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The unit-level set is precisely the unit-type reduction

Proved
QAlgorithms.Papers.Independent.Intakethird100.P93.lemma_2_10

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let t:G→Ht:G\to Ht:G→H be a topological crossed module, with continuous action α:H→Aut⁡(G)\alpha:H\to\operatorname{Aut}(G)α:H→Aut(G). Thus t(α(h)g)=ht(g)h−1t(\alpha(h)g)=ht(g)h^{-1}t(α(h)g)=ht(g)h−1 and α(t(g))k=gkg−1\alpha(t(g))k=gkg^{-1}α(t(g))k=gkg−1. Put G1=ker⁡tG_1=\ker tG1​=kert and give both G/G1G/G_1G/G1​ and H/t(G)H/t(G)H/t(G) their quotient topologies. Assume their quotient projections admit continuous local sections. All actions, structure maps and isomorphisms below are continuous, all principal bundles are locally trivial, and a bispace has a nonempty principal homogeneous underlying right space.

For an (H,G)(H,G)(H,G)-bispace (X,ψ)(X,\psi)(X,ψ) put X1={x∈X:ψ(x)=1}X_1=\{x\in X:\psi(x)=1\}X1​={x∈X:ψ(x)=1}. Then

X1≠∅  ⟺  Type⁡(X)=1.X_1\ne\varnothing\iff\operatorname{Type}(X)=1.X1​=∅⟺Type(X)=1.

If X1X_1X1​ is nonempty, it is a reduction to (1,G1)(1,G_1)(1,G1​): it is a single G1G_1G1​-orbit and its structure map has value 111. The assertion retains the possibility that X1X_1X1​ is empty and supplies the reduction only when that condition is met.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P93
Formal statement
namespace QAlgorithms.Papers.Independent.Intakethird100.P93

universe u

variable {G H : Type u} [Group G] [TopologicalSpace G] [IsTopologicalGroup G]
  [Group H] [TopologicalSpace H] [IsTopologicalGroup H]

variable (C : CrossedModule G H)

/-- Lemma 2.10, PDF p.7, both clauses. No nonemptiness is imposed at nonunit type. -/
theorem lemma_2_10 {X : Type u} [TopologicalSpace X] (B : Bispace C X) :
    ((unitLevel C B).Nonempty ↔ bispaceType C B = unitType C) ∧
    ((unitLevel C B).Nonempty →
      IsReduction C B C.t.ker (⊥ : Subgroup H) (unitLevel C B)) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P93
Source
On the existence of bibundles, arXiv:1102.4388v2, PDF p. 7, Lemma 2.10
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let G,HG,HG,H be topological groups, and let CCC consist of a continuous homomorphism t:G→Ht:G\to Ht:G→H and a homomorphism α:H→Aut⁡(G)\alpha:H\to\operatorname{Aut}(G)α:H→Aut(G) with continuous evaluation, satisfying t(αh(g))=h t(g) h−1t(\alpha_h(g))=h\,t(g)\,h^{-1}t(αh​(g))=ht(g)h−1 and αt(g)(k)=gkg−1\alpha_{t(g)}(k)=gkg^{-1}αt(g)​(k)=gkg−1 for every h∈Hh\in Hh∈H and g,k∈Gg,k\in Gg,k∈G. Assume both quotient maps G→G/ker⁡tG\to G/\ker tG→G/kert and H→H/t(G)H\to H/t(G)H→H/t(G) admit continuous right inverses on open neighborhoods of every target point. Let XXX be a topological space with a continuous right action x⋅gx\cdot gx⋅g satisfying x⋅1=xx\cdot1=xx⋅1=x and (x⋅g)⋅k=x⋅(gk)(x\cdot g)\cdot k=x\cdot(gk)(x⋅g)⋅k=x⋅(gk), such that XXX is nonempty and every orbit map g↦x⋅gg\mapsto x\cdot gg↦x⋅g is a homeomorphism G→XG\to XG→X. Let ψ:X→H\psi:X\to Hψ:X→H be continuous with ψ(x⋅g)=ψ(x)t(g)\psi(x\cdot g)=\psi(x)t(g)ψ(x⋅g)=ψ(x)t(g). Write q:H→H/t(G)q:H\to H/t(G)q:H→H/t(G) for the quotient map, let x0x_0x0​ be the point selected from the nonemptiness of XXX by classical choice in the definition of the bispace type, and put U={x∈X:ψ(x)=1}U=\{x\in X:\psi(x)=1\}U={x∈X:ψ(x)=1}. Then

U≠∅⟺q(ψ(x0))=q(1).U\ne\varnothing\quad\Longleftrightarrow\quad q(\psi(x_0))=q(1).U=∅⟺q(ψ(x0​))=q(1).

Moreover, if U≠∅U\ne\varnothingU=∅, there is an a∈Xa\in Xa∈X such that U={a⋅k:k∈ker⁡t}U=\{a\cdot k:k\in\ker t\}U={a⋅k:k∈kert}, and every x∈Ux\in Ux∈U satisfies ψ(x)∈{1}\psi(x)\in\{1\}ψ(x)∈{1}.

Human review
  • Endorsed by Shuze Chen · Nov 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Nov 1, 2026

    Confirmed by the mission captain (proposal self-audit).

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