#### Howdy, Stranger!

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

Options

# Lecture 32 - Chapter 2: Enriched Functors

edited July 2018

Whenever we study some kind of mathematical gadget, we should also study maps between gadgets of this kind. For example:

• When we study sets, we should also study functions.
• When we study preorders, we should also study monotone functions.
• When we study monoids, we should also study monoid homomorphisms.
• When we study monoidal preorders, we should also study monoidal monotones.

The reason is that mathematical structures don't live in isolation - or at least, they don't want to! They like to live in communities: they like to talk to each other.

Now we're studying enriched categories. So we should also study maps between enriched categories. These are called "enriched functors".

To get going, we need to choose a monoidal preorder $$\mathcal{V}$$. I defined $$\mathcal{V}$$-categories at the end of Lecture 30. And now I'll define maps between these:

Definition. Let $$\mathcal{X}$$ and $$\mathcal{Y}$$ be $$\mathcal{V}$$-categories. A $$\mathcal{V}$$-functor from $$\mathcal{X}$$ to $$\mathcal{Y}$$, denoted $$F\colon\mathcal{X}\to\mathcal{Y}$$, is a function

$$F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y})$$ such that

$$\mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x'))$$ for all $$x,x' \in\mathrm{Ob}(\mathcal{X})$$.

Remember, both $$\mathcal{X}(x,x')$$ and $$\mathcal{Y}(F(x),F(x'))$$ are elements of $$\mathcal{V}$$, which is a monoidal preorder. So, it makes sense to say that one is less than or equal to the other.

As always when meeting a new definition, it's good to look at examples. What's $$\mathbf{Bool}$$-functor between $$\mathbf{Bool}$$-categories?

We've seen a $$\mathbf{Bool}$$-category $$\mathcal{X}$$ is just a poset, where we say $$x \le x'$$ iff $$\mathcal{X}(x,y) = \texttt{true}$$. So, the obvious guess is that a $$\mathbf{Bool}$$-functor is just a monotone function.

Why is this obvious? Well, $$\mathbf{Bool}$$-functors go between $$\mathbf{Bool}$$-categories, which are posets, and monotone functions also go between posets... so life would be simple if these were the same thing!

And indeed it's true! Suppose we have a $$\mathbf{Bool}$$-functor $$F\colon\mathcal{X}\to\mathcal{Y}$$. This is a function

$$F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y})$$ such that

$$\mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x')).$$ But what does this last inequality mean? Remember, both $$\mathcal{X}(x,x')$$ and $$\mathcal{Y}(F(x),F(x'))$$ are elements of $$\mathbf{Bool} = \{\texttt{true}, \texttt{false} \}$$. In $$\mathbf{Bool}$$, the symbol $$\leq$$ means "implies". When we make $$\mathcal{X}$$ into a poset, $$\mathcal{X}(x,x') = \texttt{true}$$ means that $$x \le x'$$. Similarly, $$\mathcal{Y}(F(x),F(x')) = \texttt{true}$$ means that $$F(x) \le F(x')$$. So, the inequality just says

$$x \le x' \textrm{ implies } F(x) \le F(x') .$$ This says that $$F$$ is a monotone function!

Why don't you try one now? We've seen a $$\mathbf{Cost}$$-category is a Lawvere metric space.

Puzzle 92. What's a $$\mathbf{Cost}$$-functor?

To read other lectures go here.

• Options
1.

Puzzle 92. What's a $$\mathbf{Cost}$$-functor?

Definition. Let $$\mathcal{X}$$ and $$\mathcal{Y}$$ be $$\mathbf{Cost}$$-categories. A $$\mathbf{Cost}$$-functor from $$\mathcal{X}$$ to $$\mathcal{Y}$$, denoted $$F\colon\mathcal{X}\to\mathcal{Y}$$, is a function

$$F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y})$$ such that

$$\mathcal{d}(x,x') \leq \mathcal{d'}(F(x),F(x'))$$ for all $$x,x' \in\mathrm{Ob}(\mathcal{X})$$.

I would call this is a non-contracting map, since the generalized distance or cost metric may only stay the same or expand.

