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Lecture 65 - Chapter 4: Collaborative Design

Last time we reached the technical climax of this chapter: constructing the category of enriched profunctors. If you found that difficult, you'll be relieved to hear it's downhill from here on, at least in Chapter 4. We will now apply all our hard work.

One application is to collaborative design. Fong and Spivak discuss it in their the book, based on this paper by a student of Spivak:

Censi's work is based on \(\textbf{Bool}\)-enriched profunctors, also known as feasibility relations. We introduced these in Lecture 56.

Remember, a feasibility relation \( \Phi : X\nrightarrow Y \) is a monotone function

$$ \Phi : X^{\text{op}} \times Y \to \mathbf{Bool} .$$ If \( \Phi(x,y) = \text{true}\), we say \(x\) can be obtained given \(y\). The idea is that we use elements of \( X\) to describe 'requirements' - things you want - and elements of \(Y\) to describe 'resources' - things you have.

The idea of Andrea Censi's theory that we can compute the design requirements of a complex system from those of its parts using 'co-design diagrams'. These are really pictures of big complicated feasibility relations, like this:

image

This big complicated feasibility relation is built from simpler ones in various ways. Each wire in this diagram is labeled with the name of a preorder, and each little box is itself a feasibility relation between preorders. We described how to compose feasibility relations in Lecture 58, and that corresponds to feeding the outputs of one little box into another. But there are other things going on in this picture, like boxes sitting side by side, and wires that bend around backwards! This is what I need to explain - it may take a couple lectures to do this.

Instead of diving into the mathematical details today, let me quote the book's general explanation of the diagram above:

As an example, consider the design problem of creating a robot to carry some load at some velocity. The top-level planner breaks the problem into three design teams: team chassis, team motor, and team battery. Each of these teams could break up into multiple parts and the process repeated, but let's remain at the top level and consider the resources produced and the resources required by each of our three teams.

The chassis in some sense provides all the functionality—it carries the load at the velocity—but it requires some things in order to do so. It requires money, of course, but more to the point it requires a source of torque and speed. These are supplied by the motor, which in turn needs voltage and current from the battery. Both the motor and the battery cost money, but more importantly they need to be carried by the chassis: they become part of the load. A feedback loop is created: the chassis must carry all the weight, even that of the parts that power the chassis. A heavier battery might provide more energy to power the chassis, but is the extra power worth the heavier load?

In the picture, each part—chassis, motor, battery, and robot—is shown as a box with ports on the left and right. The functionalities, or resources produced by the part are on the left of the box, and the resources required by the part are on the right.

The boxes marked \(\Sigma\) correspond to summing inputs. These boxes are not to be designed, but we will see later that they fit easily into the same conceptual framework. Note also the \(\leq\)'s on each wire; they indicate that if box \(A\) requires a resource that box \(B\) produces, then \(A\)'s requirement must be less than or equal to \(B\)'s production.

Next time we'll get into more detail.

But to wrap up for today, here are few puzzles!

We've been talking about enriched functors and also enriched profunctors. How are they related? The first cool fact is that any enriched functor gives two enriched profunctors: one going each way.

Puzzle 201. Show that any \(\mathcal{V}\)-enriched functor \(F: \mathcal{X} \to \mathcal{Y}\) gives a \(\mathcal{V}\)-enriched profunctor

$$ \hat{F} \colon \mathcal{X} \nrightarrow \mathcal{Y} $$ defined by

$$ \hat{F} (x,y) = \mathcal{Y}(F(x), y ) .$$ Puzzle 202. Show that any \(\mathcal{V}\)-enriched functor \(F: \mathcal{X} \to \mathcal{Y}\) gives a \(\mathcal{V}\)-enriched profunctor

$$ \check{F} \colon \mathcal{Y} \nrightarrow \mathcal{X} $$ defined by

$$ \check{F} (y,x) = \mathcal{Y}(y,F(x)) .$$ These two constructions have funny names. \(\hat{F} \colon \mathcal{X} \nrightarrow \mathcal{Y}\) is called the companion of \(F\) and \( \check{F} \colon \mathcal{Y} \nrightarrow \mathcal{X} \), going back, is called the conjoint of \(F\).

If you have trouble remembering these, remember that a 'companion' is like a fellow traveler, going the same way as our original functor. The word 'conjoint' should remind you of 'adjoint', which means something going back the other way.

In fact there's a relationship between adjoints and conjoints!

Puzzle 203. We say a \(\mathcal{V}\)-enriched functor \(F: \mathcal{X} \to \mathcal{Y}\) is a left adjoint of a \(\mathcal{V}\)-enriched functor \(G: \mathcal{Y} \to \mathcal{X}\) if

$$ \mathcal{Y}(F(x), y) = \mathcal{X}(x,G(y)) $$ for all objects \(x\) of \(\mathcal{X}\) and \(y\) of \(\mathcal{Y}\). In this situation we also say \(G\) is the right adjoint of \(F\). Show that \(F\) is the left adjoint of \(G\) if and only if

$$ \hat{F} = \check{G} . $$ Pretty! image

To read other lectures go here.

