In the previous blog post we talked about relations. I gave an example of a thin category as a kind of relation that’s compatible with categorical structure. In a thin category, the hom-set is either an empty set or a singleton set. It so happens that these two sets form a sub-category of Set. It’s a very interesting category. It consists of the two objects — let’s give them new names o and i. Besides the mandatory identity morphisms, we also have a single morphism going from o to i, corresponding to the function we call absurd in Haskell:

absurd :: Void -> a
absurd _ = a

This tiny category is sometimes called the interval category. I’ll call it o->i.

The object o is initial, and the object i is terminal — just as the empty set and the singleton set were in Set. Moreover, the cartesian product from Set can be used to define a tensor product in o->i. We’ll use this tensor product to build a monoidal category.

Monoidal Categories

A tensor product is a bifunctor ⊗ with some additional properties. Here, in the interval category, we’ll define it through the following multiplication table:

o ⊗ o = o
o ⊗ i = o
i ⊗ o = o
i ⊗ i = i

Its action on pairs of morphisms (what we call bimap in Haskell) is also easy to define. For instance, what’s the action of ⊗ on the pair <absurd, idi>? This pair takes the pair <o, i> to <i, i>. Under the bifunctor ⊗, the first pair produces o, and the second i. There is only one morphism from o to i, so we have:

absurd ⊗ idi = absurd

If we designate the (terminal) object i as the unit of the tensor product, we get a (symmetric) monoidal category. A monoidal category is a category with a tensor product that’s associative and unital (usually, up to isomorphism — but here, strictly).

Now imagine that we replace hom-sets in our original thin category with objects from the monoidal category o->i (we’ll call them hom-objects). After all, we were only using two sets from Set. We can replace the empty hom-set with the object o, and the singleton hom-set with the object i. We get what’s called an enriched category (although, in this case, it’s more of an impoverished category).

An example of a thin category (a total order with objects 1, 2, and 3) with hom-sets replaced by hom-objects from the interval category. Think of i as corresponding to less-than-or-equal, and o as greater.

Enriched Categories

An enriched category has hom-objects instead of hom-sets. These are objects from some monoidal category V called the base category. The base category has to be monoidal because we want to define something that would replace the usual composition of morphisms. Morphisms are elements of hom-sets. However, hom-objects, in general, have no elements. We don’t know what an element of o or i is.

So to fully define an enriched category we have to come up with a sensible substitute for composition. To do that, we need to rethink composition — first in terms of hom-sets, then in terms of hom-objects.

We can think of composition as a function from a cartesian product of two hom-sets to a third hom-set:

composea b c :: C(b, c) × C(a, b) -> C(a, c)

Generalizing it, we can replace hom-sets with hom-objects (here, either o or i), the cartesian product with the tensor product, and a function with a morphism (notice: it’s a morphism in our monoidal category o->i). These composition-defining morphisms form a “composition table” for hom-objects.

As an example, take the composition of two is. Their product i ⊗ i is i again, and there is only one morphism out of i, the identity morphism. In terms of original hom-sets it would mean that the composition of two morphisms always exists. In general, we have to impose this condition when we’re defining a category, enriched or not — here it just happens automatically.

For instance (see illustration), compose0 1 2=idi:

compose0 1 2 (C(1, 2) ⊗ C(0, 1))
= compose0 1 2 (i ⊗ i)
= compose0 1 2 i
= i
= C(0, 2)

In every category we must also have identity morphisms. These are special elements in the hom-sets of the form C(a, a). We have to find a way to define their equivalent in the enriched setting. We’ll use the standard trick of defining generalized elements. It’s based on the observation that selecting an element from a set s is the same as selecting a morphism that goes from the singleton set (the terminal object in Set) to s. In a monoidal category, we replace the terminal object with the monoidal unit.

So, instead of picking an identity morphism in C(a, a), we use a morphism from the monoidal unit i:

ja :: i -> C(a, a)

Again, in the case of a thin category, there is only one morphism leaving i, and that’s the identity morphism. That’s why we are automatically guaranteed that, in a thin category, all hom-objects of the form C(a, a) are equal to i.

Composition in a category must also satisfy associativity and identity conditions. Associativity in the enriched setting translates straightforwardly to a commuting diagram, but identity is a little trickier. We have to use ja to “select” the identity from the hom-object C(a, a) while composing it with some other hom-object C(b, a). We start with the product:

i ⊗ C(b, a)

Because i is the monoidal unit, this is equal to C(b, a). On the other hand, we can tensor together two morphisms in o->i — remember, a tensor product is a bifunctor, so it also acts on morphisms. Here we’ll tensor ja and the identity at C(b, a):

ja ⊗ idC(b, a)

We act with this product on the product object i ⊗ C(b, a) to get C(a, a) ⊗ C(b, a). Then we use composition to get:

C(a, a) ⊗ C(b, a) -> C(b, a)

These two ways of getting to C(b, a) must coincide, leading to the identity condition for enriched categories.

Now that we’ve seen how the enrichment works for thin categories, we can apply the same mechanism to define categories enriched over any monoidal category V.

The important part is that V defines a (bifunctor) tensor product ⊗ and a unit object i. Associativity and unitality may be either strict or up to isomorphism (notice that a regular cartesian product is associative only up to isomorphism — (a, (b, c)) is not equal to ((a, b), c)).

Instead of sets of morphisms, an enriched category has hom-objects that are objects in V. We use the same notation as for hom-sets: C(a, b) is the hom-object that connects object a to object b. Composition is replaced by morphisms in V:

composea b c :: C(b, c) ⊗ C(a, b) -> C(a, c)

Instead of identity morphisms, we have the morphisms in V:

ja :: i -> C(a, a)

Finally, associativity and unitality of composition are imposed in the form of a few commuting diagrams.

Impoverished Yoneda

The Yoneda Lemma talks about functors from an arbitrary category to Set. To generalize the Yoneda lemma to enriched categories we first have to generalize functors. Their action on objects is not a problem; it’s the action on morphisms that needs our attention.

Enriched Functors

Since in an enriched category we no longer have access to individual morphisms, we have to define the action of functors on hom-objects wholesale. This is only possible if the hom-objects in the target category come from the same base category V as the hom-objects in the source category. In other words, both categories must be enriched over the same monoidal category. We can then use regular morphisms in V to map hom-objects.

Between any two objects a and b in C we have the hom-object C(a, b). The two objects are mapped by the functor f to f a and f b, and there is a hom-object between them, D(f a, f b). The action of f on C(a, b) is defined as a morphism in V:

C(a, b) -> D(f a, f b)

Let’s see what this means in our impoverished thin category. First of all, a functor will always map related objects to related objects. That’s because there is no morphism from i to o. A bond between two objects cannot be broken by an impoverished functor.

