> 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?