Topic 7: Using identity types

Transcription

1 Toic 7: Using identity tyes June 10, 2014 Now we would like to learn how to use identity tyes and how to do some actual mathematics with them. By now we have essentially introduced all inference rules of HoTT; the main missing ingredient is the univalence axiom. At this stage it is ossible to do a substantial amount of mathematics, and some of this we will do here. Throughout A, B U are fixed tyes. Examle: Proositional uniqueness rinciles Recall the uniqueness rincile for functions, which is an inference rule stating that f λx.f(x) for any (deendent) function f. For other tyes, we did not have such uniqueness rinciles; for examle for u 1, such a uniqueness rincile would assert that u, but we have not ostulated this and it is not derivable. However, at this oint we can show that such a uniqueness rincile holds with roositional equality in lace of judgemental equality: Theorem 1. u 1 u =. The roof of this is exemlative of many other roofs in HoTT. It is a good examle of how one can derive non-trivial equations in HoTT, although the only way to rove an equation is to reduce it in some way to reflexivity. Proof. In order to find an element w u 1 u =, we use induction on u. This reduces the roblem to finding w( ) =, which we can define to be w( ) refl. The same theorem and roof in Coq: Theorem ro_unique : forall u : unit, u = tt. Proof. intro u. induction u. reflexivity. Defined. Analogous roositional uniqueness rinciles can now be derived for many other tyes as well: Theorem 2. x A B x = (r 1 (x), r 2 (x)). Proof. In order to find an element of the tye x A B x = (r 1 (x), r 2 (x)), we use induction on x. This reduces the roblem to finding an element v (a, b) = (r 1 ((a, b)), r 2 ((a, b))). a A b B 1

2 Since r 1 ((a, b)) a and r 2 ((a, b)) b, we can take v λa.λb.refl (a,b). Theorem ro_unique_roduct ( A B : Tye ) : forall x : A * B, x = ( fst x, snd x). Proof. intro x. induction x. siml. reflexivity. Defined. One can generalize this to the deendent air tye: for any deendent tye P A U, one can similarly derive x = (r 1 (x), r 2 (x)) for any x a A P (a). We will rove a more general statement later on, when we characterize the identity tye = q for, q a A P (a). Functions acting on aths If we have x, y A together with x = y and a function f A B, does this yield an element of f(x) = f(y)? In other words, does function alication reserve roositional equality? In terms of the toological interretation, one can say that we now analyse how functions act on aths. Lemma 3. For any function f A B and x, y A, there is a function a f (x = A y) (f(x) = B f(y)) defined such that for all x A, a f (refl x ) refl f(x). Strictly seaking, we also need to consider the two elements x, y A as arguments of the function a f, whose tye therefore should be a f (x = A y) (f(x) = B f(y)). x,y A But since the first two arguments x and y are determined by the third argument x = y, we consider x and y to be imlicit arguments which are not written down exlicitly, excet for in the following roof. Proof. For any x, y A and any x = A y, we want to find a f (x, y, ) f(x) = B f(y). By induction on as an element of the (say, unbased) identity tye, it is sufficient to do so in the case that y x and refl x. But in this case, we can ut a f (x, x, refl x ) refl f(x) f(x) = B f(x). In Coq, the statement and roof look as follows: Lemma a { A B : Tye } ( f : A B) : forall x y : A, ( x = y) ( f( x) = f( y)). Proof. intros x y. induction. reflexivity. Defined. We interret Lemma 3 as saying that every function f is automatically continuous, since it mas aths to aths! So in HoTT, it is imossible to construct a function that is not continuous. 2

