Category Theory for Programmers Challenges


Part One

1 Category: The Essence of Composition

1.4 Challenges

1.4 - 1

Implement, as best as you can, the identity function in your favorite language (or the second favorite, if your favorite language happens to be Haskell).

See here.

1.4 - 2

Implement the composition function in your favorite language. It takes two functions as arguments and returns a function that is their composition.

See here.

1.4 - 3

Write a program that tries to test that your composition function respects identity.

See here.

1.4 - 4

Is the world-wide web a category in any sense? Are links morphisms?

The world-wide web is a category if we consider reachability relations as morphisms. Links are not morphisms because they can not be composited. Having a link from a to b and b to c, does not mean there is a link from a to c.

1.4 - 5

Is Facebook a category, with people as objects and friendships as morphisms?

  1. Skipped.
1.4 - 6

When is a directed graph a category?

When its transitive closure is it self.

2 Types and Functions

2.7 Challenges

2.7 - 1

Define a higher-order function (or a function object) memoize in your favorite language. This function takes a pure function f as an argument and returns a function that behaves almost exactly like f, except that it only calls the original function once for every argument, stores the result internally, and subsequently returns this stored result every time it’s called with the same argument. You can tell the memoized function from the original by watching its performance. For instance, try to memoize a function that takes a long time to evaluate. You’ll have to wait for the result the first time you call it, but on subsequent calls, with the same argument, you should get the result immediately.

See here.

2.7 - 2

Try to memoize a function from your standard library that you normally use to produce random numbers. Does it work?

No, it does not work. A function for generating random numbers needs to be able to generate different numbers for different calls, which means it is not pure. Memoizing it will cause the first generated number being remembered, then the subsequent calls will always return the same number, which changes the intended behavior.

2.7 - 3

Most random number generators can be initialized with a seed. Implement a function that takes a seed, calls the random number generator with that seed, and returns the result. Memoize that function. Does it work?

Yes it works. The output of the function is entirely determined by the seed, so it is a pure function, which means memoizing will work.

2.7 - 4

Which of these C++ functions are pure? Try to memoize them and observe what happens when you call them multiple times: memoized and not.

  1. The factorial function from the example in the text.
  2. std::getchar()
  3. bool f() {
        std::cout << "Hello!" << std::endl;
        return true;
  4. int f(int x) {
        static int y = 0;
        y += x;
        return y;
  1. Pure.
  2. Not pure, because different values could be returned depends on user input.
  3. Not pure, because it has a side effect of printing texts.
  4. Not pure, because the global variable y is accessed, which may cause the return value to be different even if the argument x is the same. For example, call f twice with 1, the first time f will return 1, and the second time f will return 2.
2.7 - 5

How many different functions are there from Bool to Bool? Can you implement them all?

There are 4 different pure functions. See here.

2.7 - 6

Draw a picture of a category whose only objects are the types Void, () (unit), and Bool; with arrows corresponding to all possible functions between these types. Label the arrows with the names of the functions.

The graph is too complex to draw, so here are the edges described in text.

Note: yes is a function that returns True for all inputs, no is a function that returns False for all inputs, and not is a function that negates a boolean value.

3 Categories Great and Small

3.6 Challenges

3.6 - 1

Generate a free category from:

  1. A graph with one node and no edges
  2. A graph with one node and one (directed) edge (hint: this edge can be composed with itself)
  3. A graph with two nodes and a single arrow between them
  4. A graph with a single node and 26 arrows marked with the letters of the alphabet: a, b, c … z.
  1. Add the identity edge.
  2. Do nothing.
  3. For each node, add one identity edge.
  4. For each string that only consists of the alphabet with length that is not 1, add one edge.
3.6 - 2

What kind of order is this?

  1. A set of sets with the inclusion relation: A is included in B if every element of A is also an element of B.
  2. C++ types with the following subtyping relation: T1 is a subtype of T2 if a pointer to T1 can be passed to a function that expects a pointer to T2 without triggering a compilation error.
  1. Partial order.
  2. Partial order.
3.6 - 3

Considering that Bool is a set of two values True and False, show that it forms two (set-theoretical) monoids with respect to, respectively, operator && (AND) and || (OR).

3.6 - 4

Represent the Bool monoid with the AND operator as a category: List the morphisms and their rules of composition.


Rules of composition:

3.6 - 5

Represent addition modulo 3 as a monoid category.


