Monads are just monoids in the category of endofunctors

2018-01-08

tl;dr: I explain the mathematical background of a joke-explanation of monads. Contains lots of math and a hasty introduction to category theory.

There is a running gag in the programming community, that newcomers will often be confused by the concept of monads (which is how sequential computations are modeled in purely functional languages) and getting the explanation "it is simple, really: Monads are just monoids in the category of endofunctors". This is not meant as an actual explanation, but rather to poke a bit of fun at the habit of functional programmers to give quite abstract and theoretical explanations at times, that are not all that helpful.

However, given my background in mathematics, I decided that I wanted to actually approach Haskell from this point of view: I am interested in how it uses math to model programming and also to, after several years of doing mostly engineering focused programming work, flex my math muscles again - as there is quite a bit of interesting math behind these concepts.

The quote is from a pretty popular book about category theory and is, in full:

All told, a monad in $$X$$ is just a monoid in the category of endofunctors of $$X$$, with product $$\times$$ replaced by composition of endofunctors and unit set by the identity endofunctor.

This, of course, is an explanation of the mathematical concept of monads, not meant for programmers. Most explanations of the quote that I found either assumed quite a bit of knowledge in Haskell or took a lot of liberties with the mathematical concepts (and relied a lot on "squinting") or both. This write up is my attempt, to walk through all the concepts needed to explain monads as a mathematical concept and how it relates to Haskell - with as little squinting as possible.

Of course, there are a couple of disclaimers, I should start with:

1. This is not the best way to understand what monads are, if you are actually interested in using them to program. In fact, it is literally the worst way. I would recommend this intro, which takes a much more practical approach.
2. This is not the best way to understand how category theory works, if you are actually interested in learning mathematics. In fact, it is literally the worst way. I would recommend the book the quote is from, it's quite good (but assumes a math audience).
3. I haven't done mathematics in years. I also don't know much Haskell either. So I might be getting a bunch of stuff wrong to varying degrees. I'm sure I will hear all about it :)
4. Even if I would understand everything correctly, there are still a lot of details, mostly of technical nature, I had to omit, to keep this "short". Not that it is short.

Originally, I intended this to be the ultimate explanation, which would teach Haskellers category theory, mathematicians Haskell and people who know neither both. Unsurprisingly, this is not what this is, at all. It ended up mostly a write up to assure myself that I understood the path myself. If anything, you can treat this as a kind of "reading companion": If you want to understand this topic of the intersection between category theory and functional programming, this post can lead you through the correct terms to search for and give you a good idea what to focus on, in the respective Wikipedia articles.

With all that out of the way, let's begin.

Categories

In mathematics, a category is (roughly) a collection of objects and a collection of arrows between them. There is not a lot of meaning behind these, but it will probably help you to think of objects as sets and arrows as mappings. Every arrow goes from an object (the domain) to an object (the codomain) and we write an arrow as $$f:X\to Y$$, where $$f$$ is the name of the arrow, $$X$$ is the domain and $$Y$$ is the codomain. Just like with mappings, there can be many arrows between any given pair of objects - or there may be none.

We do need some restrictions: First, we require a specific identity arrow $$\mathrm{id}:X\to X$$ attached to every object $$X$$, which has $$X$$ as both domain and codomain. Secondly, we require (some) arrows to be composable. That is if we have two arrows $$f:X\to Y,g:Y\to Z$$ - so, whenever the domain of $$g$$ is the codomain of $$f$$ - there should also be a composed arrow¹ $$g\circ f: X\to Z$$, that shares the domain with $$f$$ and the codomain with $$g$$.

Furthermore, the identity arrows must act as a unit for composition, that is, for every arrow $$f$$ we require $$\mathrm{id}\circ f = f = f \circ\mathrm{id}$$. We also require composition to be associative, that is $$(f\circ g)\circ h = f\circ(g\circ h)$$ (whenever all compositions exist)².

