I wrote approximately:

> **Puzzle 142.** Prove the associative law: for any natural transformations \\(\alpha: F \Rightarrow G\\), \\(\beta: G \Rightarrow H\\) and \\(\gamma: H \Rightarrow I \\) we have \\((\gamma \beta) \alpha = \gamma (\beta \alpha)\\).

Dan wrote approximately:

> In order to prove that the two composite natural transformations are equal, we show that all their components are equal, that is \\(((\gamma\circ \beta) \circ \alpha)_x = (\gamma \circ (\beta\circ \alpha))_x\\) for all objects \\(x \in \mathcal{C}\\):

> \\[
((\gamma\circ \beta)\circ \alpha)_x =
(\gamma\circ \beta)_x\circ\alpha_x =
(\gamma_x\circ\beta_x)\circ\alpha_x =
\gamma_x\circ(\beta_x\circ\alpha_x) =
\gamma_x\circ(\beta\circ \alpha)_x =
(\gamma(\beta\circ \alpha))_x.
\\]

Yes! This is one of those nice arguments one so often meets in category theory: the associativity in \\(\mathcal{D}\\) gives rise to the associativity \\(\mathcal{D}^\mathcal{C}\\), since composition in \\(\mathcal{D}\\) gives rise to composition in \\(\mathcal{D}^\mathcal{C}\\).