Scott Oswald wrote:

> I think I am not understanding something correctly about \\(\mathbf{Cost}\\)-categories. Isn't the composition rule for \\(\mathbf{Cost}\\)-categories like this: If we have objects \\(N,W,S\\), then the composition rule is \\(\mathcal{C}(S,W) + \mathcal{C}(W,N) \ge \mathcal{C}(S,N)\\) so \\(4 \ge \mathcal{C}(S,N)\\). There's another path between \\(S\\) and \\(N\\) via \\(E\\), but it gives \\(7 \ge \mathcal{C}(S,N)\\). Does this mean that \\(\mathcal{C}(S,N) = 4\\) since it's the meet of compositions that give \\(\mathcal{C}(S,N)\\)?

I explained this stuff a bit here:

* [Lecture 33 - Chapter 2: Tying up loose ends](https://forum.azimuthproject.org/discussion/2192/lecture-33-chapter-2-tying-up-loose-ends/p1).

so if you don't remember that it might help to look at it.

You're definitely on the right track! Perhaps what's confusing you is this. These pictures:

are not technically pictures of \\(\mathbf{Cost}\\)-categories; they are pictures of \\(\mathbf{Cost}\\)-weighted graphs, a concept I explained in Lecture 33. But we can _construct_ a \\(\mathbf{Cost}\\)-category from a \\(\mathbf{Cost}\\)-weighted graph, by _defining_ \\(\mathcal{C}(x,y)\\) to be the price of the least expensive route from \\(x\\) to \\(y\\), i.e. the infimum over all edge paths from \\(x\\) to \\(y\\) of the sum of the costs labelling these edges.

So, my objection to what you said is only a technical one: we are not using the rules for composition in a \\(\mathbf{Cost}\\)-category to deduce \\(\mathcal{C}(S,N) = 4\\); rather, we are using a specific recipe to construct a \\(\mathbf{Cost}\\)-category from a \\(\mathbf{Cost}\\)-weighted graph, and this recipe gives us \\(\mathcal{C}(S,N) = 4\\).

> Also, it seems to me that \\(\mathcal{V}\\)-profunctors are a way of gluing together two \\(\mathcal{V}\\)-categories such that the result is a \\(\mathcal{V}\\)-category.

This is certainly one thing one can do with a \\(\mathcal{V}\\)-profunctor! Given a \\(\mathcal{V}\\)-profunctor \\(\Psi : \mathcal{X} \nrightarrow \mathcal{Y}\\), we can glue \\(\mathcal{X}\\) and \\(\mathcal{Y}\\) together as you're imagining and get a new \\(\mathcal{V}\\)-category called their **collage.**

Simon Willerton explained this in the case \\(\mathcal{V} = \textbf{Bool}\\) back in [comment #7 to Lecture 57](https://forum.azimuthproject.org/discussion/comment/19971/#Comment_19971):

>

> In some sense, in the pictures it would be more sensible to put \\(Y\\) *on top of* \\(X\\). One can think (sometimes usefully, sometimes less usefully) of a feasability relation (or profunctor) between preorders as giving rise to a new preorder, known as the **collage**. The underlying set of the collage is the disjoint union \\(X \sqcup Y\\) and on both \\(X\\) and \\(Y\\) the new preorder restricts to the preorders you started with, but for all \\(x \in X\\) and \\(y \in Y\\) you have \\(y\not\le x\\) whilst \\(x\le y\, \iff\ \Phi(x,y)\\).

> The pictures drawn with the blue arrows are the Hasse diagrams of these collages. Nothing in \\(Y\\) ever comes before anything in \\(X\\) in this preorder, so that's why it would make sense to draw \\(Y\\) on top of \\(X\\). I'll leave it for someone to produce such a version of the examples above :-).

> I think I am not understanding something correctly about \\(\mathbf{Cost}\\)-categories. Isn't the composition rule for \\(\mathbf{Cost}\\)-categories like this: If we have objects \\(N,W,S\\), then the composition rule is \\(\mathcal{C}(S,W) + \mathcal{C}(W,N) \ge \mathcal{C}(S,N)\\) so \\(4 \ge \mathcal{C}(S,N)\\). There's another path between \\(S\\) and \\(N\\) via \\(E\\), but it gives \\(7 \ge \mathcal{C}(S,N)\\). Does this mean that \\(\mathcal{C}(S,N) = 4\\) since it's the meet of compositions that give \\(\mathcal{C}(S,N)\\)?

I explained this stuff a bit here:

* [Lecture 33 - Chapter 2: Tying up loose ends](https://forum.azimuthproject.org/discussion/2192/lecture-33-chapter-2-tying-up-loose-ends/p1).

so if you don't remember that it might help to look at it.

You're definitely on the right track! Perhaps what's confusing you is this. These pictures:

are not technically pictures of \\(\mathbf{Cost}\\)-categories; they are pictures of \\(\mathbf{Cost}\\)-weighted graphs, a concept I explained in Lecture 33. But we can _construct_ a \\(\mathbf{Cost}\\)-category from a \\(\mathbf{Cost}\\)-weighted graph, by _defining_ \\(\mathcal{C}(x,y)\\) to be the price of the least expensive route from \\(x\\) to \\(y\\), i.e. the infimum over all edge paths from \\(x\\) to \\(y\\) of the sum of the costs labelling these edges.

So, my objection to what you said is only a technical one: we are not using the rules for composition in a \\(\mathbf{Cost}\\)-category to deduce \\(\mathcal{C}(S,N) = 4\\); rather, we are using a specific recipe to construct a \\(\mathbf{Cost}\\)-category from a \\(\mathbf{Cost}\\)-weighted graph, and this recipe gives us \\(\mathcal{C}(S,N) = 4\\).

> Also, it seems to me that \\(\mathcal{V}\\)-profunctors are a way of gluing together two \\(\mathcal{V}\\)-categories such that the result is a \\(\mathcal{V}\\)-category.

This is certainly one thing one can do with a \\(\mathcal{V}\\)-profunctor! Given a \\(\mathcal{V}\\)-profunctor \\(\Psi : \mathcal{X} \nrightarrow \mathcal{Y}\\), we can glue \\(\mathcal{X}\\) and \\(\mathcal{Y}\\) together as you're imagining and get a new \\(\mathcal{V}\\)-category called their **collage.**

Simon Willerton explained this in the case \\(\mathcal{V} = \textbf{Bool}\\) back in [comment #7 to Lecture 57](https://forum.azimuthproject.org/discussion/comment/19971/#Comment_19971):

>

> In some sense, in the pictures it would be more sensible to put \\(Y\\) *on top of* \\(X\\). One can think (sometimes usefully, sometimes less usefully) of a feasability relation (or profunctor) between preorders as giving rise to a new preorder, known as the **collage**. The underlying set of the collage is the disjoint union \\(X \sqcup Y\\) and on both \\(X\\) and \\(Y\\) the new preorder restricts to the preorders you started with, but for all \\(x \in X\\) and \\(y \in Y\\) you have \\(y\not\le x\\) whilst \\(x\le y\, \iff\ \Phi(x,y)\\).

> The pictures drawn with the blue arrows are the Hasse diagrams of these collages. Nothing in \\(Y\\) ever comes before anything in \\(X\\) in this preorder, so that's why it would make sense to draw \\(Y\\) on top of \\(X\\). I'll leave it for someone to produce such a version of the examples above :-).