Let's work out what \\(\check{F} \circ \check{G}\\) is, where \\(F : \mathcal{Z} \to \mathcal{Y}, G : \mathcal{Y} \to \mathcal{X}\\).

\[(\check{F} \circ \check{G})(x, z) = \bigvee\big(\check{G}(x, y) \otimes \check{F}(y, z)\big) = \bigvee\big(\mathcal{X}(x, Gy) \otimes \mathcal{Y}(y, Fz)\big)\]

Setting \\(y = Fz\\) shows that our join \\(\geq \mathcal{X}(x, GFz) \otimes \mathcal{Y}(Fz, Fz) \geq \mathcal{X}(x, GFz) \otimes I = \mathcal{X}(x, GFz)\\)

For any \\(y\\) we have \\(\mathcal{X}(x, GFz) \geq \mathcal{X}(x, Gy) \otimes \mathcal{X}(Gy, GFz) \geq \mathcal{X}(x, Gy) \otimes \mathcal{Y}(y, Fz)\\), so we have \\(\mathcal{X}(x, GFz) \geq\\) our join.

Hence this join \\(= \mathcal{X}(x, GFz) = \check{GF}(x, z)\\). Therefore \\(\check{F} \circ \check{G} = \check{GF}\\).

(I'm wondering if we could get this result about conjoints directly from the previous one about companions by composing with cup and cap to turn a profunctor \\(\mathcal{X}\nrightarrow\mathcal{Y}\\) into a profunctor \\(\mathcal{Y}^\text{op}\nrightarrow\mathcal{X}^\text{op}\\).)