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Pure Braids, Fadell–Neuwirth, and Asphericity — Examples

1 · Prerequisites

2 · Summary

These four entries make the companion page concrete in the two lowest ranks, identify the standard generators geometrically, and delimit what the configuration braid extension can prove on its own. Each entry works at the canonical base configuration of the geometric-braids pages, with $h_{n} = \frac{1}{4 (n + 1)}$ , and inherits the Axiom-of-Choice bookkeeping of the statement it consumes.

The first example computes the rank-two group. In the short exact sequence $1 \to F_{1} \to P B_{2} \to P B_{1} \to 1$ the quotient $P B_{1}$ is trivial, so the fibre inclusion is an isomorphism: $P B_{2}$ is free on one meridian class, hence infinite cyclic, generated by $A_{12}$ as well as by its inverse. In the inverse-slicing identification $Ψ$ the positive generator $A_{12}$ is the clockwise meridian of the first puncture, of winding $- 1$ relative to the counterclockwise spine basis; the example also notes that this agrees with the general torsion-freeness theorem in rank two.

The second example computes rank three and the splitting in action. The kernel of the forgetting homomorphism $φ : P B_{3} \to P B_{2}$ is free on $A_{13}, A_{23}$ , while $P B_{2}$ is infinite cyclic. Its generator is the canonical two-strand $A_{12}$ transported to the truncated base configuration $Q^{'} = (- 1 / 8, 0)$ , where it is the image of $A_{12} \in P B_{3}$ under forgetting. With the far-right continuous section of the companion page, adjusted at the basepoint along a path in the puncture fibre so that it fixes the third point on a small representative of $A_{12}$ , the split extension reads $P B_{3} ≅ ⟨ A_{13}, A_{23} ⟩ ⋊ ⟨ A_{12} ⟩ ≅ F_{2} ⋊ Z$ . For the section chosen, and writing $w : = A_{13} A_{23}$ , the action of the positive generator on the free kernel is conjugation $x \mapsto w^{- 1} x w$ , computed directly; no direct-product decomposition and no centre statement is claimed.

The third example matches the standard generators with point pushing. For $1 \leq i < j \leq n$ the ordered slice of the word $W_{i j}$ is braid-isotopic to the motion in which only the $j$ -th marked point moves, once counterclockwise around the $i$ -th puncture along a standard meridian stem and back, the other strands returning to their base points; the base case is the explicit two-strand computation of the sibling pair, transported into the support disc of the full twist, and the inductive step slides the outer half twists off the excursion across the collar. Inverting the slicing, the class $A_{i j}$ in $P B_{n}$ is the clockwise meridian class of the fibre, and the braid–mapping-class isomorphism $Θ_{n} = Ψ_{n}^{m c} \circ (Ψ_{n}^{c o n f})^{- 1}$ of the companion page carries it to the point push of the $j$ -th point clockwise around the $i$ -th puncture; after relabelling the $j$ -th point as the terminal coordinate, this is exactly the instance of ${Push}_{n}$ computed in the sibling pair. Both inversions — the one in the configuration identification and the inverse-endpoint convention of the point-pushing definition — are displayed, and no injectivity of ${Push}_{n}$ is used.

The counterexample isolates the invalid inference that the configuration braid sequence $1 \to P B_{n} \to B_{n}^{c o n f} \to S_{n} \to 1$ would prove $B_{n}^{c o n f}$ torsion-free merely because its kernel is torsion-free and its quotient is finite. The abstract witness is $1 \to Z \to Z \times Z / 2 \to Z / 2 \to 1$ : the kernel is torsion-free, the quotient has order two, and the middle group contains the element $(0, [1]_{2})$ of order two, since $[1]_{2} \neq [0]_{2}$ while $2 [1]_{2} = [0]_{2}$ . The example makes no claim about torsion in the genuine braid group; it shows only that torsion-freeness needs the finer Fadell–Neuwirth argument of the companion page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

The two-strand pure braid group is infinite cyclic

Example

Assume the Axiom of Choice and let $Q^{(2)} = (q_{1}, q_{2})$ be the canonical base configuration of Geometric braids in the disc with setwise endpoints, so that $P B_{2} = π_{1} (F_{2} (D^{2}), Q^{(2)})$ in the convention of The pure braid group $P B_{n}$ as the fundamental group of an ordered configuration space. Then $P B_{2} ≅ Z,$ generated by the geometrically positive two-strand full twist $A_{12} = σ_{1}^{2}$ of Standard geometric pure braid generators A_ij. Under the inverse-slicing identification of The $A_{i n}$ are meridian generators of the forgetful free kernel, the generator $A_{12}$ corresponds to the clockwise meridian of $q_{1}$ inside the once-punctured disc fibre, that is to the inverse of the counterclockwise meridian class; relative to the counterclockwise spine basis its winding is $- 1$ . The group $P B_{2}$ is infinite cyclic and, in particular, torsion-free.

