#### Howdy, Stranger!

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

Options

# Lecture 62 - Chapter 4: Enriched Profunctors

2»

• Options
51.
edited July 2018

iirc the finite and cofinite subsets of $$\mathbb{N}$$ form a Heyting algebra that isn't complete (for instance the sequence $$\{0\}, \{0, 2\}, \{0,2, 4\}...$$ has no join). Setting $$\otimes$$ to be the meet would give us a closed monoidal poset that doesn't have arbitrary joins.

Comment Source:iirc the finite and cofinite subsets of \$$\mathbb{N}\$$ form a Heyting algebra that isn't complete (for instance the sequence \$$\\{0\\}, \\{0, 2\\}, \\{0,2, 4\\}...\$$ has no join). Setting \$$\otimes\$$ to be the meet would give us a closed monoidal poset that doesn't have arbitrary joins.
• Options
52.
edited July 2018

@Simon

How about $$( (-\infty,\infty), \ge, +, 0)$$?

I think that one works! In this case $$x \multimap y := y - x$$

(EDIT: I had also come up with finite-cofinite sets like Anindya, but he beat me to it!)

Comment Source:@Simon > How about \$$( (-\infty,\infty), \ge, +, 0)\$$? I think that one works! In this case \$$x \multimap y := y - x\$$ (EDIT: I had also come up with finite-cofinite sets like Anindya, but he beat me to it!)
• Options
53.

I have a feeling that if you have a closed monoidal poset $$\mathcal{V}$$, you can always 'freely throw in all joins' and get a new closed monoidal poset with all joins. Maybe Simon knows if this is true: I would try to do it by forming $$\mathbf{Bool}^{\mathcal{V}^{\text{op}}}$$: a kind of 'free cocompletion', as the experts say.

However, this may destroy the old joins you happened to have. E.g., you may throw in a new bottom element $$\bot$$, rendering the old one (if $$\mathcal{V}$$ had one) no longer the bottom.

Comment Source:I have a feeling that if you have a closed monoidal poset \$$\mathcal{V}\$$, you can always 'freely throw in all joins' and get a new closed monoidal poset with all joins. Maybe Simon knows if this is true: I would try to do it by forming \$$\mathbf{Bool}^{\mathcal{V}^{\text{op}}}\$$: a kind of 'free cocompletion', as the experts say. However, this may destroy the old joins you happened to have. E.g., you may throw in a new bottom element \$$\bot\$$, rendering the old one (if \$$\mathcal{V}\$$ had one) no longer the bottom. 
• Options
54.
edited July 2018

Simon wrote:

Or maybe even the following is true.

Suppose $$\mathcal{V}$$ is a monoidal partial order which has all joins, then $$\mathcal{V}$$ is closed if and only if the monoidal product distributes over joins.

You're really pushing Matthew hard, Simon!

Matthew proved it today, but yesterday I added a sketch of a proof to Lecture 63 because this offers two nicely different characterizations of the same structure: a quantale!

(Remember, in my lectures a quantale is a closed monoidal poset with all joins.)

Here's what I wrote:

Theorem. If $$\mathcal{V}$$ is a monoidal poset with all joins, $$\mathcal{V}$$ is a quantale if and only if

$$a \otimes \left( \bigvee_{b\in B} b\right) = \bigvee_{b \in A} (a \otimes b)$$ for every element $$a$$ and every subset $$B$$ of $$\mathcal{V}$$.

Proof Sketch. For every element $$a$$ of $$\mathcal{V}$$ there is a monotone function

$$a \otimes - \, \colon \mathcal{V} \to \mathcal{V}$$ sending each $$x \in \mathcal{V}$$ to $$a \otimes x$$ . By the Adjoint Functor Theorem for Posets, this monotone function has a right adjoint iff it preserves all joins. It a right adjoint iff $$\mathcal{V}$$ is closed, since $$\mathcal{V}$$ being closed says that

$$a \otimes x \le y \text{ if and only if } x \le a \multimap y$$ for all $$x, y$$ in $$\mathcal{V}$$, which means that

$$a \multimap - \, \colon \mathcal{V} \to \mathcal{V}$$ is a right adjoint of

$$a \otimes - \, \colon \mathcal{V} \to \mathcal{V}$$ On the other hand, $$a \otimes -$$ preserves all joins iff

$$a \otimes \left( \bigvee_{b\in B} b\right) = \bigvee_{b \in B} (a \otimes b)$$ for every element $$a$$ and every subset $$B$$ of $$\mathcal{V}$$. $$\qquad \blacksquare$$

