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}\$$.)