If the relation is a partial order, for instance less-than-or-equal, then it follows that a functor between posets preserves the ordering — it’s monotone.

A functor must also preserve composition and identity. The former can be easily expressed as a commuting diagram. Identity preservation in the enriched setting involves the use of ja. Starting from i we can use ja to get to C(a, a), which the functor maps to D(f a, f a). Or we can use jf a to get there directly. We insist that both paths be the same.

In our impoverished category, this just works because ja is the identity morphism and all C(a, a)s and D(a, a)s are equal to i.

Back to Yoneda: You might remember that we start the Yoneda construction by fixing one object a in C, and then varying another object x to define the functor:

x -> C(a, x)

This functor maps C to Set, because xs are objects in C, and hom-sets are sets — objects of Set.

In the enriched environment, the same construction results in a mapping from C to V, because hom-objects are objects of the base category V.

But is this mapping a functor? This is far from obvious, considering that C is an enriched category, and we have just said that enriched functors can only go between categories that are enriched over the same base category. The target of our functor, the category V, is not enriched. It turns out that, as long as V is closed, we can turn it into an enriched category.

Self Enrichment

Let’s first see how we would enrich our tiny category o->i. First of all, let’s check if it’s closed. Closedness means that hom-sets can be objectified — for every hom-set there is an object called the exponential object that objectifies it. The exponential object in a (symmetric) monoidal category is defined through the adjunction:

V(a⊗b, c) ≅ V(b, ca)

This is the standard adjunction for defining exponentials, except that we are using the tensor product instead of the regular product. The hom-sets are sets of morphisms between objects in V (here, in o->i).

Let’s check, for instance, if there’s an object that corresponds to the hom-set V(o, i), which we would call io. We have:

V(o⊗b, i) ≅ V(b, io)

Whatever b we chose, when multiplied by o it will yield o, so the left hand side is V(o, i), a singleton set. Therefore V(b, io) must be a singleton set too, for any choice of b. In particular, if b is i, we see that the only choice for io is:

io = i

You can check that all exponentiation rules in o->i can be obtained from simple algebra by replacing o with zero and i with one.

Every closed symmetric monoidal category can be enriched in itself by replacing hom-sets with the corresponding exponentials. For instance, in our case, we end up replacing all empty hom-sets in the category o->i with o, and all singleton hom-sets with i. You can easily convince yourself that it works, and the result is the category o->i enriched in itself.

We can now take a category C that’s enriched over a closed symmetric monoidal category V, and show that the mapping:

x -> C(a, x)

is indeed an enriched functor. It maps objects of C to objects of V and hom-objects of C to hom-objects (exponentials) of V.

An example of a functor from a total order enriched over the interval category to the interval category. This particular functor is equal to the hom-functor C(a->x) for a equal to 3.

Let’s see what this functor looks like in a poset. Given some a, the hom-object C(a, x) is equal to i if a <= x. So an x is mapped to i if it’s greater-or-equal to a, otherwise it’s mapped to o. If you think of the objects mapped to o as colored black and the ones mapped to i as colored red, you’ll see the object a and the whole graph below it must be painted red.

Enriched Natural Transformations

Now that we know what enriched functors are, we have to define natural transformations between them. This is a bit tricky, since a regular natural transformation is defined as a family of morphisms. But again, instead of picking individual morphisms from hom-sets we can work with the closest equivalent: generalized elements — morphisms going from the unit object i to hom-objects. So an enriched natural transformation between two enriched functors f and g is defined as a family of morphisms in V:

αa :: i -> V(f a, g a)

Natural transformations are very limited in our impoverished category. Let’s see what morphisms from i are at our disposal. We have one morphism from i to i: the identity morphism ida. This makes sense — we think of i as having a single element. There is no morphism from i back to o; and that makes sense too — we think of o as having no elements. The only possible generalized components of an impoverished natural transformation between two functors f and g correspond to D(f a, g a) equal to i; which means that, for every a, f a must be less-than-or-equal to g a. A natural transformation can only push a functor uphill.

When the target category is o->i, as in the impoverished Yoneda lemma, a natural transformation may never connect red to black. So once the first functor switches to red, the other must follow.

Naturality Condition

There is, of course, a naturality condition that goes with this definition of a natural transformation. The essence of it is that it shouldn’t matter if we first apply a functor and then the natural transformation α, or the other way around. In the enriched context, there are two ways of getting from C(a, b) to D(f a, g b). One is to multiply C(a, b) by i on the right:

C(a, b) ⊗ i

apply the product of g ⊗ αa to get:

D(g a, g b) ⊗ D(f a, g a)

and then apply composition to get:

D(f a, g b)

The other way is to multiply C(a, b) by i on the left:

i ⊗ C(a, b)

apply αb ⊗ f to get:

D(f b, g b) ⊗ D(f a, f b)

and compose the two to get:

D(f a, g b)

The naturality condition requires that this diagram commute.

Enriched Yoneda

The enriched version of the Yoneda lemma talks about enriched natural transformations from the functor x -> C(a, x) to any enriched functor f that goes from C to V.

Consider for a moment a functor from a poset to our tiny category o->i (which, by the way, is also a poset). It will map some objects to o (black) and others to i (red). As we’ve seen, a functor must preserve the less-than-or-equal relation, so once we get into the red territory, there is no going back to black. And a natural transformation may only repaint black to red, not the other way around.

Now we would like to say that natural transformations from x -> C(a, x) to f are in one-to-one correspondence with the elements of f a, except that f a is not a set, so it doesn’t have elements. It’s an object in V. So instead of talking about elements of f a, we’ll talk about generalized elements — morphisms from the unit object i to f a. And that’s how the enriched Yoneda lemma is formulated — as a natural bijection between the set of natural transformations and a set of morphisms from the unit object to f a.

Nat(C(a, -), f) ≅ i -> f a

In our running example, there are only two possible values for f a.

1. If the value is o then there is no morphism from i to it. The Yoneda lemma tells us that there is no natural transformation in that case. That makes sense, because the value of the functor x -> C(a, x) at x=a is i, and there is no morphism from i to o.
2. If the value is i then there is exactly one morphism from i to it — the identity. The Yoneda lemma tells use that there is just one natural transformation in that case. It’s the natural transformation whose generalized component at any object x is i->i.

