You don’t need to know anything about category theory to use Haskell as a programming language. But if you want to understand the theory *behind* Haskell or contribute to its development, some familiarity with category theory is a prerequisite.

Category theory is very easy at the beginning. I was able to explain what a category is to my 10-year old son. But the learning curve gets steeper as you go. Functors are easy. Natural transformations may take some getting used to, but after chasing a few diagrams, you’ll get the hang of it. The Yoneda lemma is usually the first serious challenge, because to understand it, you have to be able to juggle several things in your mind at once. But once you’ve got it, it’s very satisfying. Things just fall into place and you gain a lot of intuition about categories, functors, and natural transformations.

# A Teaser Problem

You are given a polymorphic function `imager`

that, for any function from `Bool`

to any type `r`

, returns a list of `r`

. Try running this code in the School of Haskell, with `colorMap`

, `heatMap`

, and `soundMap`

. You may also define your own function of `Bool`

and pass it to `imager`

.

{-# LANGUAGE ExplicitForAll #-} imager :: forall r . ((Bool -> r) -> [r]) imager = ??? data Color = Red | Green | Blue deriving Show data Note = C | D | E | F | G | A | B deriving Show colorMap x = if x then Blue else Red heatMap x = if x then 32 else 212 soundMap x = if x then C else G main = print $ imager colorMap

Can you guess the implementation of `imager`

? How many possible `imager`

s with the same signature are there? By the end of this article you should be able to validate your answers using the Yoneda’s lemma.

# Categories

A category is a bunch of objects with arrows between them (incidentally, a “bunch” doesn’t mean a *set* but a more generall *collection*). We don’t know anything about the objects — all we know is the arrows, a.k.a morphisms.

Our usual intuition is that arrows are sort of like functions. Functions are mappings between sets. Indeed, morphisms have some function-like properties, for instance composability, which is associative:

h :: a -> b g :: b -> c f :: c -> d f . (g . h) == (f . g) . h

There is also an identity morphism for every object in a category, just like the `id`

function:

id :: a -> a id . f == f . id == f

In all Haskell examples I’ll be using the category **Hask** of Haskell types, with morphisms being plain old functions. An object in **Hask** is a type, like `Int`

, `[Bool]`

, or `[a]->Int`

. Types are nothing more than just *sets* of values. `Bool`

is a two element set {`True`

, `False`

}, `Integer`

is the set of all integers, and so on.

In general, a category of all sets and functions is called **Set** .

So how good is this sets-and-functions intuition for an arbitrary category? Are all categories really like collections of sets, and morphisms are like functions from set to set? What does the word *like* even mean in this context?

# Functors

In category theory, when we say one category is “like” another category, we usually mean that there is a mapping between the two. For this mapping to be meaningful, it should preserve the structure of the category. So not only every object from one category has to be mapped into an object from another category, but also all morphisms must be mapped correctly — meaning they should preserve composition. Such a mapping has a name: it’s called a functor.

Functors in **Hask** are described by the type class `Functor`

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

A Haskell `Functor`

maps types into types and functions into functions — a type constructor does the former and `fmap`

does the latter.

A *type contructor* is a mapping from one type to another. For instance, a list type constructor takes any type `a`

and creates a list type, `[a]`

.

So instead of asking if every category is “like” the **Set** category, we can ask a more precise question: For what types of categories (if not all of them) there exist functors that map them into **Set** . Such categories are called *representable*, meaning they have a representation in **Set** .

As a physicist I had to deal a lot with groups, such as groups of spacetime rotations in various dimensions or unitary groups in complex spaces. It was very handy to *represent* these abstract groups as matrices acting on vectors. For instance, different representations of the same Lorenz group (more precisely, SL(2, C)) would correspond to physical particles with different spins. So vector spaces and matrices are to abstract groups as sets and functions are to abstract categories.

# Yoneda Embedding

One of the things Yoneda showed is that there is at least one canonical functor from any so called *locally small* category into the category of sets and functions. The construction of this functor is surpisingly easy, so let me sketch it.

This functor should map every object in category C into a set. Set of what? It doesn’t really matter, a set is a set. So how about using a set of morphisms?

How can we map any object into a set of morphisms? Easy. First, let’s arbitrarily fix one object in the category C, call it A. It doesn’t matter which object we pick, we’ll just have to hold on to it. Now, for every object X in C there is a set of morphisms (arrows) going from our fixed A to this X. We designate this set to be the image of X under the functor we are constructing. Let’s call this functor H_{A}. There is one element in the set H_{A}(X) for every morphism from A to X.

