# Product is unique up to isomorphism

https://arbital.com/p/product_is_unique_up_to_isomorphism

by Patrick Stevens Aug 28 2016 updated Aug 28 2016

If something satisfies the universal property of the product, then it is uniquely specified by that property, up to isomorphism.

[summary: Here, we prove that the Universal property of the product characterises objects uniquely up to Isomorphism. That is, if two objects both satisfy the universal property of the product of $A$ and $B$, then they are isomorphic objects.]

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 $S$. %%note: 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 $B$.%%

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