# Product is unique up to isomorphism

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 \(S\). noteI’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 \((a,b) \mapsto (b,a)\).

Parents:

- Universal property of the product
The product can be defined in a very general way, applicable to the natural numbers, to sets, to algebraic structures, and so on.