Actually, why not work this out.

Consider this row of profunctors:

\[ A \cong A \times \textbf{1} \nrightarrow A \times (A^\text{op} \times A) \cong (A \times A^\text{op}) \times A \nrightarrow \textbf{1} \times A \cong A \]

Given any objects \\(x, y\\) in \\(A\\) the value of the composite profunctor at \\((x, y)\\) is

\[ \bigvee\big(\Phi(x, (z, z', z'')) \otimes \Psi((z, z', z''), y)\big) \]

where we're joining over triples \\((z, z', z'')\\) in \\(A \times A^\text{op} \times A\\) and

\[ \Phi(x, (z, z', z'')) = \text{id}_A(x, z) \otimes \cap_A(z', z'') = \text{hom}_A(x, z) \otimes \text{hom}_A(z', z'') \]

\[ \Psi((z, z', z''), y) = \cup _{A^\mathrm{op}}(z, z') \otimes \text{id}_A(z'', y) = \text{hom} _{A^\mathrm{op}}(z', z) \otimes \text{hom}_A(z'', y) \]

(Note that we're using \\(\cup _{A^\text{op}} : A \times A^\text{op} \nrightarrow \textbf{1}\\) not \\(\cup_A : A^\text{op} \times A \nrightarrow \textbf{1}\\) here!)

So our join is

\[ \bigvee\big(\text{hom}_A(x, z) \otimes \text{hom}_A(z', z'') \otimes \text{hom} _{A^\text{op}}(z', z) \otimes \text{hom}_A(z'', y)\big) \]

Reorder the middle two terms and switch round \\(\text{hom} _{A^\text{op}}\\) to get

\[ \bigvee\big(\text{hom}_A(x, z) \otimes \text{hom}_A(z, z') \otimes \text{hom}_A(z', z'') \otimes \text{hom}_A(z'', y)\big) \]

Now for any triple \\((z, z', z'')\\) this is \\(\leq \text{hom}(x, y)\\), so the join must be \\(\leq \text{hom}(x, y)\\).

But for the triple \\((x, x, x)\\) we get the "summand" \\(\text{hom}(x, x) \otimes \text{hom}(x, x) \otimes \text{hom}(x, x) \otimes \text{hom}(x, y) \\geq I \otimes I \otimes I \otimes \text{hom}(x, y) = \text{hom}(x, y)\\)

So the join must also be \\(\geq \text{hom}(x, y)\\).

Hence the composite profunctor equals the identity profunctor. QED.