This will be a scratchpad of notes I'm taking down as I'm reading Bartosz Milewski's Category Theory for Programmers. As such, this will only contain key points, summary of ideas, and attempts at posed challenges. His posts are very detailed, the expositions are clear, presents cute illustrations, and contains exercises that will help check your understanding.

Why category theory

In this preface for this entire series, Milewski details why he is writing this series, and why he thinks a programmer should read it.

To summarize, category theory has a lot of interesting ideas for programmers. For example, the key idea in category, composition, is "the essence of programming".

He also brings up the topic of multicore and parallelism as an upcoming paradigm, which functional programming is suitable for. This new environment forces us to think in a different way and adapt, and category theory is a way to help us accomplish that.

Category: The Essence of Composition

From Bartosz Milewski's Category: The Essence of Composition

A category has objects and arrows, also called morphisms, that start from one object and end up at another object. This makes a category easy to represent in diagrams, looking like a directed graph.

Rule 1: Morphisms compose

The morphism from A to C is the composition of the previous 2 morphisms, and is usually written g ∘ f.

As a simplified and inaccurate example, in shell we have the pipe notation:

ls | wc # counts the number of files in the current directory

First there is a morphism from directory to list of files, ls, and a morphism from a list of files to a number, wc. The pipe | is the composition operator, that composes ls and wc into a morphism, ls | wc that takes directory to a number.

Rule 2: Composition is associative

An example of an associative operator is + on natural numbers.

1 + (2 + 3) = (1 + 2) + 3

In the world of category, for any three morphisms, f, g, h that can be composed, the order of composition does not matter, as long as the order of morphisms don't change.

h ∘ (g ∘f) = (h ∘ g) ∘f

Thus, we can omit parenthesis and write it simply as:

h ∘ g ∘ f

Rule 3: Every object has a unit of composition

Every object A has a morphism that goes straight to itself.

This morphism can be composed with any other morphism that starts or ends at A, giving back the same morphism.

We call this morphism the identity on A, written as idA:

f ∘ idA = f
idA ∘ f = f

At first thought, I was reminded of a monoid. A monoid requires a set of elements (morphisms), an associative operation (composition), and a identity element. In a category the identity element depends on the object to be composed with, whereas a monoid requires an identity element that works for all members in the set. So that's where the similarities break down.

As a concluding thought, this is one of my favourite quote from Bartosz Milewski in this chapter, talking about objects:

All you can ever know about it is how it relates to other object — how it connects with them using arrows. ... The moment you have to dig into the implementation of the object in order to understand how to compose it with other objects, you have lost the advantages of your programming paradigm.

I will attempt a few of the challenges he has set:

  1. Is Facebook a category, with people as objects and friendships as morphisms? No, friendships don't compose (see Rule 1); a friend (morphism from B to C) of my friend (morphism from A to B) is not immediately my friend (no morphism from A to C).
  2. When is a direct graph a category? When every node has an edge to itself (Rule 3), and there exists a transitive (if we think in terms of sets and relations) relationship (Rule 1) between nodes, i.e. if there is an edge from A to B, and an edge from B to C, there must be an edge from A to C.

Types and Functions

From Bartosz Milewski's Types and Functions

In the last chapter, we talked about objects and morphisms, which sounds very general and makes it hard to see any obvious practical use. In this section we can specialize the concepts into types and functions, something closer to what programmers deal with.

Morphisms compose, but only if the target of one morphism matches the source of the other. For example, a morphism from A to B can be composed with a morphism from B to C (the B matches) but not with a morphism from Y to Z.

This is similar to the restrictions in a programming language with a strong type system: the objects A, B, C, are types, and the morphisms are functions from one type to another.

Types can be thought of as a set of values: a Bool type is a set containing the values True and False, the Integer type can be thought of as an infinite set containing all integers (this is the case in Haskell).

This association of types with sets has some subtleties. However throughout the post Milewski works with a special category Set, in which objects are sets, and morphisms are functions. Set allows us break the rule and peek inside its objects. The goal is to generalize from this, and slowly remove the need to look into objects.

If we have an empty set, what is the corresponding type? In Haskell, this is called Void, and this type has no values. Which means that we can define a function that takes a Void, but we can never call it, because we cannot provide a value of type Void to apply the function to, because there are none!

How about the singleton set, a set with only 1 member? The type has only one value, and it is usually called unit, represented in many languages as (). If we have a function from unit to a type, say Bool, we can implement it in two ways: return a True or return a False. In fact, for any type A, there are as many functions from unit to A as there are members in A.

How about a function from A to unit? The only thing that the function can do is return unit. For any type A, there is a single function from A to unit.

Here I tackle some challenges:

  1. How many different functions are there from Bool to Bool? 4, or 2 ^ 2:

    id x = x
    not x = ! x
    true x = true
    false x = false
  2. Draw a picture of a category whose only objects are the types Void, (), and Bool; with arrows corresponding to all possible functions between these types.

    Void ---id---> Void # Rule 3
    Void -absurd-> ()
    Void -absurd-> Bool
    ()   ---id---> () # Rule 3
    ()   --true--> Bool # always returns true
    ()   --false-> Bool # always returns false
    Bool ---id---> Bool # Rule 3
    Bool ---not--> Bool # negation, turns true to false
    Bool ---not--> Bool # negation, turns false to true
    Bool --unit--> ()

Categories Great and Small

From Bartosz Milewski's Categories Great and Small

This chapter examines a few different examples of categories.

An empty category

Just like an empty set has no elements, an empty category has no objects, and thus no morphisms. This might not seem useful initially, but Milewski hints at the future, when we consider the category of categories.

Building a category from a graph

Given any directed graph, we can treat each node as an object, and edges as morphisms. Then, to build a category, we can add arrows until all the rules are satisfied.

  1. Start by adding identity arrow at each node (Rule 3),
  2. Then add composition arrows (Rule 1),
  3. More composition arrows might become required to satisfy the rules, thus
  4. GOTO 2

The category might end up with infinitely many arrows. (And Milewski assures us that it's okay, so let's trust him).

Such a category is called a free category generated by a given graph. We are adding the minimum number of items to a structure to fulfill the laws. This process is called a free construction.


Instead of plain directed graphs, imagine edges as relations: an edge exists between two nodes if they are related. Given a set of nodes and the less than or equal relation, we have a category:

The set with such a relation is called a preorder, and a preorder is a category.