3 Examle: commutativity of addition As the ossibly first examle of actual mathematics in HoTT, let us revisit the natural numbers and rove that addition is commutative. The following roof structure coincides exactly with how one would rove commutativity of addition in conventional foundations! This is true for many other things as well. Lemma 4. x,y N succ(y + x) = y + succ(x). Recall that we had defined addition x+y add(x, y) by induction over the first argument, using the definitions add(0, y) y and add(succ(x), y) succ(add(x, y)). Proof. In order to find an element of the tye succ(y + x) = y + succ(x) for any x, y N, we can use induction on y. For y 0, we have refl succ(x) succ(x) = succ(x), and by the definition of + we also have judgemental equalities succ(0 + x) succ(x) and 0 + succ(x) succ(x). Therefore, we also have refl succ(x) succ(0 + x) = 0 + succ(x). For the induction ste, we need to show that one can construct an element of the identity tye succ(succ(y) + x) = succ(y) + succ(x) from any given element v succ(y + x) = y + succ(x). Since a succ (v) succ(succ(y + x)) = succ(y + succ(x)) and we have judgemental equalities succ(succ(y) + x) succ(succ(y + x)), succ(y) + succ(x) succ(y + succ(x)) by the definition of +, we also have a succ (v) succ(succ(y) + x) = succ(y) + succ(x). Let us write add succ x,y N y + succ(x) = succ(y + x) for the element constructed in the roof. The same statement and roof in Coq: Lemma add_succ : forall x y : nat, x + S y = S ( x + y). Proof. intros x y. induction x. siml. reflexivity. Defined. siml. aly (a S). exact IHx. Theorem 5. x,y N x + y = y + x. Proof. In order to find an element of the tye x + y = y + x for any x and y, we can use induction on x. This reduces the roblem to finding elements of the following two tyes: 0 + y = y + 0, (x + y = y + x) (succ(x) + y = y + succ(x)) We start with the first. Since + was defined by induction over the first argument, we have 0 + y y, so it remains to show that y = y + 0. This in turn can be shown by induction on y as follows. For y 0, we have again by definition of +, and hence we have a judgemental equality of tyes 3

4 (0 = 0) (0 = 0 + 0), and therefore refl 0 0 = Concerning the induction ste, if we have an equality witness w y = y + 0, then we obtain a succ (w) succ(y) = succ(y + 0). Since succ(y) + 0 succ(y + 0) by the inductive definition of +, we also have a succ (w) succ(y) = succ(y) + 0. Now for the main induction ste. Suose that we have v x + y = y + x, and we want to find an element of succ(x) + y = y + succ(x). For every y, we have a succ (v) succ(x) + y = succ(y + x), where we have used succ(x + y) succ(x) + y by the inductive definition of +, and add succ(x, y) succ(y + x) = y + succ(x) thanks to Lemma 4. The claim now follows from transitivity of equality; this is what we get to next. Concatenating aths Last time, we already saw that roositional equality is symmetric, in the sense that there exists a canonical function (x = y) (y = x) which mas any ath x = y to its inverse 1 y = x. In the ath interretation, this corresonds to the fact that every ath can be traversed forwards or backwards. Recall that taking the inverse is a function 1 (x = y) (y = x) x,y A which can be defined by (based or unbased) ath induction, which reduces the roblem to defining the function on y x and refl x, in which case we define the function value to be refl x itself. Although x and y (and even A) should formally also be considered as arguments of this function, we omit them for brevity; if one knows, one can deduce x and y from the tye of, and then A as the tye of x and y. To summarize, the function 1 is defined by ath induction on, together with the stiulation that refl 1 x refl x for every x A. Similarly, it is ossible to concatenate aths: if is a ath from x to y and q is a ath from y to z, then there is a concatenated ath q from x to z. This is indeed the case: for any A U, one can construct a function (x = y) (y = z) (x = z) (1) x,y,z A as follows. By ermuting the z argument with the (x = y)-argument, it is enough to construct this function in the form (x = y) (y = z) (x = z). x,y z A Alying (unbased) ath induction lets us assume that we are dealing with the case y x and refl x x = x, in which case we simly need to construct an element of the tye (x = z) (x = z), z A for which we could simly take the identity function for every z. However, it turns out to be better to not use the identity function here, but rather to define such a function again through ath 4