When we talk about a category, we often draw diagrams like this:

$\require{AMScd} \begin{CD} X @>{f}>> Y \\ @V{g}VV @VV{p}V \\ Z @>>{q}> W \\ \end{CD}$

They show some of the objects and arrows from the category in a compact way. This particular diagram indicates that there are four objects and four arrows involved, with obvious domains and codomains. We only draw a subset of the objects and arrows, that is interesting for the point we are trying to make - for example, above diagram could also contain, of course, identity arrows and compositions $$p\circ f$$ and $$q\circ g$$), but we didn't draw them. In a square like this, we can take two paths from $$X$$ to $$W$$. If these paths are identical (that is, $$p\circ f = q\circ g$$, we say that the square commutes. A commutative diagram is a diagram, in which any square commutes, that is, it does not matter which path we take from any object to another. Most of the time, when we draw a diagram, we intend it to be commutative.

So, to summarize, to define a mathematical category, we need to:

1. Specify what our objects are
2. Specify what our arrows are, where each arrow starts and ends at a certain object
3. This collection of arrows need to include an arrow $$\mathrm{id}_X$$ for every object $$X$$, which starts and ends at $$X$$
4. And we need to be able to glue together arrows $$f:X\to Y$$ and $$g:Y\to Z$$ to an arrow $$g\circ f: X\to Z$$

1. The objects are types: Int is an object, String is an object but also Int | String, String -> Int and any other complicated type you can think of.
2. The arrows are functions: f :: a -> b is a function taking an a as an input and returning a b and is represented by an arrow f, which has a as its domain and b as its codomain. So, for example, length :: String -> Int would start at the type String and end at Int.
3. Haskell has a function id :: a -> a which gives us the identity arrow for any type a.
4. We can compose functions with the operator (.) :: (b -> c) -> (a -> b) -> (a -> c). Note, that this follows the swapped notation of $$\circ$$, where the input type of the left function is the output type of the right function.

In general, category theory is concerned with the relationship between categories, whereas in functional programming, we usually only deal with this one category. This turns out to be both a blessing and a curse: It means that our object of study is much simpler, but it also means, that it is sometimes hard to see how to apply the general concepts to the limited environment of functional programming.

Monoids

Understanding categories puts us in the position to understand monoids. A monoid is the generalized structure underlying concepts like the natural numbers: We can add two natural numbers, but we can't (in general) subtract them, as there are no negative numbers. We also have the number $$0$$, which, when added to any number, does nothing - it acts as a unit for addition. And we also observe, that addition is associative, that is, when doing a bunch of additions, the order we do them in doesn't matter.

The same properties also apply to other constructs. For example, if we take all maps from a given set to itself, they can be composed and that composition is associative and there is a unit element (the identity map).

This provides us with the following elements to define a monoid:

1. A set $$M$$
2. An operation $$\star\colon M\times M\to M$$, which "adds" together two elements to make a new one
3. We need a special unit element $$u\in M$$, which acts neutrally when added to any other element, that is $$m\star u=m=u\star m$$
4. The operation needs to be associative, that is we always require $$m\star(n\star k)=(m\star n)\star k$$

There is another way to frame this, which is closer in line with category theory. If we take $$1 := \{0\}$$ to be a 1-element set, we can see that the elements of $$M$$ are in a one-to-one correspondence to functions $$1\to M$$: Every such function chooses an element of $$M$$ (the image of $$0$$) and every element $$m\in M$$ fixes such a function, by using $$f(0) := m$$. Thus, instead of saying "we need a special element of $$M$$", we can also choose a special function $$\eta: 1\to M$$. And instead of talking about an "operation", we can talk about a function $$\mu: M\times M\to M$$. Which means, we can define a monoid via a commutative diagram like so:

