Products (but not Coproducts yet)

First some references:

• The Catsters, Lectures on Products and Coproducts 1, 2, 3, 4
• Saunders Mac Lane, Categories for the Working Mathematician Graduate Texts in Mathematics (vol 5) Springer-Verlag, Second Edition (1998);
• Jiri Adámek, Horst Herrlich, George E. Strecker Abstract and Concrete Categories: The Joy of Cats freely available online (2004)
• Serge Lang Algebra Springer-Verlag, Third Edition (2000)

We want to introduce the notion of "products" into category theory, or (equivalently) we want to categorify the notion of a product. We expect it to be such that when we work in Set we should recover the familiar Cartesian product, since that's what our intuition is based off of.

Where to start? Well, we know that with two sets $S_{1},S_{2}$ we can take their product $S_{1}\times{S_{2}}$. We can take their "projection", i.e. there are functions $\pi_{1}:S_{1}\times{S_{2}}\to{S_{1}}$ and $\pi_{2}:S_{1}\times{S_{2}}\to{S_{2}}$ which do the obvious things: they select out the first and second sets (respectively).

To generalize this notion to categories in general, we expect for some category $C$ and objects $c_{1},c_{2}\in{C}$ that their product is an object $p\in{C}$ such that we have two morphisms $\pi_{1}:p\to{c_{1}}$ and $\pi_{2}:p\to{c_{2}}$. But so what? There are probably a million such morphisms, how can pick some pair of morphisms out?

Good question, the answer is relatively intriguing. We select any third object $d\in{C}$ and demand that for every pair of morphisms $f:d\to{c_{1}}$ and $g:d\to{c_{2}}$, there needs to exist a unique function $h$ which makes the following diagram commute:

The dashed arrow indicates we assert that morphism exists, usually we have the exclamation mark "!" involved to emphasize this fact. We have chosen the product to aptly be the object labeled by $p$. Lets give the formal definition:

Definition 1. The "product of $c_{1},c_{2}$" consists of

• a mathematical object $p$

equipped with

• a morphism $\pi_{1}:p\to{c_{1}}$;
• a morphism $\pi_{2}:p\to{c_{2}}$;

such that

• for any object $d\in{C}$ and pair of morphisms $f:d\to{c_{1}}$ and $g:d\to{c_{2}}$, there exists a corresponding morphism $h:d\to{p}$ such that the following diagram commutes:
  
  

Products of Categories

We have introduced the product for arbitrary mathematical objects, and being clever mathematicians we can say "Ah a category is-a mathematical object!" So we can get to our motivation for this section: the product of categories.

Here the notation is kind of more relaxed. We consider the product of categories $B$, $C$ to be denoted by $B\times{C}$. The objects are products of objects, and the morphisms are products of morphisms. More explicitly, for any $b,b^{\prime}\in{B}$ and $c,c^{\prime}\in{C}$ and morphisms $f:b\to{b^{\prime}}$, $g:c\to{c^{\prime}}$, the objects of $B\times{C}$ are $(b,c)$, $(b^{\prime},c^{\prime})$ and the morphisms $(f,g):(b,c)\to{}(b^{\prime},c^{\prime})$. We can intuitively think that things behave componentwise.

To emphasize this notion of "componentwise-thinking", consider the following scenario for our product category: we have objects $(b,c)$, $(b^{\prime},c^{\prime})$, and $(b^{\prime\prime},c^{\prime\prime})$. We have morphisms $(f,g):(b,c)\to(b^{\prime},c^{\prime})$ and $(f^{\prime},g^{\prime}):(b^{\prime},c^{\prime})\to(b^{\prime\prime},c^{\prime\prime})$. We can compose these two morphisms "in the obvious way", i.e. $(f,g)\circ(f^{\prime},g^{\prime})=(f\circ{f^{\prime}},g\circ{g^{\prime}})$.

Now, we need projection functors for our particular situation. That is, "projection morphisms" for our product category. It's even more intuitive in our notation what they do. If we have $B\xleftarrow{P}B\times{C}\xrightarrow{Q}C$, we have these projection functors be defined by $P(f,g)=f$ and $Q(f,g)=g$ on morphisms and similarly on object $P(b,c)=b$ and $Q(b,c)=c$.

These projection functors have to satisfy a similar property as the projection morphisms, which is what we expect to happen. More explicitly, for some category $D$ and two functors $B\xleftarrow{R}D\xrightarrow{T}C$ we demand that there is a unique functor $F:D\to{B\times{C}}$ such that the following diagram commutes:

This probably shouldn't be too surprising, it's what we expect to find if we just follow our nose.

Example: Products in "Set"

This is all rather complicated, so lets first examine what products of two sets means. To be more precise, what we mean is "The product of two objects in Set is what?"

Let $P$ be the product of $C_1$, $C_2$. Let $f:\mathbf{1}\to{C_{1}}$, $g:\mathbf{1}\to{C_{2}}$. How do these functions behave? This picks out a single element of $C_1$, $C_2$ (respectively). We can identify these morphisms as "the same as" the elements. Now the product idea makes the following diagram commute:

(Such a diagram is called an "internal diagram" since it shows the guts of the object, the dots are the elements of the sets, the arrows are morphisms in Set. Category theorists love doodling these sort of diagrams, it is really enlightening when inspecting a definition in category theory.)

What this means is that $\pi_{1}\circ\langle f,g\rangle=f$ (or the morphism $\mathbf{1}\to{P}$ followed by $\pi_{1}$ is completely equivalent to $f$) and $\pi_{2}\circ\langle f,g\rangle=g$ ($\mathbf{1}\to{P}$ then $P\to{C_{2}}$ is the same as $g$). But this morphism $\langle f,g\rangle$ picks out a single element of $P$, since its domain is 1.

So what do we imagine an element of $P$ ought to be? It should be such that it can be projected into two components. This hints that the elements of $P$ are ordered pairs $(c_1,d_2)$ where $c_1\in{C_{1}}$ and $d_{2}\in{C_{2}}$.

How can we say this? Well, there are 3 elements in $C_1$ and 2 elements in $C_2$. This means that there are 2 distinct morphisms $\mathbf{1}\to{C_{2}}$ and 3 distinct morphisms in $\mathbf{1}\to{C_{1}}$. How many distinct pairs of morphisms can we pick out? There are $2\times3=6$ different ways to pick out pairs of arrows $\mathbf{1}\to{C_{1}}$ and $\mathbf{1}\to{C_{2}}$, and so 6 unique corresponding arrows from $\mathbf{1}\to{P}$. This is the number of possible ordered pairs of elements $(c_1,d_2)$. For another possible pair of arrows we could have picked, consider the following internal diagram:

For each point $f:\mathbf{1}\to{C_{1}}$, there is a corresponding pair of points $g:\mathbf{1}\to{C_{2}}$. There are 3 distinct such $f$ arrows, and each distinct arrow has two corresponding $g$ arrows, yielding 6 distinct arrows $\mathbf{1}\to{P}$. Just as for each $c_{i}\in{C_{1}}$ there is a corresponding pair of elements $d_{1},d_{2}\in{C_{2}}$; so we can write 6 possible ordered pairs $(c_i,d_j)$ where $i=1,2,3$, $j=1,2$. This seems to be exactly what we are doing with our product, which is hinted by the $\mathbf{1}\to{P}$ morphism...

Summary: from our internal diagrams, we can deduce that the product of elements of Set amounts to the Cartesian products of those sets.