Comment Source:>**Puzzle 92.** What's a \$$\mathbf{Cost}\$$-functor? **Definition.** Let \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$ be \$$\mathbf{Cost}\$$-categories. A \$$\mathbf{Cost}\$$-functor from \$$\mathcal{X}\$$ to \$$\mathcal{Y}\$$, denoted \$$F\colon\mathcal{X}\to\mathcal{Y}\$$, is a function $F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y})$ such that $\mathcal{d}(x,x') \leq \mathcal{d'}(F(x),F(x'))$ for all \$$x,x' \in\mathrm{Ob}(\mathcal{X})\$$. I would call this is a non-contracting map, since the generalized distance or cost metric may only stay the same or expand. 
• Options
2.
edited May 2018

[N/M i have been baffled by the reversed order of Cost]

Comment Source:[N/M i have been baffled by the reversed order of **Cost**]
• Options
3.
edited May 2018

Keith: don't forget the fiendish reversal of inequality in the definition of $$\mathbf{Cost}$$.

Comment Source:Keith: _don't forget the fiendish reversal of inequality in the definition of \$$\mathbf{Cost}\$$._
• Options
4.
edited May 2018

Matthew wrote:

Are we allowed to relax the definition of a $$\mathcal{V}$$-functor?

I know what you're thinking, but one important lesson from category theory is that most systematic definition of a map between mathematical gadgets is usually a map that preserves all their structure. Other definitions of map can be useful, and even tremendously important - but they're not "systematic", so it's harder to apply category theory to study them.

So, the most systematic concept of a map between metric spaces is not a continuous map! Since a metric space $$X$$ is equipped with a distance function $$d: X \times X \to \mathbb{R}$$, the most natural concept of a map between metric spaces is an isometry: a map that preserves distances.

The concept of continuous map is natural for topological spaces. When we're studying continuous maps between metric spaces, we're secretly applying a "forgetful functor" to turn these metric spaces into topological spaces, and then looking at maps between those.

This whole story has an analogue for Lawvere metric spaces, which I'm trying to probe with Puzzle 92.

The concept of $$\mathcal{V}$$-functor, introduced above, cleverly slips an inequality into the game. One might have tried demanding

$$\mathcal{X}(x,x') = \mathcal{X'}(F(x),F(x'))$$ and this would change the answers to "What is a $$\mathbf{Bool}$$-functor?" and "What is a $$\mathbf{Cost}$$-functor?" in interesting and instructive ways. It turns out that

$$\mathcal{X}(x,x') \le \mathcal{X'}(F(x),F(x'))$$ is, in general, a wiser choice... as becomes apparent when we proceed deeper into category theory.

Comment Source:Matthew wrote: > Are we allowed to relax the definition of a \$$\mathcal{V}\$$-functor? I know what you're thinking, but one important lesson from category theory is that most systematic definition of a map between mathematical gadgets is usually a map that preserves all their structure. Other definitions of map can be useful, and even tremendously important - but they're not "systematic", so it's harder to apply category theory to study them. So, the most systematic concept of a map between metric spaces is _not_ a continuous map! Since a metric space \$$X\$$ is equipped with a distance function \$$d: X \times X \to \mathbb{R}\$$, the most natural concept of a map between metric spaces is an **isometry**: a map that preserves distances. The concept of continuous map is natural for _topological spaces_. When we're studying continuous maps between metric spaces, we're secretly applying a "forgetful functor" to turn these metric spaces into topological spaces, and then looking at maps between _those_. This whole story has an analogue for Lawvere metric spaces, which I'm trying to probe with Puzzle 92. The concept of \$$\mathcal{V}\$$-functor, introduced above, cleverly slips an inequality into the game. One might have tried demanding $\mathcal{X}(x,x') = \mathcal{X'}(F(x),F(x'))$ and this would change the answers to "What is a \$$\mathbf{Bool}\$$-functor?" and "What is a \$$\mathbf{Cost}\$$-functor?" in interesting and instructive ways. It turns out that $\mathcal{X}(x,x') \le \mathcal{X'}(F(x),F(x'))$ is, in general, a wiser choice... as becomes apparent when we proceed deeper into category theory.
• Options
5.

John Baez wrote:

Keith: don't forget the fiendish reversal of inequality in the definition of $$\mathbf{Cost}$$.

Luckily, such a fix is easy.