Rules of composition:

4 Kleisli Categories

4.4 Challenge

A function that is not defined for all possible values of its argument is called a partial function. It’s not really a function in the mathematical sense, so it doesn’t fit the standard categorical mold. It can, however, be represented by a function that returns an embellished type optional:

template<class A> class optional {
    bool _isValid;
    A _value;
    optional()    : _isValid(false) {}
    optional(A v) : _isValid(true), _value(v) {}
    bool isValid() const { return _isValid; }
    A value() const { return _value; }

For example, here’s the implementation of the embellished function safe_root:

optional<double> safe_root(double x) {
    if (x >= 0) return optional<double>{sqrt(x)};
    else return optional<double>{};

Here’s the challenge:

4.4 - 1

Construct the Kleisli category for partial functions (define composition and identity).

See here.

4.4 - 2

Implement the embellished function safe_reciprocal that returns a valid reciprocal of its argument, if it’s different from zero.

See here.

4.4 - 3

Compose the functions safe_root and safe_reciprocal to implement safe_root_reciprocal that calculates sqrt(1/x) whenever possible.

See here.

5 Products and Coproducts

5.8 Challenges

5.8 - 1

Show that the terminal object is unique up to unique isomorphism.

Suppose we have two terminal objects t1 and t2. Since t1 is terminal, there must exists an unique morphism f from t1 to t2. Similarly, there must exists an unique morphism g from t2 to t1.

Now we show that fg is the identity morphism. First, since f is a morphism from t1 to t2, and g is a morphism from t2 to t1, fg must be a morphism from t1 to t1. Since t1 is a terminal object, there must be an unique morphism from t1 to t1, and since we are in a category, there must be an identity morphism form t1 to t1. So we must have fg being the identity morphism.

Similarly, gf must also be the identity morphism from t2 to t2. Now we have t1 and t2 are unique up to isomorphism. Since f and g are unique, we further have t1 and t2 are unique up to unique isomorphism.

5.8 - 2

What is a product of two objects in a poset? Hint: Use the universal construction.

A product of two objects a and b in a poset is the object c that ca and cb such that for any other object c′ that c′ ≤ a and c′ ≤ b, c′ ≤ c. In other words, the product of two objects is the largest object that is smaller than both a and b.

5.8 - 3

What is a coproduct of two objects in a poset?

A coproduct of two objects a and b in a poset is the object c that ac and bc such that for any other object c′ that ac′ and bc′, cc′. In other words, the coproduct of two objects is the smallest object that is greater than both a and b.

5.8 - 4

Implement the equivalent of Haskell Either as a generic type in your favorite language (other than Haskell).

See here.

5.8 - 5

Show that Either is a “better” coproduct than int equipped with two injections:

int i(int n) { return n; }
int j(bool b) { return b ? 0: 1; }

Hint: Define a function

int m(Either const & e);

that factorizes i and j.

See here.

5.8 - 6

Continuing the previous problem: How would you argue that int with the two injections i and j cannot be “better” than Either?

If int were better than Either, there should exist a function m from int to Either that factorizes the Left and Right function. Consider that the Left function maps 0 to Left(0) and the Right function maps true to Right(0), mi should behave like Left and mj should behave like Right. But that is not possible, because i maps 0 to 0, and j maps true to 0, there is no way for m to map 0 to both Left(0) and Right(true), so int cannot be better than Either.

5.8 - 7

Still continuing: What about these injections?

int i(int n) {
    if (n < 0) return n;
    return n + 2;

int j(bool b) { return b ? 0: 1; }

Still not better than Either. With Left, I can map std::numeric_limits<int>::max() to Left(std::numeric_limits<int>::max()) (sorry for mixing Rust and C++ grammar, but you get the idea), but i(std::numeric_limits<int>::max()) triggers signed integer overflow, which is undefined behavior. No m function can map undefined behavior to Left(std::numeric_limits<int>::max()), so Either is still better.

5.8 - 8

Come up with an inferior candidate for a coproduct of int and bool that cannot be better than Either because it allows multiple acceptable morphisms from it to Either.

Consider the (bool, i32) tuple type with the following two injections:

fn left(n: i32) -> (bool, i32) {
   (false, n)

fn right(b: bool) -> (bool, i32) {
   (true, i32::from(b))

I can have two functions from (bool, i32) to Either:

fn m1(x: (bool, i32)) -> Either<i32, bool> {
   if x.0 {
      match x.1 {
         0 => Either::Right(false),
         1 => Either::Right(true),
         _ => Either::Left(77), // <-- See here.
   } else {

fn m2(x: (bool, i32)) -> Either<i32, bool> {
   if x.0 {
      match x.1 {
         0 => Either::Right(false),
         1 => Either::Right(true),
         _ => Either::Left(88), // <-- See here.
   } else {

See here.