A functor must define a mapping of objects to objects (to sets, in our case) and morphisms to morphisms (to functions in our case). We have established the first part of the mapping. To define the second part, let’s pick an arbitrary morphism f from X to Y. We have to map it to some function from the set H_{A}(X) to the set H_{A}(Y).

Let’s define this function, we’ll call it H_{A}(f), through its action on any element of the set H_{A}(X), call it `x`

. By our construction, `x`

corresponds to some particular morphism, `u`

, from A to X. We now have at our disposal two morphisms, `u :: A -> X`

and `f :: X -> Y`

(that’s the morphism we are mapping). We can compose them. The result `f . u`

is a morphism from A to Y, so it’s a member of the set H_{A}(Y). We have just defined a function that takes an `x`

from H_{A}(X) and maps it into `y`

from H_{A}(Y), and this will be our H_{A}(f).

Of course, you have to prove that this construction of H_{A} is indeed a functor preserving composition of morphisms, but that’s reasonably easy, once the technique we have just used becomes familiar to you. Here’s the gist of this technique: Use components! When you are defining a functor from category C to category D, pick a component — an object X in C — and define its image, F(X). Then pick a morphism f in C, say from X to Y, and define its image, F(f), as a particular morphism from F(X) to F(Y).

Similarly, when defining a function from set S to T, use its components. Pick an element x of S and define its image in T. That’s exactly what we did in our construction.

Incidentally, what was that requirement that the category C be *locally small*? A category is locally small if the collection of morphisms between any two objects forms a set. This may come as a surprise but there are things in mathematics that are too big to be sets. A classic example is a collection of all sets, which cannot be a set itself, because it would lead to a paradox. A *collection* of all sets, however, is the basis of the **Set** category (which, incidentally, turns out to be locally small).

## Summary So Far

Let me summarize what we’ve learned so far. A category is just a bunch of abstract objects and arrows between them. It tells us nothing about the internal structure of objects. However, for every (locally small) category there is a structure-preserving mapping (a functor) that maps it into a category of sets. Objects in the **Set** category do have internal structure: they have elements; and morphisms are functions acting on those elements. A *representation* maps the categorie’s surface structure of morphisms into the internal structure of sets.

It is like figuring out the properties of orbitals in atoms by looking at the chemical compounds they form, and at the way valencies work. Or discovering that baryons are composed of quarks by looking at their decay products. Incidentally, no one has ever “seen” a free quark, they always live inside other particles. It’s as if physicists were studying the “category” of baryons by mapping them into sets of quarks.

## A Bar Example

This is all nice but we need an example. Let’s start with “A mathematician walks into a bar and orders a category.” The barman says, “We have this new tasty category but we can’t figure out what’s in it. All we know is that it has just one object A” — (“Oh, it’s a monoid,” the mathematician murmurs knowingly) — “…plus a whole bunch of morphisms. Of course all these morphisms go from A to A, because there’s nowhere else to go.”

What the barman doesn’t know is that the new category is just a re-packaging of the good old set of ASCII strings. The morphisms correspond to appending strings. So there is a morphism called `foo`

that apends the string `"foo"`

foo :: String -> String foo = (++"foo") main = putStrLn $ foo "Hello "

and so on.

“I can tell you what’s inside A,” says the mathematician, “but only up to an isomorphism. I’m a mathematician not a magician.”

She quickly constructs a set that contains one element for each morphism — morphisms must, by law, be listed by the manufacturer on the label. So, when she sees `foo`

, she puts an element with the label “foo”, and so on. Incidentally, there is one morphism with no name, which the mathematician maps to an empty label. (In reality this is an identity morphism that appends an empty string.)

“That’s what’s inside the object A,” she says.

“Moreover, this set comes equipped with functions that rearrange its elements. In fact there is a function for every morphism listed in the category,” she says. “Name any morphism and I’ll construct you a function.”

The barman gives her morphism `p`

, which in reality is:

p = (++"p")

“Okay,” she says, “here’s how I construct the corresponding function. Pick any element in my set.”

The barman picks “foo”.

“Okay, ‘foo’ corresponds to the morphism `foo`

,” she says, “so tell me what you call the morphism that’s the composition of `foo`

and `p`

?” (By law, the manufacturer is obliged to specify all admissible compositions of morphisms on the label.)

“It’s called `foop`

,” says the barman.

