Recall the universal property of the product:

Given objects and $B$ , we define the product to be the following collection of three objects, if it exists: $A \times B π_{A} : A \times B \to A π_{B} : A \times B \to B$ 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 a unique map $f : X \to A \times B$ such that $π_{A} \circ f = f_{A}$ and $π_{B} \circ f = f_{B}$ .

We wish to show that if the collections $(R, π_{A}, π_{B})$ and $(S, ϕ_{A}, ϕ_{B})$ satisfy the above condition, then there is an isomorphism between $R$ and $S$ . ^[1]

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 $π_{A} \circ f = f_{A}$ and $π_{B} \circ f = f_{B}$ .

Now let $f_{A} = ϕ_{A}, f_{B} = ϕ_{B}$ :

There is a unique map $ϕ : S \to R$ such that $π_{A} \circ ϕ = ϕ_{A}$ and $π_{B} \circ ϕ = ϕ_{B}$ .

Doing the same again but swapping $R$ for $S$ and $ϕ$ for $π$ (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 $π : R \to S$ such that $ϕ_{A} \circ π = π_{A}$ and $ϕ_{B} \circ π = π_{B}$ .

Now, $π \circ ϕ : S \to S$ 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 $π$ is left-inverse to $ϕ$ .

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 $ϕ_{A} \circ f = f_{A}$ and $ϕ_{B} \circ f = f_{B}$ .

Letting $f_{A} = ϕ_{A}$ and $f_{B} = ϕ_{B}$ , we obtain:

There is a unique map $f : S \to S$ such that $ϕ_{A} \circ f = ϕ_{A}$ and $ϕ_{B} \circ f = ϕ_{B}$ .

But I claim that both the identity $1_{S}$ and also $π \circ ϕ$ 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 $ϕ_{A} = ϕ_{A}$ and $ϕ_{B} = ϕ_{B}$ ;

$π \circ ϕ$ satisfies the property, since we already found that $ϕ_{A} \circ π = π_{A}$ and that $ϕ_{B} \circ π = π_{B}$ .

Therefore $π$ is left-inverse to $ϕ$ .

Now to complete the proof, we just need to repeat exactly the same steps but with $(R, π_{A}, π_{B})$ and $(S, ϕ_{A}, ϕ_{B})$ interchanged throughout. The outcome is that $ϕ$ is left-inverse to $π$ .

Hence $π$ and $ϕ$ 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 $(A \times B, π_{A}, π_{B})$ is a product of $A$ and $B$ , then so is $(B \times A, π_{A}^{'}, π_{B}^{'})$ , where $π_{A}^{'} (b, a) = a$ and $π_{B}^{'} (b, a) = b$ . (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 $(a, b) \mapsto (b, a)$ .

^{^︎}
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 $B$ .