Comment Source:Simon wrote: > Or maybe even the following is true. > > Suppose \$$\mathcal{V}\$$ is a monoidal partial order which has all joins, then \$$\mathcal{V}\$$ is closed if and only if the monoidal product distributes over joins. You're really pushing Matthew hard, Simon! Matthew proved it today, but yesterday I added a _sketch_ of a proof to [Lecture 63](https://forum.azimuthproject.org/discussion/2295/lecture-63-chapter-4-composing-enriched-profunctors/p1) because this offers two nicely different characterizations of the same structure: a quantale! (Remember, in my lectures a **quantale** is a closed monoidal poset with all joins.) Here's what I wrote: > **Theorem.** If \$$\mathcal{V}\$$ is a monoidal poset with all joins, \$$\mathcal{V}\$$ is a quantale if and only if $a \otimes \left( \bigvee\_{b\in B} b\right) = \bigvee\_{b \in A} (a \otimes b)$ for every element \$$a\$$ and every subset \$$B\$$ of \$$\mathcal{V}\$$. **Proof Sketch.** For every element \$$a\$$ of \$$\mathcal{V}\$$ there is a monotone function $a \otimes - \, \colon \mathcal{V} \to \mathcal{V}$ sending each \$$x \in \mathcal{V}\$$ to \$$a \otimes x\$$ . By the [Adjoint Functor Theorem for Posets](https://forum.azimuthproject.org/discussion/2031/lecture-16-chapter-1-the-adjoint-functor-theorem-for-posets/p1), this monotone function has a right adjoint iff it preserves all joins. It a right adjoint iff \$$\mathcal{V}\$$ is closed, since \$$\mathcal{V}\$$ being closed says that $a \otimes x \le y \text{ if and only if } x \le a \multimap y$ for all \$$x, y\$$ in \$$\mathcal{V}\$$, which means that $a \multimap - \, \colon \mathcal{V} \to \mathcal{V}$ is a right adjoint of $a \otimes - \, \colon \mathcal{V} \to \mathcal{V}$ On the other hand, \$$a \otimes -\$$ preserves all joins iff $a \otimes \left( \bigvee\_{b\in B} b\right) = \bigvee\_{b \in B} (a \otimes b)$ for every element \$$a\$$ and every subset \$$B\$$ of \$$\mathcal{V}\$$. \$$\qquad \blacksquare \$$ 
• Options
55.

Hey John,

For any poset, its Dedekind completion is the smallest complete lattice that embeds that poset.

Since the completion is an embedding, this preserves the bottom and top elements (up to isomorphism), but we can check that if you like.

Comment Source:Hey John, We talked about this a long time ago. For any poset, its *Dedekind completion* is the smallest complete lattice that embeds that poset. Since the completion is an embedding, this preserves the bottom and top elements (up to isomorphism), but we can check that if you like.
• Options
56.
edited July 2018

Interesting. I was talking about monoidal posets, not posets. If we start with a monoidal poset, does its Dedekind completion automatically get a monoidal structure extending that of our original monoidal poset? (Note: I'm not assuming the monoidal structure is $$\wedge$$.)

See, in general when you take the 'free cocompletion' of a category $$\mathcal{C}$$ you get the presheaf category $$\mathbf{Set}^{\mathcal{C}^{\text{op}}}$$, which has all (small) colimits. $$\mathcal{C}$$ sits inside $$\mathbf{Set}^{\mathcal{C}^{\text{op}}}$$ via the Yoneda embedding. And if $$\mathcal{C}$$ is monoidal, $$\mathbf{Set}^{\mathcal{C}^{\text{op}}}$$ will become monoidal too, in a manner that extends the monoidal structure on $$\mathcal{C}$$: this is called Day convolution. But the colimits that $$\mathcal{C}$$ originally had will be broken.

I was thinking it would be fun to something similar for posets, which are $$\textbf{Bool}$$-enriched categories. Colimits in posets are just joins. And the whole business I just described works for $$\mathcal{V}$$-enriched categories when $$\mathcal{V}$$ is sufficiently nice, e.g. $$\textbf{Bool}$$. (The link says what 'sufficiently nice' means.)

