What are the natural transformations induced by a category with finite products?

Question

Source: Categories for the Working Mathematician, second edition by Saunders Mac Lane.

Proposition: If a category $C$ has a terminal object $t$ and a product diagram $a\leftarrow a\times b\rightarrow b$ for any two of its objects, then $C$ has all finite products. The product objects provide, by $\langle a, b\rangle\mapsto a\times b$, a bifunctor $C\times C\to C$. For any three objects $\mathbf{a, b}$ and $\mathbf{c}$ there is an isomorphism $\mathbf{\alpha=\alpha_{a, b, c}: a\times(b\times c)\cong(a\times b)\times c}$ natural in $\mathbf{a, b}$ and $\mathbf{c}$.

My Question: I struggle with the boldface part. Take $c$, for example. I suppose for each fixed pair $\langle a, b\rangle$ the natural transformation, called it $\tau$, is between the two functors $S=a\times(b\times-)$ and $T=(a\times b)\times-: C\to C$. The object functions for $S$ and $T$ are obvious: $c\mapsto a\times(b\times c)$ and $c\mapsto(a\times b)\times c$. But what about the arrow functions? And when it comes to $\tau$, each $c$ of $C$ is mapped to which arrow of $C$?

The book can be a bit tough to follow sometimes; but I guess that's part of the charm. Any help would be greatly appreciated.

Answer 1

Here’s a diagram of the relevant morphisms (produced with quiver). The solid black arrows are the projections given by the definition of the product. The dotted red arrows are compositions of those projections. The dashed blue arrows are obtained by applying the universal properties of their target products to suitable black and red arrows.

Composing the vertical blue arrows in the centre yields morphisms from the triple products to themselves. By chasing arrows, you can check that the composition of these morphisms with the projections of the triple products again yields those projections. By uniqueness of the universal property, it follows that these morphisms are the identity morphisms on the triple products, and thus the vertical blue arrows are isomorphisms between the triple products.

Here’s a diagram for how to obtain an arrow from $(a\times b)\times c$ to $(a\times b)\times c'$ given an arrow from $c$ to $c'$. The squiggly green line is the given arrow from $c$ to $c'$. As above, the black arrows are the projections of the products, the dotted red arrow is a composition, and the dashed blue arrow is obtained by applying the universal property.

                        

And here’s the diagram for obtaining an arrow from $a\times(b\times c)$ to $a\times(b\times c')$ given an arrow from $c$ to $c'$ (same colour/texture coding as before). The coloured arrows are obtained from right to left, each using the previous one.

    

Answer 2

Some hints.

Let $T:C\times C\to C$ denote product. You then define $a\times(b\times c)$ on arrows and objects very easily as the composite functor $T\circ(1\times T):C\times C\times C\to C\times C\to C$. Similarly, $(a\times b)\times c$ stands for $T\circ(T\times 1)$. Well, that assumes you understand how $T$ is a functor. If the reader is not familiar with this, here's how it works:

For an arrow $(f,g):(x,y)\to(x',y')$ in $C\times C$, $T(f,g)$ - usually written $f\times g$ - is some arrow $x\times y\to x'\times y'$; it is the unique arrow such that $\pi_1 T(f,g)=f\pi_1:x\times y\to x\to x'$ and $\pi_2T(f,g)=g\pi_2:x\times y\to y\to y'$. Such an arrow exists and is unambiguously defined precisely by the universal property of $x'\times y'$.

This definition is an instance of the more general property that $\varprojlim$ is a functor, and $(f,g)$ defines a natural transformation between the diagrams whose limits are $x\times y$ and $x'\times y'$.

To get a map $(a\times b)\times c\to a\times(b\times c)$, use the product universal property in a similar fashion. You just need to find arrows $(a\times b)\times c\to a$ and $(a\times b)\times c\to b\times c$. The only sensible ones you can think of (if you're stuck, what are they for $C=\mathsf{Set}$?) are the correct ones.