Quick check:

p . foo == (++"p") . (++"foo") == (++"foop") foop = (++"foop")

“Okay,” she says, “the function corresponding to `p`

maps “foo” into “foop”. Hm, how curious! I bet this function will map the no-label elment into “p”, is that right?”

“Indeed, it does,” says the barman.

Quick check:

p . id == p

“I bet you this is just a string monoid,” says the mathematician.

“I think I’ll have my usual **Top** on the rocks instead.”

# Natural Transformations

We’ve seen how to construct a representation of any (locally small) category in **Set** by selecting an arbitrary object A in the category and studying morphisms originating at A. What if we choose a different object B instead? How different is the representation H_{A} from H_{B}? For that matter, what if we pick any functor F from C to **Set** ? How is it related to H_{A}? That’s what the Yoneda lemma is all about.

Let me start with a short recap.

A functor is a mapping between categories that preserves their structure. The structure of a category is defined by the way its morphisms compose. A functor `F`

maps objects into objects and morphism into morphisms in such a way that if `f . g = h`

then `F(f) . F(g) = F(h)`

.

A *natural transformation* is a mapping between functors that preserves the structure of the underlying categories.

First we have to understand how to define mappings between functors. Suppose we have functors F and G, both going from category C to category D. For a given object X in C, F will map it into F(X) in D, and G will map it into G(X) in D. A mapping Φ between functors must map object F(X) to object G(X), both in category D. We know that a mapping of objects is called a morphism. So for every object X we have to provide a morphism Φ_{X} from F(X) to G(X). This morphism is called a component of Φ at X. Or, looking at it from a different angle, Φ is a *family* of morphisms parameterized by X.

## An Example of Natural Transformation

Just to give you some Haskell intuition, consider functors on **Hask** . These are mapping of types (type constructors) such as `a -> [a]`

or `a -> Maybe a`

, with the corresponging mappings of morphisms (functions) defined by `fmap`

. Recall:

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

The mapping between Haskell functors is a family of functions parameterized by types. For instance, a mapping between the `[]`

functor and the `Maybe`

functor will map a list of `a`

, `[a]`

into `Maybe a`

. Here’s an example of such a family of functions called `safeHead`

:

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

A family of functions parameterized by type is nothing special: it’s called a polymorphic function. It can also be described as a function on both types and values, with a more verbose signature:

{-# LANGUAGE ExplicitForAll #-} safeHead :: forall a . [a] -> Maybe a safeHead [] = Nothing safeHead (x:xs) = Just x main = print $ safeHead ["One", "Two"]

Let’s go back to natural transformations. I have described what it means to define a transformation of functors in terms of objects, but functors also map morphism. It turns out, however, that the tranformation of morphisms is completely determined by the two functors. A morphism f from X to Y is transformed under F into F(f) and under G into G(f). G(f) must therefore be the image of F(f) under Φ. No choice here! Except that now we have two ways of going from F(X) to G(Y).

We can first use the morphism F(f) to take us to F(Y) and then use Φ_{Y} to get to G(Y). Or we can first take Φ_{X} to take us to G(X), and then G(f) to get to G(Y). We call Φ a *natural* transformation if these two paths result in the same morphism (the diagram *commutes*).

The best insight I can offer is that a natural transformation works on structure, while a general morphism works on contents. The naturality condition ensures that it doesn’t matter if you first rearrange the structure and then the content, or the other way around. Or, in other words, that a natural transformation doesn’t touch the content. This will become clearer in examples.

Going back to Haskell: Is `safeHead`

a natural transformation between two functors `[]`

and `Maybe`

? Let’s start with a function `f`

from some type `a`

to `b`

. There are two ways of mapping this function: one using the `fmap`

defined by `[]`

, which is the list function `map`

; and the other using the `Maybe`

‘s `fmap`

, which is defined in the `Maybe`

‘s functor instance definition:

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

The two path from `[a]`

to `Maybe b`

are:

- Apply
`fmap f`

to`[a]`

to get`[b]`

and then`safeHead`

it, or - Apply
`safeHead`

to`[a]`

and then use the`Maybe`

version of`fmap`

.

There are only two cases to consider: an empty list and a non-empty list. For an emtpy list we get `Nothing`

both ways, otherwise we get `Just`

`f`

acting on the first element of the list.

We have thus shown that `safeHead`

is a natural transformation. There are more interestig examples of natural transformations in Haskell; monadic `return`

and `join`

come to mind.

