# 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!

• Use the Yoneda lemma. These two products both have the same universal property: a map into either of them is the same thing as a triple of a map into $A$, a map into $B$, and a map into $C$. – Qiaochu Yuan Dec 14 '15 at 23:15
• I'm afraid I'm not at a level to understand (Wikipedia's version of) that lemma just yet, hence I was hoping for a more 'brute force' solution. – Alexander Heyes Dec 14 '15 at 23:49
• I don't see the relevance of the Yoneda lemma here, but the rest of Qiacho Yuan's comment captures the gist of the answer: after shuffling some brackets around $A \times B \times C$, $(A \times B) \times C$ and $A\times (B \times C)$ are all characterised by essentially the same universal property. – Rob Arthan Dec 15 '15 at 0:04
• Right, I think Awodey mentions that as well. I'm not clear on the role of $A\times B\tiems C$ - do we define it to be one of these, define it as its own thing, and if so, do we need to assume its existence over and above the existence of the other two products? – Alexander Heyes Dec 15 '15 at 0:09
• If you're willing to work through how to explicitly show that, I'd really appreciate it as an answer @Rob Arthan – Alexander Heyes Dec 15 '15 at 3:10

## 1 Answer

You can certainly construct a direct proof, as you tried, but I believe the simplest way to show it is this:

1. 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 $(A × B × C, (p_A, p_B, p_C))$.

2. 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.

3. 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, $(A × B) × C ≅ A × (B × C)$.

• Thanks very much, this is a really thorough answer! I realised last night any attempt to salvage my proof had to go through something equivalent to showing $A\times(B\times C)$ satisfied the UMP for $A\times B\times C(p_A, p_B, p_C))$ anyway. Thanks also for giving me a look at how to apply the Yoneda Lemma! Really interesting, looking forward to learning these techniques. – Alexander Heyes Dec 15 '15 at 20:17