Comments

  • 1.

    The talk of "companions" and "conjoints" reminded me of the paper 'String Diagrams For Double Categories and Equipments' by David Jaz Myers.

    The paper gives really nice string diagrammatics.

    Comment Source:The talk of "companions" and "conjoints" reminded me of the paper ['String Diagrams For Double Categories and Equipments'](https://arxiv.org/abs/1612.02762) by David Jaz Myers. The paper gives really nice string diagrammatics.
  • 2.

    Puzzle 201. Show that any \(\mathcal{V}\)-enriched functor \(F: \mathcal{X} \to \mathcal{Y}\) gives a \(\mathcal{V}\)-enriched profunctor

    $$ \hat{F} \colon \mathcal{X} \nrightarrow \mathcal{Y} $$ defined by $$ \hat{F} (x,y) = \mathcal{Y}(F(x), y) .$$

    Based on the second theorem from the previous lecture, we can prove that \(\hat{F} : \mathcal{X} \nrightarrow \mathcal{Y}\) is a \(\mathcal{V}\)-enriched profunctor by showing that

    $$ \mathcal{X}(x', x) \otimes \hat{F}(x, y) \otimes \mathcal{Y}(y, y') \leq \hat{F}(x',y') . $$ Replacing \(\hat{F}\) with its definition we have:

    $$ \mathcal{X}(x', x) \otimes \mathcal{Y}(F(x), y) \otimes \mathcal{Y}(y, y') \leq \mathcal{Y}(F(x'), y') . $$ This inequality is proved by using the facts that (i) \(F\) is a \(\mathcal{V}\)-enriched functor, and (ii) \(\mathcal{Y}\) is a \(\mathcal{V}\)-enriched category:

    $$ \begin{align} \mathcal{X}(x', x) \otimes \mathcal{Y}(F(x), y) \otimes \mathcal{Y}(y, y') & \leq \mathcal{Y}(F(x'), F(x)) \otimes \mathcal{Y}(F(x), y) \otimes \mathcal{Y}(y, y') \tag{i} \\ & \leq \mathcal{Y}(F(x'), y) \otimes \mathcal{Y}(y, y') \tag{ii} \\ & \leq \mathcal{Y}(F(x'), y') . \tag{ii} \end{align} $$


    Puzzle 202. Show that any \(\mathcal{V}\)-enriched functor \(F: \mathcal{X} \to \mathcal{Y}\) gives a \(\mathcal{V}\)-enriched profunctor

    $$ \check{F} \colon \mathcal{Y} \nrightarrow \mathcal{X} $$ defined by $$ \check{F} (y,x) = \mathcal{Y}(y,F(x)) .$$

    Similarly, to prove that \(\check{F} : \mathcal{Y} \nrightarrow \mathcal{X}\) is a \(\mathcal{V}\)-enriched profunctor we have to show that

    $$ \mathcal{Y}(y', y) \otimes \check{F}(y, x) \otimes \mathcal{X}(x, x') \leq \check{F}(y',x') $$ which is equivalent to

    $$ \mathcal{Y}(y', y) \otimes \mathcal{Y}(y, F(x)) \otimes \mathcal{X}(x, x') \leq \mathcal{Y}(y', F(x')) . $$ Again, the inequality is proved by using the facts that (i) \(F\) is a \(\mathcal{V}\)-enriched functor, and (ii) \(\mathcal{Y}\) is a \(\mathcal{V}\)-enriched category:

    $$ \begin{align} \mathcal{Y}(y', y) \otimes \mathcal{Y}(y, F(x)) \otimes \mathcal{X}(x, x') & \leq \mathcal{Y}(y', y) \otimes \mathcal{Y}(y, F(x)) \otimes \mathcal{Y}(F(x), F(x')) \tag{i} \\ & \leq \mathcal{Y}(y', F(x)) \otimes \mathcal{Y}(F(x), F(x')) \tag{ii} \\ & \leq \mathcal{Y}(y', F(x')). \tag{ii} \end{align} $$