5 induction, which allows us to ut z x and assume refl x x = x, in which case we take the value of the function to again be refl x x = x. The reason for not using the identity function, but instead using another ath induction, is mainly aesthetics: with the identity function, the resulting comutation rule for is, refl x q q for all q y = z, while with a second ath induction, the comutation rule is merely refl x refl x refl x, which we refer, since it makes both arguments behave in the same way. In logical terms, one can interret (1) as the transitivity of equality: if x is equal to y and y is equal to z, then x is also equal to z. The function (1) can be interreted as converting any two roofs x = y and q y = z into q x = z. Generally seaking, thinking in terms of aths between oints in a sace rather than roofs of equalities is the referred intuition, and so we stick with this one from now on. Theorem 6. Concatenation of aths is associative: for any w, x, y, z A and w = x, q x = y, r y = z, (q r) = ( q) r. Strictly seaking, what me mean by this theorem is that we claim to be able to construct an element of the tye (q r) = ( q) r, (2) w=x q x=y r y=z A U w,x,y,z A and this is what will be done in the roof. Proof. We use similar arguments as in the construction of : ermuting some of the arameters w, x, y, z and, q, r and then alying ath induction. Indeed, ath induction over allows us to assume x w and refl w, in which case we have q w = y and r y = z and need to rove refl w (q r) = (refl w q) r. We show this by another ath induction, this time on q, which gives y w and q refl w, so that we have reduced the roblem to finding an element of refl w (refl w r) = (refl w refl w ) r. Using the same kind of ath induction for r and alying the comutation rule refl w refl w refl w results in the assertion refl w = refl w, which has a trivial roof given by refl reflw. This result is our first coherence law. Generally seaking, a coherence law states that any two ways to construct a new object from given data are equal. In this case, it states that the concatenation of a sequence of aths does not deend on the order in which these aths are concatenated at least u to roositional equality. The element of the above tye (2) can be shown to satisfy its own coherence law! There is also a sense in which the reflexivity elements are unit elements for the concatenation of aths, in the sense that concatenating with them gives the other ath back of course, again u to roositional equality: 5

6 Theorem 7. For any x, y A and x = y, refl x =, refl y =. Finally, in the same sense, concatenating a ath with its inverse recovers the reflexivity elements: Theorem 8. For any x, y A and x = y, 1 = refl x, 1 = refl y. Both theorems can easily be roven by ath induction on. The concatenation of aths is an algebraic structure on all the identity tyes over any tye A which turns A into a so-called grouoid. Can similar things also be done with higher aths, i.e. with elements of identity tyes between aths, identities between those, and so on? Before getting to this, we briefly study how the algebraic structure on aths interacts with the a function from Lemma 3. Theorem 9. The function a has the following roerties: 1. a f ( q) = a f () a f (q), 2. a f ( 1 ) = a f () 1, 3. For f A B and g B C, we have a g f () = a g (a f ()). 4. a ida () =. Proof. All these roerties follow from ath induction on (and q). Concerning the latter two items, it would be helful to have the stronger statements that a g f = a g a f, and likewise a ida = id x=y. However, roving these statements would require to be able to conclude that two functions are equal as soon as they take on equal values on all oints; this is something which we cannot rove yet! A natural question now is, what haens if instead of f A B we consider a deendent function f x A B(x)? In this case, things are a bit more comlicated: if we have x = A y, then f(x) B(x) and f(y) B(y), so that it does not actually make sense to assert that f(x) = f(y). So, before solving this uzzle, we need to understand deendent tyes a little better; we consider this below. Tyes as higher grouoids For aths, q x = y, we can also consider a ath between these aths, α = q. More rigorously, we should indicate the base tye over which the identity tye = q is formed, and then we have to write α = x=a y q. We will also call such a ath between aths a 2-ath. Suose that we have, in addition to this data, also another ath r y = z. Geometrically, the situation looks like this: x q α y r z 6

7 Using ath induction on r, we can concatenate this data to a 2-ath of tye r = q r which we denote by α r, x r q r More recisely, there is a function which concatenates any 2-ath to any 1-ath in the sense of having the tye ( r) = (q r). A U x,y,z A,q x=y α =q r y=z Again, we only consider α and r as the arguments of this function, since the others can be deduced. Similarly, if we are given another w A and o w = x as in α r z w o x q α y then we can likewise use ath induction on o in order to define a new 2-ath o α o = o q. What oerations can we construct which concatenate two 2-aths to a new 2-ath? For one, if we have r,, q x = y and α = q and β q = r like this, x q r α β y then we already know that we can form the concatenation α β = r. Since this is only one kind of concatenation of 2-aths, it is usually called the vertical concatenation, due to how the α and β are arranged in the diagram above. On the other hand, if we have this situation, x q α y r s β z we can also construct a horizontal concatenation by defining α β (α r) (q β), which is of tye ( r) = (q s). Alternatively, it would also have been ossible to define this horizontal concatenation as ( β) (α s); the result would have been again equal, since one can find an element of the identity tye by using several consecutive ath inductions. (α r) (q β) = ( β) (α s) 7

