> Moreover, thinking about Puzzle 207, I have been wondering:
>
> 3. Is \$$\hat{F} \hat{G} = \widehat{F \circ G}\$$?
> 4. Is \$$\check{F} \check{G} = \overline{F \circ G}\$$?

**Lemma**. For two functors \$$G: \mathcal{X} \to \mathcal{Y}\$$ and \$$F: \mathcal{Y} \to \mathcal{Z}\$$, if \$$F \dashv H\$$ then \$$\hat{F} \hat{G} = \widehat{F \circ G}\$$

**Proof**. First, expanding \$$\hat{F} \hat{G}\$$ gives us

$\hat{F} \hat{G}(x,z) = \bigvee_{y} \mathcal{Y}(G(x),y) \otimes \mathcal{Z}(F(y),z)$

However, since \$$F \dashv H\$$ then \$$\mathcal{Z}(F(y),z) = \mathcal{Y}(y,H(z))\$$. Since \$$\mathcal{Y}\$$ is a \$$\mathcal{V}\$$-enriched category we have \$$\mathcal{Y}(G(x),y) \otimes \mathcal{Y}(y,H(z)) \leq \mathcal{Y}(G(x),H(z)) \$$ for all \$$y\$$. Hence:

$\bigvee_{y} \mathcal{Y}(G(x),y) \otimes \mathcal{Z}(F(y),z) \leq \mathcal{Y}(G(x),H(z)).$

Moreover, we know that since \$$I \leq \mathcal{Y}(H(z),H(z))\$$ then \$$\mathcal{Y}(G(x),H(z)) \leq \mathcal{Y}(G(x),H(z)) \otimes \mathcal{Y}(H(z),H(z))\$$, hence

$\mathcal{Y}(G(x),H(z)) \leq \bigvee_{y}\mathcal{Y}(G(x),y) \otimes \mathcal{Y}(y,H(z))$

Thus

$\bigvee_{y}\mathcal{Y}(G(x),y) \otimes \mathcal{Y}(y,H(z)) = \mathcal{Y}(G(x),H(z))$

Since \$$F \dashv H\$$, then \$$\mathcal{Y}(G(x),H(z)) = \mathcal{Y}(F(G(x)),z) = \widehat{F \circ G}(x,z)\$$. Hence

$\bigvee_{y}\mathcal{Y}(G(x),y) \otimes \mathcal{Y}(y,H(z)) = \widehat{F \circ G}(x,z)$

Which suffices to prove \$$\hat{F} \hat{G} = \widehat{F \circ G}. \quad\quad \blacksquare\$$

I can't make this lemma lemma work for conjoints.

I don't think there is a companion that answers **Puzzle 206**, so the lemma helps for **Puzzle 207**... :(