    Comment Source:> **Puzzle 201.** Show that any \\(\mathcal{V}\\)-enriched functor \\(F: \mathcal{X} \to \mathcal{Y}\\) gives a \\(\mathcal{V}\\)-enriched profunctor > > \[ \hat{F} \colon \mathcal{X} \nrightarrow \mathcal{Y} \] > > defined by > > \[ \hat{F} (x,y) = \mathcal{Y}(F(x), y) .\] Based on the second theorem from [the previous lecture](https://forum.azimuthproject.org/discussion/2298/lecture-64-chapter-5-the-category-of-enriched-profunctors/p1), we can prove that \\(\hat{F} : \mathcal{X} \nrightarrow \mathcal{Y}\\) is a \\(\mathcal{V}\\)-enriched profunctor by showing that \[ \mathcal{X}(x', x) \otimes \hat{F}(x, y) \otimes \mathcal{Y}(y, y') \leq \hat{F}(x',y') . \] Replacing \\(\hat{F}\\) with its definition we have: \[ \mathcal{X}(x', x) \otimes \mathcal{Y}(F(x), y) \otimes \mathcal{Y}(y, y') \leq \mathcal{Y}(F(x'), y') . \] This inequality is proved by using the facts that (i) \\(F\\) is a \\(\mathcal{V}\\)-enriched functor, and (ii) \\(\mathcal{Y}\\) is a \\(\mathcal{V}\\)-enriched category: \[ \begin{align} \mathcal{X}(x', x) \otimes \mathcal{Y}(F(x), y) \otimes \mathcal{Y}(y, y') & \leq \mathcal{Y}(F(x'), F(x)) \otimes \mathcal{Y}(F(x), y) \otimes \mathcal{Y}(y, y') \tag{i} \\\\ & \leq \mathcal{Y}(F(x'), y) \otimes \mathcal{Y}(y, y') \tag{ii} \\\\ & \leq \mathcal{Y}(F(x'), y') . \tag{ii} \end{align} \] --- > **Puzzle 202.** Show that any \\(\mathcal{V}\\)-enriched functor \\(F: \mathcal{X} \to \mathcal{Y}\\) gives a \\(\mathcal{V}\\)-enriched profunctor > > \[ \check{F} \colon \mathcal{Y} \nrightarrow \mathcal{X} \] > > defined by > > \[ \check{F} (y,x) = \mathcal{Y}(y,F(x)) .\] Similarly, to prove that \\(\check{F} : \mathcal{Y} \nrightarrow \mathcal{X}\\) is a \\(\mathcal{V}\\)-enriched profunctor we have to show that \[ \mathcal{Y}(y', y) \otimes \check{F}(y, x) \otimes \mathcal{X}(x, x') \leq \check{F}(y',x') \] which is equivalent to \[ \mathcal{Y}(y', y) \otimes \mathcal{Y}(y, F(x)) \otimes \mathcal{X}(x, x') \leq \mathcal{Y}(y', F(x')) . \] Again, the inequality is proved by using the facts that (i) \\(F\\) is a \\(\mathcal{V}\\)-enriched functor, and (ii) \\(\mathcal{Y}\\) is a \\(\mathcal{V}\\)-enriched category: \[ \begin{align} \mathcal{Y}(y', y) \otimes \mathcal{Y}(y, F(x)) \otimes \mathcal{X}(x, x') & \leq \mathcal{Y}(y', y) \otimes \mathcal{Y}(y, F(x)) \otimes \mathcal{Y}(F(x), F(x')) \tag{i} \\\\ & \leq \mathcal{Y}(y', F(x)) \otimes \mathcal{Y}(F(x), F(x')) \tag{ii} \\\\ & \leq \mathcal{Y}(y', F(x')). \tag{ii} \end{align} \]
  • 3.
    edited July 24

    Puzzle 203. We say a \(\mathcal{V}\)-enriched functor \(F: \mathcal{X} \to \mathcal{Y}\) is a left adjoint of a \(\mathcal{V}\)-enriched functor \(G: \mathcal{Y} \to \mathcal{X}\) if

    $$ \mathcal{Y}(F(x), y) = \mathcal{X}(x,G(y)) $$ for all objects \(x\) of \(\mathcal{X}\) and \(y\) of \(\mathcal{Y}\). In this situation we also say \(G\) is the
    right adjoint of \(F\). Show that \(F\) is the left adjoint of \(G\) if and only if $$ \hat{F} = \check{G} $$

    Suppose \(F\) is left adjoint to \(G\). Using its definition, the companion of \(F\) is \(\hat{F}(x,y) = \mathcal{Y}(F(x), y) \). Likewise the conjoint of \(G\) is \(\check{G}(x,y) = \mathcal{X}(x,G(y))\). But by the assumption of adjointness,

    $$\hat{F}(x,y) = \mathcal{Y}(F(x), y) = \mathcal{X}(x,G(y)) = \check{G}(x,y)$$ So \(\hat{F} = \check{G} \) if \(F\) is left adjoint to \(G\). The proof of the other direction is similar. Assume that \(\hat{F} = \check{G} \) given \(F,G\). The definition of the companion and conjoint lead to the same equality. \(\square\)

