> 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}\\)?

Okay, I have thought about this a little and here's what I've come up with:

**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

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


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

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**... :(