/-- The Ping-Pong-Lemma.

Given a group action of `G` on `X` so that the generators of the free groups act in specific
ways on disjoint subsets `X i` and `Y i` we can prove that `lift f` is injective, and thus the image
of `lift f` is isomorphic to the free group.

Often the Ping-Pong-Lemma is stated with regard to group elements that generate the whole group;
we generalize to arbitrary group homomorphisms from the free group to `G` and do not require the
group to be generated by the elements.
-/
theorem _root_.FreeGroup.injective_lift_of_ping_pong : Function.Injective (FreeGroup.lift a) := by
  -- Step one: express the free group lift via the free product lift
  have : FreeGroup.lift a =
      (CoprodI.lift fun i => FreeGroup.lift fun _ => a i).comp
        (@freeGroupEquivCoprodI ι).toMonoidHom := by
    ext i
    simp
  rw [this, MonoidHom.coe_comp]
  clear this
  refine' Function.Injective.comp _ (MulEquiv.injective freeGroupEquivCoprodI)
  -- Step two: Invoke the ping-pong lemma for free products
  show Function.Injective (lift fun i : ι => FreeGroup.lift fun _ => a i)
  -- Prepare to instantiate lift_injective_of_ping_pong
  let H : ι → Type _ := fun _i => FreeGroup Unit
  let f : ∀ i, H i →* G := fun i => FreeGroup.lift fun _ => a i
  let X' : ι → Set α := fun i => X i ∪ Y i
  apply lift_injective_of_ping_pong f _ X'
  show ∀ i, (X' i).Nonempty
  · exact fun i => Set.Nonempty.inl (hXnonempty i)
  show Pairwise fun i j => Disjoint (X' i) (X' j)
  · intro i j hij
    simp only
    apply Disjoint.union_left <;> apply Disjoint.union_right
    · exact hXdisj hij
    · exact hXYdisj i j
    · exact (hXYdisj j i).symm
    · exact hYdisj hij
  show Pairwise fun i j => ∀ h : H i, h ≠ 1 → f i h • X' j ⊆ X' i
  · rintro i j hij
    -- use free_group unit ≃ ℤ
    refine' FreeGroup.freeGroupUnitEquivInt.forall_congr_left'.mpr _
    intro n hne1
    change FreeGroup.lift (fun _ => a i) (FreeGroup.of () ^ n) • X' j ⊆ X' i
    simp only [map_zpow, FreeGroup.lift.of]
    change a i ^ n • X' j ⊆ X' i
    have hnne0 : n ≠ 0 := by
      rintro rfl
      apply hne1
      simp; rfl
    clear hne1
    simp only
    -- Positive and negative powers separately
    cases' (lt_or_gt_of_ne hnne0).symm with hlt hgt
    · have h1n : 1 ≤ n := hlt
      calc
        a i ^ n • X' j ⊆ a i ^ n • (Y i)ᶜ :=
          smul_set_mono ((hXYdisj j i).union_left <| hYdisj hij.symm).subset_compl_right
        _ ⊆ X i := by
          clear hnne0 hlt
          refine Int.le_induction (P := fun n => a i ^ n • (Y i)ᶜ ⊆ X i) ?_ ?_ n h1n
          · dsimp
            rw [zpow_one]
            exact hX i
          · dsimp
            intro n _hle hi
            calc
              a i ^ (n + 1) • (Y i)ᶜ = (a i ^ n * a i) • (Y i)ᶜ := by rw [zpow_add, zpow_one]
              _ = a i ^ n • a i • (Y i)ᶜ := (MulAction.mul_smul _ _ _)
              _ ⊆ a i ^ n • X i := (smul_set_mono <| hX i)
              _ ⊆ a i ^ n • (Y i)ᶜ := (smul_set_mono (hXYdisj i i).subset_compl_right)
              _ ⊆ X i := hi
        _ ⊆ X' i := Set.subset_union_left _ _
    · have h1n : n ≤ -1 := by
        apply Int.le_of_lt_add_one
        simpa using hgt
      calc
        a i ^ n • X' j ⊆ a i ^ n • (X i)ᶜ :=
          smul_set_mono ((hXdisj hij.symm).union_left (hXYdisj i j).symm).subset_compl_right
        _ ⊆ Y i := by
          refine' Int.le_induction_down (P := fun n => a i ^ n • (X i)ᶜ ⊆ Y i) _ _ _ h1n
          · dsimp
            rw [zpow_neg, zpow_one]
            

Proof state:
case refine'_1
ι : Type u_1
M : ι → Type u_2
inst✝⁴ : (i : ι) → Monoid (M i)
N : Type u_3
inst✝³ : Monoid N
inst✝² : Nontrivial ι
G : Type u_1
inst✝¹ : Group G
a : ι → G
α : Type u_4
inst✝ : MulAction G α
X Y : ι → Set α
hXnonempty : ∀ (i : ι), Set.Nonempty (X i)
hXdisj : Pairwise fun i j => Disjoint (X i) (X j)
hYdisj : Pairwise fun i j => Disjoint (Y i) (Y j)
hXYdisj : ∀ (i j : ι), Disjoint (X i) (Y j)
hX : ∀ (i : ι), a i • (Y i)ᶜ ⊆ X i
hY : ∀ (i : ι), a⁻¹ i • (X i)ᶜ ⊆ Y i
H : ι → Type := fun _i => FreeGroup Unit
f : (i : ι) → H i →* G := fun i => FreeGroup.lift fun x => a i
X' : ι → Set α := fun i => X i ∪ Y i
i j : ι
hij : i ≠ j
n : ℤ
hnne0 : n ≠ 0
hgt : n < 0
h1n : n ≤ -1
⊢ (a i)⁻¹ • (X i)ᶜ ⊆ Y i

Next tactic:
exact hY i