$\begin{CD} 1 \\ @V{\eta}VV \\ M \\ \end{CD} \hspace{1em} \begin{CD} M\times M \\ @V{\mu}VV \\ M \\ \end{CD} \hspace{1em} \begin{CD} M\times 1 @>{\mathrm{id}\times\eta}>> M\times M @<{\eta\times\mathrm{id}}<< 1\times M \\ @V{\pi_1}VV @V{\mu}VV @V{\pi_2}VV \\ M @>{\mathrm{id}}>> M @<{\mathrm{id}}<< M \\ \end{CD} \hspace{1em} \begin{CD} M\times M\times M @>{\mu\times\mathrm{id}}>> M\times M \\ @V{\mathrm{id}\times\mu}VV @V{\mu}VV \\ M\times M @>{\mu}>> M \\ \end{CD}$

$$\pi_1$$ and $$\pi_2$$ here, are the functions that project to the first or second component of a cross product respectively (that is $$\pi_1(a, b) := a, \pi_2(a, b) := b$$) and e.g. $$\mathrm{id}\times\eta$$ is the map that applies $$\mathrm{id}$$ to the first component of a cross-product and $$\eta$$ to the second: $$\mathrm{id}\times\eta(m, 0) = (m, \eta(0))$$.

There are four sub-diagrams here:

1. The first diagram just says, that we need an arrow $$\eta:1\to M$$. This chooses a unit element for us.
2. Likewise, the second diagram just says, that we need an arrow $$\mu:M\times M\to M$$. This is the operation.
3. The third diagram tells us that the chosen by $$\eta$$ should be a unit for $$\mu$$. The commutativity of the left square tells us, that it should be right-neutral, that is $\forall m\in M: m = \pi_1(m, 0) = \mu(\mathrm{id}\times\eta(m, 0)) = \mu(m, \eta(0))$ and the commutativity of the right square tells us, that it should be left-neutral, that is $\forall m\in M: m = \pi_2(0,m) = \mu(\eta\times\mathrm{id}(0, m)) = \mu(\eta(0), m)$

Thus, the first diagram is saying that the element chosen by $$\eta$$ should act like a unit. For example, the left square says

$\pi_1(m,0) = \mu((\mathrm{id}\times\eta)(m,0)) = \mu(m,\eta(0))$

Now, writing $$\mu(m,n) = m\star n$$ and $$\eta(0) = u$$, this is equivalent to saying $$m = u\star m$$.

The second diagram is saying that $$\mu$$ should be associative: The top arrow combines the first two elements, the left arrow combines the second two. The right and bottom arrows then combine the result with the remaining element respectively, so commutativity of that square means the familiar $$m\star (n\star k) = (m\star n)\star k$$.

Haskell has the concept of a monoid too. While it's not really relevant to the discussion, it might be enlightening to see, how it's modeled. A monoid in Haskell is a type-class with two (required) methods:

class Monoid a where
mempty :: a
mappend :: a -> a -> a


Now, this gives us the operation (mappend) and the unit (a), but where are the requirements of associativity and the unit acting neutrally? The Haskell type system is unable to codify these requirements, so they are instead given as a "law", that is, any implementation of a monoid is supposed to have these properties, to be manually checked by the programmer:

• mappend mempty x = x (the unit is left-neutral)
• mappend x mempty = x (the unit is right-neutral)
• mappend x (mappend y z) = mappend (mappend x y) z (the operation is associative)
Functors

I mentioned that category theory investigates the relationship between categories - but so far, everything we've seen only happens inside a single category. Functors are, how we relate categories to each other. Given two categories $$\mathcal{B}$$ and $$\mathcal{C}$$, a functor $$F:\mathcal{B}\to \mathcal{C}$$ assigns to every object $$X$$ of $$\mathcal{B}$$, an object $$F(X)$$ of $$\mathcal{C}$$. It also assigns to every arrow $$f:X\to Y$$ in $$\mathcal{B}$$ a corresponding arrow $$F(f): F(X)\to F(Y)$$ in $$\mathcal{C}$$³. So, a functor transfers arrows from one category to another, preserving domain and codomain. To actually preserve the structure, we also need it to preserve the extra requirements of a category, identities and composition. So we need, in total:

1. An object map, $$F:O_\mathcal{B} \to O_\mathcal{C}$$
2. An arrow map, $$F:A_\mathcal{B}\to A_\mathcal{C}$$, which preserves start and end object, that is the image of an arrow $$X\to Y$$ starts at $$F(X)$$ and ends at $$F(Y)$$
3. The arrow map has to preserve identities, that is $$F(\mathrm{id}_X) = \mathrm{id}_{F(X)}$$
4. The arrow map has to preserve composition, that is $$F(g\circ f) = F(g)\circ F(f)$$.

