In category theory, we attempt to avoid thinking about what an object is, and look only at how it interacts with other objects. It turns out that even if we’re not allowed to talk about the “internal” structure of an object, we can still pin down some objects just by talking about their interactions. For example, if we are not allowed to define the empty set as “the set with no elements”, we can still define it by means of a “universal property”, talking instead about the functions from the empty set rather than about the elements of the empty set.

This page is not designed to teach you any particular universal properties, but rather to convey a sense of what the idea of “universal property” is about. You are supposed to let it wash over you gently, without worrying particularly if you don’t understand words or even entire sentences.

Examples

• The empty set can be defined by a universal property. Specifically, it is an instance of the idea of an initial object, in the category of sets. The same idea captures the trivial group, the ring \(\mathbb{Z}\) of integers, and the natural number \(0\).

• The product has a universal property, generalising the set product, the product of integers, the greatest lower bound in a partially ordered set, the products of many different algebraic structures, and many other things besides.

• The free group has a universal property (we refer to this property by the unwieldy phrase “the free-group functor is left-adjoint to the forgetful functor”). The same property can be used to create free rings, the discrete topology on a set, and the free semigroup on a set. This idea of the left adjoint can also be used to define initial objects (which is the generalised version of the universal property of the empty set). noteIndeed, an initial object of category \(\mathcal{C}\) is exactly a left adjoint to the unique functor from \(\mathcal{C}\) to \(\mathbf{1}\) the one-arrow category.

The above examples show that the ideas of category theory are very general. For instance, the third example captures the idea of a “free” object, which turns up all over abstract algebra.

Definition “up to isomorphism”

explain that we only usually get things defined up to isomorphism, and what that means anyway

Universal properties might not define objects

Universal properties are often good ways to define things, but just like with any definition, we always need to check in each individual case that we’ve actually defined something coherent. There is no silver bullet for this: universal properties don’t just magically work all the time.

For example, consider a very similar universal property to that of the empty set (detailed here), but instead of working with sets, we’ll work with fields, and instead of functions between sets, we’ll work with field homomorphisms.

The corresponding universal property will turn out not to be coherent:

The initial field noteAnalogously with the empty set, but fields can’t be empty so we’ll call it “initial” for reasons which aren’t important right now. is the unique field \(F\) such that for every field \(A\), there is a unique field homomorphism from \(F\) to \(A\).

It will turn out that all we need is that there are two fields \(\mathbb{Q}\) and \(F_2\) the field on two elements. %%note: \(F_2\) has elements \(0\) and \(1\), and the relation \(1 + 1 = 0\).%%

Suppose we had an initial field \(F\) with multiplicative identity element \(1_F\); then there would have to be a field homomorphism \(f\) from \(F\) to \(F_2\). Remember, \(f\) can be viewed as (among other things) a group homomorphism from the multiplicative group \(F^*\) %%note: That is, the group whose underlying set is \(F\) without \(0\), with the group operation being “multiplication in \(F\)”.%% to \(F_2^*\).

Now \(f(1_F) = 1_{F_2}\) because the image of the identity is the identity, and so \(f(1_F + 1_F) = 1_{F_2} + 1_{F_2} = 0_{F_2}\).

But field homomorphisms are either injective or map everything to \(0\) (proof); and we’ve already seen that \(f(1_F)\) is not \(0_{F_2}\). So \(f\) must be injective; and hence \(1_F + 1_F\) must be \(0_F\) because \(f(1_F + 1_F) = 0_{F_2} = f(0_F)\).

Now examine \(\mathbb{Q}\). There is a field homomorphism \(g\) from \(F\) to \(\mathbb{Q}\). We have \(g(1_F + 1_F) = g(1_F) + g(1_F) = 1 + 1 = 2\); but also \(g(1_F + 1_F) = g(0_F) = 0\). This is a contradiction. <div><div>