Definition. Let $$\mathcal{X}$$ and $$\mathcal{Y}$$ be $$\mathbf{Cost}$$-categories. A $$\mathbf{Cost}$$-functor from $$\mathcal{X}$$ to $$\mathcal{Y}$$, denoted $$F\colon\mathcal{X}\to\mathcal{Y}$$, is a function

$$F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y})$$ such that $$\mathcal{d}(x,x') \geq \mathcal{d'}(F(x),F(x'))$$ for all $$x,x' \in\mathrm{Ob}(\mathcal{X})$$.

I would call this is a non-expanding map, since the generalized distance or cost metric may only stay the same or contract.

Comment Source:John Baez wrote: >Keith: _don't forget the fiendish reversal of inequality in the definition of \$$\mathbf{Cost}\$$._ Luckily, such a fix is easy. >**Definition.** Let \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$ be \$$\mathbf{Cost}\$$-categories. A \$$\mathbf{Cost}\$$-functor from \$$\mathcal{X}\$$ to \$$\mathcal{Y}\$$, denoted \$$F\colon\mathcal{X}\to\mathcal{Y}\$$, is a function >$F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y})$ >such that >$\mathcal{d}(x,x') \geq \mathcal{d'}(F(x),F(x'))$ >for all \$$x,x' \in\mathrm{Ob}(\mathcal{X})\$$. >I would call this is a ***non-expanding*** map, since the generalized distance or cost metric may only stay the same or ***contract***. 
• Options
6.

As I was reading, I got confused about what $$\mathcal X'$$ refers to? Should $$F: \mathcal X \to \mathcal X'$$? Instead of $$F: \mathcal X \to \mathcal Y$$?

Comment Source:As I was reading, I got confused about what \$$\mathcal X'\$$ refers to? Should \$$F: \mathcal X \to \mathcal X'\$$? Instead of \$$F: \mathcal X \to \mathcal Y\$$?
• Options
7.
edited May 2018

Sophie - sorry, $$\mathcal{X}'$$ should everywhere be $$\mathcal{Y}$$... or the other way around, but I think having two objects $$x,x' \in \mathrm{Ob}(\mathcal{X})$$ will be less confusing if there's no set $$\mathcal{X}'$$ lurking around, wishing that $$x'$$ belonged to it.

Comment Source:Sophie - sorry, \$$\mathcal{X}'\$$ should everywhere be \$$\mathcal{Y}\$$... or the other way around, but I think having two objects \$$x,x' \in \mathrm{Ob}(\mathcal{X})\$$ will be less confusing if there's no set \$$\mathcal{X}'\$$ lurking around, wishing that \$$x'\$$ belonged to it.
• Options
8.

Keith - yes, now you've got it! These days, smart people are advocating the term short map for a map $$F$$ with

$$d'(F(x),F(x')) \le d(x,x').$$ The term "nonexpansive map" is more traditional, but "short map" is descriptive and... short. People often use "contraction mapping" for a map with

$$d'(F(x),F(x')) \le k d(x,x')$$ for some constant $$k < 1$$.

Comment Source:Keith - yes, now you've got it! These days, smart people are advocating the term **short map** for a map \$$F\$$ with $d'(F(x),F(x')) \le d(x,x').$ The term ["nonexpansive map"](https://en.wikipedia.org/wiki/Metric_map) is more traditional, but "short map" is descriptive and... short. People often use ["contraction mapping"](https://en.wikipedia.org/wiki/Contraction_mapping) for a map with $d'(F(x),F(x')) \le k d(x,x')$ for some constant \$$k < 1\$$.
• Options
9.
edited May 2018

I've been thinking up some examples of $$\mathbf{Cost}$$-functors. I was inspired by John's use of the term short map and its connection to the word short cut.

Example 1

Let $$\mathcal X$$ be the $$\mathbf{Cost}$$-category where the $$\mathrm{Ob}(\mathcal{X})$$ are landmarks in my city and $$\mathcal{X}(x, x')$$ is the time it takes to walk between each landmark.

Let $$\mathcal Y$$ be the $$\mathbf{Cost}$$-category where the $$\mathrm{Ob}(\mathcal{Y})$$ are again landmarks in my city and $$\mathcal{Y}(y, y')$$ is the time it takes to bike between each landmark.

Then define $$F: \mathcal X \to \mathcal Y$$ to be the identity on objects (i.e. it maps each landmark to itself). The property

$$\mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x'))$$ is satisfied because it takes longer to walk between locations than to bike. In other words biking is a shortcut to walking!

