I wrote:

> I seem to recall Matthew had trouble proving the composite of conjoints is the conjoint of the composite:
>
> \[ \check{FG} \stackrel{?}{=} \check{F} \check{G} .\]
>
> I don't even know if this is true! It would seem strangely asymmetrical for this to be false.

Matthew wrote:

> It is asymmetrical! I don't think this is true in general.

> Instead, we have

> \[ \check{G F} = \check{F} \check{G} \]

Oh, duh! Right!



This is not asymmetrical in any 'bad' sense; it's just the natural thing. Let \\(\mathbf{Cat}_{\mathcal{V}}\\) be the category with

* \\(\mathcal{V}\\)-enriched categories as objects

* \\(\mathcal{V}\\)-enriched functors as morphisms

and let \\(\mathbf{Prof}_{\mathcal{V}}\\) be the category with

* \\(\mathcal{V}\\)-enriched categories as objects

* \\(\mathcal{V}\\)-enriched profunctors as morphisms.

Then taking companions gives an inclusion of \\(\mathbf{Cat}\_{\mathcal{V}}\\) in \\(\mathbf{Prof}\_{\mathcal{V}}\\) while taking conjoints gives an inclusion of \\(\mathbf{Cat}\_{\mathcal{V}}^{\text{op}}\\) in \\(\mathbf{Prof}_{\mathcal{V}}\\).

In other words, less technically speaking, enriched profunctors are a generalization of enriched functors where an enriched functor going _either way_ between enriched categories \\(\mathcal{X}\\) and \\(\mathcal{Y}\\) gives a profunctor going from \\(\mathcal{X}\\) to \\(\mathcal{Y}\\). The backwards way has got to be contravariant.