TheHoTTGame/1FundamentalGroup/Quest2Part0.md

Comparison maps between Ω S¹ base and ℤ

In Quest1 we have defined the map loop_times : ℤ → Ω S¹ base. Creating the inverse map is difficult without access to the entire circle. Similarly to how we used doubleCover to distinguish refl and base, the idea is to replace Bool with ℤ, allowing us to distinguish between all loops on S¹. In Part0 and Part1 we will construct one of the two comparison maps across the whole circle, called spinCount.

The plan is :

1. Define a function sucℤ : ℤ → ℤ that increases every integer by one
2. Prove that sucℤ is an isomorphism by constructing an inverse map predℤ : ℤ → ℤ.
3. Turn sucℤ into a path sucPath : ℤ ≡ ℤ using isoToPath
4. Define helix : S¹ → Type by mapping base to ℤ and a generic point loop i to sucPath i. <<<<<<< HEAD
5. Use helix and endPt to define the map base ≡ x → helix x on all x : S¹, in particular giving us Ω S¹ base → ℤ when applied to base.

In this part, we focus on 1 and 2.

5. Use helix and endPt to define the map spinCountBase : base ≡ base → ℤ Intuitively it counts how many times a path loops around S¹. a generic point loop i to sucPath i.
6. Generalize this across the circle.

In this part, we focus on 1, 2 and 3.

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sucℤ

• Setup the definition of sucℤ so that it looks of the form :

  Name : TypeOfSpace
  Name inputs = ?
  

  Compare it with our solutions in 1FundamentalGroup/Quest1.agda

• We will define sucℤ the same way we defined loop_times : by induction. Do cases on the input of sucℤ. You should have something like :

  sucℤ : ℤ → ℤ
  sucℤ pos n = ?
  sucℤ negsuc n = ?
  

• For the non-negative integers pos n we want to map to its successor. Recall that the n here is a point of the naturals ℕ whose definition is :

  data ℕ : Type where
    zero : ℕ
    suc : ℕ → ℕ
  

  Use suc to map pos n to its successor.

• The negative integers require a bit more care. Recall that annoyingly negsuc n means "- (n + 1)". We want to map - (n + 1) to - n. Try doing this. Then realise "you run out of negative integers at -(0 + 1)" so you must do cases on n and treat the -(0 + 1) case separately.

  Hint

  Do C-c C-c on n. Then map negsuc 0 to pos 0. For negsuc (suc n), map it to negsuc n.

• This completes the definition of sucℤ. Use C-c C-n to check it computes correctly. E.g. check that sucℤ (- 1) computes to pos 0 and sucℤ (pos 0) computes to pos 1.

sucℤ is an isomorphism

• The goal is to define predℤ : ℤ → ℤ which "takes n to its predecessor n - 1". This will act as the (homotopical) inverse of sucℤ. Now that you have experience from defining sucℤ, try defining predℤ.
• Imitating what we did with flipIso and give a point sucℤIso : ℤ ≅ ℤ by using predℤ as the inverse and proving <<<<<<< HEAD section sucℤ predℤ and retract sucℤ predℤ. ======= section sucℤ predℤ and retract sucℤ predℤ.

sucℤ is a path

• Imitating what we did with flipPath, upgrade sucℤIso to sucℤPath.

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