# 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.