Due to skeletality, we need to show for all inputs that \(\phi \leq U_P;\phi\) and \(U_P;\phi \leq \phi\) (the second equality to show is done similarly).
Forward direction
\(\phi(p,q) = I \otimes \phi(p,q)\)
due to unitality of \(I\) in a symmetric monoidal preorder.
\(\leq P(p,p) \otimes \phi(p,q)\)
This is because \(\forall p \in P:\) \(I \leq P(p,p)\) (a constraint of a \(\mathcal{V}\) category), the reflexivity of \(\leq\) for \(\phi(p,q)\), and the monotonicity of \(\otimes\).
\(\leq \underset{p' \in P}\bigvee(P(p,p') \otimes \phi(p',q))\)
The join is a least upper bound, and the LHS is an element of the set being joined over (the case where \(p=p'\)).
\(= (U_P;\phi)(p,q)\)
This is the profunctor composition formula, subtituting in the unit profunctor definition explicitly.
Reverse direction
Need to show \(\underset{p' \in P}\bigvee(P(p,p')\otimes \phi(p',q)) \leq \phi(p,q)\)
Show that this property holds for each \(p' \in P\) - given the join is a least upper bound, it will also be less than or equal to \(\phi(p,q)\)
\(P(p,p')\otimes\phi(p',q) = P(p,p')\otimes\phi(p',q)\otimes I\)
due to unitality of \(I\) in a symmetric monoidal preorder.
\(\leq P(p,p')\otimes \phi(p',q)\otimes Q(q,q)\)
\(\forall p:\) \(I \leq P(p,p)\) (a constraint of a \(\mathcal{V}\) category) and the monotonicity of \(\otimes\).
\(\leq\phi(p,q)\)
This was shown in Exercise 4.9
The unit profunctor is unital, i.e. for any profunctor \(P \overset{\phi}\nrightarrow Q\): \(U_P;\phi = \phi = \phi; U_Q\)