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).