Tuesday, August 11, 2009

Coproducts (but not Products!)

We introduced the notion of products as a sort of categorification of the "Cartesian Product". The product of $c_1$ and $c_2$ is an object $p$ equipped with projection morphisms $\pi_1:p\to{c_1}$ and $\pi_2:p\to{c_2}$ such that some diagram commutes.

We also introduced the duality principle, which allows us to "reverse the arrows" in diagrams to get some "dual" notion. So now we would like to use the duality principle to find the dual notion of a product, or the "co-product". First some references:

  • The Catsters, "Products and Coproducts" Lectures: 2, 3, 4
  • Saunders Mac Lane, Categories for the Working Mathematician Graduate Texts in Mathematics (vol 5) Springer-Verlag, Second Edition (1998);
  • Jiri Ad├ímek, 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 set up our diagram by reversing the arrows for the product. That is, in some category C, we have objects $c_1,c_2\in\mathbf{C}$, the product is an object $p$ such that given any object $d\in\mathbf{C}$ with morphisms $f:d\to{c_1}$, $g:d\to{c_2}$, then there exists a morphism $h:d\to{p}$ such that the following diagram commutes:

\begin{diagram}
      &                &      d        &                &      \\
      &\ldTo^{f}       & \dDashto_{!h} & \rdTo^{g}      &      \\
c_{1} & \lTo_{\pi_{1}} & p             & \rTo_{\pi_{2}} & c_{2}
\end{diagram}

Now we "turn the arrows around". We denote the coproduct of $c_1$ with $c_2$ as an object $c_1\sqcup\displaystyle{c_2}$. We want this object to be equipped with two morphisms, but they should be dual to the projection morphisms. Huh? They should have the codomain and the domain switched, so $i:c_1\to c_1\sqcup\displaystyle{c_2}$ and $j:c_2\to c_1\sqcup\displaystyle{c_2}$. We call these morphisms "injections" or "insertion maps" of the coproduct (although they are not required to be injective functions).

Now putting the diagram on its head, we end up with the following property that coproducts must satisfy: for each $d$ with morphisms $f:c_1\to{d}$ and $g:c_2\to{d}$, there exists a unique morphism $h:c_1\sqcup\displaystyle{c_2\to{d}}$ such that the following diagram commutes:

So to summarize what we've figured, lets write the formal definition of the coproduct:

Definition 1. Let C be some category and $c_{1},c_{2}\in\mathbf{C}$, the "Coproduct of $c_1$ with $c_2$" consists of
  • an object $c_{1}\sqcup\displaystyle{c_{2}\in\mathbf{C}}$
equipped with
  • a pair of morphisms $i:c_{1}\to c_{1}\sqcup\displaystyle{c_{2}}$ and $j:c_{2}\to c_{1} \sqcup\displaystyle{c_{2}}$ called "insertions"
such that
  • for any $d\in\mathbf{C}$ and pair of morphisms $f:c_1\to{d}$ and $g:c_2\to{d}$, there exists a unique morphism $\textstyle h:c_1\sqcup\displaystyle{c_2\to{d}}$ such that the following diagram commutes:

So the motivation for this construction was based off of the "dual" to the product, but we don't really know intuitively what this is (I mean, we want an intuition other than "it's related 'somehow' to the product"). What to do? We do what category theorists love to do in this situation: doodle!

We work in Set and set up the following problem:

We have only specified the sets $c_1,c_2,d$ and the morphisms $f$ and $g$. We want to figure out what $c_1\sqcup\displaystyle{c_{2}}$ is. We can use "universality", the fact that there is a unique morphism $h:c_1\sqcup\displaystyle{c_{2}\to{d}}$ which satisfies a commutativity property.

What can we deduce about $h$? Well, we see that since the image $g(c_2)$ is all of $d$, we need $h$ to map the image of $j(c_2)$ in $c_{1}\sqcup\displaystyle{c_{2}}$ to all of $d$. So in other words, we need at least two distinct elements in our coproduct. We can doodle at least part of what we know it could be:

On the other hand, since we also have to take into account $c_1$, we need to consider how this would affect the coproduct object. We see that all elements of $c_1$ are mapped to a single element of $d$, which implies our solution would be the following diagram:

Now what's going on? We see that since we chose a rather bad map $c_{1}\to{d}$, a constant map (it maps everything to a single element of $d$), we have a rather hard time getting an intuition about the situation, but we can still deduce something. It appears that $h$ is injective, which means that the coproduct would behave sort of like a disjoint union.

But, to play the Devil's advocate, how do we know this diagram is even the right one to describe our object $c_1\sqcup{c_2}$? Couldn't the following diagram also work:

Well, yes, this diagram also works, but it uniquely factors through our previous diagram. Or in other words, we can doodle an "intermediate step" using green:

Now, we can do this for any other guesses at what $c_{1}\sqcup{c_{2}}$ might be, which tells us that there is a unique object which all our guesses "factor" through. This unique object is precisely our $c_{1}\sqcup{c_{2}}$.

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