In categorical terms, a preorder will have at most one morphism going from any object a to any object b, this is known as a thin category. This example is not a thin category (omitting the identity morphism) because there are two morphisms from a to b.

a ---> b
a ---> b
b ---> a
(a != b)

If we make the less than or equal relation stronger, such that if a <= b, and b <= a, then a and b must be the same object, we have a partial order. This is not a partial order because a and b are related to each other, but they are not the same object, i.e. there are cycles.

a ---> b
b ---> a
(a != b)

If we enforce that any two objects are in a relation with each other (or equivalently, every object is related to one another), then we have a linear order or total order.


For any objects a and b in a category C, the set of morphisms from a to b is called the hom-set, written as C(a,b).

In a pre-order, every hom-set is either the singleton (definition of thin category), or the empty set (not related).

Trivially, the hom-set C(a,a) is the singleton set consisting of the identity morphism on a.

Monoid as Set and Category

A monoid can be thought of as a set with an identity element, and a binary operator that is associative. An example of a monoid is the set of strings with the empty string, "", as the identity element, and the string concatenation operator, usually ++ or ^, as the associative binary operator.

With this idea of a monoid as a set, we can try to imagine monoid as a category. To do that, we need to stop thinking of sets and elements and relations, and instead think of objects and morphisms.

Here's an idea we can try: the application of a binary operation can be thought of as the action of "shifting" things around in the set.

For example, adding 5 to a natural number, let's call this add5. It maps 0 to 5, 1 to 6. add5 is a function defined on the set of natural numbers. There are many more functions that does something similar, for example add1, add2, add10, etc. In fact there are as many of these functions as there are natural numbers, infinite! We can call each of these addN functions the "adder" of n.

Adders can compose: the composition of add5 and add1 is effectively an add6. Composing adders is equivalent to composing additions (Rule 1). The composition of adders is associative, since addition is associative (Rule 2).

This composition of adders is called mappend in Haskell

There is a special adder, add0, which is equivalent to the identity function on the set of natural numbers, it doesn't move anything around (Rule 3).

This identity of adders is called mempty in Haskell

The set of natural numbers is then a single object (no peeking into the object), with many morphisms (infinite), which are the adders (including the identity add0). So we observe that a monoid (when viewed as a category) is a single object category!

The other way holds as well, i.e. a single object category forms a monoid. From a single object category, we can extract a unique set with a binary operator. This set is the hom-set M(m, m), i.e. the set of all morphisms from m to m, where m is the single object in the category M. We can then define a binary operator for this set, call it +, such that for any two morphisms f and g, the application of this operator is the composition of these two morphisms. The identity morphism is the identity element of the hom-set under the operator +.

To put it in more concrete terms, in the world of natural numbers, if we imagine the morphisms in the category M as curried additions, such as add5, the hom-set contains these curried additions, and you can apply a binary operator to any two of these morphisms. This unique binary operator in the hom-set is defined by the composition of the curried additions in M. Notice how the binary operator is uniquely defined by the compositions. Both + and x are candidate binary operators for the set of natural numbers to form a monoid, but the morphisms defined in M requires + to be selected. If we define the morphisms differently, we end up with a different operator.

I shall attempt some of the challenges:

  1. Generate a free category from a graph with one node and no edges. Following the free construction, we need to add the identity morphism, that's it. We can try composing the identity morphism, but by Rule 3, composing identity with anything gives you identity, so we don't have any new arrows to add.
  2. Generate a free category from a graph with one node and one directed edge. Following the free construction, we first add the identity morphism. Since we have one directed edge (that is not the identity morphism), we can compose it with itself. The resulting composition can be composed with again, so we get an infinite number of morphisms that starts and ends at the node.
  3. What kind of order is this? A set of sets with the inclusion relation: A is included in B if every element of A is also an element of B. Partial order. If every element of A is in B, and every element in B is in A, then they are the same set.
  4. Considering that Bool is a set of two values True and False, show that it forms two (set-theoretical) monoids with respect to, respectively, operator && (AND) and || (OR). Monoid with respect to &&: set of natural numbers, True as identity element, && is associative. Monoid with respect to || set of natural numbers, False as identity element, || is associative.
  5. Represent the Bool monoid with the AND operator as a category: List the morphisms and their rules of composition. Two morphisms representing (AND True) and (AND False). (AND True) is also the identity morphism. The composition of morphisms follow the binary application of the boolean operators, the composition of (AND True) with any morphism f is f, the composition of (AND False) with any morphism g is (AND False)

Kleisli Categories

From Bartosz Milewski's Kleisli Categories

We start with an example of logging to motivate this section.

Suppose we have two pure functions:

toUpper :: String -> String
toWords :: String -> [String]

toUpper turns a string uppercase, and toWords breaks up a string by whitespace.

toUpper "abc" = "ABC"
toWords "a b c" = ["a", "b", "c"]

And we can define a composition of these two functions as

toUpperWords :: String -> [String]
toUpperWords = toWords . toUpper

We might want to have an audit trail (or log) of the function calls. The usual imperative way to do this might be to append to a global log, which will cause these two functions no longer be pure.

Another way is to embellish these functions, which means to add something to the return value.

toUpper :: String -> (String, String)
toWords :: String -> ([String], String)

We make the functions return a pair of values now, where the first value in the pair is the original return value, and the second value in the pair is the string we want to log.

Our original function composition no longer works, because now the types don't match.

So we have to be more explicit: we call these functions and aggregate the logs:

  1. Call toUpper on an input string to get a value and a log,
  2. Call toWords on the value from 1, to get a list of words, and another log,
  3. Return the value from 2, and append the log from 1 and 2 together
toUpperWords :: String -> ([String], String)
toUpperWords s =
    let (s1, l1) = toUpper s
        (s2, l2) = toWords s1
    in (s2, l1 ++ l2)

This becomes repetitive very quickly, and hides the original purpose, mixing the logic of logging and concatenating logs with the logic of toUpper and toWords.

We can then try to come up with a new kind of function composition that helps us to handle this embellishment:

  1. Call the first embellished function,
  2. Call the second embellished function with the first value of 1,
  3. Concatenate the second value of 1 and 2,
  4. Return a pair of the first value from 2, and the concatenated value from 3.

First, let's make up a more generic type for this sort of embellished function:

type Writer a = (a, String)
toUpper      :: String -> Writer String
toWords      :: String -> Writer [String]
toUpperWords :: String -> Writer [String]

Where Writer [String] is a type where the first component has type [String], and the second component has type String representing the log.

