Product is unique up to isomorphism

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 \(S\). 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.