In category theory, how to make sense of $(X \times Y) \times Z$ in a way that makes use of the natural projections associated with $X \times Y$? Define that a binary diagram in a category $\mathcal{C}$ is a functor $2 \rightarrow \mathcal{C},$ where $2$ is the discrete category with two objects.
Now if $X$ and $Y$ are objects of a category $\mathcal{C}$ admitting binary products, we can view $X \times Y$ as an object of a new category $\mathrm{Bin}(\mathcal{C})$ whose objects are the cones to binary diagrams in $\mathcal{C},$ and whose arrows are the usual cone homomorphisms. The problem with this point of view is that $(X \times Y) \times Z$ no longer makes sense, since $Z$ is an object of the category $\mathcal{C},$ while $X \times Y$ is an object of a different category $\mathrm{Bin}(\mathcal{C}).$
Okay that's no big deal, we can just view $X \times Y$ as an object of $\mathcal{C}$ when computing $(X \times Y) \times Z.$ But if we do this, then our computation will necessarily disregard the natural projections that $X \times Y$ comes equipped with. 
Is there a sensible way of defining $(X \times Y) \times Z$ such that those natural projections are actually taken into account?
 A: Strictly speaking you are right: $X \times Y$ is not just an object of $C$, but rather it is an object $P$ which comes equipped with two projection morphisms to $X$ resp. $Y$:
$X \times Y = (X \longleftarrow P \longrightarrow Y)$
But it is a common abuse of notation to simply write $P=X \times Y$. Then, of course, $(X \times Y) \times Z$ makes sense (and is isomorphic to $X \times Y \times Z$). If you are not happy with this abuse of notation, you should write $P \times Z$.
A: Here's an idea I just had. I'll post it here for critique (but please be gentle!)
We can generalize the notion of a binary product of objects to the notion of a binary product of diagrams.
First, we need the concept of a "disjoint union" of two diagrams. If $\mathcal{C}$ is a category and $D,D'$ are diagrams $J,J' \rightarrow \mathcal{C}$, write $D \boxplus D'$ for the diagram whose shape is the "disjoint union" of $J$ and $J',$ and which agrees with $D$ on the $J$ portion and which agrees with $D'$ on the $J'$ portion. Then we can define that $$D \times D' = \mathrm{lim}(D \boxplus D')$$
so long as the limit exists.
Importantly, the limit needs to be understood as a cone, and not just an object of $\mathcal{C}.$ So $D \times D'$ is a cone, if it exists.
Now every cone $K$ induces a diagram in a natural way (include the limit object and the natural projections). Lets denote this $\delta(K).$
Thus, if $D_0,D_1$ and $D_2$ are diagrams, we can interpret $(D_0 \times D_1) \times D_2$ as shorthand for $\delta(D_0 \times D_1) \times D_2.$
Finally, if $X$ is an object of $\mathcal{C},$ lets write $\{X\}$ for the corresponding diagram with shape $1.$ Under these definitions, we can interpret the left hand side of the following equation as shorthand for the right hand side.
$$(X \times Y) \times Z = \delta(\{X\} \times \{Y\}) \times \{Z\}$$
So that's the idea, I don't know if it really goes anywhere, but comments and suggestions would be most welcome.
One interesting observation is that if we redefine $\delta$ such that $\delta(K)$ is still the diagram associated with the cone $K$, but this time not including the limit object or the natural projections, then we essentially get the usual interpretation of $(X \times Y) \times Z.$
A: 
Okay that's no big deal, we can just view $X \times Y$ as an object of $\mathcal{C}$ when computing $(X \times Y) \times Z.$ But if we do this, then our computation will necessarily disregard the natural projections that $X \times Y$ comes equipped with. 

Why? $(X \times Y) \times Z$ itself comes with two natural projections, a projection to $X \times Y$ and a projection to $Z$. Once you have the projection to $X \times Y$, you can compose it with the projections $X \times Y \to X, X \times Y \to Y$ and get three projections to $X, Y, Z$, and then $(X \times Y) \times Z$ together with these three projections satisfies the correct universal property. 
Another way of saying this is that, if you take the perspective that $X \times Y$ is a representing object for the functor $\text{Hom}(-, X) \times \text{Hom}(-, Y)$, then $(X \times Y) \times Z$ is a representing object for the functor
$$\text{Hom}(-, X \times Y) \times \text{Hom}(-, Z)$$
which is in turn naturally isomorphic to
$$\text{Hom}(-, X) \times \text{Hom}(-, Y) \times \text{Hom}(-, Z)$$
as expected. 
