This is an exercise (I believe) to practice methods using the functoriality of limits. The exercise is as follows:
For any pair of objects $X,Y,Z$ in a category $\mathsf{C}$ with binary products, there is a unique natural isomorphism $(X \times Y) \times Z \cong X \times (Y \times Z)$ which commutes with the projections.
I should be able to do this "inside $C$" by using the universal property of the product, but I want to use the machinery we've developed in the chapter (functoriality of limits). So to show this is true, let $F,G : \bullet\bullet \to \mathsf{C}$ be the functors with images matching the factors above.
EDIT: for full clarity, the definition of $F,G : \bullet\bullet \to \mathsf{C}$ are: $$\begin{cases} F0 = (X \times Y) \\ F1 = Z \end{cases} \hphantom{1234} \begin{cases} G0 = X \\ G1 = (Y \times Z) \end{cases}$$
By picking their limits as the products above, a natural isomorphism $F \cong G$ induces a naturally-defined isomorphism $(X \times Y) \times Z \cong X \times (Y \times Z)$.
This is where I have some confusion. A natural transformation $\alpha: F \Rightarrow G$ would consist of two morphisms $\alpha_0 : X \times Y \to X$ and $\alpha_1 : Z \to Y \times Z$. We can easily take $\alpha_0 := \pi_X$, but I am at a loss as to what $\alpha_1$ should be. A morphism into $Y \times Z$ is totally determined by its projections, but are we guaranteed that there is even a morphism $Z \to Y$?
If there is no such morphism, there certainly isn't a morphism $Z \to Y \times Z$, which makes constructing a natural transformation $F \Rightarrow G$ impossible. The fact that there certainly is a natural isomorphism $(X \times Y) \times Z \to X \times (Y \times Z)$ implies that my fear is not true, but I have no idea how to show it.
If I have made some error in my reasoning, please point it out, but for now I am simply looking for clarification on how to construct $\alpha$, and where my concerns break down.
A morphism into a product breaks up into a pair of morphisms into the factors, but a morphism from a product doesn't necessarily break up like that. So you are not looking for morphisms $\alpha_0: X \times Y \to X$ and $\alpha_1: Z \to Y \times Z$. You can only break up the target. So you are looking for morphisms $\alpha_0: (X \times Y) \times Z \to X$ and $\alpha_1: (X \times Y) \times Z \to Y \times Z$. Now can you see what the morphisms are?