    Comment Source:> **Puzzle 203.** We say a \\(\mathcal{V}\\)-enriched functor \\(F: \mathcal{X} \to \mathcal{Y}\\) is a **left adjoint** of a \\(\mathcal{V}\\)-enriched functor \\(G: \mathcal{Y} \to \mathcal{X}\\) if > \[ \mathcal{Y}(F(x), y) = \mathcal{X}(x,G(y)) \] > for all objects \\(x\\) of \\(\mathcal{X}\\) and \\(y\\) of \\(\mathcal{Y}\\). In this situation we also say \\(G\\) is the **right adjoint** of \\(F\\). Show that \\(F\\) is the left adjoint of \\(G\\) if and only if > \[ \hat{F} = \check{G} \] Suppose \\(F\\) is left adjoint to \\(G\\). Using its definition, the companion of \\(F\\) is \\(\hat{F}(x,y) = \mathcal{Y}(F(x), y) \\). Likewise the conjoint of \\(G\\) is \\(\check{G}(x,y) = \mathcal{X}(x,G(y))\\). But by the assumption of adjointness, \[\hat{F}(x,y) = \mathcal{Y}(F(x), y) = \mathcal{X}(x,G(y)) = \check{G}(x,y)\] So \\(\hat{F} = \check{G} \\) if \\(F\\) is left adjoint to \\(G\\). The proof of the other direction is similar. Assume that \\(\hat{F} = \check{G} \\) given \\(F,G\\). The definition of the companion and conjoint lead to the same equality. \\(\square\\)
  • 4.

    It strikes me that some profunctors are "nice" in the sense that they are companions/conjoints of an adjoint pair of functors.

    But profunctors in general need not be "nice". Is there any way of telling which is which?

    eg a theorem along the lines of "a profunctor \(\Phi\) is 'nice' iff it satisfies [some formula involving \(\Phi\)]"

    NB by "nice" I should probably say "representable on both sides".

    EDIT to add: this is actually two problems: (i) when is a profunctor a companion? (ii) when is a profunctor a conjoint?

    Comment Source:It strikes me that some profunctors are "nice" in the sense that they are companions/conjoints of an adjoint pair of functors. But profunctors in general need not be "nice". Is there any way of telling which is which? eg a theorem along the lines of "a profunctor \\(\Phi\\) is 'nice' iff it satisfies [some formula involving \\(\Phi\\)]" NB by "nice" I should probably say "representable on both sides". EDIT to add: this is actually two problems: (i) when is a profunctor a companion? (ii) when is a profunctor a conjoint?
  • 5.
    edited July 24

    Anindya: excellent question!

    image

    At least in the case of ordinary profunctors (that is, \(\textbf{Set}\)-enriched profunctors) there's a well-known necessary condition for a profunctor to be the companion of some functor \(F\):

    Namely, it needs to have a right adjoint in the 2-category of categories, profunctors and natural transformations. And this right adjoint is just the conjoint of \(F\).

    Here I'm using a concept of 'adjoint' that make sense in any 2-category, which reduces to the usual concept of 'adjoint functor' when we apply it to the 2-category of categories, functors and natural transformations:

    This condition for a profunctor to be a companion of a functor is also sufficient if we restrict ourselves to 'Cauchy complete' categories. Every category \(\mathcal{C}\) has a Cauchy completion \(\text{Split}(\mathcal{C})\) such that profunctors out of \(\mathcal{C}\) correspond in a natural one-to-one way with profunctors out of \(\text{Split}(\mathcal{C})\). So, when doing serious work with profunctors, people often restrict attention to Cauchy complete categories.

    I've stated a necessary condition for a profunctor to be a companion, but the same result also serves to give a necessary condition for a profunctor to be a conjoint. Any profunctor that's a conjoint of a functor must have a left adjoint in the 2-category of categories, profunctors and natural transformations! And again, this necessary condition is also sufficient if we restrict attention to Cauchy complete categories.

    All this has an enriched analogue - but I have a feeling that we're already getting a bit too technical for this course! Suffice it to say that you've put your finger on an important issue.

