This post is about a problem I recently proved that seemed simple going in, but I struggled with it for quite a while. The problem statement is as follows:

Show that:

(22)2 (2 \simeq 2) \simeq 2


⚠️ This post describes exercise 2.13 of the Homotopy Type Theory book. If you are planning to attempt this problem yourself, spoilers ahead!

The following line imports some of the definitions used in this post.

open import Prelude
open Σ

The problem explained

With just that notation, the problem may not be clear. 22 represents the set with only two elements in it, which is the booleans. The elements of 22 are true\textrm{true} and false\textrm{false}.

data 𝟚 : Set where
  true : 𝟚
  false : 𝟚

In the problem statement, \simeq denotes homotopy equivalence between sets, which you can think of as basically a fancy isomorphism. For some function f:ABf : A \rightarrow B, we say that ff is an equivalence1 if:

  • there exists a g:BAg : B \rightarrow A
  • fgf \circ g is homotopic to the identity map idB:BB\textrm{id}_B : B \rightarrow B
  • gfg \circ f is homotopic to the identity map idA:AA\textrm{id}_A : A \rightarrow A

We can write this in Agda like this:

record isEquiv {A B : Set} (f : A  B) : Set where
  constructor mkEquiv
    g : B  A
    f∘g : (f  g)  id
    g∘f : (g  f)  id

Then, the expression 222 \simeq 2 simply expresses an equivalence between the boolean set to itself. Here’s an example of a function that satisfies this equivalence:

bool-id : 𝟚  𝟚
bool-id b = b

We can prove that it satisfies the equivalence, by giving it an inverse function (itself), and proving that the inverse is homotopic to identity (which is true definitionally):

bool-id-eqv : isEquiv bool-id
bool-id-eqv = mkEquiv bool-id id-homotopy id-homotopy
    id-homotopy : (bool-id  bool-id)  id
    id-homotopy _ = refl

We can package these together into a single definition that combines the function along with proof of its equivalence properties using a dependent pair2:

_≃_ : (A B : Set)  Set
A  B = Σ (A  B)  f  isEquiv f)

Hint: (click the Σ\Sigma to see how it’s defined!)

This gives us an equivalence:

bool-eqv : 𝟚  𝟚
bool-eqv = bool-id , bool-id-eqv

Great! Now what?

The space of equivalences

Turns out, the definition I gave above is just one of multiple possible equivalences between the boolean type and itself. Here is another possible definition:

bool-neg : 𝟚  𝟚
bool-neg true = false
bool-neg false = true

The remarkable thing about this is that nothing is changed by flipping true and false. Someone using booleans would not be able to tell if you gave them a boolean type where true and false were flipped. Simply put, the constructors true and false are simply names and nothing can distinguish them.

We can show this by proving that bool-neg also forms an equivalence by using itself as the inverse.

bool-neg-eqv : isEquiv bool-neg
bool-neg-eqv = mkEquiv bool-neg neg-homotopy neg-homotopy
    neg-homotopy : (bool-neg  bool-neg)  id
    neg-homotopy true = refl
    neg-homotopy false = refl

bool-eqv2 : 𝟚  𝟚
bool-eqv2 = bool-neg , bool-neg-eqv

You could imagine more complicated functions over booleans that may also exhibit this equivalence property. So how many different elements of the type 222 \simeq 2 exist?

Proving 222 \simeq 2 only has 2 inhabitants

It turns out there are only 2 equivalences, the 2 that we already found. But how do we know this for a fact? Well, if there are only 2 equivalences, that means there are only 2 inhabitants of the type 222 \simeq 2. We can prove that this type only has 2 inhabitants by establishing an equivalence between this type and another type that is known to have 2 inhabitants, such as the boolean type!

That’s where the main exercise statement comes in: “show that” (22)2(2 \simeq 2) \simeq 2, or in other words, find an inhabitant of this type.

main-theorem : (𝟚  𝟚)  𝟚

How do we do this? Well, remember what it means to define an equivalence: first define a function, then show that it is an equivalence. Since equivalences are symmetrical, it doesn’t matter which way we start first. I’ll choose to first define a function 2222 \rightarrow 2 \simeq 2. This is fairly easy to do by case-splitting:

f : 𝟚  (𝟚  𝟚)
f true = bool-eqv
f false = bool-eqv2

This maps true to the equivalence derived from the identity function, and false to the function derived from the negation function.

