Matthew and Michael - you made a lot of progress on Puzzle 168, but I have the feeling I didn't clearly explain the point of the puzzle. The main point of Puzzle 168 was not to show that this square commutes:

& & \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'} & & \\\\

As Michael notes, it obviously commutes, because when we have an adjunction we get a _natural_ isomorphism

\[ \alpha : \mathcal{B}(F(-), -) \to \mathcal{A}(-, G(-)) \]

Naturality is all about commutative squares.

The main point of the puzzle was to get a better feeling for what this naturality "really means". I wrote:

> To see what this means concretely, take a guy \\(h \in \mathcal{B}(F(a),b) \\), meaning a morphism \\(h : F(a) \to b\\), and send it around the square both ways: down and then across, or across and then down. You get two morphisms from \\( a' \\) to \\(G(b')\\); **what are they?** Since the square commutes these must be equal.

As Michael sort of noted, these two morphisms are

$$\alpha_{a',b'}(g \circ h \circ F(f))$$


$$G(g) \circ \alpha_{a,b}(h) \circ f$$

so what the naturality "really means" is that

$$\alpha_{a',b'}(g \circ h \circ F(f)) = G(g) \circ \alpha_{a,b}(h) \circ f$$

However, now we have to figure what _this_ equation "really means"! At first glance it looks pretty intimidating.

Perhaps the best approach would be to look at an example - some example of an adjunction that we understand very well. Then this equation will wind up saying something really nice and obvious, which make us feel "_of course_ we want this equation to be true."