Strong Enriched Yoneda

There is something unsatisfactory in the fact that the enriched Yoneda lemma ends up using a mapping between sets. First we try to get away from sets as far as possible, then we go back to sets of morphisms. It feels like cheating. Not to worry! There is a stronger version of the Yoneda lemma that deals with this problem. What we need is to replace the set of natural transformations with an object in V that would represent them — just like we replaced the set of morphisms with the exponential object. Such an object is defined as an end:

∫x V(f x, g x)

The strong version of the Yoneda lemma establishes the natural isomorphism:

∫x V(C(a, x), f x) ≅ f a

Enriched Profunctors

We’ve seen that a profunctor is a functor from a product category C^op × D to Set. The enriched version of a profunctor requires the notion of a product of enriched categories. We would like the product of enriched categories to also be an enriched category. In fact, we would like it to be enriched over the same base category V as the component categories.

We’ll define objects in such a category as pairs of objects from the component categories, but the hom-objects will be defined as tensor products of the component hom-objects. In the enriched product category, the hom-object between two pairs, <c, d> and <c', d'> is:

(Cop ⊗ D)(<c, d>, <c', d'>) = C(c, c') ⊗ D(d, d')

You can convince yourself that composition of such hom-objects requires the tensor product to be symmetric (at least up to isomorphism). That’s because you have to be able to rearrange the hom-objects in a tensor product of tensor products.

An enriched profunctor is defined as an enriched functor from the tensor product of two categories to the (self-enriched) base category:

Cop ⊗ D -> V

Just like regular profunctors, enriched profunctors can be composed using the coend formula. The only difference is that the cartesian product is replaced by the tensor product in V. They form a bicategory called V-Prof.

Enriched profunctors are the basis of the definition of Tambara modules, which are relevant in the application to Haskell lenses.

Conclusion

One of the reasons for using category theory is to get away from set theory. In general, objects in a category don’t have to form sets. The morphisms, however, are elements of sets — the hom-sets. Enriched categories go a step further and replace even those sets with categorical objects. However, it’s not categories all the way down — the base category that’s used for enrichment is still a regular old category with hom-sets.

Acknowledgments

I’m grateful to Gershom Bazerman for useful comments and to André van Meulebrouck for checking the grammar and spelling.

This is part 18 of Categories for Programmers. Previously: It’s All About Morphisms. See the Table of Contents.

In mathematics we have various ways of saying that one thing is like another. The strictest is equality. Two things are equal if there is no way to distinguish one from another. One can be substituted for the other in every imaginable context. For instance, did you notice that we used equality of morphisms every time we talked about commuting diagrams? That’s because morphisms form a set (hom-set) and set elements can be compared for equality.

But equality is often too strong. There are many examples of things being the same for all intents and purposes, without actually being equal. For instance, the pair type (Bool, Char) is not strictly equal to (Char, Bool), but we understand that they contain the same information. This concept is best captured by an isomorphism between two types — a morphism that’s invertible. Since it’s a morphism, it preserves the structure; and being “iso” means that it’s part of a round trip that lands you in the same spot, no matter on which side you start. In the case of pairs, this isomorphism is called swap:

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

swap happens to be its own inverse.

Adjunction and Unit/Counit Pair

When we talk about categories being isomorphic, we express this in terms of mappings between categories, a.k.a. functors. We would like to be able to say that two categories C and D are isomorphic if there exists a functor R (“right”) from C to D, which is invertible. In other words, there exists another functor L (“left”) from D back to C which, when composed with R, is equal to the identity functor I. There are two possible compositions, R ∘ L and L ∘ R; and two possible identity functors: one in C and another in D.

But here’s the tricky part: What does it mean for two functors to be equal? What do we mean by this equality:

R ∘ L = ID

or this one:

L ∘ R = IC

It would be reasonable to define functor equality in terms of equality of objects. Two functors, when acting on equal objects, should produce equal objects. But we don’t, in general, have the notion of object equality in an arbitrary category. It’s just not part of the definition. (Going deeper into this rabbit hole of “what equality really is,” we would end up in Homotopy Type Theory.)

You might argue that functors are morphisms in the category of categories, so they should be equality-comparable. And indeed, as long as we are talking about small categories, where objects form a set, we can indeed use the equality of elements of a set to equality-compare objects.

But, remember, Cat is really a 2-category. Hom-sets in a 2-category have additional structure — there are 2-morphisms acting between 1-morphisms. In Cat, 1-morphisms are functors, and 2-morphisms are natural transformations. So it’s more natural (can’t avoid this pun!) to consider natural isomorphisms as substitutes for equality when talking about functors.

So, instead of isomorphism of categories, it makes sense to consider a more general notion of equivalence. Two categories C and D are equivalent if we can find two functors going back and forth between them, whose composition (either way) is naturally isomorphic to the identity functor. In other words, there is a two-way natural transformation between the composition R ∘ L and the identity functor ID, and another between L ∘ R and the identity functor IC.

Adjunction is even weaker than equivalence, because it doesn’t require that the composition of the two functors be isomorphic to the identity functor. Instead it stipulates the existence of a one way natural transformation from ID to R∘L, and another from L∘R to IC. Here are the signatures of these two natural transformations:

η :: ID -> R ∘ L
ε :: L ∘ R -> IC

η is called the unit, and ε the counit of the adjunction.

Notice the asymmetry between these two definitions. In general, we don’t have the two remaining mappings:

R ∘ L -> ID -- not necessarily
IC -> L ∘ R -- not necessarily

Because of this asymmetry, the functor L is called the left adjoint to the functor R, while the functor R is the right adjoint to L. (Of course, left and right make sense only if you draw your diagrams one particular way.)

The compact notation for the adjunction is:

L ⊣ R

To better understand the adjunction, let’s analyze the unit and the counit in more detail.

Let’s start with the unit. It’s a natural transformation, so it’s a family of morphisms. Given an object d in D, the component of η is a morphism between I d, which is equal to d, and (R ∘ L) d; which, in the picture, is called d':

ηd :: d -> (R ∘ L) d

Notice that the composition R∘L is an endofunctor in D.

This equation tells us that we can pick any object d in D as our starting point, and use the round trip functor R ∘ L to pick our target object d'. Then we shoot an arrow — the morphism ηd — to our target.

By the same token, the component of of the counit ε can be described as:

εc :: (L ∘ R) c -> c

It tells us that we can pick any object c in C as our target, and use the round trip functor L ∘ R to pick the source c' = (L ∘ R) c. Then we shoot the arrow — the morphism εc — from the source to the target.