6 Simple Algebraic Data Types

6.5 Challenges

6.5 - 1

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

See here.

6.5 - 2

Here’s a sum type defined in Haskell:

data Shape = Circle Float
           | Rect Float Float

When we want to define a function like area that acts on a Shape, we do it by pattern matching on the two constructors:

area :: Shape -> Float
area (Circle r) = pi * r * r
area (Rect d h) = d * h

Implement Shape in C++ or Java as an interface and create two classes: Circle and Rect. Implement area as a virtual function.

See here.

6.5 - 3

Continuing with the previous example: We can easily add a new function circ that calculates the circumference of a Shape. We can do it without touching the definition of Shape:

circ :: Shape -> Float
circ (Circle r) = 2.0 * pi * r
circ (Rect d h) = 2.0 * (d + h)

Add circ to your C++ or Java implementation. What parts of the original code did you have to touch?

See here.

I have to update the definition of the Shape interface and the implementation of Shape for both Circle and Rect types.

6.5 - 4

Continuing further: Add a new shape, Square, to Shape and make all the necessary updates. What code did you have to touch in Haskell vs. C++ or Java? (Even if you’re not a Haskell programmer, the modifications should be pretty obvious.)

See here.

6.5 - 5

Show that a + a = 2 × a holds for types (up to isomorphism). Remember that 2 corresponds to Bool, according to our translation table.

See here.

7 Functors

7.4 Challenges

7.4 - 1

Can we turn the Maybe type constructor into a functor by defining:

fmap _ _ = Nothing

which ignores both of its arguments? (Hint: Check the functor laws.)

No, because identity is not preserved. For example, fmap id (Just 4) = Nothing, while id (Just 4) = Just 4.

7.4 - 2

Prove functor laws for the reader functor. Hint: it’s really simple.

7.4 - 3

Implement the reader functor in your second favorite language (the first being Haskell, of course).

See here.

7.4 - 4

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

8 Functoriality

8.9 Challenges

8.9 - 1

Show that the data type:

data Pair a b = Pair a b

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

See here.

instance Bifunctor Pair where
    bimap f g (a, b) = (f a, g b)
    first f (a, b) = (f a, b)
    second g (a, b) = (a, g b)
8.9 - 2

Show the isomorphism between the standard definition of Maybe and this desugaring:

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

Hint: Define two mappings between the two implementations. For additional credit, show that they are the inverse of each other using equational reasoning.

See here.

maybeToEither :: Maybe a -> Either (Const () a) (Identity a)
maybeToEither Nothing = Left (Const ())
maybeToEither (Just a) = Right a

eitherToMaybe :: Either (Const () a) (Identity a) -> Maybe a
eitherToMaybe Left (Const ()) = Nothing
eitherToMaybe Right a = Just a

Proofs of maybeToEither and eitherToMaybe are the inverse of each other.

8.9 - 3

Let’s try another data structure. I call it a PreList because it’s a precursor to a List. It replaces recursion with a type parameter b.

data PreList a b = Nil | Cons a b

You could recover our earlier definition of a List by recursively applying PreList to itself (we’ll see how it’s done when we talk about fixed points).

Show that PreList is an instance of Bifunctor.

See here.

8.9 - 4

Show that the following data types define bifunctors in a and b:

data K2 c a b = K2 c

data Fst a b = Fst a

data Snd a b = Snd b

For additional credit, check your solutions against Conor McBride’s paper Clowns to the Left of me, Jokers to the Right.

See here.

8.9 - 5

Define a bifunctor in a language other than Haskell. Implement bimap for a generic pair in that language.

See here.

8.9 - 6

Should std::map be considered a bifunctor or a profunctor in the two template arguments Key and T? How would you redesign this data type to make it so?

Personal opinion: intuitively, a map is like a function that maps the key type to the value type, so like a function, it would seem that map should be considered a profunctor. The problem is that the map type from the standard library is a mutable data structure, and I am not sure what the semantic of mutating mapped value should be, such as insertion and removing.

The best I can come up with is this, only a getter is implemented.