Facts & Assumptions

Given: the Axiom of Choice, the canonical base configuration $Q^{(2)} = (q_{1}, q_{2})$ of interior points of the disc, and the truncation notation of The Fadell-Neuwirth short exact sequence for pure braids.

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

In ZF, AC implies DC (AC implies DC implies countable choice).

[F2]

Assume AC and let $n \geq 2$ . With the base configuration $Q^{(n)}$ and its truncation $q^{'}$ , the forgetting map $φ : P B_{n} \to P B_{n - 1}$ sits in the short exact sequence $1 \to F_{n - 1} \overset{κ}{\to} P B_{n} \overset{φ}{\to} P B_{n - 1} \to 1$ with $κ$ injective, $φ$ surjective and $im κ = \ker φ$ , and $P B_{1}$ is trivial; here $F_{n - 1} = π_{1} (int D^{2} ∖ {q_{1}, \dots, q_{n - 1}}, q_{n})$ is free on the $n - 1$ positively oriented meridian classes of the punctures $q_{1}, \dots, q_{n - 1}$ (The Fadell-Neuwirth short exact sequence for pure braids).

[F3]

Assume AC and let $n \geq 2$ . Under the identification $F_{n - 1} ≅ \ker φ$ given by $κ$ , the elements $A_{1 n}, \dots, A_{n - 1, n}$ of Standard geometric pure braid generators A_ij are a free basis of $\ker φ$ , and the $i$ -th of them corresponds to the clockwise meridian of the $i$ -th puncture, the inverse of the counterclockwise spine-basis class (The $A_{i n}$ are meridian generators of the forgetful free kernel).

[F4]

The free group on $X = {γ_{1}}$ has the reduced-word model, in which each element has a unique reduced word in $γ_{1}^{\pm 1}$ (Reduced words form the free group on an alphabet). Since there is only one generator, a reduced word contains only $γ_{1}$ or only $γ_{1}^{- 1}$ , so every element is uniquely $γ_{1}^{k}$ for some $k \in Z$ .

[F5]

Assume AC. For every $n \geq 0$ the pure braid group $P B_{n}$ is torsion-free (Pure braid groups are torsion-free).

Verification

technique · direct

1.1A1F1F2

The choice deduction and the exact sequence at $n = 2$ . By [F1] the Axiom of Choice [A1] yields DC, so the sequence of [F2] is available at $n = 2$ : $1 \to F_{1} \overset{κ}{\to} P B_{2} \overset{φ}{\to} P B_{1} \to 1$ with $κ$ injective, $im κ = \ker φ$ , and $F_{1} = π_{1} (int D^{2} ∖ {q_{1}}, q_{2})$ . Since $P B_{1}$ is trivial by [F2], $φ$ has trivial codomain, so $\ker φ = P B_{2}$ and $im κ = P B_{2}$ : the map $κ$ is a group isomorphism $F_{1} \to P B_{2}$ .

1.2F2F3

The fibre and its meridian basis. By [F2] the fibre group $F_{1} = π_{1} (int D^{2} ∖ {q_{1}}, q_{2})$ is free on the single positively oriented meridian class of $q_{1}$ , written $γ_{1}$ , and by the case $n = 2$ of [F3] the corresponding generator $A_{12}$ corresponds under $κ$ to the inverse class $γ_{1}^{- 1}$ , the clockwise meridian; that is, $κ (γ_{1}^{- 1}) = A_{12}$ and $κ (γ_{1}) = A_{12}^{- 1}$ .

2.1step 1.1step 1.2F2F4

$P B_{2}$ is infinite cyclic. By [F2] and [F4], every element of the free group $F_{1}$ on ${γ_{1}}$ is uniquely $γ_{1}^{k}$ for $k \in Z$ . Concatenation followed by free reduction adds exponents, so $ν : F_{1} \to (Z, +)$ , $ν (γ_{1}^{k}) = k$ , is a group isomorphism. Composing the negation automorphism $k \mapsto - k$ of $Z$ with $ν^{- 1}$ and then with $κ$ of step 1.1 gives the isomorphism $Z \to P B_{2}$ , $k \mapsto κ (γ_{1}^{- k}) = A_{12}^{k}$ , by step 1.2. Thus $P B_{2}$ is infinite cyclic, generated by $A_{12}$ , and its inverse sends $A_{12}$ to $1$ , as claimed.