Example 2

This example is a throwback to our discussions about pie! Consider the set of ingredients needed to make a pie $$S = \{\textrm{flour}, \textrm{water}, \textrm{butter}, \textrm{crust}, \textrm{filling}, \textrm{pie}\}$$. And suppose that it takes 30 minutes to turn flour, water, and butter into a crust, and 60 minutes to turn crust and filling into pie.

Let $$\mathcal X$$ be the $$\mathbf{Cost}$$-category where the objects are the elements of $$\mathbb N[S]$$ and $$\mathcal X (x, x')$$ is the time it takes to turn the list of ingredients described by $$x$$ into the list of ingredients described by $$x'$$ or $$\infty$$ if this is impossible. Some examples:

$$\mathcal{X}( [\textrm{flour}] + [\textrm{water}] + [\textrm{butter}], [\textrm{crust}]) = 30$$ $$\mathcal{X}( [2\textrm{flour}] +2 [\textrm{water}] + 2[\textrm{butter}] + [\textrm{filling}] , [\textrm{crust}] + [\textrm{pie}]) = 120$$ $$\mathcal{X}([\textrm{pie}], [\textrm{pie}]) = 0$$ $$\mathcal{X}([\textrm{water}], [\textrm{pie}]) = \infty$$ Now suppose I am a busy person who prefers to buy my pie crust from the store instead of making it from scratch. Now the set of ingredients I need to make a pie is $$T = \{\textrm{crust}, \textrm{filling}, \textrm{pie}\}$$. I can create another $$\mathbf{Cost}$$-category $$\mathcal{Y}$$ where the objects are the elements of $$\mathbb{N}[T]$$ and $$\mathcal{Y}(y, y')$$ is again given by the time it takes to make transform the ingredients of $$y$$ into the ingredients of $$y'$$.

Lastly we need to define a $$\mathbf{Cost}$$-functor $$F: \mathcal{X} \to \mathcal{Y}$$. On objects define $$F$$ so that the ingredients needed to make a crust maps to a whole crust; crusts, fillings, and pies map to themselves; and any extra ingredients disappear. Formally:

$$F (a [\textrm{flour}] + b [\textrm{water}] + c [\textrm{butter}] + d [\textrm{crust}] + e [\textrm{filling}] + f[\textrm{pie}]) = (\textrm{min}(a,b,c) + d) [\textrm{crust}] + e [\textrm{filling}] + f[\textrm{pie}])$$ $$F$$ satisfies the property

$$\mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x'))$$ because it takes longer to make a pie from scratch than it does to buy pre-made ingredients from the store!