The intuition behind natural transformations is that they deal with structure, not contents. `safeHead`

couldn’t care less about what’s stored in a list, but it understands the structure of the list: things like the list being empty, or having a first element. The type of this element doesn’t matter. In Haskell, natural transformations are polymorphic functions that can, like `safeHead`

be typed using `forall`

:

safeHead :: forall a . [a] -> Maybe a

# Yoneda Lemma

Going back to the Yoneda lemma, it states that for any functor from C to **Set** there is a natural transformation from our canonical representation H_{A} to this functor. Moreover, there are *exactly* as many such natural transformations as there are elements in F(A).

That, by the way, answers our other question about the dependence on the choice of A in the Yoneda embedding. The Yoneda lemma tells us that there are natural transformations both ways between H_{A} and H_{B}.

Amazingly, the proof of the Yoneda lemma, at least in one direction, is quite simple. The trick is to first define the natural transformation Φ on one special element of H_{A}(A): the element that corresponds to the identity morphism on A (remember, there is always one of these for every object). Let’s call this element p. Its image under Φ_{A} will be in F(A), which is a set. You can pick *any element* of this set and it will define a different but equally good Φ. Let’s call this element q. So we have fixed Φ_{A}(p) = q.

Now we have to define the action of Φ on an arbitrary element in the image of H_{A}. Remember that the functor H_{A} transforms objects in C into sets. So let’s take an arbitrary object X and its image H_{A}(X). The elements in H_{A}(X) correspond to morphisms from A to X. So let’s pick one such morphism and call it f. Its image is an element r in H_{A}(X). The question is, what does r map into under Φ? Remember, it’s image must be an element of F(X).

To figure that out, let’s consider the F route. F being a functor transforms our morphism f into F(f) — which is a morphism from F(A) to F(X). But, as you may remember, we have selected a special element in F(A) — our q. Now apply F(f) to q and you get an element in F(X), call it s. (Remember, F(f) is just a regular function between two sets, F(A) and F(X).)

There’s nothing more natural than picking Φ_{X}(r) to be this s! We have thus defined a natural transformation Φ for any X and r.

The straightforward proof that this definition of Φ is indeed *natural* is left as an exercise to the user.

## A Haskell Example

I’ve been very meticulous about distinguishing between morphisms from A to X in C and the corresponding set elements in H_{A}(X). But in practice it’s more convenient to skip the middle man and define natural transformations in the Yoneda lemma as going directly from these morphisms to F(X). Keeping this in mind, the Haskell version of the Yoneda lemma is ofter written as follows:

forall r . ((a -> r) -> f r) ~ f a

where the (lowercase) `f`

is the functor (think of it as a type constructor and its corresponding `fmap`

), `(a -> r)`

is a function corresponding to the morphism from A to X in our orginal formulation. The Yoneda’s natural transformation maps this morphism into the image of `r`

under `f`

— the F(X) in the original formulation. The `forall r`

means that the function `((a -> r) -> f r)`

works for any type `r`

, as is necessary to make it a natural transformation.

The lemma states that the type of this function, `forall r . ((a -> r) -> f r)`

is equivalent to the much simpler type `f a`

. If you remember that types are just sets of values, you can interpret this result as stating that there is one-to-one correspondence between natural transformations and values of the type `f r`

.

Remember the example from the beginning of this article? There was a function `imager`

with the following signature:

imager :: forall r . ((Bool -> r) -> [r])

This looks very much like a natural transformation from the Yoneda lemma with the type `a`

fixed to `Bool`

and the functor, the list functor `[]`

. (I’ll call the functions `Bool->r`

*iffies*.)

The question was, how many different implementations of this signature are there?

The Yoneda lemma tells us exactly how to construct such natural transformations. It instructs us to start with an identity iffie: `idBool :: Bool -> Bool`

, and pick any element of `[Bool]`

to be its image under our natural transformation. We can, for instance, pick `[True, False, True, True]`

. Once we’ve done that, the action of this natural transformation on *any* iffie `h`

is fixed. We just map the morphism `h`

using the functor (in Haskell we `fmap`

the iffie), and apply it to our pick, `[True, False, True, True]`

.

Therefore, all natural transformations with the signature:

forall r . ((Bool -> r) -> [r])

are in one-to-one correspondence with different lists of `Bool`

.

