[summary: Here, we prove that the Universal property of the product characterises objects uniquely up to Isomorphism. That is, if two objects both satisfy the universal property of the product of $~$A$~$ and $~$B$~$, then they are isomorphic objects.]

Recall the Universal property of the product:

Given objects $~$A$~$ and $~$B$~$, we define the

productto be the following collection of three objects, if it exists: $$~$A \times B \\ \pi_A: A \times B \to A \\ \pi_B : A \times B \to B$~$$ with the requirement that for every object $~$X$~$ and every pair of maps $~$f_A: X \to A, f_B: X \to B$~$, there is auniquemap $~$f: X \to A \times B$~$ such that $~$\pi_A \circ f = f_A$~$ and $~$\pi_B \circ f = f_B$~$.

We wish to show that if the collections $~$(R, \pi_A, \pi_B)$~$ and $~$(S, \phi_A, \phi_B)$~$ satisfy the above condition, then there is an Isomorphism between $~$R$~$ and $~$S$~$. %%note: I'd write $~$A \times_1 B$~$ and $~$A \times_2 B$~$ instead of $~$R$~$ and $~$S$~$, except that would be really unwieldy. Just remember that $~$R$~$ and $~$S$~$ are both standing for products of $~$A$~$ and $~$B$~$.%%

# Proof

The proof follows a pattern which is standard for these things.

Since $~$R$~$ is a product of $~$A$~$ and $~$B$~$, we can let $~$X = S$~$ in the universal property to obtain:

For every pair of maps $~$f_A: S \to A, f_B: S \to B$~$ there is a unique map $~$f: S \to R$~$ such that $~$\pi_A \circ f = f_A$~$ and $~$\pi_B \circ f = f_B$~$.

Now let $~$f_A = \phi_A, f_B = \phi_B$~$:

There is a unique map $~$\phi: S \to R$~$ such that $~$\pi_A \circ \phi = \phi_A$~$ and $~$\pi_B \circ \phi = \phi_B$~$.

Doing the same again but swapping $~$R$~$ for $~$S$~$ and $~$\phi$~$ for $~$\pi$~$ (basically starting over with the line "Since $~$S$~$ is a product of $~$A$~$ and $~$B$~$, we can let $~$X = R$~$…"), we obtain:

There is a unique map $~$\pi: R \to S$~$ such that $~$\phi_A \circ \pi = \pi_A$~$ and $~$\phi_B \circ \pi = \pi_B$~$.

Now, $~$\pi \circ \phi: S \to S$~$ is a map which we wish to be the identity on $~$S$~$; that would get us halfway to the answer, because it would tell us that $~$\pi$~$ is left-inverse to $~$\phi$~$.

But we can use the universal property of $~$S$~$ once more, this time looking at maps into $~$S$~$:

For every pair of maps $~$f_A: S \to A, f_B: S \to B$~$ there is a unique map $~$f: S \to S$~$ such that $~$\phi_A \circ f = f_A$~$ and $~$\phi_B \circ f = f_B$~$.

Letting $~$f_A = \phi_A$~$ and $~$f_B = \phi_B$~$, we obtain:

There is a unique map $~$f: S \to S$~$ such that $~$\phi_A \circ f = \phi_A$~$ and $~$\phi_B \circ f = \phi_B$~$.

But I claim that both the identity $~$1_S$~$ and also $~$\pi \circ \phi$~$ satisfy the same property as $~$f$~$, and hence they're equal by the uniqueness of $~$f$~$. Indeed,

- $~$1_S$~$ certainly satisfies the property, since that would just say that $~$\phi_A = \phi_A$~$ and $~$\phi_B = \phi_B$~$;
- $~$\pi \circ \phi$~$ satisfies the property, since we already found that $~$\phi_A \circ \pi = \pi_A$~$ and that $~$\phi_B \circ \pi = \pi_B$~$.

Therefore $~$\pi$~$ is left-inverse to $~$\phi$~$.

Now to complete the proof, we just need to repeat *exactly* the same steps but with $~$(R, \pi_A, \pi_B)$~$ and $~$(S, \phi_A, \phi_B)$~$ interchanged throughout.
The outcome is that $~$\phi$~$ is left-inverse to $~$\pi$~$.

Hence $~$\pi$~$ and $~$\phi$~$ are genuinely inverse to each other, so they are both isomorphisms $~$R \to S$~$ and $~$S \to R$~$ respectively.

# The characterisation is not unique

To show that we can't do better than "characterised up to isomorphism", we show that the product is not characterised *uniquely*.
Indeed, if $~$(A \times B, \pi_A, \pi_B)$~$ is a product of $~$A$~$ and $~$B$~$, then so is $~$(B \times A, \pi'_A, \pi'_B)$~$, where $~$\pi'_A(b, a) = a$~$ and $~$\pi'_B(b, a) = b$~$.
(You can check that this does satisfy the universal property for a product of $~$A$~$ and $~$B$~$.)

Notice, though, that $~$A \times B$~$ and $~$B \times A$~$ are isomorphic as guaranteed by the theorem. The isomorphism is the map $~$A \times B \to B \times A$~$ given by $~$(a,b) \mapsto (b,a)$~$.