I claim that a possible product of $a$ and $b$ is:
$\begin{aligned}
(&p = \{ 1, 2, 3, 4\}, \\
&\pi_a = 1 \mapsto \alpha, 2 \mapsto \alpha, 3 \mapsto \beta, 4 \mapsto \beta,\\
&\pi_b = 1 \mapsto \gamma, 2 \mapsto \delta, 3 \mapsto \gamma, 4 \mapsto \delta)
\end{aligned}$
p πa a
1------------*--->{α}
| 2--------─┘ { }
| | 3-------*--->{β}
| | | 4----─┘
+----* |
πb | | |
| *---+
| |
v v
b {γ δ}
Now given a 3-tuple $(q, \pi'_a \in Hom(q, a), \pi'_b \in Hom(q, b))$, we construct
the factorization map $f: q \rightarrow p$, where:
$f: q \rightarrow p; \quad
f(x) \equiv
\begin{cases}
1 & \pi'_a(x) = \alpha \land \pi'_b(x) = \gamma \\
2 & \pi'_a(x) = \alpha \land \pi'_b(x) = \delta \\
3 & \pi'_a(x) = \beta \land \pi'_b(x) = \gamma \\
4 & \pi'_a(x) = \beta \land \pi'_b(x) = \delta \\
\end{cases}$
That is, we build $f(x)$ such that on composing with $\pi_a$ and $\pi_b$, we will
get the right answer. For example, if $\pi'_a(x) = \alpha$, then we know that
we should map $x$ to an element $y \in p$ such that $\pi(y) = \alpha$. So
we need $y = 1 \lor y = 2$. Then looking at $\pi'_b(x)$ allows us to pick a
unique y. There is zero choice in the
construction of $f$. We need exactly 4 elements to cover all possible ways
in which $q$ could map into $a$ and $b$ through $\pi'_a, \pi'_b$ such that
we cover all possibilities with no redundancy.
This choice of product is goldilocks : it does not have too few elements such
that some elements are not representable; it also does not have too many
elements such that some elements are redundant.
§ Non uniqueness: product as $\{10, 20, 30, 40\}$
Note that instead of using $p = \{1, 2, 3 4\}$, I could have used
$p = \{10, 20, 30, 40 \}$
and nothing would have changed.
I have never depended on using the values $1, 2, 3, 4$.
Rather, I've only used them as \emph{labels}.
§ Non uniqueness: product as $\{ (\alpha, \gamma), (\alpha, \delta), (\beta, \gamma), (\beta, \delta) \}$
Indeed, our usual idea of product also satisfies the product, since it is
in unique isomorphism to the set $\{1, 2, 3, 4\}$ we had considered previously.
But this might be strange: Why is it that the categorical definition of product
allows for so many other "spurious" products? Clearly this product of tuples
is the best one, is it not?
Of course not! Inside the category of sets, anything with the same cardinality
is isomorphic. So nothing inside the category can distinguish between the
sets $\{1, 2, 3, 4\}$ and $\{ (\alpha, \gamma), (\alpha, \delta), (\beta, \gamma), (\beta, \delta) \}$.
Hence, the usual product we are so used to dealing with is not privileged.
This should make us happy, not sad. We have removed the (un-necessary)
privilege we were handing to this one set because it "felt" like it was
canonical, and have instead identified what actually makes the product of sets
tick: the fact that their cardinality is the product of the cardinalities of
the individual sets!
§ How to think about the product
Since we are specifying the data of $(p, \pi_a, \pi_b)$, we can simply think of
elements of $p$ as being "pre-evaluated", as $(x \in p, \pi_a(x), \pi_b(x))$.
So in our case, we can simplify the previous situation with $(p = \{ 10, 20, 30, 40 \}, \pi_a, \pi_b)$
by writing the set as
$p = \{ (10, \alpha, \gamma), (20, \alpha, \delta), (30, \beta, \gamma), (40, \beta, \delta) \}$.
This tells us "at a glance" that every element of $a \times b$ is represented,
as well as what element it is represented by.
§ Proof of uniqueness upto unique isomorphism
- Assume we have two products $(p, \pi_a, \pi_b)$ and $(q, \pi_a, \pi_b)$, which are products of $a$ and $b$
- By the universality of $p$ wrt $q$, we get a unique! map $q2p$ such that the diagaram commutes.
- By the universality of $q$ wrt $p$, we get a unique! map $p2q$
- We get a map $q2p \cdot p2q : p \rightarrow p$. by the universality of $p$ with respect to $p$, we get a unique! map that makes the diagram commute.But we have two such maps: $id_p$ as well as $q2p . p2q$. Hence we must have $id_p = q2p \cdot p2q$. In pictures:
∃!id(p)
+---+
| |
| v
+---------ppppp--------+
| πa | ^ πb |
v ∃!p2q | v
a | | b
^ | ∃!q2p ^
| π'b v | π'b |
+---------qqqqq--------+
- The full diagram commutes.
- By definition of identity/commutativity of the diagram, $\pi_a \circ id_p = \pi_a$.
- By definition of identity/commutativity of the diagram, $\pi_b \circ id_p = \pi_b$.
- By the commutativity of the diagram, we have $\pi_a \circ (q2p \circ p2q) = \pi_a$.
- By the commutativity of the diagram, we have $\pi_b \circ (q2p \circ p2q) = \pi_b$.
- We can consider the universal property of the product, where we have $(p, \pi_a, \pi_b)$ as one product, and $(p, \pi_a , \pi_b)$ as another product.
- This gives us a unique map $h$ such that $\pi_a \circ h = \pi_a$, and $\pi_b \circ h = \pi_b$.
- We have two candidates for $h$: $h = id_p$ and $h = p2q \circ q2p$. Hence, by the uniqueness of $h$, we have that $id_p = p2q \circ q2p$.
§ Wrapping up