
4.

Sorry, this exercise was not well defined! Thanks for spotting the errors. We've now written solutions to all exercises, and they'll be part of the next release, along with amendments to impossible exercises such as this one. This exercise now reads:

The concepts of opposite, dagger, and skeleton extend from preorders to $$\mathcal{V}$$-categories. The opposite of a $$\mathcal{V}$$-category $$\mathcal{X}$$ is denoted $$\mathcal{X}^{op}$$ and is defined by

(i) $$Ob\mathcal{X}^{op} = Ob\mathcal{X}$$$, and (ii) for $$x,y \in \mathcal{X}$$, we have $$\mathcal{X}^{op}(x,y)=\mathcal{X}(y,x)$$. A $$\mathcal{V}$$-category $$\mathcal{X}$$ is a dagger $$\mathcal{V}$$-category if the identity function is a $$\mathcal{V}$$-functor $$\dagger\colon\mathcal{X}\to\mathcal{X}^{op}$$. And a skeletal $$\mathcal{V}$$-category is one in which if $$I \le \mathcal{X}(x,y)$$ and $$I \le \mathcal{X}(y,x)$$, then $$x=y$$. Recall that an extended metric space $$(X,d)$$ is a Lawvere metric space with two extra properties; see properties (b) and (c) in Definition 2.49. 1. Show that a skeletal dagger Cost-category is an extended metric space. 2. Use Exercise 1.69 to make sense of the following analogy: "preorders are to sets as Lawvere metric spaces are to extended metric spaces.'' Comment Source:Sorry, this exercise was not well defined! Thanks for spotting the errors. We've now written solutions to all exercises, and they'll be part of the next release, along with amendments to impossible exercises such as this one. This exercise now reads: --- The concepts of opposite, dagger, and skeleton extend from preorders to \$$\mathcal{V} \$$-categories. The *opposite* of a \$$\mathcal{V} \$$-category \$$\mathcal{X} \$$ is denoted \$$\mathcal{X}^{op} \$$ and is defined by (i) \$$Ob\mathcal{X}^{op} = Ob\mathcal{X} \$$$, and (ii) for \$$x,y \in \mathcal{X} \$$, we have \$$\mathcal{X}^{op}(x,y)=\mathcal{X}(y,x) \$$. A \$$\mathcal{V} \$$-category \$$\mathcal{X} \$$ is a *dagger* \$$\mathcal{V} \$$-category if the identity function is a \$$\mathcal{V} \$$-functor \$$\dagger\colon\mathcal{X}\to\mathcal{X}^{op} \$$. And a *skeletal* \$$\mathcal{V} \$$-category is one in which if \$$I \le \mathcal{X}(x,y)\$$ and \$$I \le \mathcal{X}(y,x)\$$, then \$$x=y\$$. Recall that an extended metric space \$$(X,d) \$$ is a Lawvere metric space with two extra properties; see properties (b) and (c) in Definition 2.49. 1. Show that a skeletal dagger **Cost**-category is an extended metric space. 2. Use Exercise 1.69 to make sense of the following analogy: "preorders are to sets as Lawvere metric spaces are to extended metric spaces.'' ---