3.1step 1.2step 2.1F5

The winding sign and torsion-freeness. By step 1.2 the generator $A_{12}$ corresponds under $κ$ to $γ_{1}^{- 1}$ , the clockwise meridian of $q_{1}$ , while the counterclockwise spine-basis class is $γ_{1}$ itself; the identification of step 1.2 therefore assigns to $A_{12}$ the winding $- 1$ relative to the counterclockwise basis, as claimed. The isomorphism in step 2.1 shows that $P B_{2}$ is infinite cyclic and generated by $A_{12}$ ; it is torsion-free by the general theorem [F5].

∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

PB_3 as F_2 by Z, with its section action

Example

Assume the Axiom of Choice and let, with the standard pure braid generators of Standard geometric pure braid generators A_ij for $n = 3$ ,

$a : = A_{12}, b : = A_{13}, c : = A_{23} \in P B_{3} .$

Then $⟨ b, c ⟩$ is the kernel of the forgetting homomorphism $φ : P B_{3} \to P B_{2}$ , it is free on $b, c$ , and $P B_{2} = ⟨ a ⟩$ is infinite cyclic. With the far-right section of A choice-free continuous section of planar coordinate forgetting — adjusted at the basepoint along a path in the fibre, so that on a small representative of $a$ it fixes the third point — the extension splits and

$P B_{3} ≅ ⟨ b, c ⟩ ⋊ ⟨ a ⟩ ≅ F_{2} ⋊ Z .$

Writing $w : = b c$ , the action of the positive generator $a$ on the free kernel is

$x ⟼ w^{- 1} x w (x \in F_{2}),$

for this section and the first-under-second convention. This does not assert a direct-product decomposition.

Facts & Assumptions

Given: the Axiom of Choice, the canonical base configuration $Q = (q_{1}, q_{2}, q_{3})$ of The elementary geometric half twist, its support disc, and its opposite, so that $h = \frac{1}{16}$ , $q_{1} = - \frac{1}{8}$ , $q_{2} = 0$ and $q_{3} = \frac{1}{8}$ ; the half twists $σ_{1}, σ_{2}$ and the geometric braid group $G_{3}$ of The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism; the open and closed ordered configuration spaces $F_{3} (int D^{2}) \subseteq F_{3} (D^{2})$ and $F_{2} (int D^{2}) \subseteq F_{2} (D^{2})$ at the base configurations $Q$ and $Q^{'} : = (q_{1}, q_{2})$ ; the pure braid groups $P B_{3}$ , $P B_{2}$ and the forgetting homomorphism $φ : P B_{3} \to P B_{2}$ of The Fadell-Neuwirth short exact sequence for pure braids.

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

In ZF, AC implies DC (AC implies DC implies countable choice).

[F2]

The standard generators are $A_{i j} = [W_{i j}]$ with $W_{i j} = σ_{j - 1} \dots σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} \dots σ_{j - 1}^{- 1}$ under first-under-second stacking, and the same letters denote their images under the isomorphism $Ψ$ in $P B_{n}$ ; for $n = 3$ this gives $A_{12} = [σ_{1}^{2}]$ , $A_{13} = [σ_{2} σ_{1}^{2} σ_{2}^{- 1}]$ and $A_{23} = [σ_{2}^{2}]$ (Standard geometric pure braid generators A_ij).

[F3]

The geometric three strand relation holds: $[σ_{1}] [σ_{2}] [σ_{1}] = [σ_{2}] [σ_{1}] [σ_{2}]$ in $G_{3}$ (The geometric three strand braid relation).

[F4]

$G_{3}$ is a group with product induced by stacking, $[γ] [β] = [γ ⋆ β]$ , and the classes $[σ_{1}], [σ_{2}]$ generate $G_{3}$ ; the endpoint permutation $π_{g e o} : G_{3} \to S_{3}$ is a homomorphism with $π_{g e o} ([σ_{i}])$ the transposition of $i$ and $i + 1$ , and $G_{3}^{p u r e} = \ker π_{g e o}$ (The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, The Artin presentation surjects onto the geometric braid group, The elementary geometric half twist, its support disc, and its opposite).

