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LeanPool.WhiteheadTheorem.HEP.Retract

LeanPool.WhiteheadTheorem.HEP.Retract #

Imported Lean Pool material for LeanPool.WhiteheadTheorem.HEP.Retract.

structure StrongDeformRetr (X : Type u_1) [TopologicalSpace X] (A : Set X) :
Type u_1

A strong deformation retraction from X to its subset A

Instances For
    theorem Cube.inclToBot_discardLast_eq_of {n : } (y : Fin (n + 1)unitInterval) (hz : y (Fin.last n) = 0) :

    (y₀, y₁, …, yₙ₋₁, yₙ) ↦ (y₀, y₁, …, yₙ₋₁) ↦ (y₀, y₁, …, yₙ₋₁, 0) does nothing if yₙ = 0.

    noncomputable def Cube.projToBoundary {n : } :
    (Fin (n + 1)unitInterval)(boundary (Fin (n + 1)))

    Project the $(n+1)$-dimensional cube to its boundary ∂I^(n+1). This fixes every point on the boundary, and maps other points to the default value 0. This (non-continuous) function is convenient for diagram-chasing when applying the homotopy extension property in retrToBoundaryJar. This function is noncomputable due to Real.decidableEq.

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      theorem Cube.projToBoundary_eq_of_mem_boundary {n : } (y : Fin (n + 1)unitInterval) (hy : y boundary (Fin (n + 1))) :
      (projToBoundary y) = y

      projToBoundary y does nothing if y is already in the boundary.

      noncomputable def Cube.projToSides {n : } :
      (Fin (n + 1 + 1)unitInterval)(boundary (Fin (n + 1))) × unitInterval

      Project the $(n+2)$-dimensional cube to its sides (∂I^(n+1)) × I (the closure of the complement of the top and bottom faces of the boundary). Since ∂I^0 = ∅, the cube shoule be at least 2-dimensional.

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        theorem Cube.inclToSides_projToSides_eq_of {n : } (y : Fin (n + 1 + 1)unitInterval) (hs : (splitAtLast y).2 boundary (Fin (n + 1))) :
        noncomputable def Cube.retrToBoundaryJar (n : ) :
        { r' : C(Fin (n + 1)unitInterval, (boundaryJar (n + 1))) // yboundaryJar (n + 1), (r' y) = y }

        A retraction from I^n to ⊔I^n, for n ≥ 1.

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          noncomputable def Cube.strongDeformRetrToBoundaryJar (n : ) :

          A strong deformation retraction from I^n to ⊔I^n, for n ≥ 1. I'm writing down the formula for each coordinate because:

          • failed to synthesize ContinuousSMul (↑I) (Fin (n + 1) → ↑I)
          • Using Convex requires going to a subset of ℝ^(n+1) and back (although Convex ℝ I is very helpful)
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          • One or more equations did not get rendered due to their size.
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            def Cube.deformRetrToBot {n : } :
            C(unitInterval × (Fin (n + 1)unitInterval), Fin (n + 1)unitInterval)

            Deformation retraction from the cube to its bottom face

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            • One or more equations did not get rendered due to their size.
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              theorem Cube.deformRetrToBot_apply_bot {n : } (t : unitInterval) {y : Fin (n + 1)unitInterval} (hy : y (Fin.last n) = 0) :

              If y is in the bottom face of the cube, then deformRetrToBot ⟨t, y⟩ is in the bottom face for all t.

              theorem Cube.deformRetrToBot_apply_sides {n : } (t : unitInterval) {y : Fin (n + 1)unitInterval} (hy : (splitAtLast y).2 boundary (Fin n)) :

              If y is in the sides (the closure of the complement of the top and bottom faces of the boundary) of the cube, then deformRetrToBot ⟨t, y⟩ is in the sides for all t.

              Deformation retraction from ⊔I^(n + 1) to its bottom face

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              • One or more equations did not get rendered due to their size.
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                hequivBoundaryJar

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                  Deformation retraction from the cube to the point 0.

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                  • One or more equations did not get rendered due to their size.
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