> 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**.