[F5]

The map $Ψ : G_{n}^{p u r e} \to P B_{n}$ , $Ψ ([β]) = (ι_{*}^{F} [z_{β}])^{- 1}$ , is a group isomorphism, where $z_{β}$ is the coordinate path of $β$ and $ι_{*}^{F}$ is the open-to-closed isomorphism on fundamental groups of configuration spaces (Pure geometric braids and ordered configuration loops, The interior-disc and closed-disc configuration spaces are homotopy equivalent).

[F6]

Under AC the sequence $1 \to F_{2} \overset{κ}{\to} P B_{3} \overset{φ}{\to} P B_{2} \to 1$ is short exact with $im κ = \ker φ$ , and under the identification of $κ$ the elements $A_{13}, A_{23}$ are a free basis of $\ker φ$ , each represented by the clockwise meridian of the corresponding puncture (The Fadell-Neuwirth short exact sequence for pure braids, The $A_{i n}$ are meridian generators of the forgetful free kernel).

[F7]

At the canonical two-point configuration $Q^{(2)} = (- 1 / 12, 1 / 12)$ , $P B_{2}$ is infinite cyclic generated by $A_{12}^{(2)}$ (The two-strand pure braid group is infinite cyclic). Here the quotient is instead $P B_{2} = π_{1} (F_{2} (D^{2}), Q^{'})$ , with $Q^{'} = (- 1 / 8, 0)$ . The path $η (r) = ((1 + r / 3) q_{1} + r / 12, (1 + r / 3) q_{2} + r / 12)$ runs from $Q^{'}$ to $Q^{(2)}$ . Basepoint transport gives an isomorphism $τ : π_{1} (F_{2} (D^{2}), Q^{(2)}) \to π_{1} (F_{2} (D^{2}), Q^{'})$ , $τ ([ℓ]) = [η * ℓ * \overset{ˉ}{η}]$ (Conjugating loop classes by a path is an isomorphism of fundamental groups, Higher homotopy basepoint transport and moving homotopies). Write $a^{'} : = τ (A_{12}^{(2)})$ for the quotient generator; occurrences of $a$ in the cyclic quotient factor of the Example mean $a^{'}$ . Step 1.3 checks that forgetting sends $a \in P B_{3}$ to $a^{'}$ , using the actual coordinate-forgetting map of The Fadell-Neuwirth short exact sequence for pure braids.

[F8]

The extension of [F6] splits: there is a homomorphic section $s : P B_{2} \to P B_{3}$ of $φ$ , and then $P B_{3} ≅ F_{2} ⋊ P B_{2}$ compatibly with $κ$ and $φ$ , the action of $h \in P B_{2}$ on the free kernel being $h \cdot x = s (h) x s (h)^{- 1}$ ; the section is the based version of the far-right explicit section, and for a section the action depends on that section (The pure braid extension splits as a semidirect product, Splitting lemma for groups: a section, a complement, and a semidirect-product decomposition are equivalent).

[F9]

The far-right section of the planar forgetful map is $s (z_{1}, \dots, z_{n - 1}) = (z_{1}, \dots, z_{n - 1}, 1 + ∣ z_{1} ∣ + \dots + ∣ z_{n - 1} ∣)$ for $F_{n} (C) \to F_{n - 1} (C)$ , transported to the disc by the radial homeomorphism $h (w) = w / (1 + ∣ w ∣)$ with inverse $h^{- 1} (z) = z / (1 - ∣ z ∣)$ ; fixing a base configuration $q$ and a path $α$ in the fibre from $q$ to $s^{'} (q^{'})$ , the formula $σ ([β]) = [(α * (s^{'} \circ β)) * \overset{ˉ}{α}]$ defines a homomorphism $σ : π_{1} (F_{n - 1} (int D^{2}), q^{'}) \to π_{1} (F_{n} (int D^{2}), q)$ with ${\tilde{p}}_{*} \circ σ = id$ , and the section $s$ of [F8] is $ι_{*}^{F} \circ σ \circ (ι_{*}^{F})^{- 1}$ (A choice-free continuous section of planar coordinate forgetting).

[F10]

For $i = 1, 2$ the support disc of $σ_{i}$ is $U_{i} = B (m_{i}, \frac{3}{2} h)$ with $m_{1} = - \frac{1}{16}$ and $m_{2} = \frac{1}{16}$ ; the two strands of the braid $σ_{1}^{2}$ stay in $U_{1}$ at every time, and $U_{1} \subseteq B (0, \frac{5}{2} h) = B (0, \frac{5}{32})$ , while $q_{3} = \frac{1}{8} \notin U_{1} \cup {q_{1}, q_{2}}$ and the real interval $[\frac{1}{8}, \frac{37}{64}]$ is disjoint from $U_{1} \cup {q_{1}, q_{2}}$ (The elementary geometric half twist, its support disc, and its opposite).