Then, the composition of embellished functions is:

embellished_compose :: (a -> Writer b) ->
                       (b -> Writer c) ->
                       (a -> Writer c)
embellished_compose f g s =
    let (s1, l1) = f s
        (s2, l2) = g s1
    in (s2, l1 + l2)

And thus, toUpperWords looks like this:

toUpperWords :: String -> Writer [String]
toUpperWords s = embellished_compose toUpper toWords

Now we can look at this new composition from a categorical point of view. The types are still objects, and the morphisms are the embellished functions. The important point is that the morphism is still considered an arrow between String and String, or String and [String], although the function actually returns a pair.

In this new category, embellished_compose is composition, and the identity morphism will take a type to the same type, and not add anything to the log, implemented as the return function:

return :: a -> Writer a
return s = (s, "")

We can check that this forms a category by checking all the rules:

  1. Morphisms compose, as shown by embellished_compose, satisfying Rule 1,
  2. Composition is associative, because at the first part of the pair we have function composition (which is associative) and on the second part we have concatenation, which is associative, satisfying Rule 2,
  3. Every object has a unit of composition, the return function, satisfying Rule 3.

The log here is actually monoidal, "" is mempty, and + is mappend.

Kleisli Categories

This category we have constructed is actually an example of a Kleisli category, a category based on a monad. The objects of a Kleisli category are the types of the underlying programming language, and morphisms from A to B are functions from the types A to an embellished type B.

The morphisms in the above example are from the objects String to String, and String to [String], whereas the functions were from the types String to Writer String, and String to Writer [String].

Each Kleisli category defines it's own embellishment, composition of morphisms, and identity morphisms. The example Kleisli category above is based on a writer monad, which has the embellishments, composition, and identity has described.

Previously when we thought of programming languages as categories, we modelled types as objects and functions as morphisms, and defined categorical compositions as function compositions, which merely passes the output of one function as the input to another. Here we defined a different category, where the categorical compositions are not the usual function compositions: it does more than passing output as input, and (in this example) can append to a log as well.

Here I attempt the challenges (I don't write C++ so the answers probably won't run, but will hint at the solution):

  1. Construct the Kleisli category for partial functions (define composition and identity). The composition of two embellished function f and g first calls f with the input. If it is not valid, return invalid. Otherwise pass the return value of f to g, and returns the result. The identity is a function that a valid optional with value of type A.

    auto const compose = [](auto m1, auto m2) {
        return [m1, m2](auto x) {
            auto p1 = m1(x);
            if (p1.isValid()) return m2(p1.value())
            // this is optional<B>, where B is the return type of m2
            else return optional()
    template<class A>
    optional<A> identity(A x) {
      return optional(x);
  2. Implement the embellished function safe_reciprocal that returns a valid reciprocal of its argument, if it’s different from zero.

    optional<double> safe_reciprocal(double x) {
      if (x == 0) return optional<double>{};
      else return optional<double>{1/x};
  3. Compose safe_root and safe_reciprocal to implement safe_root_reciprocal that calculates sqrt(1/x) whenever possible.

    optional<double> safe_root_reciprocal(double x) {
      return compose(safe_reciprocal, safe_root)(x);

Products and Coproducts

From Bartosz Milewski's Products and Coproducts

Universal construction

In a category, we only have objects and morphisms. An object can be singled out by describing its relationships with other objects based on morphisms. We can pick a pattern of relationships, a shape, and try to find occurrences of this pattern in the category. We can then rank these occurrences and find the best-fit. This is the universal construction.

Initial object

Let's start with the pattern of the single object. There are as many of these as there are objects in a category. We can rank these objects using an initial trait. An object A is more initial than B if there is a morphism from A to B. Then, the most initial object has morphisms to every other object.

This object might not exist, or there may be more than one such object. Then, if we restrict the category to be ordered, such as a preorder, we are able to find the initial object, which has one and only one morphism going to any object in the category.

This restriction to ordered categories still does not guarantee uniqueness of the initial object, only up to isomorphism, explained later.

For example, the set of natural numbers has an initial object, 0, with the less than or equal relation. The set of integers however, is infinite, and has no initial object. In the category of types and function, the initial object is Void, the empty set of values. The morphism from Void to any other type is this absurd :: Void -> a.

Terminal object

If we think of the initial object as the bottom, then maybe we can have a top, called the terminal object. We define this by tweaking our ranking slightly: an object A is more terminal than B if there is a morphism from B to A. The most terminal object is then the object which has one and only one morphism coming to it from any object in the category. Note that this does not say anything about morphisms going out from A.

Similar to the initial object, the terminal object might not exist, or there might be more than one such object, again up to isomorphism.

For example, the set of integers does not have a terminal object. In the category of types and functions, the type which has one and only one arrow going to it from any other type is unit, or (), and the function is unit : a -> ().


There is a symmetry between the definition of the initial object and terminal object. The difference is direction of the morphism, and hence the ranking. In fact, for any category, we can define the opposite category with all the morphisms flipped, and the compositions flipped, and identity remaining identical.

This duality means that for every pattern in category theory, you get the opposite one for free: product and coproduct, monad and comonad, etc.

So, the terminal object can be thought of as the initial object in the opposite category.


Isomorphic objects look the same, the structure have a one-to-one mapping. An isomorphism is a pair of morphisms, one being the inverse of the other. So given objects A and B, the morphism f from A to B and the morphism g from B to A, if f and g are the inverse of each other then they form an isomorphism.

We can examine isomorphisms in the context of composition and identity:

f ∘ g = id
g ∘ f = id

The composition of isomorphisms is the identity morphism, since they are the inverse of each other.

Looking back at how we defined initial objects, any two initial objects a and b must be isomorphic.

  1. Since a is an initial object, there is one and only one morphism f from a to b
  2. Since b is an initial object, there is one and only one morphism g from b to a
  3. Thus, f ∘ g must be a morphism from a to a
  4. Since a is initial, there is only one morphism from a to a
  5. Since a is an object in a category, it must have an identity morphism
  6. Therefore, f ∘ g must be the identity morphism and is unique

A symmetrical proof of g ∘ f can be made. And by duality, this proof can be used for showing uniqueness up to isomorphism for terminal objects as well.

Observe that because of the initial object conditions, f and g are unique as well. Thus we actually have uniqueness up to unique isomorphism, i.e. not only is the initial object unique up to isomorphism, the isomorphisms are themselves unique.

What we have done so far is to use universal construction to find the initial object (unique up to isomorphism) and the terminal object (unique up to isomorphism). Let's spot more patterns


The cartesian product of two sets is a set of pairs, but we shouldn't be able to peek into the objects, so let's think in terms of morphisms and compositions.

