And on a similar note:

Given a profunctor \\(\Phi : \mathcal{X} \nrightarrow \mathcal{Y}\\), define \\(\Phi^\text{op} : \mathcal{Y}^\text{op} \nrightarrow \mathcal{X}^\text{op}\\) by \\(\Phi^\text{op}(y, x) = \Phi(x, y)\\). It's easy to prove \\(\Phi^\text{op}\\) is indeed a profunctor.

Note that \\((\Phi^\text{op})^\text{op} = \Phi\\), \\((\Psi\circ\Phi)^\text{op} = \Phi^\text{op} \circ \Psi^\text{op}\\) and \\((\Phi\otimes\Psi)^\text{op} = \Phi^\text{op} \otimes \Psi^\text{op}\\).

More interesting is this (which I'll leave as a puzzle to prove):

\[\Phi^\text{op} = (\cap\_\mathcal{Y} \otimes 1_{\mathcal{X}^\text{op}})\circ(1_{\mathcal{Y}^\text{op}} \otimes \Phi \otimes 1_{\mathcal{X}^\text{op}})\circ(1_{\mathcal{Y}^\text{op}} \otimes \cup\_ \mathcal{X})\]

Or in pictures: