INTRODUCTION TO HOMOTOPY TYPE THEORY

EUTypes School, Ohrid 2019 

NOTES AND EXERCISES FOR LECTURE ONE
31 August 2019

Chapter 1: What is (a) dependent type theory?

 * (the basis for) a programming language
 * (the basis for) a proof assistant
 * a foundation of maths
 * a language for higher categories (more precisely: infinity-groupoids)

Chapter 2: Basic components of dependent type theory

(see also 1.4-1.7 of the HoTT book)

 * products, A*B
 * coproducts, A+B
 * Sigma types, Sigma(x:A).B(x)
 * Pi types, Pi(x:A).B(x)

Chapter 3: Equality types

(see also 1.12 of the HoTT book)

Notation:
  Pi(x y z:A). ...
means the same as 
  Pi(x:A). Pi(y:A). Pi(z:A). ...

The rules for the equality type are as follows:

formation: If x:A and y:A, then x=y is a type

introduction: If x:A, then refl : x=x

elimination: given
  P : Pi(x y:A). (x=y) -> Type
  d : Pi(x:A) P x x refl
then we get
  J(P,d) : Pi(x y:A). Pi(q:x=y). P x y q

computation: given P,d as above,
  J(P,d) x x refl
computes to
  d x

EXERCISES:

(1) Using the rules above, construct a term
  trans : Pi(x y z:A). x=y -> y=z -> x=z

(2) Show associativity, i.e. construct a term
  assoc : Pi(x y z w:A). Pi(p:x=y). Pi(q:y=z). Pi(s:z=w). trans x y w p (trans y z w q s) = trans x z w (trans x y z p q) s

(3) Construct a term
  sym : Pi(x y:A). x=y -> y=x

(4) Show:
  selfinv : Pi(x y:A). Pi(p:x=y). sym y x (sym x y p) = p

SOLUTIONS:

(1) choose
  P x y p := Pi(z:A). y=z -> x=z
  d x := \z.\q.q     --- \ is lambda
then define
  trans x y z p q := J(P,d) x y p z q

(2) We use the names P2, d2 instead of P, d to avoid confusion with the P, d of (1).
We choose
  P2 x y p := Pi(z w:A). Pi(q:y=z). Pi(s:z=w). trans x y w p (trans x z w q s) = trans x z w (trans x y z p q) s
Let us check what the type of 
  d2 : Pi(x:A). P x x refl
is, by unfolding P2. We get:
  d2 : Pi(x:A). Pi(z w:A). Pi(q:x=z). Pi(s:z=w). trans x x w refl (trans x z w q s) = trans x z w (trans x x z refl q) s
-- (note how we have simply replaced (y,p) by (x,refl)!)
Since the term trans is defined using J, the expression 'trans x x ... refl ...' computes to 'd x ...', which gives:
  d2 : Pi(x:A). Pi(z w:A). Pi(q:x=z). Pi(s:z=w). trans x z w q s = trans x z w q s
This is easy:
  d2 x := \z. \w. \q. \s. refl
Thus, we can define:
  assoc x y z w p q s := J(P2,d2) x y p z w q s

(3) choose
  P3 x y p := y=x
  d3 x := refl

(4) choose
  P4 x y p := sym y x (sym x y p) = p
  d4 x := refl