Comment Source:I've been thinking up some examples of \$$\mathbf{Cost}\$$-functors. I was inspired by John's use of the term **short map** and its connection to the word **short cut**. **Example 1** Let \$$\mathcal X \$$ be the \$$\mathbf{Cost}\$$-category where the \$$\mathrm{Ob}(\mathcal{X})\$$ are landmarks in my city and \$$\mathcal{X}(x, x')\$$ is the time it takes to walk between each landmark. Let \$$\mathcal Y\$$ be the \$$\mathbf{Cost}\$$-category where the \$$\mathrm{Ob}(\mathcal{Y})\$$ are again landmarks in my city and \$$\mathcal{Y}(y, y')\$$ is the time it takes to bike between each landmark. Then define \$$F: \mathcal X \to \mathcal Y\$$ to be the identity on objects (i.e. it maps each landmark to itself). The property $\mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x'))$ is satisfied because it takes longer to walk between locations than to bike. In other words biking is a shortcut to walking! **Example 2** This example is a throwback to our discussions about pie! Consider the set of ingredients needed to make a pie \$$S = \\{\textrm{flour}, \textrm{water}, \textrm{butter}, \textrm{crust}, \textrm{filling}, \textrm{pie}\\}\$$. And suppose that it takes 30 minutes to turn flour, water, and butter into a crust, and 60 minutes to turn crust and filling into pie. Let \$$\mathcal X\$$ be the \$$\mathbf{Cost}\$$-category where the objects are the elements of \$$\mathbb N[S]\$$ and \$$\mathcal X (x, x')\$$ is the time it takes to turn the list of ingredients described by \$$x\$$ into the list of ingredients described by \$$x'\$$ or \$$\infty\$$ if this is impossible. Some examples: $\mathcal{X}( [\textrm{flour}] + [\textrm{water}] + [\textrm{butter}], [\textrm{crust}]) = 30$ $\mathcal{X}( [2\textrm{flour}] +2 [\textrm{water}] + 2[\textrm{butter}] + [\textrm{filling}] , [\textrm{crust}] + [\textrm{pie}]) = 120$ $\mathcal{X}([\textrm{pie}], [\textrm{pie}]) = 0$ $\mathcal{X}([\textrm{water}], [\textrm{pie}]) = \infty$ Now suppose I am a busy person who prefers to buy my pie crust from the store instead of making it from scratch. Now the set of ingredients I need to make a pie is \$$T = \\{\textrm{crust}, \textrm{filling}, \textrm{pie}\\}\$$. I can create another \$$\mathbf{Cost}\$$-category \$$\mathcal{Y}\$$ where the objects are the elements of \$$\mathbb{N}[T]\$$ and \$$\mathcal{Y}(y, y')\$$ is again given by the time it takes to make transform the ingredients of \$$y\$$ into the ingredients of \$$y'\$$. Lastly we need to define a \$$\mathbf{Cost}\$$-functor \$$F: \mathcal{X} \to \mathcal{Y}\$$. On objects define \$$F\$$ so that the ingredients needed to make a crust maps to a whole crust; crusts, fillings, and pies map to themselves; and any extra ingredients disappear. Formally: $F (a [\textrm{flour}] + b [\textrm{water}] + c [\textrm{butter}] + d [\textrm{crust}] + e [\textrm{filling}] + f[\textrm{pie}]) = (\textrm{min}(a,b,c) + d) [\textrm{crust}] + e [\textrm{filling}] + f[\textrm{pie}])$ \$$F\$$ satisfies the property $\mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x'))$ because it takes longer to make a pie from scratch than it does to buy pre-made ingredients from the store! 
• Options
10.
edited May 2018

These days, smart people are advocating the term short map for a map $$F$$ with

$$d'(F(x),F(x')) \le d(x,x').$$

This is in line with my intuition - these are exactly "pairwise Lipschitz continuous functions with constant 1".

Sorry I screwed up the order previously.

The concept of $$\mathcal{V}$$-functor, introduced above, cleverly slips an inequality into the game. One might have tried demanding

$$\mathcal{X}(x,x') = \mathcal{X'}(F(x),F(x'))$$ and this would change the answers to "What is a $$\mathbf{Bool}$$-functor?" and "What is a $$\mathbf{Cost}$$-functor?" in interesting and instructive ways.

To answer "What is a $$\mathbf{Bool}$$-functor?", it is an elementary embedding of one preorder into another.

To answer "What is a $$\mathbf{Cost}$$-functor?", it is a generalization of an isometry like you mentioned.

Comment Source:> These days, smart people are advocating the term **short map** for a map \$$F\$$ with > > $d'(F(x),F(x')) \le d(x,x').$ This is in line with my intuition - these are exactly "pairwise [Lipschitz continuous](https://en.wikipedia.org/wiki/Lipschitz_continuity) functions with constant 1". Sorry I screwed up the order previously. > The concept of \$$\mathcal{V}\$$-functor, introduced above, cleverly slips an inequality into the game. One might have tried demanding > > $\mathcal{X}(x,x') = \mathcal{X'}(F(x),F(x'))$ > > and this would change the answers to "What is a \$$\mathbf{Bool}\$$-functor?" and "What is a \$$\mathbf{Cost}\$$-functor?" in interesting and instructive ways. To answer "What is a \$$\mathbf{Bool}\$$-functor?", it is an [elementary embedding](https://ncatlab.org/nlab/show/elementary+embedding) of one preorder into another. To answer "What is a \$$\mathbf{Cost}\$$-functor?", it is a generalization of an [isometry](https://en.wikipedia.org/wiki/Isometry) like you mentioned.
• Options
11.
edited May 2018

To answer "What is a $$\mathbf{Bool}$$-functor?", it is an elementary embedding of one preorder into another.

I tend to associate the term "elementary embedding" with more fancy contexts from logic... but maybe this is right.

In simpler terms, it's a map between posets $$f : X \to Y$$ such that $$x \le x'$$ if and only if $$f(x) \le f(x')$$.

