This is a note based on the lecture about functoriality and bifunctors. The same topic is also discussed in the Functoriality chapter from the book Category Theory for Programmers.

What is discussed in this lecture?

This lecture discussed two main topics:

• What is a bifunctor
• product and coproduct are bifunctors, in both Hask and general categories

What is a bifunctor?

A bifunctor maps two objects \(a\) and \(b\), from categories \(C\) and \(D\), respectively, to a new object \(e\) in category \(E\).

Instead of defining a new concept from scratch, we reuse the concept of functors when defining bifunctors. We could define a new category \(C \times D\) whose objects are pairs of pairs of objects from \(C\) and \(D\). As a result, \(a\) and \(b\) form a pair \((a, b)\) in \(C \times D\).

A functor maps not only the objects but also the morphisms between objects. Similarly, a bifunctor should also somehow map the morphisms. Suppose we have another pair of objects \(a'\) and \(b'\) which are represented by an object \((a', b')\) in \(C \times D\), and we have \(f\) goes from \(a\) to \(a'\) and \(g\) from \(b\) to \(b'\), we could thus define a new morphism called \((f, g)\) in \(C \times D\) which maps the first component of \((a, b)\) using its first component \(f\), and the second component using \(g\). Note that every morphism from \((a, b)\) to \((a', b')\) is constructed by picking one morphism from the morphisms between \(a\) and \(a'\), and another one from the morphisms between \(b\) and \(b'\). Therefore, if there are \(m\) morphisms between \(a\) and \(a'\), and \(n\) between \(b\) and \(b'\), there will be \(m \times n\) morphisms between \((a, b)\) and \((a', b')\).

A bifunctor, as a result, is defined as a functor \(F\) from \(C \times D\) to \(E\), which maps \((a, b)\) to \(F a b\) and \((a', b')\) to \(F a' b'\). It also preserves the morphism structure by mapping \((f, g)\) to a morphism \(F (f, g)\) between \(F a b\) and \(F a' b'\).

However, we usually deal with a single category when working on Hask, so actually both \(D\) and \(E\) are the same as \(C\). Therefore, we could think of a bifunctor as one which takes two objects \(a\) and \(b\) and maps them to another object \(F a b\). The bifunctor also maps two morphisms, \(f\) and \(g\), to \(F f g\).

Products and coproducts are bifunctors

Hask

A product in Hask is a bifunctor which maps \(a\) and \(b\) to \((a, b)\), and \(f\) and \(g\) to \((f, g)\). Or if we define it in terms of the product category (right side), it’s just an \(id\) mapping.

Similarly, a coproduct in Hask is also a bifunctor which maps \(a\) and \(b\) to \(a | b\). What is interesting is how it maps the morphisms. The coproduct bifunctor takes two morphisms \(f\) and \(g\), and, based on whether the source object is \(a\) or \(b\), apply \(f\) or \(g\) conditionally.

General categories

We have two objects \(a\) and \(b\) whose product is \(a * b\), which means projection morphisms \(p\) and \(q\) from \(a * b\) to \(a\) and \(b\), respectively. We also have \(a'\), \(b'\) and \(a' * b'\). Now we want to map the morphisms \(f\) and \(g\) to some other morphism in the category which maps \(a * b\) to \(a' * b'\).

We could consider \(a * b\) to be another candidate of the product of \(a'\) and \(b'\). By the definition of product, there must be a unique morphism \(h\) from \(a * b\) to \(a' * b'\) such that \(p' \cdot h = f \cdot p\) and \(q' \cdot h = g \cdot q\). Let’s define \(f * g\) to be this \(h\), and here is how the product bifunctor maps morphisms.

Similarly, coproducts are also bifunctors, see the figure below.