Another way of looking at unit and counit is that unit lets us introduce the composition R ∘ L anywhere we could insert an identity functor on D; and counit lets us eliminate the composition L ∘ R, replacing it with the identity on C. That leads to some “obvious” consistency conditions, which make sure that introduction followed by elimination doesn’t change anything:

L = L ∘ ID -> L ∘ R ∘ L -> IC ∘ L  = L

R = ID ∘ R -> R ∘ L ∘ R -> R ∘ IC = R

These are called triangular identities because they make the following diagrams commute:

These are diagrams in the functor category: the arrows are natural transformations, and their composition is the horizontal composition of natural transformations. In components, these identities become:

ε L d ∘ L η d = id L d
R ε c ∘ η R c = id R c

We often see unit and counit in Haskell under different names. Unit is known as return (or pure, in the definition of Applicative):

return :: d -> m d

and counint as extract:

extract :: w c -> c

Here, m is the (endo-) functor corresponding to R∘L, and w is the (endo-) functor corresponding to L∘R. As we’ll see later, they are part of the definition of a monad and a comonad, respectively.

If you think of an endofunctor as a container, the unit (or return) is a polymorphic function that creates a default box around a value of arbitrary type. The counit (or extract) does the reverse: it retrieves or produces a single value from a container.

We’ll see later that every pair of adjoint functors defines a monad and a comonad. Conversely, every monad or comonad may be factorized into a pair of adjoint functors — this factorization is not unique, though.

In Haskell, we use monads a lot, but only rarely factorize them into pairs of adjoint functors, primarily because those functors would normally take us out of Hask.

We can however define adjunctions of endofunctors in Haskell. Here’s part of the definition taken from Data.Functor.Adjunction:

class (Functor f, Representable u) =>
      Adjunction f u | f -> u, u -> f where
  unit         :: a -> u (f a)
  counit       :: f (u a) -> a

This definition requires some explanation. First of all, it describes a multi-parameter type class — the two parameters being f and u. It establishes a relation called Adjunction between these two type constructors.

Additional conditions, after the vertical bar, specify functional dependencies. For instance, f -> u means that u is determined by f (the relation between f and u is a function, here on type constructors). Conversely, u -> f means that, if we know u, then f is uniquely determined.

I’ll explain in a moment why, in Haskell, we can impose the condition that the right adjoint u be a representable functor.

Adjunctions and Hom-Sets

There is an equivalent definition of the adjunction in terms of natural isomorphisms of hom-sets. This definition ties nicely with universal constructions we’ve been studying so far. Every time you hear the statement that there is some unique morphism, which factorizes some construction, you should think of it as a mapping of some set to a hom-set. That’s the meaning of “picking a unique morphism.”

Furthermore, factorization can be often described in terms of natural transformations. Factorization involves commuting diagrams — some morphism being equal to a composition of two morphisms (factors). A natural transformation maps morphisms to commuting diagrams. So, in a universal construction, we go from a morphism to a commuting diagram, and then to a unique morphism. We end up with a mapping from morphism to morphism, or from one hom-set to another (usually in different categories). If this mapping is invertible, and if it can be naturally extended across all hom-sets, we have an adjunction.

The main difference between universal constructions and adjunctions is that the latter are defined globally — for all hom-sets. For instance, using a universal construction you can define a product of two select objects, even if it doesn’t exist for any other pair of objects in that category. As we’ll see soon, if the product of any pair of objects exists in a category, it can be also defined through an adjunction.

Here’s the alternative definition of the adjunction using hom-sets. As before, we have two functors L :: D->C and R :: C->D. We pick two arbitrary objects: the source object d in D, and the target object c in C. We can map the source object d to C using L. Now we have two objects in C, L d and c. They define a hom-set:

C(L d, c)

Similarly, we can map the target object c using R. Now we have two objects in D, d and R c. They, too, define a hom set:

D(d, R c)

We say that L is left adjoint to R iff there is an isomorphism of hom sets:

C(L d, c) ≅ D(d, R c)

that is natural both in d and c.

Naturality means that the source d can be varied smoothly across D; and the target c, across C. More precisely, we have a natural transformation φ between the following two (covariant) functors from C to Set. Here’s the action of these functors on objects:

c -> C(L d, c)
c -> D(d, R c)

The other natural transformation, ψ, acts between the following (contravariant) functors:

d -> C(L d, c)
d -> D(d, R c)

Both natural transformations must be invertible.

It’s easy to show that the two definitions of the adjunction are equivalent. For instance, let’s derive the unit transformation starting from the isomorphism of hom-sets:

C(L d, c) ≅ D(d, R c)

Since this isomorphism works for any object c, it must also work for c = L d:

C(L d, L d) ≅ D(d, (R ∘ L) d)

We know that the left hand side must contain at least one morphism, the identity. The natural transformation will map this morphism to an element of D(d, (R ∘ L) d) or, inserting the identity functor I, a morphism in:

D(I d, (R ∘ L) d)

We get a family of morphisms parameterized by d. They form a natural transformation between the functor I and the functor R ∘ L (the naturality condition is easy to verify). This is exactly our unit, η.

Conversely, starting from the existence of the unit and co-unit, we can define the transformations between hom-sets. For instance, let’s pick an arbitrary morphism f in the hom-set C(L d, c). We want to define a φ that, acting on f, produces a morphism in D(d, R c).

There isn’t really much choice. One thing we can try is to lift f using R. That will produce a morphism R f from R (L d) to R c — a morphism that’s an element of D((R ∘ L) d, R c).

What we need for a component of φ, is a morphism from d to R c. That’s not a problem, since we can use a component of ηd to get from d to (R ∘ L) d. We get:

φf = R f ∘ ηd

The other direction is analogous, and so is the derivation of ψ.

Going back to the Haskell definition of Adjunction, the natural transformations φ and ψ are replaced by polymorphic (in a and b) functions leftAdjunct and rightAdjunct, respectively. The functors L and R are called f and u:

class (Functor f, Representable u) =>
      Adjunction f u | f -> u, u -> f where
  leftAdjunct  :: (f a -> b) -> (a -> u b)
  rightAdjunct :: (a -> u b) -> (f a -> b)

The equivalence between the unit/counit formulation and the leftAdjunct/rightAdjunct formulation is witnessed by these mappings:

  unit           = leftAdjunct id
  counit         = rightAdjunct id
  leftAdjunct f  = fmap f . unit
  rightAdjunct f = counit . fmap f

It’s very instructive to follow the translation from the categorical description of the adjunction to Haskell code. I highly encourage this as an exercise.

We are now ready to explain why, in Haskell, the right adjoint is automatically a representable functor. The reason for this is that, to the first approximation, we can treat the category of Haskell types as the category of sets.