9 Function Types

10 Natural Transformations

10.6 Challenges

10.6 - 1

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

See here.

10.6 - 2

Define at least two different natural transformations between Reader () and the list functor. How many different lists of () are there?

See here.

There are infinite countable number of lists of ().

10.6 - 3

Continue the previous exercise with Reader Bool and Maybe.

See here.

10.6 - 4

Show that horizontal composition of natural transformation satisfies the naturality condition (hint: use components). It’s a good exercise in diagram chasing.

We need to prove that for any

(βα) is a natural transformation from (GF) to (G′ ∘ F′), that is:

\[((G' ∘ F') f) ∘ (β ∘ α)_a = (β ∘ α)_b ∘ ((G ∘ F) f).\]

Let X = GF, Y = G′ ∘ F′, and γ = βα, we need to prove that:

\[(Y f) ∘ γ_a = γ_b ∘ X f\text{.}\]


First, We apply \((Y f) ∘ γ_a\) to X a:

\(((Y f) ∘ γ_a) (X a)\)
= \((Y f) (γ_a (X a))\)
= \((Y f) (Y a)\)
= \(Y b\).

Then, we also apply \(γ_b ∘ (X f)\) to X a:

\((γ_b ∘ (X f)) (X a)\)
= \(γ_b ((X f) (X a))\)
= \(γ_b (X b)\)
= \(Y b\).

So indeed \((Y f) ∘ γ_a = γ_b ∘ X f\), we have proved that composition of natural transformations satisfies the naturality condition.

10.6 - 5

Write a short essay about how you may enjoy writing down the evident diagrams needed to prove the interchange law.


10.6 - 6

Create a few test cases for the opposite naturality condition of transformations between different Op functors. Here’s one choice:

op :: Op Bool Int
op = Op (\x -> x > 0)


f :: String -> Int
f x = read x

See here.

Part Two

11 Declarative Programming

12 Limits and Colimits

12.5 Challenges

12.5 - 1

How would you describe a pushout in the category of C++ classes?

Assume morphisms are inheritance relation from subclasses to superclasses, if we have a diagram of class A inheriting from both B and C, then the pushout is a class D that for any class E that is the superclass of both B and C, it is the super class of D, that is, D is the maximally common superclass of B and C.

12.5 - 2

Show that the limit of the identity functor Id :: CC is the initial object.

The image of Id is C itself, which contains all objects in C, so the limit is a natural transformation, there exist a morphism from the limit to each object in C, since the natural transformation is universal, the morphisms from the limit to each object is unique, which satisfied the condition of the initial object.

12.5 - 3

Subsets of a given set form a category. A morphism in that category is defined to be an arrow connecting two sets if the first is the subset of the second. What is a pullback of two sets in such a category? What’s a pushout? What are the initial and terminal objects?

The pullback of two sets is their intersection set.

The pushout of two sets is their union set.

The initial object is the empty set.

The terminal object is the union set of all sets.

12.5 - 4

Can you guess what a coequalizer is?

The coequalizer is a colimit of a two-element category with two parallel morphisms going between them.

12.5 - 5

Show that, in a category with a terminal object, a pullback towards the terminal object is a product.

The pullback of objects a and c with two morphisms:

is the object d with two morphisms:

that for any object object d′ with two morphisms:

where fp′ = gq′, there is a unique morphism h: d′ → d, so that:

The trick is that b is a terminal object, both fp′ and gq′ must be the unique morphism from d′ → b, so the fp′ = gq′ condition is always satisfied. Without the constraint, d′ can be any object with any two morphisms:

Using the unique h morphism as the factoring morphism, we can satisfy the product condition, with d as the product of a and c.

12.5 - 6

Similarly, show that a pushout from an initial object (if one exists) is the coproduct.

The pushout of objects a and c with two morphisms:

is the object d with two morphisms:

that for any object object d′ with two morphisms:

where p′ ∘ f = q′ ∘ g, there is a unique morphism h: dd′, so that:

The trick is that b is an initial object, both p′ ∘ f and q′ ∘ g must be the unique morphism from bd′, so the p′ ∘ f = q′ ∘ g condition is always satisfied. Without the constraint, d′ can be any object with any two morphisms:

Using the unique h morphism as the factoring morphism, we can satisfy the coproduct condition, with d as the coproduct of a and c.

13 Free Monoids

13.3 Challenges

13.3 - 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.

