Michael Hong wrote:
>**Puzzle 168**
>
> So the 2 morphisms are :
>
> $$\alpha_{a',b'}(F(f) \circ h \circ g)$$
> $$f \circ \alpha_{a,b}(h) \circ G(g)$$

Are you sure? I was thinking that if \\(x \in \mathcal{B}(F(a),b)\\) and \\(y \in \mathcal{A}(a,G(b))\\) then:

\[ \begin{align}
\mathcal{B}(F(f),g)(x) & = g \circ x \circ F(f) \\\\
\mathcal{A}(f,G(g))(y) & = G(g) \circ y \circ f
\end{align}
\]

Here is my go at a proof:

> **Puzzle 168.** Suppose \\(F\\) is a left adjoint of \\(G\\), as above. If \\(a \in \mathrm{Ob}(\mathcal{A})\\) and \\(b \in \mathrm{Ob}(\mathcal{B})\\) we get a bijection
>
> \[ \alpha_{a,b} : \mathcal{B}(F(a),b) \to \mathcal{A}(a,G(b)). \]
>
> Similarly, if \\(a' \in \mathrm{Ob}(\mathcal{A})\\) and \\(b' \in \mathrm{Ob}(\mathcal{B})\\) we get a bijection
>
> \[ \alpha_{a',b'} : \mathcal{B}(F(a'),b') \to \mathcal{A}(a',G(b')). \]
>
> Show that if \\(f: a' \to a\\) and \\(g: b \to b'\\) we get a commutative square
>
> \[
\begin{matrix}
& & \alpha_{a,b} & & \\\\
& \mathcal{B}(F(a),b) & \rightarrow & \mathcal{A}(a,G(b)) &\\\\
\mathcal{B}(F(f),g) & \downarrow & & \downarrow & \mathcal{A}(f,G(g)) \\\\
& \mathcal{B}(F(a'),b') & \rightarrow & \mathcal{A}(a',G(b')) &\\\\
& & \alpha_{a',b'} & & \\\\
\end{matrix}
\]

Assume \\(x \in \mathcal{B}(F(a),b)\\). To show the diagram commutes, we must show:

\[
\alpha_{a',b'}(\mathcal{B}(F(f),g)(x)) = \mathcal{A}(f,G(g))(\alpha_{a,b}(x))
\]

In other words we must show

\[
\alpha_{a',b'} \circ \mathcal{B}(F(f),g) = \mathcal{A}(f,G(g)) \circ \alpha_{a,b}
\]

However, this is exactly the [naturality condition](https://en.wikipedia.org/wiki/Natural_transformation) on \\(\alpha\\) from the [hom-set definition of an adjunction](https://en.wikipedia.org/wiki/Adjoint_functors#Hom-set_adjunction).