When the right category D is Set, the right adjoint R is a functor from C to Set. Such a functor is representable if we can find an object rep in C such that the hom-functor C(rep, _) is naturally isomorphic to R. It turns out that, if R is the right adjoint of some functor L from Set to C, such an object always exists — it’s the image of the singleton set () under L:

rep = L ()

Indeed, the adjunction tells us that the following two hom-sets are naturally isomorphic:

C(L (), c) ≅ Set((), R c)

For a given c, the right hand side is the set of functions from the singleton set () to R c. We’ve seen earlier that each such function picks one element from the set R c. The set of such functions is isomorphic to the set R c. So we have:

C(L (), -) ≅ R

which shows that R is indeed representable.

Product from Adjunction

We have previously introduced several concepts using universal constructions. Many of those concepts, when defined globally, are easier to express using adjunctions. The simplest non-trivial example is that of the product. The gist of the universal construction of the product is the ability to factorize any product-like candidate through the universal product.

More precisely, the product of two objects a and b is the object (a × b) (or (a, b) in the Haskell notation) equipped with two morphisms fst and snd such that, for any other candidate c equipped with two morphisms p::c->a and q::c->b, there exists a unique morphism m::c->(a, b) that factorizes p and q through fst and snd.

As we’ve seen earlier, in Haskell, we can implement a factorizer that generates this morphism from the two projections:

factorizer :: (c -> a) -> (c -> b) -> (c -> (a, b))
factorizer p q = \x -> (p x, q x)

It’s easy to verify that the factorization conditions hold:

fst . factorizer p q = p
snd . factorizer p q = q

We have a mapping that takes a pair of morphisms p and q and produces another morphism m = factorizer p q.

How can we translate this into a mapping between two hom-sets that we need to define an adjunction? The trick is to go outside of Hask and treat the pair of morphisms as a single morphism in the product category.

Let me remind you what a product category is. Take two arbitrary categories C and D. The objects in the product category C×D are pairs of objects, one from C and one from D. The morphisms are pairs of morphisms, one from C and one from D.

To define a product in some category C, we should start with the product category C×C. Pairs of morphism from C are single morphisms in the product category C×C.

It might be a little confusing at first that we are using a product category to define a product. These are, however, very different products. We don’t need a universal construction to define a product category. All we need is the notion of a pair of objects and a pair of morphisms.

However, a pair of objects from C is not an object in C. It’s an object in a different category, C×C. We can write the pair formally as <a, b>, where a and b are objects of C. The universal construction, on the other hand, is necessary in order to define the object a×b (or (a, b) in Haskell), which is an object in the same category C. This object is supposed to represent the pair <a, b> in a way specified by the universal construction. It doesn’t always exist and, even if it exists for some, might not exist for other pairs of objects in C.

Let’s now look at the factorizer as a mapping of hom-sets. The first hom-set is in the product category C×C, and the second is in C. A general morphism in C×C would be a pair of morphisms <f, g>:

f :: c' -> a
g :: c'' -> b

with c'' potentially different from c'. But to define a product, we are interested in a special morphism in C×C, the pair p and q that share the same source object c. That’s okay: In the definition of an adjunction, the source of the left hom-set is not an arbitrary object — it’s the result of the left functor L acting on some object from the right category. The functor that fits the bill is easy to guess — it’s the diagonal functor from C to C×C, whose action on objects is:

Δ c = <c, c>

The left-hand side hom-set in our adjunction should thus be:

(C×C)(Δ c, <a, b>)

It’s a hom-set in the product category. Its elements are pairs of morphisms that we recognize as the arguments to our factorizer:

(c -> a) -> (c -> b) ...

The right-hand side hom-set lives in C, and it goes between the source object c and the result of some functor R acting on the target object in C×C. That’s the functor that maps the pair <a, b> to our product object, a×b. We recognize this element of the hom-set as the result of the factorizer:

... -> (c -> (a, b))

We still don’t have a full adjunction. For that we first need our factorizer to be invertible — we are building an isomorphism between hom-sets. The inverse of the factorizer should start from a morphism m — a morphism from some object c to the product object a×b. In other words, m should be an element of:

C(c, a×b)

The inverse factorizer should map m to a morphism <p, q> in C×C that goes from <c, c> to <a, b>; in other words, a morphism that’s an element of:

(C×C)(Δ c, <a, b>)

If that mapping exists, we conclude that there exists the right adjoint to the diagonal functor. That functor defines a product.

In Haskell, we can always construct the inverse of the factorizer by composing m with, respectively, fst and snd.

p = fst ∘ m
q = snd ∘ m

To complete the proof of the equivalence of the two ways of defining a product we also need to show that the mapping between hom-sets is natural in a, b, and c. I will leave this as an exercise for the dedicated reader.

To summarize what we have done: A categorical product may be defined globally as the right adjoint of the diagonal functor:

(C × C)(Δ c, <a, b>) ≅ C(c, a×b)

Here, a×b is the result of the action of our right adjoint functor Product on the pair <a, b>. Notice that any functor from C×C is a bifunctor, so Product is a bifunctor. In Haskell, the Product bifunctor is written simply as (,). You can apply it to two types and get their product type, for instance:

(,) Int Bool ~ (Int, Bool)

Exponential from Adjunction

The exponential ba, or the function object a⇒b, can be defined using a universal construction. This construction, if it exists for all pairs of objects, can be seen as an adjunction. Again, the trick is to concentrate on the statement:

For any other object z with a morphism

g :: z × a -> b

there is a unique morphism

h :: z -> (a⇒b)

This statement establishes a mapping between hom-sets.

In this case, we are dealing with objects in the same category, so the two adjoint functors are endofunctors. The left (endo-)functor L, when acting on object z, produces z × a. It’s a functor that corresponds to taking a product with some fixed a.

The right (endo-)functor R, when acting on b produces the function object a⇒b (or ba). Again, a is fixed. The adjunction between these two functors is often written as:

- × a ⊣ (-)a

The mapping of hom-sets that underlies this adjunction is best seen by redrawing the diagram that we used in the universal construction.

Notice that the eval morphism is nothing else but the counit of this adjunction:

(a⇒b) × a -> b

where:

(a⇒b) × a = (L ∘ R) b

I have previously mentioned that a universal construction defines a unique object, up to isomorphism. That’s why we have “the” product and “the” exponential. This property translates to adjunctions as well: if a functor has an adjoint, this adjoint is unique up to isomorphism.