Let's define a pattern using two morphism from a product, c, to its constituents, a and b. p is a morphism from c to a, and q is a morphism from c to b. In this example, p and q can also be called projections.

p :: c -> a
q :: c -> b

Again, there might be lots of candidates c that fits into this pattern. For example, for the constituent types Int and Bool, a candidate for their product is Int. Then,

p :: Int -> Int
p x = x

q :: Int -> Bool
q _ = True

This candidate seems too small, q doesn't look sensible because it always returns the same thing.

Another candidate is a triple of (Int, Int, Bool):

p :: (Int, Int, Bool) -> Int
p (x, _, _) = x

q :: (Int, Int, Bool) -> Bool
q (_, _, b) = b

This candidate seems too big, the second Int is never used.

We can say that a candidate c is better than another candidate c' if there is a morphism m from c' to c. We also want the projections p and q of c, to be better than the projections p' and q' of c'. Better here means that one can be constructed in terms of another:

p' = p . m
q' = q . m

Another way to say this is that m factorizes p' and q'.

Let's have another candidate for c, (Int, Bool), and show that this is the best candidate:

p :: (Int, Bool) -> Int
p (x, _) = x

q :: (Int, Bool) -> Bool
q (_, b) = b

First we construct a morphism m from c', Int, to c, (Int, Bool):

m :: Int -> (Int, Bool)
m x = (x, True)

And observe that p' and q' is defined in terms of p, q and m:

p' x = (p . m) x = x
q' x = (q . m) x = True

To show that it s better than (Int, Int, Bool), we construct a morphism m from c', (Int, Int, Bool), to c, (Int, Bool):

m :: (Int, Int, Bool) -> (Int, Bool)
m (x, _, b) = (x, b)

And observe that p' and q' is defined in terms of p, q and m:

p' (x, y, b) = (p . m) (x, y, b) = x
q' (x, y, b) = (q . m) (x, y, b) = b

Thus, (Int, Bool) is better than the first two candidates we came up with. How about the other way round?

For the first candidate, Int, we lose information: we will never be able to get back the value of the Bool, since it always returns True.

For the second candidate, there are more than one way to factorize p' and q'. If we try to define an m' from (Int, Bool) to (Int, Int, Bool), we can do it in many ways:

m' (x, b) = (x, x, b)
m' (x, b) = (x, 42, b)
m' (x, b) = (x, 99, b)

From our examples, we see that for any type c', with projections p' and q', there is a unique m from c' to the cartesian product (a, b), that factorizes them. Thus, this makes the cartesian product the best match.

Categorically, a product of two objects a and b, is the object c with two projections, p from c to a and q from c to b, such that for any other object c' with two projections p' from c' to a and q' from c' to b', there is a unique morphism m from c' to c, such that m factorizes p' and q'.


By duality, the product has a dual. When we reverse all the arrows in a product, we get a coproduct. Define two morphisms i and j from objects a and b to an object c. i and j are also called injections.

i :: a -> c
j :: b -> c

Suppose there is another object c', and there are two morphisms i' from a to c' and j' from b to c'.

We say that c is better than c' if there is a morphism m from c to c' (when we flip the arrows, the ranking is flipped too), that factorizes the injections.

i' = m . i
j' = m . j

The best candidate c, is called a coproduct, and if exists, is unique up to unique isomorphism.

In the category of types (as sets) and functions, the coproduct is the disjoint union of two sets. In the example below, the type Toss is the coproduct, the disjoint union of the sets Heads and Tails. Heads and Tails also act as the morphisms i and j.

data Toss = Heads | Tails

Here I attempt some challenges:

  1. Show that the terminal object is unique up to unique isomorphism. The terminal object, has one and only one morphism from any object to it. Assume we have two terminal objects a and b, there must be a morphism f from a to b, and a morphism g from b to a. The composition of these two morphisms f ∘ g is a morphism from b to b. Because b is a terminal object, there can only be one morphism from b to b. And because b is an object in a category, it has an identity morphism. Thus the morphism from b to b is the identity morphism and must be unique. The same argument can be made for the composition g ∘ f. Therefore we have unique isomorphisms between a and b, and the terminal object is unique up to isomorphism.
  2. What is a product of two objects in a poset? Hint: Use the universal construction. Let c be an object with projections p to an object a, and q to object b. When this is a poset, it means that c is related to a and b. Let c' be an object with projections p' and q'. Then there exists a unique morphism m from c' to c, which means that c' is related to c. So the product of two objects in a poset is the greatest (defined by the ordering) object that is related to both a and b. Using less than equal relation as an example, the product of two numbers, is the smaller number. Using sets and subset relation as an example, the product of two sets is the intersection.
  3. What is a coproduct of two objects in a poset? Let c be an object with injections i from a, and j from b. Let c' be an object with injections i' from a and j' from b'. For c to be a coproduct of a and b, there must be a morphism m from c to c'. In a poset this means that c must be ordered less than c'. For example, with the less than or equal relation, the coproduct of two objects is the bigger object. For example, with the subset relation, the coproduct of two objects is the union of the sets.

Simple Algebraic Data Types

From Bartosz Milewski's Simple Algebraic Data Types

Product types

In many languages, a product of two types is implemented as a pair. Pairs are not commutative.

For example, (Bool, Int) is not the same as (Int, Bool), although they carry the same information. The isomorphism between these two types is the swap function, which is it's own inverse.

swap :: (a, b) -> (b, a)
swap (x, y) = (y, x)

Pairs can be nested as well, for example ((a, b), c), a pair of two types, the first value being a pair (a, b), and the second value a c.

This is isomorphic to the nested pair (a, (b, c)).

If you view the , as a binary operation to create a pair out of two types, then , looks like an associative operator.

Also, the pair of any type a with unit, (), is isomorphic to a itself.

In this sense, the category of sets (in programming languages), is a monoidal category. It is a category, that's also a monoid. The binary associative operator is the product, and the identity element is the unit type, ().

Coproduct types (Sum types)

A sum type can be expressed in Haskell as:

data Either a b = Left a | Right b

Similar to pairs, Either is commutative up to isomorphism, and can be nested.

The category of sets is also a monoidal category with the disjoint sum as the binary operator, and Void as the identity element.

Algebra of Types

We can try to view types as numbers, Void maps to zero, () to one. Then product can be thought of as multiplication, and coproduct as addition.

Previously we defined a pair (a, ()) and said that it was isomorphic to a. In this sense, a * 1 = a.

