Show that a \(\mathcal{V}\) profunctor is the same as a function \(Ob(\mathcal{X})\times Ob(\mathcal{Y}) \xrightarrow{\phi} V\) such that, \(\forall x,x' \in \mathcal{X}, y,y' \in \mathcal{Y}\), the following holds in \(\mathcal{V}\):
\(\mathcal{X}(x',x)\otimes \phi(x,y) \otimes \mathcal{Y}(y,y') \leq \phi(x',y')\)
A \(\mathcal{V}\) profunctor must be a function satisfying the following constraint, according to the \(\mathcal{V}\) functor definition:
\(Z((x,y),(x',y')) \leq\) \(\mathcal{V}(\phi((x,y)),\phi((x',y')))\)
where \(Z = \mathcal{X}^{op}\times \mathcal{Y}\)
Unpacking the definition of a product \(\mathcal{V}\) category, we obtain
\(\mathcal{X}^{op}(x,x') \otimes \mathcal{Y}(y,y') \leq \mathcal{V}(\phi((x,y)),\phi((x',y')))\)
And applying opposite category definition: \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y') \leq \mathcal{V}(\phi((x,y)),\phi((x',y')))\)
Noting the definition of \(\multimap\) for a \(\mathcal{V}\) category enriched in itself:
\(\mathcal{V}(v,w)=v\multimap w\), so now we have: \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y') \leq \phi((x,y)) \multimap \phi((x',y'))\)
From the constraint of a hom-element of a symmetric monoidal preorder \(\mathcal{V}\), i.e. \(a \leq (v \multimap w)\) iff \((a \otimes v) \leq w\), we see that the first case pattern matches with:
\(a \mapsto\) \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y')\)
\(v \mapsto\) \(\phi((x,y))\)
\(w \mapsto\) \(\phi((x',y'))\)
So using the iff we can rewrite as \((a \otimes v) \leq w\), and use the commutativity of \(\otimes\) to obtain the desired expression.