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}\$$.