Numbers Types





a * 1

(a, ())

a + b

Either a b = Left a | Right b

In natural numbers, we have distribution of multiplication over addition:

a * (b + c) = a * b + a * c

The left hand side translates, at the types level, to:

(a, Either b c)

And the right hand side to:

Either (a, c) (b, c)

To show that these two are equal up to isomorphism, we have to find a morphism f from the left hand side to the right hand side, and the inverse morphism g:

f :: (a, Either b c) -> Either (a, c) (b, c)
f (a, Left b) = Left a b
f (a, Right c) = Right b c

g :: Either (a, c) (b, c) -> (a, Either b c)
g Left (a, b) = (a, Left b)
g Right (a, c) = (a, Right c)

A structure with two intertwined monoids of these sort is called a semiring.

Specifically, a semiring is a structure with an associative, commutative addition operator with an identity 0, a associative multiplication operator with an identity 1, the multiplication left and right distributes over addition, and multiplication by 0 gives 0.

Recursive types

The List type is commonly defined as a coproduct of Nil, indicating the end of the list, and Cons a (List a), indicating an element of type a and the rest of the list:

List a = Nil | Cons a (List a)

In terms of an algebraic equation, letting x = List a:

x = 1 + (a * x)

Note that Nil is isomorphic to (), and hence 1. And Cons is isomorphic to ,, or the pair operator, and so it's the * of two types, a and List a, which we defined to be x.

We can try to solve this algebraic equation by expansion:

x = 1 + (a * x)
x = 1 + (a * (1 + a * x)) = 1 + a + a * a * x
x = 1 + a + a * a + ... + a * a * a * a...

Which can be interpreted as, x is either Nil (empty), or a singleton (a), or a pair of values (a * a), or three values (a * a * a), ...

I previously wrote a post that goes a bit more into the relationship between numbers and types.

There is also a relationship between types and intuitionistic logic, specifically Void and false, () and true, product with and, coproduct with or. This relationship is more commonly known as the Curry-Howard isomorphism, and also popularized as propositions as types.

Here I attempt some challenges:

  1. Show the isomorphism between Maybe a and Either () a.

    f :: Maybe a -> Either () a
    f Nothing = Left ()
    f Just x = Right x
    g :: Either () a -> Maybe a
    g Left _ = Nothing
    g Right x = Just x
  2. Show that a + a = 2 * a holds for types (up to isomorphism). Remember that 2 corresponds to Bool, according to our translation table.

    f :: Either a a -> (Bool, a)
    f Left x = (True, x)
    f Right x = (False, x)
    g :: (Bool, a) -> Either a a
    g True x = Left x
    g False x = Right x


From Bartosz Milewski's Functors

A functor is a mapping between categories. A functor F from category C to category D maps objects and morphisms in C to D.

Given an object a in C, the functor F maps a to the object F a in D. We call F a the image of a in D.

Given a morphism f from a to b in C, the functor F will map f to a morphism F f from F a to F b.

A functor preserves the source and destinations of a morphism under F, and also preserves composition of morphism.

If h is a composition of morphisms f and g, h = g . f, then the image under F is a composition F h = F g . F f.

And the functor also preserves the identity morphisms: for an object A, it maps the identity morphism idA from A to A in C to F idA in D, which is also the identity morphism idFA from F a to F a in D.

Functors in our programming language

In the category of types and functions in programming, we can talk about functors that map this category onto itself. A functor where the source and target category are the same is called an endofunctor. A functor in the this category then maps types to types, and functions to functions.

The mapping of types to types has been seen previously, it's the case of building a type parameterised by another type, such as a Maybe a, or an Either a b.

data Maybe a = Nothing | Just a
data Either a b = Left a | Right b

Maybe is a type constructor that maps types to types. Give it a type Bool and you will get a type Maybe Bool.

We can try making Maybe a functor. It already maps types to types, what remains is to map functions to functions. Given a function f :: a -> b, a functor should map f to a function f' :: Maybe a -> Maybe b. Such a function f' can be defined as such:

f' :: Maybe a -> Maybe b
f' Nothing = Nothing
f' (Just x) = Just (f x)

We can also define a higher order function to map the function f to f'::

fmap :: (a -> b) -> (Maybe a -> Maybe b)
fmap _ Nothing = Nothing
fmap f (Just x) = Just (f x)

fmap can be said to lift a function from the type of f to the type of f', so now the lifted function can work on values of type Maybe a

The final steps to show that Maybe (which maps types to types), and fmap (which maps morphisms to morphisms) form a function, we need to prove the functor laws, namely that they preserve composition and identity.

Let's start with identity, we need to prove that fmap preserves identity morphisms:

fmap id = id
-- case one, Nothing
fmap id Nothing = Nothing
                = id Nothing
-- case two, Just x
fmap id (Just x) = (Just id x)
                 = (Just x)
                 = id (Just x)

Then show that fmap preserves composition:

fmap (g . f) = fmap g . fmap f
-- case one, Nothing
fmap (g . f) Nothing = Nothing
                     = fmap g Nothing
                     = fmap g (fmap f Nothing)
                     = (fmap g . fmap f) Nothing
-- case two, Just x
fmap (g . f) (Just x) = Just ((g . f) x)
                 = Just (g (f x))
                 = fmap g (Just (f x))
                 = fmap g (fmap f (Just x))
                 = (fmap g . fmap f) (Just x)

Thus we show that Maybe and fmap forms a functor.


Haskell uses a typeclass mechanism to specify a functor. It specifies that types adhere to a common interface. For example we can specify a Functor typeclass.

class Functor f where
    fmap :: (a -> b) -> f a -> f b

This defines that f is a Functor if it supports an fmap function.

And to declare that a type is a functor, for example Maybe, we declare the type as an instance of a typeclass:

instance Functor Maybe where
    fmap _ Nothing = Nothing
    fmap f (Just x) = Just (f x)

This implementation of fmap for Maybe is exactly what we came up with earlier.

List functor

We can try to spot more functors in our programming language. Good candidates for functors are type constructors, since they map types to types, such as List.

The List type constructor maps a type a to a type List a, and its fmap function looks takes a a -> b to List a -> List b.

We can make List an instance of a Functor as such:

instance Functor List where
    fmap _ Nil = Nil
    fmap f (Cons x tl) = Cons (f x) (fmap f tl)

What's happening here is that we are applying f to the head of the list, and recursively using fmap to apply f to the rest of the list (the tail).

Let's talk about the type of functions. Whenever we are defining a function, we use -> in the type signature, such as f :: a -> b. We can think of -> as an infix type constructor that takes a two types, a and b, to a type a -> b.

This is different from our previous examples, where the type constructors only take one type.