So, for any poset $$\mathcal{C}$$ we can form a poset $$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}$$ whose elements are monotone maps from $$\mathcal{C}^{\text{op}}$$ to $$\textbf{Bool}$$. This poset should contain $$\mathcal{C}$$ in some obvious way, and it should have all joins. And if $$\mathcal{C}$$ is monoidal, so will be $$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}$$.

I'm not sure how this construction is related to the Dedekind completion. Nor am I sure that the joins in $$\mathcal{C}$$ are no longer joins in $$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}$$!

One way to think about a monotone map from $$\mathcal{C}^{\text{op}}$$ to $$\textbf{Bool}$$ is to look at the set of elements that map to $$\text{true}$$; this is an upper set in $$\mathcal{C}^{\text{op}}$$, or a lower set in $$\mathcal{C}$$. This seems a bit Dedekind-y; right now I don't even remember how the Dedekind completion of a poset works.

Comment Source:Interesting. I was talking about monoidal posets, not posets. If we start with a monoidal poset, does its Dedekind completion automatically get a monoidal structure extending that of our original monoidal poset? (Note: I'm not assuming the monoidal structure is \$$\wedge\$$.) See, in general when you take the 'free cocompletion' of a category \$$\mathcal{C}\$$ you get the presheaf category \$$\mathbf{Set}^{\mathcal{C}^{\text{op}}}\$$, which has all (small) colimits. \$$\mathcal{C}\$$ sits inside \$$\mathbf{Set}^{\mathcal{C}^{\text{op}}}\$$ via the Yoneda embedding. And if \$$\mathcal{C}\$$ is monoidal, \$$\mathbf{Set}^{\mathcal{C}^{\text{op}}}\$$ will become monoidal too, in a manner that extends the monoidal structure on \$$\mathcal{C}\$$: this is called [Day convolution](https://ncatlab.org/nlab/show/Day+convolution#ForMonoidalCategories). But the colimits that \$$\mathcal{C}\$$ originally had will be broken. I was thinking it would be fun to something similar for posets, which are \$$\textbf{Bool}\$$-enriched categories. Colimits in posets are just _joins_. And the whole business I just described works for \$$\mathcal{V}\$$-enriched categories when \$$\mathcal{V}\$$ is sufficiently nice, e.g. \$$\textbf{Bool}\$$. (The link says what 'sufficiently nice' means.) So, for any poset \$$\mathcal{C}\$$ we can form a poset \$$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}\$$ whose elements are monotone maps from \$$\mathcal{C}^{\text{op}}\$$ to \$$\textbf{Bool}\$$. This poset should contain \$$\mathcal{C}\$$ in some obvious way, and it should have all joins. And if \$$\mathcal{C}\$$ is monoidal, so will be \$$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}\$$. I'm not sure how this construction is related to the Dedekind completion. Nor am I sure that the joins in \$$\mathcal{C}\$$ are no longer joins in \$$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}\$$! One way to think about a monotone map from \$$\mathcal{C}^{\text{op}}\$$ to \$$\textbf{Bool}\$$ is to look at the set of elements that map to \$$\text{true}\$$; this is an [upper set](https://en.wikipedia.org/wiki/Upper_set) in \$$\mathcal{C}^{\text{op}}\$$, or a lower set in \$$\mathcal{C}\$$. This seems a bit Dedekind-y; right now I don't even remember how the Dedekind completion of a poset works. 
• Options
57.
edited July 2018

Oh, I just noticed something.

Our friend $$\leq$$, being somewhat of a Boolean hom-functor, is adjoint to the Boolean Yoneda embedding,

$\leq : \mathcal{P}^{op} \times \mathcal{P} \to \mathbf{Bool} \\ \Leftrightarrow \\ よ : \mathcal{P} \to \mathbf{Bool}^{\mathcal{P}^{op}}.$

Edit: The symbol よ ["yo"] is Japanese hiragana, coming from the Japanese surname ねだ /米田 ["yoneda"], of the Japanese mathematician Nobuo Yoneda. Interestingly, the symbols 米田 can also be pronounced as めた ["meta"] in Japanese, which is sort of what the Yoneda lemma and embedding are.