Challenges

1. Derive the naturality square for ψ, the transformation between the two (contravariant) functors:

   a -> C(L a, b)
   a -> D(a, R b)

2. Derive the counit ε starting from the hom-sets isomorphism in the second definition of the adjunction.
3. Complete the proof of equivalence of the two definitions of the adjunction.
4. Show that the coproduct can be defined by an adjunction. Start with the definition of the factorizer for a coproduct.
5. Show that the coproduct is the left adjoint of the diagonal functor.
6. Define the adjunction between a product and a function object in Haskell.

Next: Free/Forgetful Adjunctions.

Acknowledgments

I’d like to thank Edward Kmett and Gershom Bazerman for checking my math and logic, and André van Meulebrouck, who has been volunteering his editing help throughout this series of posts.
Follow @BartoszMilewski

This is part 16 of Categories for Programmers. Previously: The Yoneda Lemma. See the Table of Contents.

We’ve seen previously that, when we fix an object a in the category C, the mapping C(a, -) is a (covariant) functor from C to Set.

x -> C(a, x)

(The codomain is Set because the hom-set C(a, x) is a set.) We call this mapping a hom-functor — we have previously defined its action on morphisms as well.

Now let’s vary a in this mapping. We get a new mapping that assigns the hom-functor C(a, -) to any a.

a -> C(a, -)

It’s a mapping of objects from category C to functors, which are objects in the functor category (see the section about functor categories in Natural Transformations). Let’s use the notation [C, Set] for the functor category from C to Set. You may also recall that hom-functors are the prototypical representable functors.

Every time we have a mapping of objects between two categories, it’s natural to ask if such a mapping is also a functor. In other words whether we can lift a morphism from one category to a morphism in the other category. A morphism in C is just an element of C(a, b), but a morphism in the functor category [C, Set] is a natural transformation. So we are looking for a mapping of morphisms to natural transformations.

Let’s see if we can find a natural transformation corresponding to a morphism f :: a->b. First, lets see what a and b are mapped to. They are mapped to two functors: C(a, -) and C(b, -). We need a natural transformation between those two functors.

And here’s the trick: we use the Yoneda lemma:

[C, Set](C(a, -), F) ≅ F a

and replace the generic F with the hom-functor C(b, -). We get:

[C, Set](C(a, -), C(b, -)) ≅ C(b, a)

This is exactly the natural transformation between the two hom-functors we were looking for, but with a little twist: We have a mapping between a natural transformation and a morphism — an element of C(b, a) — that goes in the “wrong” direction. But that’s okay; it only means that the functor we are looking at is contravariant.

Actually, we’ve got even more than we bargained for. The mapping from C to [C, Set] is not only a contravariant functor — it is a fully faithful functor. Fullness and faithfulness are properties of functors that describe how they map hom-sets.

A faithful functor is injective on hom-sets, meaning that it maps distinct morphisms to distinct morphisms. In other words, it doesn’t coalesce them.

A full functor is surjective on hom-sets, meaning that it maps one hom-set onto the other hom-set, fully covering the latter.

A fully faithful functor F is a bijection on hom-sets — a one to one matching of all elements of both sets. For every pair of objects a and b in the source category C there is a bijection between C(a, b) and D(F a, F b), where D is the target category of F (in our case, the functor category, [C, Set]). Notice that this doesn’t mean that F is a bijection on objects. There may be objects in D that are not in the image of F, and we can’t say anything about hom-sets for those objects.

The Embedding

The (contravariant) functor we have just described, the functor that maps objects in C to functors in [C, Set]:

a -> C(a, -)

defines the Yoneda embedding. It embeds a category C (strictly speaking, the category C^op, because of contravariance) inside the functor category [C, Set]. It not only maps objects in C to functors, but also faithfully preserves all connections between them.

This is a very useful result because mathematicians know a lot about the category of functors, especially functors whose codomain is Set. We can get a lot of insight about an arbitrary category C by embedding it in the functor category.

Of course there is a dual version of the Yoneda embedding, sometimes called the co-Yoneda embedding. Observe that we could have started by fixing the target object (rather than the source object) of each hom-set, C(-, a). That would give us a contravariant hom-functor. Contravariant functors from C to Set are our familiar presheaves (see, for instance, Limits and Colimits). The co-Yoneda embedding defines the embedding of a category C in the category of presheaves. Its action on morphisms is given by:

[C, Set](C(-, a), C(-, b)) ≅ C(a, b)

Again, mathematicians know a lot about the category of presheaves, so being able to embed an arbitrary category in it is a big win.

Application to Haskell

In Haskell, the Yoneda embedding can be represented as the isomorphism between natural transformations amongst reader functors on the one hand, and functions (going in the opposite direction) on the other hand:

forall x. (a -> x) -> (b -> x) ≅ b -> a

(Remember, the reader functor is equivalent to ((->) a).)

The left hand side of this identity is a polymorphic function that, given a function from a to x and a value of type b, can produce a value of type x (I’m uncurrying — dropping the parentheses around — the function b -> x). The only way this can be done for all x is if our function knows how to convert a b to an a. It has to secretly have access to a function b->a.

Given such a converter, btoa, one can define the left hand side, call itfromY, as:

fromY :: (a -> x) -> b -> x
fromY f b = f (btoa b)

Conversely, given a function fromY we can recover the converter by calling fromY with the identity:

fromY id :: b -> a

This establishes the bijection between functions of the type fromY and btoa.

An alternative way of looking at this isomorphism is that it’s a CPS encoding of a function from b to a. The argument a->x is a continuation (the handler). The result is a function from b to x which, when called with a value of type b, will execute the continuation precomposed with the function being encoded.

The Yoneda embedding also explains some of the alternative representations of data structures in Haskell. In particular, it provides a very useful representation of lenses from the Control.Lens library.

Preorder Example

This example was suggested by Robert Harper. It’s the application of the Yoneda embedding to a category defined by a preorder. A preorder is a set with an ordering relation between its elements that’s traditionally written as <= (less than or equal). The “pre” in preorder is there because we’re only requiring the relation to be transitive and reflexive but not necessarily antisymmetric (so it’s possible to have cycles).

A set with the preorder relation gives rise to a category. The objects are the elements of this set. A morphism from object a to b either doesn’t exist, if the objects cannot be compared or if it’s not true that a <= b; or it exists if a <= b, and it points from a to b. There is never more than one morphism from one object to another. Therefore any hom-set in such a category is either an empty set or a one-element set. Such a category is called thin.

It’s easy to convince yourself that this construction is indeed a category: The arrows are composable because, if a <= b and b <= c then a <= c; and the composition is associative. We also have the identity arrows because every element is (less than or) equal to itself (reflexivity of the underlying relation).