If we provide only one type to ->, we get r ->, or as a prefix (->) r, which is a whole family of type constructors parameterized by r.

To satisfy our functor laws, we want to lift a -> b to (r -> a) -> (r -> b).

instance Functor ((->) r) where
    fmap :: (a -> b) -> (r -> a) -> (r -> b)
    fmap f g = f . g

The type constructor ((->) r) and the fmap implementation forms the reader functor.

Const functor

Another functor is the Const functor, which ignores it's second argument, so the implementation of fmap can ignore the function argument too:

data Const c a = Const c

instance Functor (Const c) where
    fmap _ (Const v) = Const v

Functor composition

Functors between categories compose as well, just like morphisms compose.

When you compose two functors, the object mappings compose, and the morphisms compose too: identity morphisms and composition of morphisms in the initial category are still identity morphisms and composition of morphism in the final category.

Like morphisms in a category, functor composition is associative (since the mapping of object and morphisms is associative), and there is also an identity functor in every category (maps every object and morphism to itself).

So functors seem to act like morphisms, if that's the case, what category is it? A category of categories, where the objects are categories and morphisms are functors. A category of all categories would get us into trouble, leading us to Russell's Paradox. Instead there is a category of all small categories called Cat, which itself is big, and so does not contain itself.

Here I attempt some challenges:

  1. Can we turn the Maybe type constructor into a functor by defining: One of the functor laws is that of preserving identity morphisms. However, this definition of fmap does not do that:

    fmap id (Just x) = Nothing
                    != Just x
                     = id (Just x)
  2. Prove functor laws for the reader functor. Hint: it’s really simple.

    fmap id = id
    fmap id r = id . r
              = r
              = id r
    fmap (f . g) r = (f . g) . r
                   = f . (g . r)
                   = fmap f (g . r)
                   = fmap f (fmap g r)
                   = (fmap f . fmap g) r
  3. Prove the functor laws for the list functor. Assume that the laws are true for the tail part of the list you’re applying it to (in other words, use induction).

    fmap id ls = id ls
    fmap id Nil = Nil
                = id Nil
    fmap id (Cons x t) = Cons (id x) (fmap id t)
                       = Cons x t -- inductive hypothesis, fmap id t = t
                       = id (Const x t)
    fmap (f . g) Nil = Nil
                     = fmap f Nil
                     = fmap f (fmap g Nil)
                     = (fmap f . fmap g) Nil
    fmap (f . g) (Cons x t) = Cons ((f . g) x) (fmap (f . g) t)
                            -- induction hypothesis fmap (f . g) t = (fmap f . fmap g) t
                            = Cons (f (g x)) ((fmap f . fmap g) t)
                            = Cons (f (g x)) (fmap f (fmap g t))
                            = fmap f (Cons (g x) (fmap g t))
                            = fmap f (fmap g (Const x t))
                            = (fmap f . fmap g) (Const x t)


From Bartosz Milewski's Functoriality


A functor maps objects and morphisms from one category to another. A functor with two arguments is called a bifunctor.

A bifunctor maps every pair of objects, one from category C, one from category D, to an object in category E. In other words, it is mapping the cartesian product of C and D to E.

Also, it has to map a pair of morphisms, one from C and one from D, to E. It maps the cartesian product of morphisms in C and D to E. These morphisms in E can also be composed:

(f , g) . (f', g') = (f . f', g . g')

This composition of pairs of morphisms is associative and has an identity (id, id)

If you have a mapping from a pair of categories to a third category, and it can be proven that it acts like a functor on each argument separately, then the mapping is a bifunctor.

A functor in Haskell is a typeclass that takes one argument, a bifunctor in Haskell is then a typeclass that takes two arguments.

class Bifunctor f where
    bimap :: (a -> c) -> (b -> d) -> f a b -> f c d
    bimap g h = first g . second h
    first :: (a -> c) -> f a b -> f c b
    first g = bimap g id
    second :: (b -> d) -> f a b -> f a d
    second = bimap id

bimap lifts a pair of functions, a -> c and b -> d to a f a b -> f c d, and can be defined in terms of first and second, which lifts a function from each argument of the bifunctor.

Product and Coproduct Bifunctors

An example of a bifunctor is the categorical product.

instance Bifunctor (,) where
    bimap f g (x, y) = (f x, g y)

It applies the first function to the first value of the pair, and the second function to the second value of the pair.

By duality, a coproduct is defined too, and is a bifunctor.

instance Bifunctor Either where
    bimap f _ (Left x) = Left (f x)
    bimap _ g (Right y) = Right (g y)

It was mentioned earlier that the category of types and functions is a monoidal category in two ways, with cartesian product as the operator and () as identity element, and with disjoint union as the operator and Void as the identity element. The operator also needs to be a bifunctor (and in this category they are).

In the previous chapter we defined functor instances for several data types, such as Maybe and Either. In fact they are built up from products and coproducts. Since we have seen that both product and coproduct are functorial and that functors compose. If we can show that that basic elements in an algebraic data types are functorial, we can show that parametrized algebraic data types are functorial too.

When building up our types, there are broadly two kinds of type constructors: the first do not take in any type arguments, such as Nothing in Maybe, the second takes an argument, like Just in Maybe.

The first kind is equivalent to a Const functor, which ignores the type parameter, and the second is equivalent to an Identity functor.

data Const c a = Const c

instance Functor (Const c) where
    fmap _ (Const v) = Const v

data Identity a = Identity a

instance Functor Identity where
    fmap f (Identity x) = Identity (f x)

Then Maybe can be defined, up to isomorphism, in terms of functors:

type Maybe a = Either (Const () a) (Identity a)

Writer functor

In an earlier chapter talking about the Kleisli category, we had an embellished type Writer:

tyep Writer a = (a, String)

The embellishment in the Kleisli category is always a functor. Kleisli category has a >=> composition operator of embellished type, and an identity morphism:

(>=>) :: (a -> Writer b) -> (b -> Writer c) -> (a -> Writer c)
return :: a -> Writer a

How can we define a functor for this embellishment?

fmap :: a -> b -> Writer a -> Writer b
fmap f = id >=> (\b -> return (f b))

Reader functor

The Reader functor is based on the partially applied arrow type constructor: (->) r.

type Reader r a = r -> a

We can try to make it functorial in the first argument, by defining a synonym:

type Op r a = a -> r

We would then need to implement an fmap with this signature:

fmap :: (a -> b) -> (a -> r) -> (b -> r)

However this is impossible, we won't know what to do with b! What we can try to do is to get b -> a, by somehow inverting the first argument to fmap. But we cannot do that easily. But we can go to an opposite category.