Now we need another function in the other direction. We can’t case-split on functions, but we can certainly case-split on their output. Specifically, we can differentiate id from neg by their behavior when being called on true:

  • id(true):true\textrm{id}(\textrm{true}) :\equiv \textrm{true}
  • neg(true):false\textrm{neg}(\textrm{true}) :\equiv \textrm{false}
g : (𝟚  𝟚)  𝟚
g eqv = (fst eqv) true
  -- same as this:
  -- let (f , f-is-equiv) = eqv in f true

Hold on. If you know about the cardinality of functions, you’ll observe that in the space of functions f:22f : 2 \rightarrow 2, there are 22=42^2 = 4 equivalence classes of functions. So if we mapped 4 functions into 2, how could this be considered an equivalence?

A key observation here is that we’re not mapping from all possible functions f:22f : 2 \rightarrow 2. We’re mapping functions f:22f : 2 \simeq 2 that have already been proven to be equivalences. This means we can count on them to be structure-preserving, and can rule out cases like the function that maps everything to true, since it can’t possibly have an inverse.

We’ll come back to this later.

First, let’s show that gfidg \circ f \sim \textrm{id}. This one is easy, we can just case-split. Each of the cases reduces to something that is definitionally equal, so we can use refl.

g∘f : (g  f)  id
g∘f true = refl
g∘f false = refl

Now comes the complicated case: proving fgidf \circ g \sim \textrm{id}.


Since Agda’s comment syntax is --, the horizontal lines in the code below are simply a visual way of separating out our proof premises from our proof goals.

module f∘g-case where
  goal :
    (eqv : 𝟚  𝟚)
     f (g eqv)  id eqv

  f∘g : (f  g)  id
  f∘g eqv = goal eqv

Now our goal is to show that for any equivalence eqv:22\textrm{eqv} : 2 \simeq 2, applying fgf ∘ g to it is the same as not doing anything. We can evaluate the g(eqv)g(\textrm{eqv}) a little bit to give us a more detailed goal:

  goal2 :
    (eqv : 𝟚  𝟚)
     f ((fst eqv) true)  eqv

  -- Solving goal with goal2, leaving us to prove goal2
  goal eqv = goal2 eqv

The problem is if you think about equivalences as encoding some rich data about a function, converting it to a boolean and back is sort of like shoving it into a lower-resolution domain and then bringing it back. How can we prove that the equivalence is still the same equivalence, and as a result proving that there are only 2 possible equivalences?

Note that the function ff ultimately still only produces 2 values. That means if we want to prove this, we can case-split on ff‘s output. In Agda, this uses a syntax known as with-abstraction:

  goal3 :
    (eqv : 𝟚  𝟚)  (b : 𝟚)  (p : fst eqv true  b)
     f b  eqv

  -- Solving goal2 with goal3, leaving us to prove goal3
  goal2 eqv with b(fst eqv) true in p = goal3 eqv b p

We can now case-split on bb, which is the output of calling ff on the equivalence returned by gg. This means that for the true case, we need to show that f(b)=bool-eqvf(b) = \textrm{bool-eqv} (which is based on id) is equivalent to the equivalence that generated the true.

Let’s start with the id case; we just need to show that for every equivalence ee where running the equivalence function on true also returned true, ef(true)e \equiv f(\textrm{true}).

Unfortunately, we don’t know if this is true unless our equivalences are mere propositions, meaning if two functions are identical, then so are their equivalences.

isProp(P):x,y:P(xy) \textrm{isProp}(P) :\equiv \prod_{x, y: P}(x \equiv y)

Definition 3.3.1 from the HoTT book

Applying this to isEquiv(f)\textrm{isEquiv}(f), we get the property:

f:AB(e1,e2:isEquiv(f)e1e2) \sum_{f : A → B} \left( \prod_{e_1, e_2 : \textrm{isEquiv}(f)} e_1 \equiv e_2 \right)

This proof is shown later in the book, so I will use it here directly without proof3:

    equiv-isProp : {A B : Set}
       (e1 e2 : A  B)  (Σ.fst e1  Σ.fst e2)
       e1  e2

Now we’re going to need our key observation that we made earlier, that equivalences must not map both values to a single one. This way, we can pin the behavior of the function on all inputs by just using its behavior on true, since its output on false must be different.

