> For any \\(\mathcal{V}\\)-functor, \\(F\\), define a \\(\mathcal{V}\\)-profunctor \\(\Phi\\) in \\(\mathcal{V}\\) using it's closed structure,
>\[ \Phi := x \multimap F(x)\]
> I believe this defines a \\(\mathcal{V}\\)-profunctor.

I am still not seeing how this works, Keith.

For a \\(\mathcal{V}\\)-functor \\(F : \mathcal{X} \rightarrow \mathcal{Y}\\), we know from the definition that

\[\mathcal{X}(a,b) \leq \mathcal{Y}(F(a),F(b))\]

As far as I know, \\(F\\) acts like a function \\(\mathrm{Obj}(\mathcal{X}) \to \mathrm{Obj}(\mathcal{Y})\\), so we can't use it in an expression like \\(x \multimap F(x)\\) unless \\(\mathrm{Obj}(\mathcal{X}) = \mathrm{Obj}(\mathcal{Y}) = \mathrm{Obj}(\mathcal{V})\\).

Is there something I am missing?