In algebra we are interested in thinking about objects. Objects can be many things including numbers, sets, and many more complicated constructions. Category theory is a powerful tool which analyzes these objects and how they are related.

When defining a category we define what the objects are and what the “morphisms” between those objects are. Morphisms are however we choose objects to be related. There are some additional restrictions which we can touch on later.

Let’s look at a simple category: In this category, the objects are positive integers and morphisms compare the value of these numbers. We’ll say the morphism points towards the bigger number. More rigorously, there exists a morphism from \(a\) to \(b\) if and only if \(a\leq b\). We can visualize this with the following diagram:

We can make some obvious statements about morphisms in this category. It is always true that \(a\leq a\). We can also say that if \(a\leq b\) and \(b\leq c\) then \(a\leq c\). These are the additional constrictions on morphisms: every object must have a morphism to itself (called the identity morphism) and that composing morphisms in a row implies the existence of a morphism connecting the ends:

So now we understand a basic category and how it works.

Let’s look at a more interesting category. The objects are still positive integers and now the morphisms are divisibility operators. This means that there is a morphism from \(a\) to \(b\) if and only if \(a\) divides \(b\) (written as \(a | b\)). For example: \(3|9\), \(2|e\) if \(e\) is even.

Let’s take a look at our first Universal Property. Say we have two numbers \(a\) and \(b\) which both divide \(z\). What number \(w\) has the property that no matter what \(z\) is, as long as \(a\) and \(b\) both divide \(w\), it is also true that \(w\) divides \(z\)? Observe the diagram

Take a second to understand the question. For **any** \(z\) which is divisible by both \(a\) and \(b\), we want to find a \(w\) which is also divisible by both \(a\) and \(b\) which must **necessarily** divide **every** choice of \(z\). For every multiple of both \(a\) and \(b\) (called \(z\)), we want to find a multiple of \(a\) and \(b\) (called \(w\)) which always divides it.

Let’s consider some examples. Let \(a=4\), \(b=6\). What are some common multiples of these? There’s \(6\times 4 =24\) but there’s also \(6\times 2 = 4\times 3 = 12\) and \(6\times 100 = 4\times 150 = 600\). These are a few of our candidates for \(w\).

So then, what is a reasonable choice? Can we choose a huge number like \(600\)? No, because \(600\) doesn’t divide \(24\). So, we are looking for the *smallest* of these values which divides every possible \(z\). A *least common multiple*, if you will. That would be \(12\). This works for every \(z\) value i.e. every common multiple of \(4\) and \(6\). Any number which is a common multiple, is itself a multiple of the *least* common multiple. We can choose \(z=12\) or \(z=600\) and it doesn’t matter. \(w=12\) will always work.

Let’s look at what we just did there. We took this diagram, which admittedly we kind of just conjured out of thin air, but we used it to rigorously define the least common multiple. A number is the least common multiple if and only if it satisfies this diagram in this category. This is a rigorous definition. This quality of how the least common multiple interacts with arbitrary numbers is its universal property.

Later, we will discuss at greater length how these kinds of structures can be applied to sets. Even though with the least common multiple there might be simpler ways to define it, we often find in Algebra that the best way to define objects is by their universal property.

As an exercise to the adventurous and curious reader, consider the the following diagram in the same category. What object (number) \(w\) satisfies the universal property:

Remember, solid lines are given. The dotted line is what you are trying to prove exists for any choice of \(z\).

(Hint: \(z\) must satisfy the properties indicated by the arrows to \(a\) and \(b\))

Contributors: David Staudinger, Joseph Brewster