Hi -

> 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).

Yes; we can actually reduce the problem of constructing the collage of a \\(\mathcal{V}\\)-profunctor to the problem of getting a \\(\mathcal{V}\\)-category from a \\(\mathcal{V}\\)-weighted graph, which I'll assume you know how to solve.

Notice that any \\(\mathcal{V}\\)-category \\(\mathcal{C}\\) gives a \\(\mathcal{V}\\)-weighted graph with an edge from any object \\(c\\) to any object \\(c'\\), weighted by \\(\mathcal{C}(c,c')\\). If we take this \\(\mathcal{V}\\)-weighted graph and use our recipe for turning it into a \\(\mathcal{V}\\)-category, we get \\(\mathcal{C}\\) back!

Next notice that any \\(\mathcal{V}\\)-profunctor \\(\Phi : \mathcal{C} \nrightarrow \mathcal{D}\\) gives a \\(\mathcal{V}\\)-weighted graph! This has:

* an edge from any object \\(c\\) to any object \\(c'\\) of \\(\mathcal{C}\\), weighted by \\(\mathcal{C}(c,c')\\).

* an edge from any object \\(d\\) to any object \\(d'\\) of \\(\mathcal{D}\\), weighted by \\(\mathcal{D}(d,d')\\).

* an edge from any object \\(c\\) of \\(\mathcal{C}\\) to any object \\(d\\) of \\(\mathcal{D}\\), weighted by \\(\Phi(c,d)\\).

If we take this \\(\mathcal{V}\\)-weighted graph and use our recipe for turning it into a \\(\mathcal{V}\\)-category, we get the **collage** of \\(\Phi\\).

By the way, when you have a \\(\mathcal{V}\\)-enriched category \\(\mathcal{Z}\\) and two objects \\(a\\) and \\(b\\) in it, we call \\(\mathcal{Z}(a,b) \in \mathcal{V}\\) the **hom-object** from \\(a\\) to \\(b\\). This is like when we have an ordinary category \\(\mathcal{Z}\\): then \\(\mathcal{Z}(a,b)\\) is a set, called the **hom-set** from \\(a\\) to \\(b\\) and often written \\(\text{hom}(a,b)\\). In the enriched case \\(\mathcal{Z}(a,b) \in \mathcal{V}\\) is not analogous to a morphism; it's analogous to a set of morphisms.