The Universal property

Category theory is about structures, not about details. It doesn’t care about what you look like, but rather who and how you interact with. Category theory takes a bird’s eye view of mathematics.

One of the most important concept in category theory is that of universal property. They are ubiquitous in all the branches in mathematics, often in disguise. To get a better understanding of universal property, let’s start from some examples.

As a first example, let’s consider sets. Let $1$ denote the set with only one elements. It doesn’t matter what the element really is, may it be a real number, a complex number, or even a person. It again shows the “taking a bird’s view” nature of category theory. Given a set $1$, for all sets $X$, there exists a unique map from $X$ to $1$. In this context, the words “map”, “arrow” and “function” are all the same thing.

The key word here is unique. Given any map $X$, there is indeed only one possible way to map all and every elements of $X$ to $1$, namely the constant map, $f(x)=1$.

Phrases of the form “there exists a unique such-and-such satisfying some condition” are common in category theory. The phrase means that there is only one such-and-such satisfying the condition. To prove it, we need to prove that i) there exists at least one such-and-such and ii) there exists at most 1 such-and-such. If there are two things satisfying the condition, they must be the same.

This kind of properties are called universal properties, since it describes how the object being described (in our example, the set $1$) relates to all the other objects in the universe it lives in (the universe of sets). In our example, the property starts with “for all sets $X$”, and therefore says something about the relationship between $1$ and the entire universe of sets.

As the second example, consider the universe of rings. In our note rings are always assume to contain a multiplicative identity $1$. Recall that homomorphisms between rings are maps that preserves addition and multiplication, and of course identity $1$.

The ring of integers $\mathbb{Z}$ is special among rings in the sense that, for all rings $R$, there exists a unique homomorphism $f:\mathbb{Z}\to R$, which maps number $1$ to the multiplicative identity of $R$. Note this is again of the form “there exists a unique such-and-such that blahblahblah”. Any number can be written as the summation of certain number of $1$, for example $5$ is $1+1+1+1+1$, then the homomorphism $f$ maps number $5$ to an element in $R$ which is the summation of five multiplicative identity. In this way, the homomorphism is unique.

To prove the uniqueness, we don’t need to rely on any specific properties of rings! Let $\phi:\mathbb{Z}\to R$ be another homomorphism, we want to prove that $\phi$ is the same as $f$. Now, two maps between sets are equal when they map every element to the same element. So we need to look at how $f$ and $\phi$ acts on elements of $\mathbb{Z}$. Since homomorphism preserves identity, we have $\phi(1)=1=f(1)$. Since homomorphism preserves addition, we have

\[\phi(n) = \phi(1+\dots+1) = \phi(1)+\dots+\phi(1)=1+\dots+1=f(1)+\dots+f(1)=f(1+\dots+1)=f(n).\]

Since homomorphism preserves zero, we have $f(0)=0=\phi(0)$. Thus $f(z)=\phi(z)$ for all $z\in\mathbb{Z}$.

Crucially, there can be essentially only one object object that satisfying a given universal property. The word “essentially” means the two objects satisfying the same universal property need not literally be equal, but only equal up to isomorphism. Recall that isomorphism between $A$ and $B$ means that there are two arrows $A\to B$ and $A\leftarrow B$, whose compositions are identity on both ends. In our example of rings, the uniqueness of such objects says:

Let $A$ be a ring with the following property: for all rings $R$, there exists a unique homomorphism $A\to R$. Then $A\cong B$.

Before proving it, let me mention that a ring with this property is called an initial object, since there is a arrow coming out from this object to any other objects. We could say that $A$ is initial in the category of rings (I haven’t said what a category is yet, for now just think of category as a collection of rings).

Now the lemma we want to prove has become, if $A$ is initial, and we already know that $\mathbb{Z}$ is initial, prove that $A\cong \mathbb{Z}$.

Since $A$ is initial, there is a unique map from $A$ to all the other rings, including $\mathbb{Z}$, call it $\phi:A\to \mathbb{Z}$. Since $\mathbb{Z}$ is also initial, there exists a unique map $\phi’:\mathbb{Z}\to A$ as well. Now $\phi’\circ\phi:A\to A$ is a homomorphism (the composition of two homomorphisms is itself a homomorphism), but so is the identity map $1{A}:A\to A$. Since $A$ is initial, there is a unique map from $A\to A$ as well! It means that $\phi’\circ\phi$ can’t be different from $1{A}$, otherwise the map $A\to A$ won’t be unique. Thus $\phi’\circ\phi\cong {1}{A}$. In the same way we can prove that $\phi\circ\phi’\cong 1{\mathbb{Z}}$. Thus $A$ is isomorphic to $\mathbb{Z}$.

The proof has very little to do with the properties of rings (addition, multiplication are defined, etc.), it belongs to a higher level of generality.

Since example are so important to understanding, in the following we list some more examples of universal property.

Given a set $S$, we can make it into a topology space by equipping $S$ with the discrete topology: all subsets of $S$, including $\emptyset$ and $S$ itself. The now topological space is denoted $D(S)$. It’s one of most trivial way to bestow on $S$ a topology structure. With discrete topology, any map from $S$ to a space $X$ is continuous.

Given a map $f:S\to X$, we can regard it as a map from $D(S)$ to $X$, since $D(S)$ and $S$ have same elements. Then we would have a continuous function $\overline{f}:D(S)\to X$. $\overline{f}$ is just $f$ with a different perspective. To generalize it, for all functions $f$ from $S$ to all spaces $X$, there exists a unique (denoted by $\exists !$) map $\overline{f}$ which is continuous. This is another example for universal property.

This statement feels more or less trivial, but still it tells us something about the discrete topology. We mentioned that there can be only one (up to isomorphism) object satisfying a universal property, here the unique object being the discrete topology. The universal properties only holds for discrete topology. If we equip the set with indiscrete topology, for example, where the only closed sets are $S$ and $\emptyset$, the universal property will not hold.

As our last example, consider the collection of vector spaces. Given three vector spaces $U,V$ and $W$, a bilinear map $f:U\times V\to W$ is a function $f$ that is linear in each variable:

\[\begin{align} f(u,v_{1}+\lambda v_{2})&=f(u)+\lambda f(v_{2}),\\ f(\lambda u_{1}+u_{2},v)&=\lambda f(u_{1})+f(u_{2}). \end{align}\]

There turns out to be a universal map out of $U\times V$. In other words, there exists a certain vector space $T$ and a certain bilinear map $b:U\times V\to T$ such that every bilinear map from $f:U\times V\to W$ induces a unique bilinear map $\overline{f}:T\to W$, satisfying

\[b\circ\overline{f}=f.\]

Roughly speaking, the property says that bilinear maps out of $U\times V$ corresponds 1-2-1 with linear maps out of a single vector space $T$. The vector space $T$ is called the tensor product of $U$ and $V$.

Along the study of category theory, we will develop different ways of understanding universal property.