8 Theorem 10. (Exchange law) In a situation like this, o r x q α β y s t γ δ z we have (α β) (γ δ) = (α γ) (β δ). Proof. As usual, we aly ath induction, this time on α, β, γ and δ, which reduces the roblem to this case: s x refl refl y s s refl s refl s z In this case, the desired equality is (refl refl ) (refl s refl s ) = (refl refl s ) (refl refl s ). Since we have refl refl refl, and similar for refl s and (refl refl s ), this simlifies to refl refl s = refl refl s, which is true again by reflexivity. There are many more oerations on higher-dimensional aths which can be constructed by ath induction. We will sto our short discussion of this asect by noting that all these higher oerations can be regarded as one ossible definition of an algebraic structure called an -grouoid. Deendent tyes as fibrations We already learnt that we can think of a tye A as a sace with elements x A as oints and equalities x = y as aths. We also just realized that a function f A B can be interreted as a continuous ma from the sace A to the sace B. Now suose that we aly this in the case B U. So, the universe U is itself a gigantic sace whose elements are themselves other saces. (Such a sace of saces is often also called a moduli sace.) So, a deendent tye over A or tye family over A is a function D A U; intuitively, this means that D assigns to every x A another sace D(x) U in a way which deends continuously on x. Conventionally, such a deendent sace is known as a fibration; so, a tye family is the HoTT analogue of the notion of fibration. As an examle, if we start with the base sace A being a circle, and attach to each x A a coy of the interval, such that as we go once around the circle, the interval comes back with a twist, then we obtain the Möbius stri of Figure 1. What is deicted is not the function D A U, but rather the total sace x A D(x), which is the union of 8

9 Figure 1: A Möbius stri all the individual D(x). One can identify the circle S 1 with the central line that winds around the stri; for every x, the sace D(x) then consists of the interval erendicular to that circle. Higher inductive tyes will let us ermit the definition of a tye modelling the circle and a tye modelling the interval, and then the Möbius stri will be an actual tye family that we can construct. Until then, it is nothing but an illustration. So, the deendent sum x A D(x) corresonds to the total sace of the fibration D A U; but what about the deendent roduct x A D(x)? Intuitively, an element s x A D(x) corresonds to an assignment x s(x) with s(x) D(x) such that the deendence on x is continuous. Under the corresondence to fibrations, such an s is conventionally known as a section. The Möbius stri is very secial in that it ermits a section which embeds the circle as the base into the total sace x A D(x); in general, this is not the case. With this intuition in mind, we can now move on to discussing the behavior of aths with resect to such fibrations. Theorem 11. Given a tye family D A U over A and a ath x = A y, there exists a function transort D () D(x) D(y). Proof. Path induction on, using the definition transort D (refl x ) id D(x). Whenever the tye family D is clear, then we also use the shorthand notation D(x) D(y) in lace of transort D (). If D A U is interreted as a roosition deending on x A, then we can say that x A has roerty D if D(x) is inhabited. Therefore, the theorem tells us that if x is equal to y, and x has roerty D, then so does y. More recisely, any witness or roof of the statement x has roerty D can be transorted into a witness or roof of the statement y has roerty D. In the toological theory of fibrations, the transort function can be interreted as a connection, which is a fundamental structure in differential geometry. In this sense, a tye family is more than just a fibration: it is a fibration equied with a connection. Imortant roerties of the transort function are the following, which are all easily roven by the usual ath induction: 9

