Why is the Associativity Isomorphism Natural in a Category Closed Under Finite Products? From Categories for the Working Mathematician:

Question: I am concerned only with the associativity isomorphism $\alpha$, and understand everything in this theorem except the very last claim, that $\alpha$ is natural in $a$, $b$, and $c$ (which the author doesn't justify). Why is $\alpha$ natural?
I'm assuming the functors involved are each from $C \times C \times C$ to $C$ with $F(a,b,c) \mapsto a \times (b \times c)$ and $G(a,b,c) \mapsto (a \times b) \times c$.
 A: Here is a sketch of the argument: you can explicitly compute $\alpha=\alpha^{-1}_2\circ \alpha_1$ by showing using UMP of the product(s) that there are isos  $a\times (b\times c)\overset{\alpha_1}{\rightarrow}a\times b\times c$ and $(a\times b)\times c\overset{\alpha_2}{\rightarrow}a\times b\times c.$ Naturality will follow by chasing the appropriate square.
For example, with $F=-\times (b\times c),\ G=(-\times b)\times c,$ and $f:a\to a',$  you need to check that the following square commutes:
$\require{AMScd}
\begin{CD}
a\times(b\times c) @>{\alpha_a}>> (a\times b)\times c\\
@VVV @VVV \\
a'\times(b\times c) @>{\alpha_{a'}}>> (a'\times b)\times c
\end{CD}$
That is, that $\alpha_{a'}\circ Ff=Gf\circ \alpha_{a'}.$ Explicitly, we require $\alpha_{a'}\circ (f\times (b\times c))=((f\times b)\times c)\circ \alpha_{a}.$
This result follows by construction of $\alpha$ and the definition of $F$ and $G$ on arrows. 
Similarly, you can check naturality in the other arguments by considering the appropriate functors and squares. Naturality in the three arguments then follows by the functoriality of the trifunctors $(-\times-)\times -$ and $-\times(-\times -).$