We can now apply the co-Yoneda embedding to a preorder category. In particular, we’re interested in its action on morphisms:

[C, Set](C(-, a), C(-, b)) ≅ C(a, b)

The hom-set on the right hand side is non-empty if and only if a <= b — in which case it’s a one-element set. Consequently, if a <= b, there exists a single natural transformation on the left. Otherwise there is no natural transformation.

So what’s a natural transformation between hom-functors in a preorder? It should be a family of functions between sets C(-, a) and C(-, b). In a preorder, each of these sets can either be empty or a singleton. Let’s see what kind of functions are there at our disposal.

There is a function from an empty set to itself (the identity acting on an empty set), a function absurd from an empty set to a singleton set (it does nothing, since it only needs to be defined for elements of an empty set, of which there are none), and a function from a singleton to itself (the identity acting on a one-element set). The only combination that is forbidden is the mapping from a singleton to an empty set (what would the value of such a function be when acting on the single element?).

So our natural transformation will never connect a singleton hom-set to an empty hom-set. In other words, if x <= a (singleton hom-set C(x, a)) then C(x, b) cannot be empty. A non-empty C(x, b) means that x is less or equal to b. So the existence of the natural transformation in question requires that, for every x, if x <= a then x <= b.

for all x, x ≤ a ⇒ x ≤ b

On the other hand, co-Yoneda tells us that the existence of this natural transformation is equivalent to C(a, b) being non-empty, or to a <= b. Together, we get:

a ≤ b if and only if for all x, x ≤ a ⇒ x ≤ b

We could have arrived at this result directly. The intuition is that, if a <= b then all elements that are below a must also be below b. Conversely, when you substitute a for x on the right hand side, it follows that a <= b. But you must admit that arriving at this result through the Yoneda embedding is much more exciting.

Naturality

The Yoneda lemma establishes the isomorphism between the set of natural transformations and an object in Set. Natural transformations are morphisms in the functor category [C, Set]. The set of natural transformation between any two functors is a hom-set in that category. The Yoneda lemma is the isomorphism:

[C, Set](C(a, -), F) ≅ F a

This isomorphism turns out to be natural in both F and a. In other words, it’s natural in (F, a), a pair taken from the product category [C, Set] × C. Notice that we are now treating F as an object in the functor category.

Let’s think for a moment what this means. A natural isomorphism is an invertible natural transformation between two functors. And indeed, the right hand side of our isomorphism is a functor. It’s a functor from [C, Set] × C to Set. Its action on a pair (F, a) is a set — the result of evaluating the functor F at the object a. This is called the evaluation functor.

The left hand side is also a functor that takes (F, a) to a set of natural transformations [C, Set](C(a, -), F).

To show that these are really functors, we should also define their action on morphisms. But what’s a morphism between a pair (F, a) and (G, b)? It’s a pair of morphisms, (Φ, f); the first being a morphism between functors — a natural transformation — the second being a regular morphism in C.

The evaluation functor takes this pair (Φ, f) and maps it to a function between two sets, F a and G b. We can easily construct such a function from the component of Φ at a (which maps F a to G a) and the morphism f lifted by G:

(G f) ∘ Φa

Notice that, because of naturality of Φ, this is the same as:

Φb ∘ (F f)

I’m not going to prove the naturality of the whole isomorphism — after you’ve established what the functors are, the proof is pretty mechanical. It follows from the fact that our isomorphism is built up from functors and natural transformations. There is simply no way for it to go wrong.

Challenges

1. Express the co-Yoneda embedding in Haskell.
2. Show that the bijection we established between fromY and btoa is an isomorphism (the two mappings are the inverse of each other).
3. Work out the Yoneda embedding for a monoid. What functor corresponds to the monoid’s single object? What natural transformations correspond to monoid morphisms?
4. What is the application of the covariant Yoneda embedding to preorders? (Question suggested by Gershom Bazerman.)
5. Yoneda embedding can be used to embed an arbitrary functor category [C, D] in the functor category [[C, D], Set]. Figure out how it works on morphisms (which in this case are natural transformations).

Next: It’s All About Morphisms.

Acknowledgments

I’d like to thank Gershom Bazerman for checking my math and logic.
Follow @BartoszMilewski

This is part 13 of Categories for Programmers. Previously: Limits and Colimits. See the Table of Contents.

Monoids are an important concept in both category theory and in programming. Categories correspond to strongly typed languages, monoids to untyped languages. That’s because in a monoid you can compose any two arrows, just as in an untyped language you can compose any two functions (of course, you may end up with a runtime error when you execute your program).

We’ve seen that a monoid may be described as a category with a single object, where all logic is encoded in the rules of morphism composition. This categorical model is fully equivalent to the more traditional set-theoretical definition of a monoid, where we “multiply” two elements of a set to get a third element. This process of “multiplication” can be further dissected into first forming a pair of elements and then identifying this pair with an existing element — their “product.”

What happens when we forgo the second part of multiplication — the identification of pairs with existing elements? We can, for instance, start with an arbitrary set, form all possible pairs of elements, and call them new elements. Then we’ll pair these new elements with all possible elements, and so on. This is a chain reaction — we’ll keep adding new elements forever. The result, an infinite set, will be almost a monoid. But a monoid also needs a unit element and the law of associativity. No problem, we can add a special unit element and identify some of the pairs — just enough to support the unit and associativity laws.

Let’s see how this works in a simple example. Let’s start with a set of two elements, {a, b}. We’ll call them the generators of the free monoid. First, we’ll add a special element e to serve as the unit. Next we’ll add all the pairs of elements and call them “products”. The product of a and b will be the pair (a, b). The product of b and a will be the pair (b, a), the product of a with a will be (a, a), the product of b with b will be (b, b). We can also form pairs with e, like (a, e), (e, b), etc., but we’ll identify them with a, b, etc. So in this round we’ll only add (a, a), (a, b) and (b, a) and (b, b), and end up with the set {e, a, b, (a, a), (a, b), (b, a), (b, b)}.

In the next round we’ll keep adding elements like: (a, (a, b)), ((a, b), a), etc. At this point we’ll have to make sure that associativity holds, so we’ll identify (a, (b, a)) with ((a, b), a), etc. In other words, we won’t be needing internal parentheses.

You can guess what the final result of this process will be: we’ll create all possible lists of as and bs. In fact, if we represent e as an empty list, we can see that our “multiplication” is nothing but list concatenation.

This kind of construction, in which you keep generating all possible combinations of elements, and perform the minimum number of identifications — just enough to uphold the laws — is called a free construction. What we have just done is to construct a free monoid from the set of generators {a, b}.