Consider a functor F from the opposite category of C, Cop, to D. The morphism from b to a in C, now becomes a to b in Cop. Consider another mapping called G, which is not a functor, from C to D. It maps objects like F, but reverses all morphisms. It maps an morphism b to a in C and maps it to the possible morphism g from a to b. Then G is a contravariant functor because it inverts the direction of the morphisms.


A profunctor is a mapping that is contravariant in the first argument and covariant in the second, and the arrow operator is an instance of a profunctor. It takes the cartesian product of the opposite category of C, Cop, and D to an object in E:

class Profunctor p where
  dimap :: (a -> b) -> (c -> d) -> p b c -> p a d
  dimap f g = lmap f . rmap g
  lmap :: (a -> b) -> p b c -> p a c
  lmap f = dimap f id
  rmap :: (b -> c) -> p a b -> p a c
  rmap = dimap id

Here I attempt some challenges:

  1. Show that the data type:

    data Pair a b = Pair a b

    is a bifunctor. For additional credit implement all three methods of Bifunctor and use equational reasoning to show that these definitions are compatible with the default implementations whenever they can be applied. This data type is equivalent to ,, which is a bifunctor.

    data Pair a b = Pair a b
    instance Bifunctor Pair where
        bimap :: (a -> c) -> (b -> d) -> Pair a b -> Pair c d
        bimap f g (Pair a b) = Pair (f a) (g b)
    -- default implementation of bimap
    bimap g h (Pair a b) = (first g . second h) (Pair a b)
                         = first g (second h (Pair a b))
                         = first g (bimap id h (Pair a b))
                         = first g (Pair (id a) (h b))
                         = first g (Pair a (h b))
                         = bimap g id (Pair a (h b))
                         = Pair (g a) (id (h b))
                         = Pair (g a) (h b)
    -- default implementation of first
    first g (Pair a b) = bimap g id (Pair a b)
                       = Pair (g a) (id b)
                       = Pair (g a) b
    -- default implementation of second
    second g (Pair a b) = bimap id g (Pair a b)
                        = Pair (id a) (g b)
                        = Pair a (g b)
  2. Show the isomorphism between the standard definition of Maybe and this desugaring:

    type Maybe' a = Either (Const () a) (Identity a)
    f :: Maybe a -> Maybe' a
    f Nothing = Left ()
    f (Just a) = Right (Identity a)
    g :: Maybe' a -> Maybe a
    g (Left _) = Nothing
    g (Right (Identity a)) = Just a
    -- to show that f is the inverse of g, we show that both left and right
    -- composition of f and g are equivalent to id
    (g . f) Nothing = g (Left ())
                    = Nothing
    (g . f) (Just a) = g (Right (Identity a))
                     = Just a
    (f . g) (Left ()) = f Nothing
                      = Left ()
    (f . g) (Right (Identity a)) = f (Just a)
                                 = Right (Identity a)
  3. Show that PreList is an instance of Bifunctor. We can show that this definition is functorial in both arguments by redefining it in terms of basic functorial building blocks.

    data PreList a b = Nil | Cons a b
    data Prelist a b = Either (Const () a) (Identity (a ,b))
    instance Bifunctor Prelist where
        bimap :: (a -> c) -> (b -> d) -> Prelist a b -> Prelist c d
        bimap f g Nil = Nil
        bimap f g (Cons a b) = Cons (f a) (g b)

Function Types

From Bartosz Milewski's Function Types

When working in the category of types and functions, the objects are sets representing types. For example the object Bool is a set with two elements. A function type a -> b must exist in the set as well, but what does it contain? It contains the morphisms from a to b: it's a hom-set.

It is not always the case that a hom-set exists in the same category as the objects it was built from. If the hom-set exists and is in the same category, it is an internal hom-set. If the hom-set exists and is in a different category, it is an external hom-set.

Universal Construction

We can try to construct an internal hom-set from scratch. We can think of A function type as a composite type, much like product and coproduct types. The product and coproduct types were constructed using by looking for patterns in the category, using universal construction. The same strategy can be used to find function types.

In this case an object z contains functions that maps the argument type a, to the return type b. This pattern represents function application

If we peek inside z, we can find functions f which maps a value x of type a to a value f x of type b. A pair of (f, a) will map to b. However we want to avoid peeking, so instead of looking at an individual function, we look at the entire object z. Then we can build a morphism g between the cartesian product of z and a, to b. In the category of types and functions, g will then be a function that maps every pair (f, x) to f x.

How do we rank candidates for z? An object z with a morphism g from z * a to b, is better than an object z' with morphism g' from z * a' to b, if there is a morphism h from z' to z that factorizes g. This is not completely sufficient, because we have the products z' * a and z * a, but only a morphism h from z' to z. What we would like is a morphism h' from z' * a to z * a. Since we already know that a product is functorial, the product of morphisms is also functorial. So the factorization is:

g' = g . (h * id)

The best candidate in this case will be

  1. an object a=>b with
  2. a morphism eval :: ((a=>b) * a) -> b
  3. such that any other candidate z with a morphism g :: z * a -> b
  4. there is a unique morphism h :: z -> (a=>b)
  5. that factorizes g through eval

This object a=>b might not exist for any pair of objects a, b, in a category, but in the category of types and functions it always does. In fact, in this category, the object a=>b is isomorphic to the hom-set of a and b.


The morphism g can be seen as a function of two variables, z and a. The morphism h can be seen as a function of one variable, z, returning a function from a to b. This makes h a higher ordered function. h is the curried version of g.


The function object is often called the exponential, denoted by b^a, where a is the argument type.

A function from Bool to Int accepts two values as it's argument, True and False, and for each value can return any of the values in Int. So the number of possible functions from Bool to Int is Int * Int, or Int ^ 2.

In this sense, functions can be turned into table lookups. This is the equivalence between function types as objects, and functions as morphisms.

Cartesian Closed Categories

So far, we have been using the category of Set as a model for types and functions. There is a larger family of categories that can be used as well. These categories are called cartesian closed. The category of Set is an example.

The requirements of a cartesian closed category is that it must contain:

  1. The terminal object
  2. Product of any pair of objects
  3. Exponential of any pair of objects

Exponential can be seen as a product of arbitrary many terms, and the terminal object as a product of zero objects, or zeroth power of an object.

The cartesian closed category provide models for the simply typed lambda calculus.

The terminal object and product have their duals, the terminal object and coproduct. A cartesian closed category that also supports these two constructions, and that product can be distributed over coproduct, is called a bicartesian closed category, which Set is.