Comment Source:Oh, I just noticed something. Our friend \$$\leq\$$, being somewhat of a Boolean hom-functor, is adjoint to the Boolean Yoneda embedding, \$\leq : \mathcal{P}^{op} \times \mathcal{P} \to \mathbf{Bool} \\\\ \Leftrightarrow \\\\ よ : \mathcal{P} \to \mathbf{Bool}^{\mathcal{P}^{op}}. \$ Edit: The symbol よ ["yo"] is Japanese hiragana, coming from the Japanese surname **よ**ねだ /米田 ["**yo**neda"], of the Japanese mathematician Nobuo Yoneda. Interestingly, the symbols 米田 can also be pronounced as めた ["meta"] in Japanese, which is sort of what the Yoneda lemma and embedding are. 
• Options
58.
edited July 2018

Very nice observation, Keith!

Yes, in this lecture we saw that whenever $$\mathcal{V}$$ is a commutative quantale, every $$\mathcal{V}$$-enriched category $$\mathcal{C}$$ has a hom-functor

$$\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}$$ In Lecture 64 we saw this gives a $$\mathcal{V}$$-enriched profunctor which is just the identity

$$1_{\mathcal{C}} : \mathcal{C} \nrightarrow \mathcal{C} .$$ We haven't talked about functor categories for enriched categories. But they work okay in this context. We can feed the $$\mathcal{V}$$-enriched functor

$$\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}$$ an object of $$\mathcal{C}$$ and it will give back a functor from $$\mathcal{C}^{\text{op}}$$ to $$\mathcal{V}$$, which is an object of $$\mathcal{V}^{ \mathcal{C}^{\text{op}} }$$.

So, one can check, we get a functor

$$Y : \mathcal{C} \to \mathcal{V}^{ \mathcal{C}^{\text{op}} }$$ which is called the Yoneda embedding.

So the Yoneda embedding, the identity profunctor

$$1_{\mathcal{C}} : \mathcal{C} \nrightarrow \mathcal{C}$$ and the hom-functor

$$\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}$$ are all just different views of the same thing!

Comment Source:Very nice observation, Keith! Yes, in this lecture we saw that whenever \$$\mathcal{V}\$$ is a commutative quantale, every \$$\mathcal{V}\$$-enriched category \$$\mathcal{C}\$$ has a hom-functor $\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}$ In [Lecture 64](https://forum.azimuthproject.org/discussion/2298/lecture-64-chapter-5-the-category-of-enriched-profunctors/p1) we saw this gives a \$$\mathcal{V}\$$-enriched profunctor which is just the identity $1_{\mathcal{C}} : \mathcal{C} \nrightarrow \mathcal{C} .$ We haven't talked about functor categories for enriched categories. But they work okay in this context. We can feed the \$$\mathcal{V}\$$-enriched functor $\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}$ an object of \$$\mathcal{C}\$$ and it will give back a functor from \$$\mathcal{C}^{\text{op}} \$$ to \$$\mathcal{V} \$$, which is an object of \$$\mathcal{V}^{ \mathcal{C}^{\text{op}} } \$$. So, one can check, we get a functor $Y : \mathcal{C} \to \mathcal{V}^{ \mathcal{C}^{\text{op}} }$ which is called the **Yoneda embedding**. So the Yoneda embedding, the identity profunctor $1_{\mathcal{C}} : \mathcal{C} \nrightarrow \mathcal{C}$ and the hom-functor $\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}$ are all just different views of the same thing! 
• Options
59.

Interesting. I was talking about monoidal posets, not posets. If we start with a monoidal poset, does its Dedekind completion automatically get a monoidal structure extending that of our original monoidal poset?

So, for any poset $$\mathcal{C}$$ we can form a poset $$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}$$ whose elements are monotone maps from $$\mathcal{C}^{\text{op}}$$ to $$\textbf{Bool}$$. This poset should contain $$\mathcal{C}$$ in some obvious way, and it should have all joins. And if $$\mathcal{C}$$ is monoidal, so will be $$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}$$.

I'm not sure how this construction is related to the Dedekind completion. Nor am I sure that the joins in $$\mathcal{C}$$ are no longer joins in $$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}$$!

One way to think about a monotone map from $$\mathcal{C}^{\text{op}}$$ to $$\textbf{Bool}$$ is to look at the set of elements that map to $$\text{true}$$; this is an upper set in $$\mathcal{C}^{\text{op}}$$, or a lower set in $$\mathcal{C}$$. This seems a bit Dedekind-y; right now I don't even remember how the Dedekind completion of a poset works.

