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.