# 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 $$(a,b) \mapsto (b,a)$$.

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