Exponentials and Algebraic Data Types

The function type from 0 to any type a is equivalent to the type 1, there is only one function from Void to any type a, absurd, the hom-set C(0,a) is a singleton set.

a ^ 0 = 1

The function type from a to 1, is equivalent to 1, there is only one function from any type a to the type (), unit,

1^a = 1

Morphism from the terminal object is isomorphic to the object itself:

a^1 = a

The exponential form of a coproduct of two objects, is isomorphic to a product of two exponentials. When writing a function that takes in a coproduct b + c, it is sufficient to write a pair of functions dealing with each type in the coproduct separately.

a ^ (b+c) = a^b * a^c

The exponential of an exponential is the product of the exponents, which is a way to express uncurrying.

(a^b)^c = a^(b*c)

A function returning a pair is equivalent to a pair of functions returning one element of the pair each.

(a * b)^c = a^c * b^c

Curry-Howard isomorphism

Algebraic data types and logic have a certain correspondence described above. In particular, the new function types we have learned corresponds to implication: the function type a -> b is the logical implication that, a implies b.

The Curry-Howard isomorphism states that every type can be viewed as a proposition, and the proposition is true if the type can be inhabited, false if not. A logical implication is true if the argument and return type is inhabited. Thus an implementation of a function is then a proof of the proposition, that is the type signature of the function.

Natural Transformations

From Bartosz Milewski's Natural Transformations

Functors are mappings between categories that preserve structure (composition, identity). It embeds one category into another.

There can be many functors from a source category C to a target category D. Natural transformations allows us to compare different functors, they are mapping functors that preserver the functorial nature of functors.

Given two functors F and G, both from a category C to another category D, an object a in a C can be mapped to two objects in D, F a or G a. Then a mapping of functors will map F a to G a.

F a and G a are objects in the same category D, and the mapping between them will be a morphism in D. A natural transformation α is a selection of morphisms, for every a, select a morphism from F a to G a. This morphism is called the component of α at a.

α_a :: F a -> G a

If for some a, there is no morphism between F a and G a, then there is no natural transformation between F and G.

Functors also map morphisms, so natural transformations have to map these mapped morphisms as well. F f must be mapped to G f.

F f :: F a -> F b
G f :: G a -> G b

The natural transformation then provides an additional morphisms:

α_a :: F a -> G a
α_b :: F b -> G b

And we impose the naturality condition that for any f:

G f . α_a  = α_b  . F f

This ensures that if we use α_a first to get from F a to G a, then use G f to get from G a to G b, we get the same result as if we use F f to get from F a to F b, then use α_b to get from F b to G b.

This requirement is pretty stringent, and so functors related by natural transformations are not easy to find. But if found, they can tell you a lot about the structure of the categories.

You can see natural transformation as mapping objects to morphisms (maps an object a to to a morphism α_a from F a to G a). It also maps morphisms to commuting squares (the morphism f from a to b is mapped to a square that shows two ways of getting from F a to G b).

Natural transformations can be used to define isomorphisms of functors. Natural isomorphism is defined as a natural transformation whose components are all isomorphisms.

Polymorphic functors

In Haskell, functors map types to types, and lifts functions using fmap. One way to think about functors is that they are generalized containers. Then, natural transformations are a way to repackage the contents of one container into another container.

One example of a natural transformation is between the List functor and the Maybe functor.

safeHead :: [a] -> Maybe a
safeHead [] = Nothing
safeHead (x:xs) = Just x

To show that this satisfies the naturality condition, we need to show that:

G f . α_a  = α_b  . F f
-- G f is fmap for G
-- α_a s the component at a
-- F f is fmap for F
-- α_b s the component at b
-- In the case of safeHead, G is Maybe, and F is List

fmap f safeHead = safeHead . fmap f

fmap f (safeHead []) = fmap f Nothing
                     = Nothing
                     = safeHead []
                     = safeHead (fmap f [])
fmap f (safeHead (x:xs)) = fmap f (Just x)
                         = Just (f x)
                         = safeHead (f x : fmap f xs)
                         = safeHead (fmap f (x : xs))

Another functor is the Reader functor:

newtype Reader e a = Reader (e -> a)
instance Functor (Reader e) where
    fmap f (Reader g) = Reader (\x -> f (g x))

This functor is parameterized by two types e and a, and functorial only in a.

For every type e, you can define a natural transform from Reader e to a functor F. The Yoneda lemma states that the members of this family are always in one to one correspondence of the elements of F e. For example the functor Reader () takes any type a and maps it a function type () -> a. We can try to build a natural transformation from this functor to Maybe.

α_a :: Reader () a -> Maybe a
dumb (Reader _) -> Nothing
obvious (Reader g) -> Just (g ())

Natural transformations feel like morphisms, they map a functor to another functor. And in fact, functors form a category. Objects in this category are functors from C to D, and morphisms are the natural transformations between those functors.

If we have a category, we need composition. Composition of natural transformations is the composition of the morphisms that the natural transformations select.

Given a natural transformation α from F to G, the component at a is a morphism:

α_a :: F a -> G a

If we have another natural transformation β from G to H, the component at a is:

β_a :: G a -> H a

Thus the composition of the natural transformations at a is:

. α)_a = β_a . α_a

And the result of this composition is a natural transformation from F to H.

-- naturality condition
G f . α_a  = α_b  . F f
-- rewritten in terms of the composition
H f .. α)_a =. α)_b . F f

For each functor F, there is an identity natural transformation:

id_fa :: F a -> F a

Here I attempt some challenges:

  1. Define a natural transformation from the Maybe functor to the list functor. Prove the naturality condition for it.

    g :: Maybe a -> [a]
    g Nothing = []
    g Just a = [a]
    -- G f . α_a  = α_b  . F f
    fmap f (g Nothing) = []
                       = g Nothing
                       = g (fmap f Nothing)
                       = (g . fmap f) Nothing
    fmap f (g (Just a)) = fmap f [a]
                        = g (Just (f a))
                        = g (fmap f (Just a))
                        = (g . fmap f) (Just a)
  2. Define at least two different natural transformations between Reader () and the list functor. How many different lists of () are there?

    α_a :: Reader () a -> [A]
    dumb    (Reader _) = []
    obvious (Reader a) = [a ()]
  3. Continue the previous exercise with Reader Bool and Maybe.

    α_a :: Reader Bool a -> Maybe a
    alpha (Reader _) = Nothing
    beta (Reader f) = Just (f True)
    beta (Reader f) = Just (f False)