Is the product bifunctor uniquely determined by how it acts on objects?

Question

Saunders MacLane states that "The product objects provide, by $<a, b> \mapsto a \times b$, a bifunctor $C \times C \to C.$ " I know that this is not in fact a bifunctor, as the product is unique only up to isomorphism. However, if one defines the object function of the bifunctor to send each object $<a,b>$ to a particular product, say $(a \times b)_1$, is there a unique arrow function that satisfies the functor laws?$\\$ Edit: changed coproduct to product.

Answer 1

Just a minor note: you talk about coproducts while Mac Lane talks about products.

That being said, yes, he assumes that we have a choice of product for each pair of objects. In general this would require the axiom of choice (class sized choice, possibly), but in many categories we have a canonical choice of product. For example, in the category of sets the product of two sets $A$ and $B$ is just given by their cartesian product $A \times B$.

Once we have a choice of product for any two objects $A$ and $B$, this assignment naturally extends to a functor. Namely for an arrow $(f, g): (A, B) \to (C, D)$ in $\mathcal{C} \times \mathcal{C}$, there is a unique arrow $f \times g: A \times B \to C \times D$ defined as follows. Let $\pi_A: A \times B \to A$ and $\pi_B: A \times B \to B$ be the projections. Then we have arrows $f \pi_A: A \times B \to C$ and $g \pi_B: A \times B \to D$, so the universal property of the product gives a unique arrow $\langle f \pi_A, g \pi_B \rangle: A \times B \to C \times D$. The notation $f \times g$ is just a shorthand for $\langle f \pi_A, g \pi_B \rangle$. So no more choice is required for the arrows.

This is not saying that this is the only bifunctor that sends $(A, B)$ to $A \times B$, as noted in another answer. This is just the natural way to extend to a binfunctor, and it is what Mac Lane means in the sentence you quoted.

Answer 2

Assuming you have a fixed object $(a\times b, \pi^{a,b}_a, \pi^{a,b}_b)$ for each pair $a,b$, then there is a unique bifunctor $P:C\times C\to C$ such that :

1- For all $a,b\in C, P(a,b)=a\times b$

2- Let $I_1 : C\times C\to C$ denote the projection-onto-the-first-factor functor, and $I_2$ the other one, then $(\pi^{a,b}_a)_{(a,b)\in C\times C}$ is a natural transformation $P\to I_1$ and $(\pi^{a,b}_b)_{(a,b)\in C\times C}$ is a natural transformation $P\to I_2$.

It is not enough in general to just ask that $P(a,b) = a\times b$, I'll let you find some counterexamples

The proof is easy : if you have $f:a\to c, g: b\to d$, there is a unique map $f\times g : a\times b \to c\times d$ such that the following diagram commutes :

$\require{AMScd}\begin{CD} a @>f>> c \\ @A\pi^{a,b}_aAA @A\pi^{c,d}_cAA \\ a\times b @>>> c\times d \\ @V\pi^{a,b}_bVV @V\pi^{c,d}_cVV \\ b @>g>> d \end{CD}$

If you think about it, you'll see that this is what we'd want to represent $(x,y)\mapsto (f(x),g(y))$

You can obviously (given the proof) replace 2. by the condition that $P(f,g)$ be the map defined above

In any case, that's what is meant by "the product bifunctor"