    Comment Source:Anindya: excellent question! <img width = "100" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> At least in the case of ordinary profunctors (that is, \\(\textbf{Set}\\)-enriched profunctors) there's a well-known _necessary_ condition for a profunctor to be the companion of some functor \\(F\\): * Wikipedia, [Profunctor: lifting functors to profunctors](https://en.wikipedia.org/wiki/Profunctor#Properties). Namely, it needs to have a right adjoint in the 2-category of categories, profunctors and natural transformations. And this right adjoint is just the conjoint of \\(F\\). Here I'm using a concept of 'adjoint' that make sense in any 2-category, which reduces to the usual concept of 'adjoint functor' when we apply it to the 2-category of categories, functors and natural transformations: * nLab, [Adjunction: definition](https://ncatlab.org/nlab/show/adjunction#definition). This condition for a profunctor to be a companion of a functor is also _sufficient_ if we restrict ourselves to 'Cauchy complete' categories. Every category \\(\mathcal{C}\\) has a [Cauchy completion](https://en.wikipedia.org/wiki/Karoubi_envelope) \\(\text{Split}(\mathcal{C})\\) such that profunctors out of \\(\mathcal{C}\\) correspond in a natural one-to-one way with profunctors out of \\(\text{Split}(\mathcal{C})\\). So, when doing serious work with profunctors, people often restrict attention to Cauchy complete categories. I've stated a necessary condition for a profunctor to be a companion, but the same result also serves to give a necessary condition for a profunctor to be a conjoint. Any profunctor that's a conjoint of a functor must have a _left_ adjoint in the 2-category of categories, profunctors and natural transformations! And again, this necessary condition is also sufficient if we restrict attention to Cauchy complete categories. All this has an enriched analogue - but I have a feeling that we're already getting a bit too technical for this course! Suffice it to say that you've put your finger on an important issue.
  • 6.
    edited July 24

    Nice solution to Puzzles 201 and 202, Dan Oneata! It's neat how you used that theorem from last time. I guess this way of checking that something is an enriched profunctor by checking a single inequality is pretty efficient!

    Comment Source:Nice [solution to Puzzles 201 and 202](https://forum.azimuthproject.org/discussion/comment/20303/#Comment_20303), Dan Oneata! It's neat how you used that theorem from last time. I guess this way of checking that something is an enriched profunctor by checking a single inequality is pretty efficient!
  • 7.

    Elegant solution to Puzzle 203, Scott Oswald! It becomes clear that left and right adjoints, companions and conjoints fit together in a neat package.

    Comment Source:Elegant [solution to Puzzle 203](https://forum.azimuthproject.org/discussion/comment/20305/#Comment_20305), Scott Oswald! It becomes clear that left and right adjoints, companions and conjoints fit together in a neat package.
  • 8.

    @John – glad to hear this question was Important But Difficult, I've been hurtling up various dead ends trying to get a grasp on it today... I'll have a think tomorrow about what a "natural transformation" of profunctors might be.

    fwiw I was wondering that given \(\Phi\) maybe we could construct an \(F\) such that \(\hat{F} = \Phi\) by some sort of colimit construction – then I realised I had no idea what colimits in a \(\mathcal{V}\)-category might mean.

    I got this far: the coproduct of objects \(a, b\) is an object \(a + b\) such that for all \(c\)

    $$\mathcal{X}(a + b, c) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c)$$ Two problems arise. First, there could be several objects that meet this criterion. We would expect them to all be isomorphic, however. But what does it mean for two objects in a \(\mathcal{V}\)-category to be isomorphic? After a bit of fiddling I came up with this definition, which people might want to check out:

    $$ a, b \in \text{Ob}(\mathcal{X}) \text{ isomorphic} \iff I \leq \mathcal{X}(a, b) \text{ and } I \leq \mathcal{X}(b, a) $$ The second issue, which stumped me completely, is this: how do we define an infinite coproduct? We can't just infinitely \(\otimes\) a bunch of stuff. So where on earth would we get the coproduct from?

    Comment Source:@John – glad to hear this question was Important But Difficult, I've been hurtling up various dead ends trying to get a grasp on it today... I'll have a think tomorrow about what a "natural transformation" of profunctors might be. fwiw I was wondering that given \\(\Phi\\) maybe we could construct an \\(F\\) such that \\(\hat{F} = \Phi\\) by some sort of colimit construction – then I realised I had no idea what colimits in a \\(\mathcal{V}\\)-category might mean. I got this far: the coproduct of objects \\(a, b\\) is an object \\(a + b\\) such that for all \\(c\\) \[\mathcal{X}(a + b, c) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c)\] Two problems arise. First, there could be several objects that meet this criterion. We would expect them to all be isomorphic, however. But what does it mean for two objects in a \\(\mathcal{V}\\)-category to be isomorphic? After a bit of fiddling I came up with this definition, which people might want to check out: \[ a, b \in \text{Ob}(\mathcal{X}) \text{ isomorphic} \iff I \leq \mathcal{X}(a, b) \text{ and } I \leq \mathcal{X}(b, a) \] The second issue, which stumped me completely, is this: how do we define an infinite coproduct? We can't just infinitely \\(\otimes\\) a bunch of stuff. So where on earth would we get the coproduct from?
  • 9.
    edited July 25

    image

    Anindya wrote:

    I'll have a think tomorrow about what a "natural transformation" of profunctors might be.