I know in lattice theory, for any poset $$(X, \le)$$ then if we consider the set $$X^\uparrow$$ of $$X$$'s upper sets, this set forms a complete lattice under arbitrary joins and meets. This lattice is called an Alexandrov Topology. In fact $$\bullet^\uparrow$$ extends to a functor. A monotone function $$f: X \to Y$$ can be lifted to $$f^\uparrow : X^\uparrow \to Y^\uparrow$$ as follows: if $$u \in X^\uparrow$$ then $$f^\uparrow (u) := f_!(u)$$ following the notation from Lecture 9 in chapter 1. In fact $$f_!$$ is a continuous function between the two topologies. We also have the map $$\uparrow \lbrace x\rbrace := \lbrace y \mid x \le y \rbrace$$ is an injective monotone function.

This might be the free cocompletion you are thinking of, or perhaps its dual.

The Dedekind-MacNeil completion is similar. Say $$\uparrow X := \lbrace y \mid \forall x \in X. a \leq y\rbrace$$ and $$\downarrow X := \lbrace y \mid \forall x \in X. y \leq x\rbrace .$$ A set $$A \subseteq P$$ is in the Dedekind-MacNeil completion of $$P$$ if and only if $$A = \downarrow (\uparrow A) .$$

The question now is if these things are preserving a monoidal poset.

• If $$A$$ and $$B$$ are upper sets, then is $$A \bar{\otimes} B = \lbrace a \otimes b \mid a \in A \text{ and } b \in B \rbrace$$ an upper set?

• If $$A$$ and $$B$$ are in the Dedekind-MacNeil completion of $$P$$, then is $$A \bar{\otimes} B$$ also in the completion too?

Hopefully I can find the time tomorrow to dig into this.

Comment Source:> Interesting. I was talking about monoidal posets, not posets. If we start with a monoidal poset, does its Dedekind completion automatically get a monoidal structure extending that of our original monoidal poset? I don't know the answer. > So, for any poset \$$\mathcal{C}\$$ we can form a poset \$$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}\$$ whose elements are monotone maps from \$$\mathcal{C}^{\text{op}}\$$ to \$$\textbf{Bool}\$$. This poset should contain \$$\mathcal{C}\$$ in some obvious way, and it should have all joins. And if \$$\mathcal{C}\$$ is monoidal, so will be \$$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}\$$. > I'm not sure how this construction is related to the Dedekind completion. Nor am I sure that the joins in \$$\mathcal{C}\$$ are no longer joins in \$$\mathbf{Bool}^{\mathcal{C}^{\text{op}}}\$$! > One way to think about a monotone map from \$$\mathcal{C}^{\text{op}}\$$ to \$$\textbf{Bool}\$$ is to look at the set of elements that map to \$$\text{true}\$$; this is an [upper set](https://en.wikipedia.org/wiki/Upper_set) in \$$\mathcal{C}^{\text{op}}\$$, or a lower set in \$$\mathcal{C}\$$. This seems a bit Dedekind-y; right now I don't even remember how the Dedekind completion of a poset works. I know in lattice theory, for any poset \$$(X, \le)\$$ then if we consider the set \$$X^\uparrow\$$ of \$$X\$$'s upper sets, this set forms a complete lattice under arbitrary joins and meets. This lattice is called an [Alexandrov Topology](https://en.wikipedia.org/wiki/Alexandrov_topology). In fact \$$\bullet^\uparrow\$$ extends to a functor. A monotone function \$$f: X \to Y\$$ can be lifted to \$$f^\uparrow : X^\uparrow \to Y^\uparrow\$$ as follows: if \$$u \in X^\uparrow\$$ then \$$f^\uparrow (u) := f_!(u)\$$ following the notation from [Lecture 9](https://forum.azimuthproject.org/discussion/1931/lecture-9-chapter-1-adjoints-and-the-logic-of-subsets/p1) in chapter 1. In fact \$$f_!\$$ is a *continuous* function between the two topologies. We also have the map \$$\uparrow \lbrace x\rbrace := \lbrace y \mid x \le y \rbrace\$$ is an injective monotone function. This might be the free cocompletion you are thinking of, or perhaps its dual. The Dedekind-MacNeil completion is similar. Say \$$\uparrow X := \lbrace y \mid \forall x \in X. a \leq y\rbrace \$$ and \$$\downarrow X := \lbrace y \mid \forall x \in X. y \leq x\rbrace .\$$ A set \$$A \subseteq P \$$ is in the Dedekind-MacNeil completion of \$$P\$$ if and only if \$$A = \downarrow (\uparrow A) .\$$ The question now is if these things are preserving a monoidal poset. • If \$$A\$$ and \$$B\$$ are upper sets, then is \$$A \bar{\otimes} B = \lbrace a \otimes b \mid a \in A \text{ and } b \in B \rbrace \$$ an upper set? • If \$$A\$$ and \$$B\$$ are in the Dedekind-MacNeil completion of \$$P\$$, then is \$$A \bar{\otimes} B\$$ also in the completion too? Hopefully I can find the time tomorrow to dig into this.
• Options
60.