To answer "What is a $$\mathbf{Cost}$$-functor?", it is a generalization of an isometry like you mentioned.

Yes, if we change the inequality to an equation in the definition of $$\mathbf{Cost}$$-functor we get an isometry between Lawvere metric spaces: a map that preserves distances.

Comment Source:> To answer "What is a \$$\mathbf{Bool}\$$-functor?", it is an [elementary embedding](https://ncatlab.org/nlab/show/elementary+embedding) of one preorder into another. I tend to associate the term "elementary embedding" with more fancy contexts from logic... but maybe this is right. In simpler terms, it's a map between posets \$$f : X \to Y\$$ such that \$$x \le x'\$$ if and only if \$$f(x) \le f(x')\$$. > To answer "What is a \$$\mathbf{Cost}\$$-functor?", it is a generalization of an [isometry](https://en.wikipedia.org/wiki/Isometry) like you mentioned. Yes, if we change the inequality to an equation in the definition of \$$\mathbf{Cost}\$$-functor we get an isometry between Lawvere metric spaces: a map that preserves distances.
• Options
12.

I tend to associate the term "elementary embedding" with more fancy contexts from logic... but maybe this is right.

In simpler terms, it's a map between posets $$f : X \to Y$$ such that $$x \le x'$$ if and only if $$f(x) \le f(x')$$.

Elementary embedding is not right. It's merely an embedding, rather than an elementary embedding.

Here's an example of an embedding that is not elementary. We can embed the singleton poset Unit (the singleton poset) into Bool with $$f$$ where $$f(\ast) = \mathtt{false}$$. While $$\mathbf{Unit} \models \forall x. x \leq \ast$$, it is not the case that $$\mathbf{Bool} \models \forall x. x \leq f(\ast)$$.

Thanks for catching my mistake.

Comment Source:> I tend to associate the term "elementary embedding" with more fancy contexts from logic... but maybe this is right. > > In simpler terms, it's a map between posets \$$f : X \to Y\$$ such that \$$x \le x'\$$ if and only if \$$f(x) \le f(x')\$$. *Elementary embedding* is not right. It's merely an [*embedding*](https://en.wikipedia.org/wiki/Embedding#Universal_algebra_and_model_theory), rather than an elementary embedding. Here's an example of an embedding that is not elementary. We can embed the singleton poset **Unit** (the singleton poset) into **Bool** with \$$f\$$ where \$$f(\ast) = \mathtt{false}\$$. While \$$\mathbf{Unit} \models \forall x. x \leq \ast\$$, it is not the case that \$$\mathbf{Bool} \models \forall x. x \leq f(\ast)\$$. Thanks for catching my mistake.
• Options
13.

If we have a pair $$f \vdash g: \mathcal{X} \to \mathcal{Y}$$ of adjoint contraction mappings then can we construct an equivalence between $$\mathcal{X}$$ and $$\mathcal{Y}$$?

Comment Source:If we have a pair \$$f \vdash g: \mathcal{X} \to \mathcal{Y}\$$ of adjoint contraction mappings then can we construct an equivalence between \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$?
• Options
14.

Hey Chris,

I think you mean:

If we have a pair $$f \dashv g: \mathcal{X} \to \mathcal{Y}$$ of adjoint short-maps then can we construct an equivalence between $$\mathcal{X}$$ and $$\mathcal{Y}$$?

Contraction mappings are Lipschitz continuous functions with constants $$k < 1$$, while short-maps correspond to $$\mathbf{Cost}$$-functors.

I don't know the answer to this.

Comment Source:Hey Chris, I think you mean: > If we have a pair \$$f \dashv g: \mathcal{X} \to \mathcal{Y}\$$ of adjoint short-maps then can we construct an equivalence between \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$? Contraction mappings are Lipschitz continuous functions with constants \$$k < 1\$$, while short-maps correspond to \$$\mathbf{Cost}\$$-functors. I don't know the answer to this.
• Options
15.

I think you might need the flexibility of the constant to get adjoint maps that aren't inverses.

Comment Source:I think you might need the flexibility of the constant to get adjoint maps that aren't inverses.
• Options
16.

I think you might need the flexibility of the constant to get adjoint maps that aren't inverses.

I still don't see the proof. Can you write it up?

Comment Source:> I think you might need the flexibility of the constant to get adjoint maps that aren't inverses. I still don't see the proof. Can you write it up?
• Options
17.