Conversely, if you want to find out what list of `Bool`

is hidden in a given implementation of `imager`

, just pass it an identity iffie. Try it:

{-# LANGUAGE ExplicitForAll #-} imager :: forall r . ((Bool -> r) -> [r]) imager iffie = fmap iffie [True, False, True, True] data Color = Red | Green | Blue deriving Show data Note = C | D | E | F | G | A | B deriving Show colorMap x = if x then Blue else Red heatMap x = if x then 32 else 212 soundMap x = if x then C else G idBool :: Bool -> Bool idBool x = x main = print $ imager idBool

Remember, this application of the Yoneda lemma is only valid if `imager`

is a natural transformation — its naturality square must commute. The two functors in the `imager`

naturality diagram are the Yoneda embedding and the list functor. Naturality of `imager`

translates into the requirement that any function `f :: a -> b`

modifying an iffie could be pulled out of the `imager`

:

imager (f . iffie) == map f (imager iffie)

Here’s an example of such a function translating colors to strings commuting with the application of imager:

{-# LANGUAGE ExplicitForAll #-} imager :: forall r . ((Bool -> r) -> [r]) imager iffie = fmap iffie [True, False, True, True] data Color = Red | Green | Blue deriving Show colorMap x = if x then Blue else Red f :: Color -> String f = show main = do print $ imager (f . colorMap) print $ map f (imager colorMap)

# The Structure of Natural Transformations

That brings another important intuition about the Yoneda lemma in Haskell. You start with a type signature that describes a natural transformation: a particular kind of polymorphic function that takes a probing function as an argument and returns a type that’s the result of a functor acting on the result type of the probing function. Yoneda tells us that the structure of this natural transformation is tightly constrained.

One of the strengths of Haskell is its very strict and powerful type system. Many Haskell programers start designing their programs by defining type signatures of major functions. The Yoneda lemma tells us that type signatures not only restrict how functions can be combined, but also how they can be implemented.

As an extreme, there is one particular signature that has only one implementation: `a->a`

(or, more explicitly, `forall a. a -> a`

). The only *natural* implementation of this signature is the identity function, `id`

.

Just for fun, let me sketch the proof using the Yoneda lemma. If we pick the source type as the singleton unit type, `()`

, then the Yoneda embedding consists of all functions taking unit as an argument. A function taking unit has only one return value so it’s really equivalent to this value. The functor we pick is the identity functor. So the question is, how many natural tranformation of the the following type are there?

forall a. ((() -> a) -> a)

Well, there are as many as there are elements in the image of `()`

under the identity functor, which is exactly one! Since a function `()->a`

can be identified with `a`

, it means we have only one natural transformation with the following signature:

forall a. (a -> a)

Moreover, by Yoneda construction, this function is defined by `fmap`

ping the function `()->a`

over the element `()`

using the identity functor. So our natural transformation, when probed with a value of the type `a`

will return the same value. But that’s just the definition of the identity function. (In reality things are slightly more complicated because every Haskell type must include `undefined`

, but that’s a different story.)

Here’s an exercise for the reader: Show that the naturality square for this example is equivalent to `id`

commuting with any function: `f . id == id . f`

.

# Conclusion

I hope I provided you with enough background information and intuition so that you’ll be able to easily read more advanced blog posts, like this one:

Reverse Engineering Machines with the Yoneda Lemma by Dan Piponi, or GADTs by Gabriel Gonzales.

# Acknowledgments

I’d like to thank Gabriel Gonzales for providing useful comments and John Wiegley, Michael Sloan, and Eric Niebler for many interesting conversations.

May 16, 2013 at 1:14 am

Thank you. An accessible introduction to category theory is something I’ve been looking for. Wikipedia gets a little dense and this helps a lot.

June 12, 2013 at 2:17 pm

In a paragraph following Fig. 6, beginning:

To gain some insight, ….

the last sentence seems to be at bit garbled.

This is an amazing presentation! With almost no background, I’ve been able to make great progress. This kind of introductory presentation of advanced topics is very rare. Many thanks!

June 12, 2013 at 4:52 pm

Thank you guys. I really appreciate the kind words. They give me incentive to keep writing.

(I fixed the paragraph.)

June 13, 2013 at 5:31 am

I don’t see the fix yet. I was referring to the last sentence in this paragraph:

“To gain some insight, imagine that a natural transformation expresses the idea that F(X) looks, from a certain angle, the same as G(X). Similarly, F(Y) looks the same as G(Y). If that’s true, then a morphism from X to Y must map into something that looks like a single morphism from the concept that combines F(X) and G(X) to the concept that combines F(Y) and G(Y). This way F(f) looks the same and G(f) and the same as (ΦY . F(f)) and the same as (G(f) . ΦX).”