A trivial example of a functor is the identity functor (which we will call $$I$$), which assigns each object to itself and each arrow to itself - that is, it doesn't change the category at all.

A simple example is the construction of the free monoid, which maps from the category of sets to the category of monoids. The Free monoid $$S^*$$ on a set $$S$$ is the set of all finite length strings of elements of $$S$$, with concatenation as the operation and the empty string as the unit. Our object map then assigns to each set $$S$$ its free monoid $$S^*$$. And our arrow map assigns to each function $$f:S\to T$$ the function $$f^*:S^*\to T^*$$, that applies $$f$$ to each element of the input string.

There is an interesting side note here: Mathematicians love to abstract. Categories arose from the observation, that in many branches of mathematics we are researching some class of objects with some associated structure and those maps between them, that preserve this structure. It turns out that category theory is a branch of mathematics that is researching the objects of categories, with some associated structure (identity arrows and composition) and maps (functors) between them, that preserve that structure. So it seems obvious that we should be able to view categories as objects of a category, with functors as arrows. Functors can be composed (in the obvious way) and every category has an identity functor, that just maps every object and arrow to itself.

Now, in Haskell, Functors are again a type class:

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


This looks like our arrow map: It assigns to each function g :: a -> b a function fmap g :: f a -> f b. The object map is implicit: When we write f a, we are referring to a new type, that depends on a - so we "map" a to f a .

Again, there are additional requirements the type system of Haskell can not capture. So we provide them as laws the programmer has to check manually:

• fmap id == id (preserves identities)
• fmap (f . g) == fmap f . fmap g (preserves composition)

There is one thing to note here: As mentioned, in Haskell we only really deal with one category, the category of types. That means that a functor always maps from the category of types to itself. In mathematics, we call such a functor, that maps a category to itself, an endofunctor. So we can tell, that in Haskell, every functor is automatically an endofunctor.

Natural transformations

We now understand categories and we understand functors. We also understand, that we can look at something like the category of categories. But the definition of a monad given to us talks about the category of endofunctors. So we seem to have to step up yet another level in the abstraction hierarchy and somehow build this category. As objects, we'd like to have endofunctors - and arrows will be natural transformations, which take one functor to another, while preserving its internal structure (the mapping of arrows). If that sounds complicated and abstract, that's because it is.

We need two functors $$F,G:\mathcal{B}\to \mathcal{C}$$ of the same "kind" (that is, mapping to and from the same categories). A natural transformation $$\eta:F\dot\to G$$ assigns an arrow $$\eta_X: F(X)\to G(X)$$ (called a component of $$\eta$$) to every object in $$\mathcal{B}$$. So a component $$\eta_X$$ describes, how we can translate the action of $$F$$ on $$X$$ into the action of $$G$$ on $$X$$ - i.e. how to translate their object maps. We also have to talk about the translation of the arrow maps. For that, we observe that for any arrow $$f:X\to Y$$ in $$\mathcal{B}$$, we get four new arrows in $$\mathcal{C}$$:

$\begin{CD} X \\ @V{f}VV \\ Y \\ \end{CD} \hspace{1em} \begin{CD} F(X) @>{\eta_X}>> G(X) \\ @V{F(f)}VV @VV{G(f)}V \\ F(Y) @>>{\eta_Y}> G(Y) \\ \end{CD}$

For a natural transformation, we require the resulting square to commute.

So, to recap: To create a natural transformation, we need

1. Two functors $$F,G:\mathcal{B}\to\mathcal{C}$$
2. For every object $$X$$ in $$\mathcal{B}$$, an arrow $$\eta_X: F(X)\to G(X)$$
3. The components need to be compatible with the arrow maps of the functors: $$\eta_Y\circ F(f) = G(f)\circ \eta_X$$.

