Triple Products are Isomorphic I am currently working through Awodey's Introduction to Category Theory, and I'm learning how to move around complicated diagrams. 
I want to show that $A\times(B\times C)\cong(A\times B)\times C$; but I want to do so quite explicitly, preferably by constructing the isomorphism. This is so I understand the different techniques involved in moving around such diagrams, and how to best use Universal Mapping Properties. In some ways, I would prefer a less-elegant proof!
Details and Working:
I am using Awodey's definition for products via the Universal Mapping Property: 
A product $A\times B$ is an object with two arrows $p_A:A\times B\rightarrow A$ (and similarly for $p_B$) such that, given any other object $X$ with a pair of arrows $x_A:X\rightarrow A$ (and similarly $x_B$), there exists a unique arrow $u:X\rightarrow A\times B$ such that the obvious diagram commutes ($p_A\circ u=x_A$ and similarly $p_B$).
So far I have constructed various unique arrows 
$$f:A\times(B\times C)\rightarrow A\times B$$ $$g:(A\times B)\times C\rightarrow B\times C$$ that make the following diagram commute as follows:
$$af=\alpha\qquad bf=\beta\delta$$
$$\gamma g=c\qquad \beta g=bd$$

Then using $f, g$, I construct
 $$i:A\times (B\times C)\rightarrow (A\times B)\times C$$ $$j:(A\times B)\times C\rightarrow A\times (B\times C)$$
satisfying
$$di=f\qquad ci=\gamma\delta$$
$$\delta j=g\qquad \alpha j=ad$$
These should then be the required iso's, but I can't seem to prove that they have to be left and right inverses. For example, I have shown 
$$\alpha ji=adi=af=\alpha$$
I know that if I can show $$\delta ji=\delta$$ then by uniqueness of $1_{(A\times B)\times C}$ as the only map that should satisfy those two relations, 
$$ji=1_{(A\times B)\times C}$$
but I can't seem to get there!
 A: You can certainly construct a direct proof, as you tried, but I believe the simplest way to show it is this:


*

*Define what a product of 3 (or $n$, or $κ$) objects is. This is just a minor generalization of the binary case.
To be explicit, let's write the data of a product of $A$, $B$, $C$ as $(A × B × C, (p_A, p_B, p_C))$.

*Show that for any two products $(P, (p_A, p_B, p_C))$ and $(Q, (q_A, q_B, q_C))$ of $A$, $B$ and $C$, $P$ and $Q$ must be isomorphic.
As a bonus, don't forget that there is in fact a unique isomorphism $f : P → Q$ which commutes with the projection maps, ie. $q_Af = p_A$ etc.
This is best done by noticing that $(A × B × C, (p_A, p_B, p_C))$ is the terminal object in an appropriate category.

*If $(A × B, (p_A, p_B))$ is a product of $A$ and $B$, and $((A × B) × C, (q, q_C))$ of $A × B$ and $C$, show that $((A × B) × C, (qp_A, qp_B, q_C))$ is a product of $A$, $B$ and $C$, ie. that it satisfies its universal property.
Now obviously the same can be done for $A × (B × C)$, so by the previous part, we have $A × (B × C) ≅ (A × B) × C$. And again, don't forget you also have uniqueness in the appropriate sense.
A shorter proof moves everything to $\mathrm{Set}$, where the isomorphism is obvious. To do it properly you need the Yoneda lemma, but to get the gist of it you just need to check that for every object $X$, $\mathrm{Hom}(X, A × B) ≅ \mathrm{Hom}(X, A) × \mathrm{Hom}(X, B)$ -- this is just a (very useful btw.) restatement of the universal property. Anyway, here's the proof:
\begin{align}
\mathrm{Hom}(X, (A × B) × C)) &≅ \mathrm{Hom}(X, A × B) × \mathrm{Hom}(X, C) \\
&≅ (\mathrm{Hom}(X, A) × \mathrm{Hom}(X, B)) × \mathrm{Hom}(X, C) \\
&≅ \mathrm{Hom}(X, A) × (\mathrm{Hom}(X, B) × \mathrm{Hom}(X, C)) \\
&≅ \mathrm{Hom}(X, A) × \mathrm{Hom}(X, B × C) \\
&≅ \mathrm{Hom}(X, A × (B × C))
\end{align}
naturally in X, so by the Yoneda lemma, $(A × B) × C ≅ A × (B × C)$.