We can use a proof that true≢false\textrm{true} \not\equiv \textrm{false} that I’ve shown previously.

  true≢false : true  false
      -- read: true ≡ false → ⊥
  true≢false p = bottom-generator p
      map : 𝟚  Set
      map true = 
      map false = 

      bottom-generator : true  false  
      bottom-generator p = transport map p tt

Let’s transform this into information about ff‘s output on different inputs:

  observation : (f : 𝟚  𝟚)  isEquiv f  f true  f false
                                 -- read: f true ≡ f false → ⊥
  observation f (mkEquiv g f∘g g∘f) p =
      -- Given p : f true ≡ f false
      -- Proof strategy is to call g on it to get (g ∘ f) true ≡ (g ∘ f) false
      -- Then, use our equivalence properties to reduce it to true ≡ false
      -- Then, apply the lemma true≢false we just proved
      step1 = ap g p
      step2 = sym (g∘f true)  step1  (g∘f false)
      step3 = true≢false step2
    in step3

For convenience, let’s rewrite this so that it takes in an arbitrary bb and returns the version of the inequality we want:

  observation' : (f : 𝟚  𝟚)  isEquiv f  (b : 𝟚)  f b  f (bool-neg b)
                                                  -- ^^^^^^^^^^^^^^^^^^^^
  observation' f eqv true p = observation f eqv p
  observation' f eqv false p = observation f eqv (sym p)

Then, solving the true case is simply a matter of using function extensionality (functions are equal if they are point-wise equal) to show that just the function part of the equivalences are identical, and then using this property to prove that the equivalences must be identical as well.

  goal3 eqv true p = equiv-isProp bool-eqv eqv functions-equal
      e = fst eqv

      pointwise-equal : (x : 𝟚)  e x  x
      pointwise-equal true = p
      pointwise-equal false with efe false in eq = helper ef eq
          helper : (ef : 𝟚)  e false  ef  ef  false
          helper true eq = rec-⊥ (observation' e (snd eqv) false (eq  sym p))
          helper false eq = refl

      -- NOTE: fst bool-eqv = id, definitionally
      functions-equal : id  e
      functions-equal = sym (funExt pointwise-equal)

The false case is proved similarly.

  goal3 eqv false p = equiv-isProp bool-eqv2 eqv functions-equal
      e = fst eqv

      pointwise-equal : (x : 𝟚)  e x  bool-neg x
      pointwise-equal true = p
      pointwise-equal false with efe false in eq = helper ef eq
          helper : (ef : 𝟚)  e false  ef  ef  true
          helper true eq = refl
          helper false eq = rec-⊥ (observation' e (snd eqv) true (p  sym eq))

      functions-equal : bool-neg  e
      functions-equal = sym (funExt pointwise-equal)

Putting this all together, we get the property we want to prove:

open f∘g-case

-- main-theorem : (𝟚 ≃ 𝟚) ≃ 𝟚
main-theorem = g , mkEquiv f g∘f f∘g

Now that Agda’s all happy, our work here is done!

Going through all this taught me a lot about how the basics of equivalences work and how to express a lot of different ideas into the type system. Thanks for reading!


  1. There are other ways to define equivalences. As we’ll show, an important property that is missed by this definition is that equivalences should be mere propositions. The reason why this definition falls short of that requirement is shown by Theorem 4.1.3 in the HoTT book.

  2. A dependent pair (or Σ\Sigma-type) is like a regular pair x,y\langle x, y\rangle, but where yy can depend on xx. For example, x,isPrime(x)\langle x , \textrm{isPrime}(x) \rangle. In this case it’s useful since we can carry the equivalence information along with the function itself. This type is rather core to Martin-Löf Type Theory, you can read more about it here.

  3. This is shown in Theorem 4.2.13 in the HoTT book. I might write a separate post about that when I get there, it seems like an interesting proof as well!