[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 product to 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 a unique map $f: X \to A \times B$ 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 $(a,b) \mapsto (b,a)$.