I believe they do preserve monoidal structure.

Upper sets preserve joins, while down sets preserve meets (Or maybe it's backward).

Basically, upper and down sets are like presheaves and copresheaves in my head (or at least what I think are (co)presheaves).

In fact, given John Baez's earlier comments, I'm pretty sure a decategorized presheaf gives a quantale.

Comment Source:I believe they do preserve monoidal structure. Upper sets preserve joins, while down sets preserve meets (Or maybe it's backward). Basically, upper and down sets are like presheaves and copresheaves in my head (or at least what I think are (co)presheaves). In fact, given John Baez's earlier comments, I'm pretty sure a decategorized presheaf gives a quantale.
• Options
61.

@John wrote:

So the Yoneda embedding, the identity profunctor $$1_{\mathcal{C}} : \mathcal{C} \nrightarrow \mathcal{C}$$ and the hom-functor $$\mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}$$ are all just different views of the same thing!

We can generalise this surely: any $$\mathcal{V}$$-profunctor $$\Phi : \mathcal{C} \nrightarrow \mathcal{D}$$ corresponds to a unique $$\mathcal{V}$$-functor $$\mathcal{D} \to \mathcal{V}^{C^\text{op}}$$, and vice versa. Which I suppose gives another perspective on what profunctors "are".

Comment Source:@John wrote: > So the Yoneda embedding, the identity profunctor \$$1_{\mathcal{C}} : \mathcal{C} \nrightarrow \mathcal{C}\$$ and the hom-functor \$$\mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V}\$$ are all just different views of the same thing! We can generalise this surely: any \$$\mathcal{V}\$$-profunctor \$$\Phi : \mathcal{C} \nrightarrow \mathcal{D}\$$ corresponds to a unique \$$\mathcal{V}\$$-functor \$$\mathcal{D} \to \mathcal{V}^{C^\text{op}}\$$, and vice versa. Which I suppose gives another perspective on what profunctors "are".
• Options
62.
edited July 2018

@Keith

I believe they do preserve monoidal structure.

Upper sets preserve joins, while down sets preserve meets (Or maybe it's backward).

Basically, upper and down sets are like presheaves and copresheaves in my head (or at least what I think are (co)presheaves).

I appreciate the vote of confidence, Keith, but having slept on this I don't think $$A \bar{\otimes} B = \lbrace a \otimes b \mid a \in A \text{ and } b \in B \rbrace$$ works.

Consider any bounded poset $$P$$ with more than two elements. Call the initial object $$\bot$$ and the terminal object $$\top$$. Define:

$$a \bullet b := \begin{cases} a & \text{if } b = \top \\ b & \text{if } a = \top \\ \bot & \text{otherwise} \end{cases}$$ We can see $$(P, \bullet, \top)$$ forms a monoid. We also need to check that $$(P, \leq, \bullet, \top)$$ is a monoidal partial order. This boils down to checking $$a \leq b$$ and $$c \leq d$$ then $$a \bullet c \leq b \bullet d$$. This could be done by checking the 16 cases where $$a = b = c = d = \top$$, $$a = b = c = \top$$ and $$d \neq \top$$, etc.

If $$A$$ and $$B$$ are upper sets, then $$A\; \bar{\bullet}\; B$$ may not be an upper set. Consider the case where $$P = \mathcal{P}(\lbrace a,b \rbrace)$$, ordered by inclusion. We have $$X = \lbrace \lbrace a \rbrace, \lbrace a, b \rbrace \rbrace$$ is an upper set. But $$X\; \bar{\bullet}\; X = \lbrace \varnothing, \lbrace a \rbrace, \lbrace a, b \rbrace \rbrace$$ which is not an upper set.