An isomorphism between monoids A and B is a pair of morphisms

so that:

If f and g preserve multiplication, we have: for all a, b in A, f a × f b = f (a × b).

Assume e is the unit element in A, we need to prove that f e is the unit element in B, which means for all c in B, we need to have:

f e × c
= f e × (fg) c (Since fg is the identity morphism on B.)
= f e × f (g c)
= f (e × g c) (Since multiplication is preserved.)
= f (g c)
= (fg) c
= c (Since fg is the identity morphism on B.)

Similarly, we can prove that c × f e = c, so we provide that f e is the unit element in B, so unit is preserved.

13.3 - 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?

  1. The image of the empty list [] is 1 since homomorphism preserves unit.
  2. The image of [1, 2, 3, 4] is 24 since homomorphism preserves multiplication.
  3. There lists map to integer 12 are lists that have total product of 12. There are infinite countable number of lists that have product of 12, since you can add 1 to the list any times you want without changing the product.
  4. We can also map lists of integers concatenation to integers with addition.

13.3 - 3

What is the free monoid generated by a one-element set? Can you see what it’s isomorphic to?

It is lists that only contain that one-element zero or more times. It’s isomorphic to natural numbers with addition where we map list to the length of the list.

14 Representable Functors

14.3 Challenges

14.3 - 1

Show that the hom-functors map identity morphisms in C to corresponding identity functions in Set.

For functor C(a, -), it maps a morphism f to a function that precomposes f on its argument:

(C(a, -) f) g = fg

So when applied to the identity morphism:

(C(a, -) id) g = idg = g

So the identity morphism is mapped to a function that just returns the argument, which is the identity function.

For contravariant functor C(-, a), it maps a morphism f to a function that composes f to its argument:

(C(-, a) f) g = gf

So when applied to the identity morphism:

(C(-, a) id) g = gid = g

So the identity morphism is mapped to a function that just returns the argument, which is the identity function.

14.3 - 2

Show that Maybe is not representable.

For Maybe to be representable, we need for create a function ax from a Maybe x value for some representing type a. The problem is that a Maybe x value could be Nothing, from which we are not able to create a function that returns an x value, so Maybe is not representable.

14.3 - 3

Is the Reader functor representable?

Yes, trivially.

14.4 - 4

Using Stream representation, memoize a function that squares its argument.

Not sure if I understand the challenge correctly, but here is my attempt:

See here.

14.4 - 5

Show that tabulate and index for Stream are indeed the inverse of each other. (Hint: use induction.)

Wwe prove that both tabulateindex and indextabulate are the identity function.

Let f = tabulate . index, we have:

f (Cons x y)
= (tabulate . index) (Cons x y)
= tabulate (index (Cons x y))
= Cons (index (Cons x y) 0) (tabulate ((index (Cons x y)) . (+1)))
= Cons x (tabulate ((index (Cons x y)) . (+1)))
= Cons x (tabulate ((index (Cons x y)) . (+1)))
= Cons x (tabulate ((\n -> if n == 0 then x else index y (n - 1)) . (+1)))
= Cons x (tabulate (\n -> if n + 1 == 0 then x else index y (n + 1 - 1)))
= Cons x (tabulate (\n -> index y n))
= Cons x (tabulate (index y))
= Cons x (f y)

we know f will preserve the first value, thus by induction on f y, we know that it will also preserve the second value, and also the third value, etc. So f will preserve all values in a Stream object, which means it is the identity function.

Let g = index . tabulate, we need to prove for any h, g h = h, which means for any x, g h x = h x. We use proof by induction on x:

Base case, if x = 0:

g h 0
= ((index . tabulate) h) 0
= index (tabulate h) 0
= index (Cons (h 0) (tabulate (h . (+1)))) 0
= (h 0)

Inductive cases, if x > 0:

g h x
= (index . tabulate) h x
= index (tabulate h) x
= index (Cons (h 0) (tabulate (h . (+1)))) x
= index (tabulate (h . (+1))) (x - 1) (since x > 0)
= (index . tabulate) (h . (+1)) (x - 1)
= g (h . (+1)) (x - 1)
= (h . (+1)) (x - 1) (by induction)
= h ((+1) (x - 1))
= h x

So for any x, g h x = h x, thus for any h, g h = h, which means g is the identity function.

14.4 - 6

The functor:

Pair a = Pair a a

is representable. Can you guess the type that represents it? Implement tabulate and index.

The bool type represents the Pair functor.

See here.