Well, I'll write up a counter example this evening. >. < That's sort of like a proof.

Comment Source:Well, I'll write up a counter example this evening. >. < That's sort of like a proof.
• Options
18.

Sophie, thank you for continually providing excellent examples for this course's concepts. There's definitely a Cost-functor for the time it takes me to navigate the interdependent concepts of this course without -> with your examples.

Comment Source:Sophie, thank you for continually providing excellent examples for this course's concepts. There's definitely a Cost-functor for the time it takes me to navigate the interdependent concepts of this course without -> with your examples.
• Options
19.
edited May 2018

So consider the classic metric space $$\{A,B,C, D\}$$ with d (x,y)= if x = y then 0 else 1, and the pair f = rotate obj clock wise, and g flip A with B.

Hmm. Now what i actually meant to mean was that there are no adjunct f and g that are not inverses. And i do think that is true, but am not sure.

Comment Source:So consider the classic metric space \$$\\{A,B,C, D\\} \$$ with d (x,y)= if x = y then 0 else 1, and the pair f = rotate obj clock wise, and g flip A with B. Hmm. Now what i actually meant to mean was that there are no adjunct f and g that are not inverses. And i do think that is true, but am not sure. 
• Options
20.
edited May 2018

Matthew and Christopher wrote:

If we have a pair $$f \dashv g: \mathcal{X} \to \mathcal{Y}$$ of adjoint short maps then can we construct an equivalence between $$\mathcal{X}$$ and $$\mathcal{Y}$$?

That's a nice question! But the first thing to note is that we haven't officially discussed adjoint functors between enriched categories! That's just due to a shortage of time.

So, the first thing to do is figure out the definition of adjoint $$\mathcal{V}$$-functors between $$\mathcal{V}$$-categories when $$\mathcal{V}$$ is a symmetric monoidal preorder. Or look it up, but figuring it out is much better. We know what $$\mathcal{V}$$-functors are - I explained it in this lecture. So we just need to make up a concept of adjoint $$\mathcal{V}$$-functors that generalizes the concept of adjoint monotone maps between preorders, which is the special case $$\mathcal{V} = \mathbf{Bool}$$.

Once we see the definition, so we know what the question actually is, I bet we can answer the question without a huge amount of work.

So:

Definition. Given $$\mathcal{V}$$-categories $$\mathcal{X}$$ and $$\mathcal{Y}$$, we say a $$\mathcal{V}$$-functor $$F : \mathcal{X} \to \mathcal{Y}$$ is the left adjoint of a $$\mathcal{V}$$-functor $$G : \mathcal{Y} \to \mathcal{X}$$ if....

Comment Source:Matthew and Christopher wrote: > If we have a pair \$$f \dashv g: \mathcal{X} \to \mathcal{Y}\$$ of adjoint short maps then can we construct an equivalence between \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$? That's a nice question! But the first thing to note is that we haven't officially discussed adjoint functors between enriched categories! That's just due to a shortage of time. So, the first thing to do is figure out the definition of adjoint \$$\mathcal{V}\$$-functors between \$$\mathcal{V}\$$-categories when \$$\mathcal{V}\$$ is a symmetric monoidal preorder. Or look it up, but figuring it out is much better. We know what \$$\mathcal{V}\$$-functors are - I explained it in this lecture. So we just need to make up a concept of adjoint \$$\mathcal{V}\$$-functors that generalizes the concept of adjoint monotone maps between preorders, which is the special case \$$\mathcal{V} = \mathbf{Bool}\$$. Once we see the definition, so we know what the question actually _is_, I bet we can answer the question without a huge amount of work. So: **Definition.** Given \$$\mathcal{V}\$$-categories \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$, we say a \$$\mathcal{V}\$$-functor \$$F : \mathcal{X} \to \mathcal{Y}\$$ is the **left adjoint** of a \$$\mathcal{V}\$$-functor \$$G : \mathcal{Y} \to \mathcal{X}\$$ if....
• Options
21.
edited May 2018

Nice examples in comment #9, Sophie! Now we can study enriched flour using enriched functors! :P

Comment Source:Nice examples in comment #9, Sophie! Now we can study [enriched flour](https://en.wikipedia.org/wiki/Enriched_flour) using enriched functors! :P