Free Monoid in Haskell

A two-element set in Haskell is equivalent to the type Bool, and the free monoid generated by this set is equivalent to the type [Bool] (list of Bool). (I am deliberately ignoring problems with infinite lists.)

A monoid in Haskell is defined by the type class:

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

This just says that every Monoid must have a neutral element, which is called mempty, and a binary function (multiplication) called mappend. The unit and associativity laws cannot be expressed in Haskell and must be verified by the programmer every time a monoid is instantiated.

The fact that a list of any type forms a monoid is described by this instance definition:

instance Monoid [a] where
    mempty  = []
    mappend = (++)

It states that an empty list [] is the unit element, and list concatenation (++) is the binary operation.

As we have seen, a list of type a corresponds to a free monoid with the set a serving as generators. The set of natural numbers with multiplication is not a free monoid, because we identify lots of products. Compare for instance:

2 * 3 = 6
[2] ++ [3] = [2, 3] // not the same as [6]

That was easy, but the question is, can we perform this free construction in category theory, where we are not allowed to look inside objects? We’ll use our workhorse: the universal construction.

The second interesting question is, can any monoid be obtained from some free monoid by identifying more than the minimum number of elements required by the laws? I’ll show you that this follows directly from the universal construction.

Free Monoid Universal Construction

If you recall our previous experiences with universal constructions, you might notice that it’s not so much about constructing something as about selecting an object that best fits a given pattern. So if we want to use the universal construction to “construct” a free monoid, we have to consider a whole bunch of monoids from which to pick one. We need a whole category of monoids to chose from. But do monoids form a category?

Let’s first look at monoids as sets equipped with additional structure defined by unit and multiplication. We’ll pick as morphisms those functions that preserve the monoidal structure. Such structure-preserving functions are called homomorphisms. A monoid homomorphism must map the product of two elements to the product of the mapping of the two elements:

h (a * b) = h a * h b

and it must map unit to unit.
For instance, consider a homomorphism from lists of integers to integers. If we map [2] to 2 and [3] to 3, we have to map [2, 3] to 6, because concatenation

[2] ++ [3] = [2, 3]

becomes multiplication

2 * 3 = 6

Now let’s forget about the internal structure of individual monoids, and only look at them as objects with corresponding morphisms. You get a category Mon of monoids.

Okay, maybe before we forget about internal structure, let us notice an important property. Every object of Mon can be trivially mapped to a set. It’s just the set of its elements. This set is called the underlying set. In fact, not only can we map objects of Mon to sets, but we can also map morphisms of Mon (homomorphisms) to functions. Again, this seems sort of trivial, but it will become useful soon. This mapping of objects and morphisms from Mon to Set is in fact a functor. Since this functor “forgets” the monoidal structure — once we are inside a plain set, we no longer distinguish the unit element or care about multiplication — it’s called a forgetful functor. Forgetful functors come up regularly in category theory.

We now have two different views of Mon. We can treat it just like any other category with objects and morphisms. In that view, we don’t see the internal structure of monoids. All we can say about a particular object in Mon is that it connects to itself and to other objects through morphisms. The “multiplication” table of morphisms — the composition rules — are derived from the other view: monoids-as-sets. By going to category theory we haven’t lost this view completely — we can still access it through our forgetful functor.

To apply the universal construction, we need to define a special property that would let us search through the category of monoids and pick the best candidate for a free monoid. But a free monoid is defined by its generators. Different choices of generators produce different free monoids (a list of Bool is not the same as a list of Int). Our construction must start with a set of generators. So we’re back to sets!

That’s where the forgetful functor comes into play. We can use it to X-ray our monoids. We can identify the generators in the X-ray images of those blobs. Here’s how it works:

We start with a set of generators, x. That’s a set in Set.

The pattern we are going to match consists of a monoid m — an object of Mon — and a function p in Set:

p :: x -> U m

where U is our forgetful functor from Mon to Set. This is a weird heterogeneous pattern — half in Mon and half in Set.

The idea is that the function p will identify the set of generators inside the X-ray image of m. It doesn’t matter that functions may be lousy at identifying points inside sets (they may collapse them). It will all be sorted out by the universal construction, which will pick the best representative of this pattern.

We also have to define the ranking among candidates. Suppose we have another candidate: a monoid n and a function that identifies the generators in its X-ray image:

q :: x -> U n

We’ll say that m is better than n if there is a morphism of monoids (that’s a structure-preserving homomorphism):

h :: m -> n

whose image under U (remember, U is a functor, so it maps morphisms to functions) factorizes through p:

q = U h . p

If you think of p as selecting the generators in m; and q as selecting “the same” generators in n; then you can think of h as mapping these generators between the two monoids. Remember that h, by definition, preserves the monoidal structure. It means that a product of two generators in one monoid will be mapped to a product of the corresponding two generators in the second monoid, and so on.

This ranking may be used to find the best candidate — the free monoid. Here’s the definition:

We’ll say that m (together with the function p) is the free monoid with the generators x if and only if there is a unique morphism h from m to any other monoid n (together with the function q) that satisfies the above factorization property.

Incidentally, this answers our second question. The function U h is the one that has the power to collapse multiple elements of U m to a single element of U n. This collapse corresponds to identifying some elements of the free monoid. Therefore any monoid with generators x can be obtained from the free monoid based on x by identifying some of the elements. The free monoid is the one where only the bare minimum of identifications have been made.

We’ll come back to free monoids when we talk about adjunctions.

Challenges

1. You might think (as I did, originally) that the requirement that a homomorphism of monoids preserve the unit is redundant. After all, we know that for all a

   h a * h e = h (a * e) = h a

   So h e acts like a right unit (and, by analogy, as a left unit). The problem is that h a, for all a might only cover a sub-monoid of the target monoid. There may be a “true” unit outside of the image of h. Show that an isomorphism between monoids that preserves multiplication must automatically preserve unit.

2. Consider a monoid homomorphism from lists of integers with concatenation to integers with multiplication. What is the image of the empty list []? Assume that all singleton lists are mapped to the integers they contain, that is [3] is mapped to 3, etc. What’s the image of [1, 2, 3, 4]? How many different lists map to the integer 12? Is there any other homomorphism between the two monoids?
3. What is the free monoid generated by a one-element set? Can you see what it’s isomorphic to?

Next: Representable Functors.

Acknowledgments

I’d like to thank Gershom Bazerman for checking my math and logic, and André van Meulebrouck, who has been volunteering his editing help throughout this series of posts.
Follow @BartoszMilewski