I am not sure if this is broken for the Dedekind-MacNeil completion, but I suspect it may be.

Comment Source:@Keith > I believe they do preserve monoidal structure. > > Upper sets preserve joins, while down sets preserve meets (Or maybe it's backward). > > Basically, upper and down sets are like presheaves and copresheaves in my head (or at least what I think are (co)presheaves). I appreciate the vote of confidence, Keith, but having slept on this I don't think \$$A \bar{\otimes} B = \lbrace a \otimes b \mid a \in A \text{ and } b \in B \rbrace \$$ works. Consider any bounded poset \$$P\$$ with more than two elements. Call the initial object \$$\bot\$$ and the terminal object \$$\top\$$. Define: $a \bullet b := \begin{cases} a & \text{if } b = \top \\\\ b & \text{if } a = \top \\\\ \bot & \text{otherwise} \end{cases}$ We can see \$$(P, \bullet, \top)\$$ forms a monoid. We also need to check that \$$(P, \leq, \bullet, \top)\$$ is a monoidal partial order. This boils down to checking \$$a \leq b\$$ and \$$c \leq d\$$ then \$$a \bullet c \leq b \bullet d\$$. This could be done by checking the 16 cases where \$$a = b = c = d = \top\$$, \$$a = b = c = \top\$$ and \$$d \neq \top\$$, etc. If \$$A\$$ and \$$B\$$ are upper sets, then \$$A\; \bar{\bullet}\; B\$$ may not be an upper set. Consider the case where \$$P = \mathcal{P}(\lbrace a,b \rbrace)\$$, ordered by inclusion. We have \$$X = \lbrace \lbrace a \rbrace, \lbrace a, b \rbrace \rbrace\$$ is an upper set. But \$$X\; \bar{\bullet}\; X = \lbrace \varnothing, \lbrace a \rbrace, \lbrace a, b \rbrace \rbrace\$$ which is *not* an upper set. I am not sure if this is broken for the Dedekind-MacNeil completion, but I suspect it may be.
• Options
63.

A question Matt: Why are you defining the monoidal product in that way?

It seems unnecessary.

Comment Source:A question Matt: Why are you defining the monoidal product in that way? It seems unnecessary. 
• Options
64.

A question Matt: Why are you defining the monoidal product in that way?

I defined $$a \bullet b$$ as a counter example to the idea that the upper set embedding and Dedekind-MacNeil embedding preserve monoidal products.

Regarding the definition of $$A \; \bar{\otimes} \; B = \lbrace a \otimes b \mid a \in A \text{ and } b \in B \rbrace$$, it seems natural. Maybe you can think of something else?

If we want to embed some mathematical object $$\mathfrak{M} \hookrightarrow \mathfrak{N}$$, then if $$\mathfrak{M}$$ has a tensor $$\otimes$$ (or an order relation, or whatever) then $$\mathfrak{N}$$ has to have one too.

Comment Source:> A question Matt: Why are you defining the monoidal product in that way? I defined \$$a \bullet b\$$ as a counter example to the idea that the upper set embedding and Dedekind-MacNeil embedding preserve monoidal products. Regarding the definition of \$$A \; \bar{\otimes} \; B = \lbrace a \otimes b \mid a \in A \text{ and } b \in B \rbrace\$$, it seems natural. Maybe you can think of something else? If we want to embed some mathematical object \$$\mathfrak{M} \hookrightarrow \mathfrak{N}\$$, then if \$$\mathfrak{M}\$$ has a tensor \$$\otimes\$$ (or an order relation, or whatever) then \$$\mathfrak{N}\$$ has to have one too.
• Options
65.
edited July 2018

I honestly don't know why this matters, but I'll give you the benefit of the doubt.

However, it's clear to me now that downsets preserve meets ($$\land$$), and uppersets preserve joins ($$\lor$$).

Proof:

A downset is defined to be a set $\downarrow_{x} := \lbrace y \mid y \leq x \rbrace$

Given two downsets, $$\downarrow_{\lbrace x \rbrace},\downarrow_{\lbrace y \rbrace}$$ we will say $$\downarrow_{\lbrace x \rbrace} \subseteq \downarrow_{\lbrace y \rbrace}$$ if, and only if, $$x \leq y$$.

