> 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\\)

Of course, I think I forgot about that.

> 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):

First, I like the word "collage", it means "gluing" in French.

I am trying to see if there's a way to take that intuition and turn it into a general formula for combining \\(\mathcal{V}\\)-categories given a \\(\mathcal{V}\\)-profunctor. But when \\(\mathcal{V} = \mathbf{Cost}\\), we are only required to have \\( \mathcal{C}(x,y) + \mathcal{C}(y,z) \ge \mathcal{C}(x,z)\\) meaning that for \\(a \in \mathrm{ob}\left(\mathcal{X}\right)\\) and \\(b \in \mathrm{ob}\left(\mathcal{Y}\right)\\), the morphism (am I allowed to say morphism for enriched categories?) \\(\mathcal{Z}(a,b)\\) is only required to be less than the least expensive route from \\(a\\) to \\(b\\). We can choose that the least expensive route defines \\(\mathcal{Z}(a,b)\\) (like we do for building \\(\mathbf{Cost}\\)-categories from weighted graphs).

I don't know, maybe it depends on the answer to **Puzzle 185**.

> 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\\)

Of course, I think I forgot about that.

> 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):

First, I like the word "collage", it means "gluing" in French.

I am trying to see if there's a way to take that intuition and turn it into a general formula for combining \\(\mathcal{V}\\)-categories given a \\(\mathcal{V}\\)-profunctor. But when \\(\mathcal{V} = \mathbf{Cost}\\), we are only required to have \\( \mathcal{C}(x,y) + \mathcal{C}(y,z) \ge \mathcal{C}(x,z)\\) meaning that for \\(a \in \mathrm{ob}\left(\mathcal{X}\right)\\) and \\(b \in \mathrm{ob}\left(\mathcal{Y}\right)\\), the morphism (am I allowed to say morphism for enriched categories?) \\(\mathcal{Z}(a,b)\\) is only required to be less than the least expensive route from \\(a\\) to \\(b\\). We can choose that the least expensive route defines \\(\mathcal{Z}(a,b)\\) (like we do for building \\(\mathbf{Cost}\\)-categories from weighted graphs).

I don't know, maybe it depends on the answer to **Puzzle 185**.