In Haskell, we can define a natural transformation like so:

class (Functor f, Functor g) => Transformation f g where
eta :: f a -> g a


f and g are functors and a natural transformation from f to g provides a map f a -> g a for every type a. Again, the requirement of compatibility with the actions of the functors is not expressible as a type signature, but we can require it as a law:

• eta (fmap fn a) = fmap fn (eta a)

This, finally, puts us in the position to define monads. Let's look at our quote above:

All told, a monad in $$X$$ is just a monoid in the category of endofunctors of $$X$$, with product $$\times$$ replaced by composition of endofunctors and unit set by the identity endofunctor.

It should be clear, how we can compose endofunctors. But it is important, that this is a different view of these things than if we'd look at the category of categories - there, objects are categories and functors are arrows, while here, objects are functors and arrows are natural transformations. That shows, how composition of functors can take the role of the cross-product of sets: In a set-category, the cross product makes a new set out of two other set. In the category of endofunctors, composition makes a new endofunctor out of two other endofunctors.

When we defined monoids diagrammatically, we also needed a cross product of mappings, that is, given a map $$f:X_1\to Y_1$$ and a map $$g:X_2\to Y_2$$, we needed the map $$f\times g: X_1\times X_2\to Y_1\times Y_2$$, which operated on the individual constituents. If we want to replace the cross product with composition of endofunctors, we need an equivalent for natural transformations. That is, given two natural transformations $$\eta:F\to G$$ and $$\epsilon:J\to K$$, we want to construct a natural transformation $$\eta\epsilon:J\circ F\to K\circ G$$. This diagram illustrates how we get there (working on components):

$\begin{CD} F(X) @>{\eta_X}>> G(X) @. \\ @V{J}VV @VV{J}V @. \\ J(F(X)) @>{J(\eta_X)}>> J(G(X)) @>{\epsilon_{G(X)}}>> K(G(X)) \\ \end{CD}$

As we can see, we can build an arrow $$\epsilon_{G(X)}\circ J(\eta_X): J(F(X)) \to K(G(X))$$, which we can use as the components of our natural transformation $$\eta\epsilon:J\circ F\to K\circ G$$. This construction is called the horizontal composition of natural transformations. We should verify that this is indeed a natural transformation - for now, let's just accept that it follows from the naturality of $$\eta$$ and $$\epsilon$$.

Lastly, there is an obvious natural transformation taking a functor to itself; each component being just the identity arrow. We call that natural transformation $$\iota$$, staying with the convention of using Greek letters for natural transformations.

With this, we can redraw the diagram we used to define monoids above, the replacements indicated by the quotes:

$\begin{CD} I \\ @V{\eta}VV \\ M \\ \end{CD} \hspace{1em} \begin{CD} M\circ M \\ @V{\mu}VV \\ M \\ \end{CD} \hspace{1em} \begin{CD} M\circ I @>{\iota\ \eta}>> M\circ M @<{\eta\ \iota}<< I\circ M \\ @VVV @V{\mu}VV @VVV \\ M @>{\iota}>> M @<{\iota}<< M \\ \end{CD} \hspace{1em} \begin{CD} M\circ M\circ M @>{\mu\ \iota}>> M\circ M \\ @V{\iota\ \mu}VV @V{\mu}VV \\ M\circ M @>{\mu}>> M \\ \end{CD}$

The vertical arrows in the middle diagram now simply apply the composition of functors, using that the identity functor is a unit.

These diagrams encode these conditions on our natural transformations:

• $$\mu\circ\eta\iota = \mu = \iota\eta\circ\mu$$, that is $$\eta$$ serves as a unit
• $$\mu\circ\mu\iota = \mu\circ\iota\mu$$, that is $$\mu$$ is associative

To recap, a monad, in category theory, is

• An endofunctor $$M$$
• A natural transformation $$\eta: I\to M$$, which serves as an identity for horizontal composition.
• A natural transformation $$\mu: M\circ M\to M$$, which is associative in respect to horizontal composition.

First, what is the identity functor in Haskell? As we pointed out above, the object function of functors is implicit, when we write f a instead of a. As such, the identity functor is simply a - i.e. we map any type to itself. fmap of that functor would thus also just be the identity fmap :: (a -> a) -> (a -> a).