Then clearly meets are preserved, since if

$x_1 \leq y$

and,

$x_1 \leq y$

and,

$x_1 \land x_2$ is the meet of $$x_1$$ and $$x_2$$, ie

$x_1 \land x_2 \leq x_1$

and, $x_1 \land x_2 \leq x_2$

then,

$x_1 \land x_2 \leq y$

and,

$x_1 \land x_2 \leq x_1 \leq y \Rightarrow \\ \downarrow_{\lbrace x_1 \land x_2 \rbrace} \subseteq \downarrow_{\lbrace x_1 \rbrace} \subseteq \downarrow_{\lbrace y \rbrace}$

and,

$x_1 \land x_2 \leq x_2 \leq y \Rightarrow \\ \downarrow_{\lbrace x_1 \land x_2 \rbrace} \subseteq \downarrow_{\lbrace x_2 \rbrace} \subseteq \downarrow_{\lbrace y \rbrace}$

furthermore, this is monoidal with,

$\downarrow_{\lbrace x_1 \land x_2 \ \rbrace} = \downarrow_{\lbrace x_1 \rbrace} \cap \downarrow_{\lbrace x_2 \rbrace}.$

$$\blacksquare$$

The proof with upper sets and joins proceeds similarity.

Comment Source:I honestly don't know why this matters, but I'll give you the benefit of the doubt. However, it's clear to me now that downsets preserve meets (\$$\land\$$), and uppersets preserve joins (\$$\lor\$$). Proof: A downset is defined to be a set \$\downarrow\_{x} := \lbrace y \mid y \leq x \rbrace \$ Given two downsets, \$$\downarrow\_{\lbrace x \rbrace},\downarrow\_{\lbrace y \rbrace}\$$ we will say \$$\downarrow\_{\lbrace x \rbrace} \subseteq \downarrow\_{\lbrace y \rbrace} \$$ if, and only if, \$$x \leq y\$$. Then clearly meets are preserved, since if \$x\_1 \leq y \$ and, \$x\_1 \leq y \$ and, \$x\_1 \land x\_2 \$ is the meet of \$$x\_1\$$ and \$$x\_2\$$, ie \$x\_1 \land x\_2 \leq x\_1 \$ and, \$x\_1 \land x\_2 \leq x\_2 \$ then, \$x\_1 \land x\_2 \leq y \$ and, \$x\_1 \land x\_2 \leq x\_1 \leq y \Rightarrow \\\\ \downarrow\_{\lbrace x\_1 \land x\_2 \rbrace} \subseteq \downarrow\_{\lbrace x\_1 \rbrace} \subseteq \downarrow\_{\lbrace y \rbrace} \$ and, \$x\_1 \land x\_2 \leq x\_2 \leq y \Rightarrow \\\\ \downarrow\_{\lbrace x\_1 \land x\_2 \rbrace} \subseteq \downarrow\_{\lbrace x\_2 \rbrace} \subseteq \downarrow\_{\lbrace y \rbrace} \$ furthermore, this is monoidal with, \$\downarrow\_{\lbrace x\_1 \land x\_2 \ \rbrace} = \downarrow\_{\lbrace x\_1 \rbrace} \cap \downarrow\_{\lbrace x\_2 \rbrace}. \$ \$$\blacksquare\$$ The proof with upper sets and joins proceeds similarity. 
• Options
66.

Keith,

You are only preserving $$\wedge$$, but a monoidal preorder can have a different monoidal operation such as $$+$$ in $$\mathbf{Cost}$$ or my example of $$\bullet$$.

This is exactly what John was concerned about when he said to me:

Interesting. I was talking about monoidal posets, not posets. If we start with a monoidal poset, does its Dedekind completion automatically get a monoidal structure extending that of our original monoidal poset? (Note: I'm not assuming the monoidal structure is $$\wedge$$.)

Comment Source:Keith, You are only preserving \$$\wedge\$$, but a monoidal preorder can have a different monoidal operation such as \$$+\$$ in \$$\mathbf{Cost}\$$ or my example of \$$\bullet\$$. This is exactly what John was concerned about when he said to me: > Interesting. I was talking about monoidal posets, not posets. If we start with a monoidal poset, does its Dedekind completion automatically get a monoidal structure extending that of our original monoidal poset? (Note: I'm not assuming the monoidal structure is \$$\wedge\$$.)