Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Smooth representatives of configuration loops

Statement

Let Qn=(q1,…,qn) be the fixed base configuration of Boundary-fixed mapping class group of a punctured disk. Every based loop α:I→Cn(int⁡D2) at the basepoint [Qn] is path homotopic relative to {0,1} to a based loop β whose unique ordered lift z:I→Fn(int⁡D2) from Qn consists of coordinate paths z1,…,zn that are smooth, pairwise collision-free (zi(t)≠zj(t) for i≠j), take values in int⁡D2, and are constant on [0,ε) and on (1−ε,1] for some ε>0. The construction uses no choice principle.

Facts & Assumptions

Given: The based loop α:I→Cn(int⁡D2) with α(0)=α(1)=[Qn].

[L1]

For every f∈C([0,1],R) and ε>0 there is a polynomial p with sup⁡x∈[0,1]∣p(x)−f(x)∣<ε (Polynomials are uniformly dense in C([0,1],R)).

[L2]

The standard smooth step function σ(t)=β(t)/(β(t)+β(1−t)) is smooth, equals 0 for t≤0 and equals 1 for t≥1 (The standard smooth step function).

[L3]

The quotient p:Fn(int⁡D2)→Cn(int⁡D2) is a covering map with n!-element fibres, and both spaces are path-connected (Ordered configuration spaces cover the unordered ones regularly with deck group Sn).

[L4]

A covering map has a unique path lift through any prescribed starting point: if α~(0)=e0 and p∘α~=α, then α~ is unique (Existence and uniqueness of path lifts through a covering map).

[L5]

Fn(X) consists of the tuples with pairwise distinct coordinates and Cn(X)=Fn(X)/Sn with quotient map p(x)=[x] (Ordered configuration spaces Fn(X), Unordered configuration spaces Cn(X)).

Proof

technique · direct

If n=0, the unique based loop represents itself and has the unique empty ordered lift; the claim is immediate. Assume n≥1 below.

1.1L3L4L5

The ordered lift and its margin. Let p:Fn(int⁡D2)→Cn(int⁡D2) be the quotient covering map of [L3]. By [L4] there is a unique path u:I→Fn(int⁡D2) with u(0)=Qn and p∘u=α; its coordinates u1,…,un are continuous and satisfy ui(t)≠uj(t) for i≠j and ui(t)∈int⁡D2. Compactness and finiteness give a positive boundary margin d:=min⁡i,t(1−∥ui(t)∥2)>0; if n≥2, also put c:=min⁡i<j,t∥ui(t)−uj(t)∥2>0, and if n=1 put c:=1. Then m:=min⁡(c,d)>0 bounds every pairwise separation and every boundary margin from below (with the pairwise condition vacuous for n=1).

1.2L1L2

Smooth approximation with fixed endpoints and flat time ends. Write ui=(ui,1,ui,2). For each of the finitely many functions ui,k apply [L1] with ε1>0 to obtain a polynomial Pi,k with sup⁡t∣Pi,k(t)−ui,k(t)∣<ε1, and put Ri,k(t):=Pi,k(t)+(1−t)(ui,k(0)−Pi,k(0))+t(ui,k(1)−Pi,k(1)), so that Ri,k(0)=ui,k(0), Ri,k(1)=ui,k(1) and sup⁡t∣Ri,k(t)−ui,k(t)∣<2ε1. Now choose a small ε∈(0,14) and use [L2] to define the smooth time change λ(t):=σ((t−ε)/ε)σ((1−ε−t)/ε)t+(1−σ((1−ε−t)/ε)); it is smooth, equals 0 on [0,ε], equals 1 on [1−ε,1], satisfies 0≤λ(t)≤1 and ∣λ(t)−t∣≤4ε on I, and equals t on [2ε,1−2ε]. Set ri(t):=(Ri,1(t),Ri,2(t)) and zi:=ri∘λ. Then each zi is smooth, is constant on [0,ε] and on [1−ε,1], and because the finitely many ui are uniformly continuous there is a modulus of continuity ω for all of them on I with ∥ui(λ(t))−ui(t)∥2≤ω(4ε). Choosing ε1 and ε so small that 3ε1+ω(4ε)<m/4, we obtain ∥zi(t)−ui(t)∥2≤∥Ri(λ(t))−ui(λ(t))∥2+∥ui(λ(t))−ui(t)∥2<m/4 for every i and t.

2.1L5step 1.1step 1.2

The approximating tuple is collision-free, interior, and based. For i≠j and all t∈[0,1] the estimates of step 1.2 give ∥zi(t)−zj(t)∥2≥m−2(m/4)=m/2>0 and 1−∥zi(t)∥2≥m−m/4>0, so all zi(t) lie in int⁡D2 and are pairwise distinct. Moreover zi(0)=ri(λ(0))=ri(0)=ui(0)=qi and zi(1)=ri(1)=ui(1), so [z(1)]=[u(1)]=α(1)=[Qn]: the terminal tuple z(1) is a permutation of Qn, and z:I→Fn(int⁡D2) is an ordered path from Qn to that permutation, while p∘z is a based loop at [Qn].

3.1L3L5step 1.1step 1.2step 2.1

A relative homotopy to the smooth representative. For s,t∈[0,1] put γ(s,t):=(1−s)u(t)+s z(t), computed coordinatewise in R2n. The map γ is continuous, and by the estimates of steps 1.2 and 2.1 every γ(s,t) again has pairwise distinct coordinates at distance at least m/2 and lies in int⁡D2: the interpolation moves each point by at most m/4 from ui(t). Hence γ lands in Fn(int⁡D2), and its composition with the quotient map p of [L3] is a continuous map H:I×I→Cn(int⁡D2), H(s,t):=[γ(s,t)], with H(0,t)=α(t) and H(1,t)=[z(t)]; the identities γ(s,0)=u(0)=Qn and γ(s,1)=u(1), together with [u(1)]=α(1)=[Qn], show that H(s,0)=H(s,1)=[Qn] for every s, so that H is a path homotopy relative to {0,1}.

4.1L3L4step 1.2step 2.1step 3.1∎

The lift of the representative is z itself. The path [z]:=p∘z is a based loop at [Qn] by step 2.1, and z is a lift of it with z(0)=Qn; by the uniqueness clause [L4] the unique ordered lift of [z] from Qn is exactly z. Together with steps 1.2, 2.1 and 3.1 this exhibits the required smooth collision-free lift with flat time ends and the path homotopy α≃[z] rel {0,1}; all constructions used only the given loop, fixed polynomials and the fixed step function, so no choice principle is spent.

Remarks

  • The time change λ is built from the published smooth step so that the approximating path is stationary near both ends of the interval; this is what later allows the motion to be extended across the endpoints by constancy.
  • The estimate is uniform in t and uses only finitely many continuous functions on the compact interval, so no selection from infinitely many approximations is made.

Depends on

Used by

Dependency tree · two levels

49 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources