OK, recapitulating the composition of conjoints rule one last time using opposite profunctors...

We've shown that \\(\check{F}(y, z) = \widehat{F^\text{op}}(z, y)\\). This is equivalent to \\((\check{F})^\text{op} = \widehat{F^\text{op}}\\).

From which we get:

\[(\check{GF})^\text{op} = \widehat{(GF)^\text{op}} = \widehat{G^\text{op}F^\text{op}} = \widehat{G^\text{op}}\circ\widehat{F^\text{op}} = (\check{G})^\text{op} \circ (\check{F})^\text{op} = (\check{F}\circ\check{G})^\text{op}\]

Hence \\(\check{GF} = \check{F}\circ\check{G}\\). QED