    It's not too hard for ordinary profunctors, since a profunctor \(F : \mathcal{C} \nrightarrow \mathcal{D}\) is really a functor \(F : \mathcal{C}^{\text{op}} \times \mathcal{D} \to \mathbf{Set} \) and we know what natural transformations between functors are.

    What about the enriched case? Well, we haven't talked about natural transformations between \(\mathcal{V}\)-enriched functors, but we could. Our restriction to having \(\mathcal{V}\) be a commutative quantale instead of symmetric monoidal closed category with all colimits tends to make some aspects of enriched category theory trivial or near-trivial, and this is one. Fong and Spivak make this restriction because it makes life incredibly easy, but it also tends to make things boring, and it means that our work on enriched categories doesn't even include ordinary categories as a special case... because this restriction prevents us from taking \(\mathcal{V}\) to be \(\mathbf{Set}\).

    I got this far: the coproduct of objects \(a, b\) is an object \(a + b\) such that for all \(c\)

    $$\mathcal{X}(a + b, c) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c)$$

    That doesn't sound right. For example, \(\mathbf{Vect}\) is enriched over itself and it has coproducts, but it doesn't obey the above. It's probably wise to guide yourself with some examples.

    We can't just infinitely ⊗ a bunch of stuff.

    I don't think we should ever want to. I think any 'infinitary' stuff we want to do with \(\mathcal{V}\) should arise from it having all colimits and perhaps also all limits.

    By the way, all the questions you're raising are answered in the bible of enriched category theory:

    But beware - this is not an easy read! I've never managed to fully penetrate it.

    You may take a look at Kelly's book and quickly decide it's easier to figure things out yourself... but I can't resist giving you a warning: people working on enriched categories eventually learned through experience that in the enriched context colimits should be generalized to weighted colimits. This is one reason Kelly's book is hard: enriched category theory seems very much like ordinary category theory until one reaches the theory of colimits (and limits), and then it looks very weird unless one carefully thinks about examples.

    Another reason Kelly's book is hard is that he's very macho, not friendly and gentle like me.

    Comment Source:<img width = "100" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> Anindya wrote: > I'll have a think tomorrow about what a "natural transformation" of profunctors might be. It's not too hard for ordinary profunctors, since a profunctor \\(F : \mathcal{C} \nrightarrow \mathcal{D}\\) is really a functor \\(F : \mathcal{C}^{\text{op}} \times \mathcal{D} \to \mathbf{Set} \\) and we know what natural transformations between functors are. What about the enriched case? Well, we haven't talked about natural transformations between \\(\mathcal{V}\\)-enriched functors, but we could. Our restriction to having \\(\mathcal{V}\\) be a commutative quantale instead of symmetric monoidal closed category with all colimits tends to make some aspects of enriched category theory trivial or near-trivial, and this is one. Fong and Spivak make this restriction because it makes life incredibly easy, but it also tends to make things boring, and it means that our work on enriched categories doesn't even include ordinary categories as a special case... because this restriction prevents us from taking \\(\mathcal{V}\\) to be \\(\mathbf{Set}\\). > I got this far: the coproduct of objects \\(a, b\\) is an object \\(a + b\\) such that for all \\(c\\) > \[\mathcal{X}(a + b, c) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c)\] That doesn't sound right. For example, \\(\mathbf{Vect}\\) is enriched over itself and it has coproducts, but it doesn't obey the above. It's probably wise to guide yourself with some examples. > We can't just infinitely ⊗ a bunch of stuff. I don't think we should ever want to. I think any 'infinitary' stuff we want to do with \\(\mathcal{V}\\) should arise from it having all colimits and perhaps also all limits. By the way, all the questions you're raising are answered in the bible of enriched category theory: * Max Kelly, _[Basic Concepts of Enriched Category Theory](http://www.tac.mta.ca/tac/reprints/articles/10/tr10.pdf)_. But beware - this is not an easy read! I've never managed to fully penetrate it. You may take a look at Kelly's book and quickly decide it's easier to figure things out yourself... but I can't resist giving you a warning: people working on enriched categories eventually learned through experience that in the enriched context _colimits_ should be generalized to _[weighted colimits](https://ncatlab.org/nlab/show/weighted+colimit)_. This is one reason Kelly's book is hard: enriched category theory seems very much like ordinary category theory _until_ one reaches the theory of colimits (and limits), and then it looks very weird unless one carefully thinks about examples. Another reason Kelly's book is hard is that he's very macho, not friendly and gentle like me.
  • 10.
    • Max Kelly, Basic Concepts of Enriched Category Theory.

    But beware - this is not an easy read! I've never managed to fully penetrate it.

    It's like the joke that any math textbook with the word "Advanced" in the title for high schoolers, maybe undergrads while the word "Elementary" means the book is intended for grad students

    Comment Source:> + Max Kelly, Basic Concepts of Enriched Category Theory. > But beware - this is not an easy read! I've never managed to fully penetrate it. It's like the joke that any math textbook with the word "Advanced" in the title for high schoolers, maybe undergrads while the word "Elementary" means the book is intended for grad students
  • 11.

    Yes, some of us were talking about that recently on Twitter. Jacobson's Basic Algebra, Serre's A Course in Arithmetic...

    Comment Source:Yes, some of us were talking about that recently on Twitter. Jacobson's _Basic Algebra_, Serre's _A Course in Arithmetic_...
  • 12.

    There is a joke somewhere here that mathematicians come up with bad naming schemes.

    However, mathematicians probably use the word "joke" in a more esoteric, yet precise manner than how the general public would.

    Also, it takes a Ph.D. dissertation to first even properly define a math joke, and prove the correctness of the set-up to the joke, which is followed up with the creation of 20 or so follow up papers arguing that under certain circumstances the punchline to the joke is indeed funny.

    Comment Source:There is a joke somewhere here that mathematicians come up with bad naming schemes. However, mathematicians probably use the word "joke" in a more esoteric, yet precise manner than how the general public would. Also, it takes a Ph.D. dissertation to first even properly define a math joke, and prove the correctness of the set-up to the joke, which is followed up with the creation of 20 or so follow up papers arguing that under certain circumstances the punchline to the joke is indeed funny.
  • 13.

    surely there must be a joke in here somewhere about how co-design is the dual of design

    Comment Source:surely there must be a joke in here somewhere about how co-design is the dual of design
  • 14.

    @John wrote re coproducts in \(\mathcal{V}\)-categories:

    That doesn't sound right. For example, \(\textbf{Vect}\) is enriched over itself and it has coproducts, but it doesn't obey the above. It's probably wise to guide yourself with some examples.

    OK I've tried this but not got very far.

    My original thinking was that \((a, b)\mapsto a + b\) should be left adjoint to the diagonal map \(c\mapsto (c, c)\). Hence we'd have

    $$ \mathcal{X}(a + b, c) = (\mathcal{X\times X})((a, b), (c, c)) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c) $$ But I've just realised that the diagonal map is not necessarily a \(\mathcal{V}\)-functor!

    $$ (\mathcal{X\times X})((c, c), (c', c')) = \mathcal{X}(c, c')\otimes\mathcal{X}(c, c') $$ and there's no reason to think the right hand side is always \(\geq \mathcal{X}(c, c')\).

    So let's try a few examples. If \(\mathcal{V} = \textbf{Bool}\) then a \(\mathcal{V}\)-category \(\mathcal{X}\) is a \(\textbf{Bool}\)-category, which is a preorder. So the coproduct should be a join in that preorder, ie

    $$ (a + b \leq_\mathcal{X} c) \text{ if and only if } (a \leq_\mathcal{X} c \text{ and } b \leq_\mathcal{X} c) $$ which is the same as

    $$ \mathcal{X}(a + b, c) = \mathcal{X}(a, c)\wedge \mathcal{X}(b, c) $$ And if we consider \(\textbf{Vect}\) enriched over itself we have

    $$ \textbf{Vect}(A \oplus B, C) = \textbf{Vect}(A, C) \oplus \textbf{Vect}(B, C) $$ and again we have a binary "meet" (ie \(\oplus\) considered as a binary product in \(\textbf{Vect}\)) on the right hand side.

    So my current guess is that we should be using meets/products in \(\mathcal{V}\) (which unlike \(\otimes\) do admit infinitary generalisations) to define coproducts in \(\mathcal{V}\)-categories.

    I'll take a look at that Kelly book to see if any of this is barking up the right tree.

    Comment Source:@John wrote re coproducts in \\(\mathcal{V}\\)-categories: > That doesn't sound right. For example, \\(\textbf{Vect}\\) is enriched over itself and it has coproducts, but it doesn't obey the above. It's probably wise to guide yourself with some examples. OK I've tried this but not got very far. My original thinking was that \\((a, b)\mapsto a + b\\) should be left adjoint to the diagonal map \\(c\mapsto (c, c)\\). Hence we'd have \[ \mathcal{X}(a + b, c) = (\mathcal{X\times X})((a, b), (c, c)) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c) \] But I've just realised that the diagonal map is *not* necessarily a \\(\mathcal{V}\\)-functor! \[ (\mathcal{X\times X})((c, c), (c', c')) = \mathcal{X}(c, c')\otimes\mathcal{X}(c, c') \] and there's no reason to think the right hand side is always \\(\geq \mathcal{X}(c, c')\\). So let's try a few examples. If \\(\mathcal{V} = \textbf{Bool}\\) then a \\(\mathcal{V}\\)-category \\(\mathcal{X}\\) is a \\(\textbf{Bool}\\)-category, which is a preorder. So the coproduct should be a join in that preorder, ie \[ (a + b \leq_\mathcal{X} c) \text{ if and only if } (a \leq_\mathcal{X} c \text{ and } b \leq_\mathcal{X} c) \] which is the same as \[ \mathcal{X}(a + b, c) = \mathcal{X}(a, c)\wedge \mathcal{X}(b, c) \] And if we consider \\(\textbf{Vect}\\) enriched over itself we have \[ \textbf{Vect}(A \oplus B, C) = \textbf{Vect}(A, C) \oplus \textbf{Vect}(B, C) \] and again we have a binary "meet" (ie \\(\oplus\\) considered as a binary product in \\(\textbf{Vect}\\)) on the right hand side. So my current guess is that we should be using meets/products in \\(\mathcal{V}\\) (which unlike \\(\otimes\\) do admit infinitary generalisations) to define coproducts in \\(\mathcal{V}\\)-categories. I'll take a look at that Kelly book to see if any of this is barking up the right tree.
  • 15.
    edited July 26