15 The Yoneda Lemma

15.3 Challenges

15.3 - 1

Show that the two functions phi and psi that form the Yoneda isomorphism in Haskell are inverses of each other.

phi :: (forall x . (a -> x) -> F x) -> F a
phi alpha = alpha id
psi :: F a -> (forall x . (a -> x) -> F x)
psi fa h = fmap h fa

We need to show that both (phi . psi) and (psi . phi) are identity functions:

(phi . psi) fa
= phi (psi fa)
= phi (\h -> fmap h fa)
= (\h -> fmap h fa) id
= fmap id fa
= fa

So (phi . psi) is an identity function.

(psi . phi) alpha
= psi (phi alpha)
= psi (alpha id)
= \h -> fmap h (alpha id)
= \h -> (fmap h . alpha) id
= \h -> (alpha . fmap h) id (by naturality condition)
= \h -> alpha (fmap h id)
= \h -> alpha (h . id)
= \h -> alpha h
= alpha

So (psi . phi) is also an identity function, we conclude that phi and psi are inverses of each other.

15.3 - 2

A discrete category is one that has objects but no morphisms other than identity morphisms. How does the Yoneda lemma work for functors from such a category?


15.3 - 3

A list of units [()] contains no other information but its length. So, as a data type, it can be considered an encoding of integers. An empty list encodes zero, a singleton [()] (a value, not a type) encodes one, and so on. Construct another representation of this data type using the Yoneda lemma for the list functor.

According to the Yoneda lemma, [()] is equivalent to type forall x . (() -> x) -> [x].


16.5 Challenges

16.5 - 1

Express the co-Yoneda embedding in Haskell.


16.5 - 2

Show that the bijection we established between fromY and btoa is an isomorphism (the two mappings are the inverse of each other).

Let m be the funcion that converts btoa to fromY and n be the function that converts fromY to btoa, we have: m = \x -> \f -> \b -> f (x b) and n = \x -> x id.

(m . n) x
= m (n x)
= m (x id)
= \f -> \b -> f ((x id) b)
= \f -> \b -> (f . (x id)) b
= \f -> f . (x id)
= \f -> fmap f (x id) (since fmap f = \x -> f . x)
= \f -> ((fmap f) . x) id
= \f -> (x . (fmap f)) id (by naturality condition)
= \f -> x (fmap f id)
= \f -> x (f . id)
= \f -> x f
= x

(n . m) x
= n (m x)
= n (\f -> \b -> f (x b))
= (\f -> \b -> f (x b)) id
= \b -> id (x b)
= \b -> x b
= x

So both m . n and n . m are identity functions, we conclude that fromY and btoa are isomorphic.

16.5 - 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?

Let M be the category, and m be the single object, Yoneda embedding maps m to the functor M(m, -).

Since M only has a single object m, all morphisms in M are in the hom-set M(m, m), they will be mapped to natural transforms between functor M(m, -) and it self.

16.5 - 4

What is the application of the covariant Yoneda embedding to preorders? (Question suggested by Gershom Bazerman.)


16.5 - 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).

Let F and G be two objects (functors) in [C, D], and α be a morphism (natural transforms) from F to G, Yoneda embedding maps α to a natural transform from [C, D](G, -) to [C, D](F, -).


17.8 Challenges

17.8 - 1

Consider some degenerate cases of a naturality condition and draw the appropriate diagrams. For instance, what happens if either functor F or G map both objects a and b (the ends of fab) to the same object, e.g., F a = F b or G a = G b? (Notice that you get a cone or a co-cone this way.) Then consider cases where either F a = G a or F b = G b. Finally, what if you start with a morphism that loops on itself — faa?

  1. Either functor F or G map both objects a and b (the ends of fab) to the same object:

     Ff        Gf
     ┌─┐       ┌─┐
     │ ↓   α   │ ↓
    Fa/Fb ──→ Ga/Gb

    G fα = αF f.

  2. Either F a = G a or F b = G b:

     αa         αb
     ┌─┐        ┌─┐
     │ ↓   Ff   │ ↓
    Fa/Ga ───→ Fb/Gb
      │          ↑

    G f ∘ \(α_a\) = \(α_b\) ∘ F f.

  3. A morphism that loops on itself:

     Ff          Gf
    ┌──┐        ┌───┐
    │  ↓   αa   │   ↓
    └─ Fa ────→ Ga ←┘

    G f ∘ \(α_a\) = \(α_a\) ∘ F f.