Thank you.

June 13, 2013 at 12:05 pm

Charles: Oh, that paragraph. You’re right, it might be confusing. I’ll try something else.

October 8, 2013 at 11:58 am

[…] function? I’ve seen this pattern in the Yoneda lemma (for a refresher, see my tutorial: Understanding Yoneda). In Haskell we usually see a slightly narrower application of Yoneda. It says that a certain set […]

January 14, 2014 at 10:40 am

[…] functor (the Reader functor). I said it had something to do with the Yoneda lemma. I wrote a whole blog about the Yoneda lemma, so I’m not going to repeat it here — just translate it into the container […]

May 8, 2014 at 12:46 pm

[…] But what does it mean that the isomorphism is natural? What are the two functors that are mapped by it? One functor maps V into Hom(V, U), and the other maps V into Nat(ΔV, F). Both Hom(V, U) and Nat(ΔV, F) are sets in the category of sets. So these are functors from C to Set. You might recognize the first one from the Yoneda lemma. […]

September 29, 2014 at 2:54 am

[…] that should be mentioned. So let me recap what Yoneda is in the more traditional setting (see my blog post about it). It’s about functors from an arbitrary category to a much simpler category of sets — a […]

October 9, 2014 at 8:52 am

Hello Bartosz,

I’m trying to understand Yoneda lemma and your article are pretty nice and getting me closer to that point! And here are some questions I want to ask:

Consider a monoid viewed as a category. By taking Yoneda embedding it turned into a Set, if what I have observed above is true, then is that happens to be the case that for each element of the monoid, I can find both a corresponding element in the only resulting object of Set and a corresponding morphism in Set. And both those correspondences are one-to-one? (got little confused in the bar example between morphisms from the unknown category and elements from the set that mathematician has constructed)

For the bar example, can I interpret it this way: the morphism “p” is like a mapping X -> Y, and “foo” A -> X, by asking what is “foo . p” you actually get a mapping A -> Y. And here the trick is to replace “foo” with the identity so you can figure out (up to isomorphism?) what really “p” is?

Also for the bar example, I don’t get it when mathematician says “this set comes equipped with functions that rearrange its elements.”, how come can elements of an unorderred structure(in this case, a set) be rearranged? I think I’m taking this sentence wrong.

Thank you!

October 9, 2014 at 1:26 pm

@Javran:

A monoid taken as a category is just one object (mono means one) with no internal structure. It has no “elements”. But it does have morphisms, so its hom-set may be quite rich. The elements of the hom set are what you would normally call the elements of a monoid. You have one element corresponding to identity, and multiplication of elements is given by the composition of original morphisms. But a set also has its own morphisms, which are just functions. These functions can be seen as (a) images of original morphisms of the monoid under the functor, or (b) as functions induced by multiplication in the set. These are just two ways of looking at the same thing.

and 4. Functions that map a set to itself can be thought of as rearrangements of its elements. The identity function maps each element to itself. Some functions permute the elements. Others map multiple elements into one, and so on.

Think of a set of strings. There is a morphism that maps “foo” to “foop” and “bur” to “burp” and “” to “p”. That’s the morphism that was called p by the manufacturer. What the mathematician did was to guess the simple naming convention used by the manufacturer. Of course the manufacturer could have called it q instead, and it would have worked too. It would just introduce a non-trivial isomorphism into the picture.

October 30, 2014 at 8:18 am

Hm. I do not quite understand what you mean to illustrate with the Bar Example.

Surely, you do not want to say that all monoids are isomorphic to some (free) monoid of all strings over some alphabet?

There are finite monoids, for example!

April 13, 2015 at 4:20 am

Hi,

This post was very helpful. There were a couple of places I got stuck, and clarifying these may make this post more helpful to people with background similar to mine:

“Its image will be in F(A), which is a set.” -> since the symbol p, which is also in context, is a morphism; it may be useful to clarify that you’re talking about the image of id_A under PHI_A.

“Now apply F(f) to q and you get an element in F(X), call it s.” -> it may be helpful to explicitly state that F(f) is an arrow in

theset category and hence is a function, and we can talk about “apply” it like a normal set-theoretic function (did I understand this correctly)?Thank you for writing this!

April 13, 2015 at 1:45 pm

@Sanjoy: I added some clarifications. Thanks for your feedback.

June 12, 2017 at 1:27 pm

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