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 $(S,{\varphi }_A,{\varphi }_B)$ 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 ${\pi }_A\circ f=f_A$ and ${\pi }_B\circ f=f_B$.

Now let $f_A={\varphi }_A,f_B={\varphi }_B$:

There is a unique map $\varphi :S\to R$ such that ${\pi }_A\circ \varphi ={\varphi }_A$ and ${\pi }_B\circ \varphi ={\varphi }_B$.

Doing the same again but swapping $R$ for $S$ and $\varphi$ 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 ${\varphi }_A\circ \pi ={\pi }_A$ and ${\varphi }_B\circ \pi ={\pi }_B$.

Now, $\pi \circ \varphi :S\to S$ 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 $\varphi$.

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 ${\varphi }_A\circ f=f_A$ and ${\varphi }_B\circ f=f_B$.

Letting $f_A={\varphi }_A$ and $f_B={\varphi }_B$, we obtain:

There is a unique map $f:S\to S$ such that ${\varphi }_A\circ f={\varphi }_A$ and ${\varphi }_B\circ f={\varphi }_B$.

But I claim that both the identity $1_S$ and also $\pi \circ \varphi$ 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 ${\varphi }_A={\varphi }_A$ and ${\varphi }_B={\varphi }_B$;
• $\pi \circ \varphi$ satisfies the property, since we already found that ${\varphi }_A\circ \pi ={\pi }_A$ and that ${\varphi }_B\circ \pi ={\pi }_B$.

Therefore $\pi$ is left-inverse to $\varphi$.

Now to complete the proof, we just need to repeat exactly the same steps but with $(R,{\pi }_A,{\pi }_B)$ and $(S,{\varphi }_A,{\varphi }_B)$ interchanged throughout. The outcome is that $\varphi$ is left-inverse to $\pi$.

Hence $\pi$ and $\varphi$ 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 $(B\times A,{\pi }_A^{\prime },{\pi }_B^{\prime })$, where ${\pi }_A^{\prime }(b,a)=a$ and ${\pi }_B^{\prime }(b,a)=b$. (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 $(a,b)\mapsto (b,a)$.

1. ^{^︎}

   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 $B$.