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.