So, what would our natural transformation $$\eta$$ look like? As we said, a natural transformation between two functors is just a map f a -> g a. So (if we call our endofunctor m) the identity transformation of our monoid is eta :: a -> m a mapping the identity functor to m. We also need our monoidal operation, which should map m applied twice to m: mu :: m (m a) -> m a.

Now, Haskellers write return instead of eta and write join instead of mu, giving us the type class

class (Functor m) => Monad where
return :: a -> m a
join :: m (m a) -> m a


As a last note, it is worth pointing out that you usually won't implement join, but instead a different function, called "monadic bind":

(>>=) :: m a -> (a -> m b) -> m b


The reason is, that this more closely maps to what monads are actually used for in functional programming. But we can move between join and >>= via

(>>=) :: m a -> (a -> m b) -> m b
v >>= f = join ((fmap f) v)

join :: m (m a) -> m a
join v = v >>= id

Conclusion

This certainly was a bit of a long ride. It took me much longer than anticipated both to understand all the steps necessary and to write them down. I hope you found it helpful and I hope I didn't make too many, too glaring mistakes. If so (either), feel free to let me know on Twitter, reddit or Hacker News - but please remember to be kind :)

I want to thank Tim Adler and mxf+ for proof-reading this absurdly long post and for making many helpful suggestions for improvements

[1] It is often confusing to people, that the way the arrows point in the notation and the order they are written seems to contradict each other: When writing $$f:X\to Y$$ and $$g:Y\to Z$$ you might reasonably expect their composite to work like $$f\circ g: X\to Z$$, that is, you glue together the arrows in the order you are writing them.

The fact that we are not doing that is a completely justified criticism, that is due to a historical accident - we write function application from right to left, that is we write $$f(x)$$, for applying $$f$$ to $$x$$. Accordingly, we write $$g(f(x))$$, when applying $$g$$ to the result of applying $$f$$ to $$x$$. And we chose to have the composite-notation be consistent with that, instead of the arrow-notation.

I chose to just eat the unfortunate confusion, as it turns out Haskell is doing exactly the same thing, so swapping things around would just increase the confusion.

Sorry.

[2] Keep in mind that this is a different notion from the ones for monoids, which we come to a bit later: While the formulas seem the same and the identities look like a unit, the difference is that only certain arrows can be composed, not all. And that there are many identity arrows, not just one. However, if we would have only one object, it would have to be the domain and codomain of every arrow and there would be exactly one identity arrow. In that case, the notions would be the same and indeed, "a category with exactly one object" is yet another way to define monoids.

[3] It is customary, to use the same name for the object and arrow map, even though that may seem confusing. A slight justification of that would be, that the object map is already given by the arrow map anyway: If $$F$$ is the arrow map, we can define the object map as $$X\mapsto \mathrm{dom}(F(\mathrm{id}_X))$$. So, given that they are always occurring together and you can make one from the other, we tend to just drop the distinction and save some symbols.

What was that? Oh, you thought Mathematicians where precise? Ha!

[4] It is important to note, that this is not really a function. Functions operate on values of a given type. But here, we are operating on types and Haskell has no concept of a "type of types" built in that a function could operate on. There are constructs operating on types to construct new types, like data, type, newtype or even deriving. But they are special syntactical constructs that exist outside of the realm of functions.

This is one of the things that was tripping me up for a while: I was trying to figure out, how I would map types to other types in Haskell or even talk about the object map. But the most useful answer is "you don't".

[5] An important note here, is that the $$\eta_X$$ are arrows. Where the object map of a functor is just a general association which could look anything we like, the components of a natural transformation need to preserve the internal structure of the category we are working in.

[6] You will often see these conditions written differently, namely written e.g. $$\mu M$$ instead of $$\mu\iota$$. You can treat that as a notational shorthand, it really means the same thing.

[7] There is a technicality here, that Haskell also has an intermediate between functor and monad called "applicative". As I understand it, this does not have a clear category theoretical analogue. I'm not sure why it exits, but I believe it has been added into the hierarchy after the fact.