    image

    Anindya wrote:

    My original thinking was that \((a, b)\mapsto a + b\) should be left adjoint to the diagonal map \(c\mapsto (c, c)\). Hence we'd have

    $$ \mathcal{X}(a + b, c) = (\mathcal{X\times X})((a, b), (c, c)) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c) $$ But I've just realised that the diagonal map is
    not necessarily a \(\mathcal{V}\)-functor!

    Right; this is one reason people switch from colimits to weighted colimits when working in enriched categories. The nLab article:

    motivates weighted colimits as follows:

    In enriched category theory, where one considers categories \(C\) enriched in a "nice" monoidal category \(V\) (generally one where \(V\) is a complete, cocomplete closed symmetric monoidal category) there is in general no \(V\)-enriched diagonal functor \(\Delta: C \to C^J\) to speak of. For example, when \(V\) is the category \(\textbf{Ab}\) of abelian groups we have \(C \simeq C^I\) where \(I\) is the unit \(V\)-category having one object \(1\) for which \(\text{hom}(1, 1) = \mathbb{Z}\), but then for a general \(\textbf{Ab}\)-enriched category \(J\), there is no enriched functor \(J \to I\) to pull back along (or, there may be many, but none stand out as canonical). This shows that the usual notion of colimit doesn't adapt particularly well to the general enriched setting.

    The more flexible notion of weighted colimit (also called an indexed colimit in some of the older accounts) was introduced by Borceux (and Kelly?) as giving the right notion of colimit for enriched category theory.

    It's possible the nLab article will get you started faster than Kelly's book. They also refer to a "very nice description" by a guy named Baez, here. He's talking about "indexed colimits", but these are just the same as weighted colimits.

    Comment Source:<img width = "100" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> Anindya wrote: > My original thinking was that \\((a, b)\mapsto a + b\\) should be left adjoint to the diagonal map \\(c\mapsto (c, c)\\). Hence we'd have > \[ \mathcal{X}(a + b, c) = (\mathcal{X\times X})((a, b), (c, c)) = \mathcal{X}(a, c)\otimes\mathcal{X}(b, c) \] > But I've just realised that the diagonal map is *not* necessarily a \\(\mathcal{V}\\)-functor! Right; this is one reason people switch from colimits to weighted colimits when working in enriched categories. The nLab article: * nLab, [weighted colimit](https://ncatlab.org/nlab/show/weighted+colimit) motivates weighted colimits as follows: > In enriched category theory, where one considers categories \\(C\\) enriched in a "nice" monoidal category \\(V\\) (generally one where \\(V\\) is a complete, cocomplete closed symmetric monoidal category) there is in general no \\(V\\)-enriched diagonal functor \\(\Delta: C \to C^J\\) to speak of. For example, when \\(V\\) is the category \\(\textbf{Ab}\\) of abelian groups we have \\(C \simeq C^I\\) where \\(I\\) is the unit \\(V\\)-category having one object \\(1\\) for which \\(\text{hom}(1, 1) = \mathbb{Z}\\), but then for a general \\(\textbf{Ab}\\)-enriched category \\(J\\), there is no enriched functor \\(J \to I\\) to pull back along (or, there may be many, but none stand out as canonical). This shows that the usual notion of colimit doesn't adapt particularly well to the general enriched setting. > The more flexible notion of weighted colimit (also called an _indexed colimit_ in some of the older accounts) was introduced by Borceux (and Kelly?) as giving the _right_ notion of colimit for enriched category theory. It's possible the nLab article will get you started faster than Kelly's book. They also refer to a "very nice description" by a guy named Baez, [here](https://golem.ph.utexas.edu/category/2007/02/day_on_rcfts.html#c007688). He's talking about "indexed colimits", but these are just the same as weighted colimits.
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