10 Theorem 12. If q y = z is another ath, then for all t D(x). ( q) (t) = q ( (t)) So the transort ma has the exected behaviour uon concatenation of aths; note the similarity to the revious theorem a f (q ) = a f (q) a f (). Using transort, we can not only transort elements around, we can also transort aths themselves, a transort D (s = t) ( (s) = (t)). (3) s,t D(x) The behaviour of transort with resect to changing the base tye A is like this: Theorem 13. If f B A is a function, then D f B U is again a tye family, and for all x = B y with x, y B, and t D(f(x)). transort D f (, t) = transort D (a f (), t). Finally, we also note the behaviour of transort with resect to changing the tye family, while keeing the base A fixed: Theorem 14. Given any two tye families D, E A U and a family of deendent functions f x A D(x) E(x) and any x = A y and t D(x), we have transort E (, f(x, t)) = f(y, transort D (, t)). We can exress this as saying that transort commutes with the alication of a deendent function. In terms of category theory, f is a transformation between functors, and the theorem shows that this transformation is automatically natural. This is similar to how Lemma 3 shows that every function is automatically continuous. If we aly transort in the case of a non-deendent tye, it acts trivially: Theorem 15. If D λx.b A U for some fixed B U, then for any x, y A and x = A y there is a family of aths transortconst B () transort λx.b (, t) = t t B We are now in a situation to generalize the function a (x = y) (f(x) = f(y)) to the case in which f is a deendent function, by simly transorting f(x) from D(x) to D(y) before comaring it to f(y): Theorem 16. Given any deendent function f x A D(x), there exists a function ad f (f(x)) = D(y) f(y). x=y Again we consider the elements x, y A to be imlicit arguments of ad f for which we omit mention. The roof of the theorem is again straightforward by ath induction on. While the revious a was a function (x = y) (f(x) = f(y)), the new ad f for deendent f is itself a deendent function. Also ad f has the exected behaviour with resect to concatenation of aths, and more generally there are more comlicated analogues of Theorem 9, but we omit these. When f is a non-deendent function, we would exect ad f to reduce to a f, assuming that we identify f A B with the corresonding element of x A B. This is indeed the case: 10

11 Theorem 17. For all f A B and x = A y, we have ad f () = transortconst B (, f(x)) a f (). The is well-tyed since the ath on the left goes from transort λx.b (, f(x)) to f(y), while the one on the right is a concatenation of a ath from transort λx.b (, f(x)) to f(x) and a ath from f(x) to f(y). As usual, the roof is by ath induction on, in which case both sides reduce to refl f(x). Path lifting The revious considerations involving transort were about turning a ath x = y on the base A into a function transort D () D(x) D(y). But can we also lift a ath in the base to a ath in D? In order to make sense of this, we need to consider aths in the total sace x A D(x), and then we have: Theorem 18 (Path lifting roerty). For any x A and t D(x), every ath x = A y gives a ath and this ath lies over in the sense that lift(t, ) (x, t) = x A D(x) (y, (t)), a r1 (lift(t, )) =. Proof. Using ath induction on reduces the roof of both statements to the case y x and refl x, for which they are trivial. Omniotent ath induction? The revious results may suggest that ath induction is a roof method with which one can rove almost anything, since it always reduces the roof to a trivial case. However, this is by no means the case: for examle, the statement = refl x (4) x A x=x is not derivable by ath induction. In fact, ath induction does not aly here since it would require at least one endoint of to be free based ath induction alies to statements of the form x A x=y C(x, y, ) only! We will see soon that together with the univalence axiom, one can show (4) to be false. Actually, (4) is closely related to Axiom K, which is a commonly imosed axiom resulting in so-called extensional tye theory, in which two things can be equal in at most one way. Due to the incomatibility with the univalence axiom, and the realization that univalence is a more owerful rincile resulting in a foundation of mathematics close to mathematical ractice, we do not imose Axiom K. 11

12 Examle: discriminating the elements of Bool In order to illustrate the use of the results derived here, we show that the two canonical elements of Bool are not equal: Theorem 19. (0 = 1) 0. Proof. Define the tye family D Bool U by induction over Bool using the values D(0) 0, D(1) 1 on the base cases. We know that D(1) is inhabited by D(1). Hence for any utative equation 0 = 1, we have ( 1 ) ( ) D(0), which is an element of 0 since D(0) 0. 12