Verification

technique · direct

1.1A1F1F2F6F8

Choice bookkeeping and the standing identifications. By [F1] the Axiom of Choice [A1] yields DC, so the short exact sequence and the splitting of [F6] and [F8] are available. Throughout, $a, b, c$ denote the elements of $P B_{3}$ named in the statement, that is $a = Ψ ([σ_{1}^{2}])$ , $b = Ψ ([σ_{2} σ_{1}^{2} σ_{2}^{- 1}])$ and $c = Ψ ([σ_{2}^{2}])$ , The quotient generator is $a^{'} = τ (A_{12}^{(2)})$ of [F7]; only the letter $a$ is reused for that cyclic factor, and $b, c$ remain kernel elements in $P B_{3}$ .

1.2F2F3F4F5

The product $a b c$ is central. Put $s : = [σ_{1}]$ and $t : = [σ_{2}]$ in $G_{3}$ . By [F2] the geometric classes of the three words are $s^{2}$ , $t s^{2} t^{- 1}$ and $t^{2}$ , so in the group $G_{3}$ of [F4] the bracketing is immaterial and $s^{2} \cdot t s^{2} t^{- 1} \cdot t^{2} = s^{2} t s^{2} t = s (s t s) s t = s (t s t) s t = (s t)^{3},$ where the middle step uses the relation $t s t = s t s$ of [F3]; put $Δ : = s t s$ . The same relation gives $Δ = t s t$ , hence $Δ s = t Δ$ and $Δ t = s Δ$ , and therefore $Δ^{2} s = Δ (t Δ) = (Δ t) Δ = (s Δ) Δ = s Δ^{2}, Δ^{2} t = Δ (s Δ) = (Δ s) Δ = (t Δ) Δ = t Δ^{2} .$ So $Δ^{2}$ commutes with $s$ and $t$ ; since $s$ and $t$ generate $G_{3}$ by [F4], $Δ^{2}$ is central in $G_{3}$ . Also $π_{g e o} (Δ^{2}) = π_{g e o} (s t)^{3} = i d$ because $π_{g e o} (s t)$ is a product of the two distinct transpositions $(1 2)$ and $(2 3)$ , a three-cycle whose cube is the identity; hence $Δ^{2} \in G_{3}^{p u r e}$ and $z : = Ψ ([Δ^{2}])$ is defined. By [F5] the map $Ψ$ is a homomorphism, so with the identification of [F2] $a b c = Ψ ([σ_{1}^{2}]) Ψ ([σ_{2} σ_{1}^{2} σ_{2}^{- 1}]) Ψ ([σ_{2}^{2}]) = Ψ ([σ_{1}^{2} σ_{2} σ_{1}^{2} σ_{2}^{- 1} \cdot σ_{2}^{2}]) = Ψ ([Δ^{2}]) = z .$ Since $Ψ$ is an isomorphism onto $P B_{3}$ and $Δ^{2}$ is central in $G_{3}$ , the element $z = a b c$ is central in $P B_{3}$ .

1.3F5F6F7F10

Forgetting and the transported quotient generator. Let $z = (z_{1}, z_{2})$ be the two moving coordinates of the rank-three word $σ_{1}^{2}$ , a loop at $Q^{'}$ . Then $φ (a) = (ι_{*}^{F} [z])^{- 1}$ by [F5] and the naturality of coordinate forgetting in [F6]. Define $g_{r} (v) = (1 + r / 3) v + r / 12$ and $H (t, r) = (g_{r} (z_{1} (t)), g_{r} (z_{2} (t)))$ . The map $g_{r}$ is injective, so these coordinates stay distinct. By [F10], $∣ z_{i} (t) ∣ \leq 5 / 32$ , whence $∣ g_{r} (z_{i} (t)) ∣ \leq (4 / 3) (5 / 32) + 1 / 12 = 7 / 24 < 1$ . Thus $H$ is a homotopy through ordered configurations with basepoint track $η$ from [F7]. At $r = 1$ the midpoint $- 1 / 16$ is sent to $0$ , while the relative diamond displacement $ρ$ is multiplied by $4 / 3$ , changing its scale from $1 / 16$ to $1 / 12$ . Therefore $H (-, 1)$ is exactly the raw coordinate loop of the canonical rank-two full twist. The moving-basepoint identity of [F7], and its compatibility with open-to-closed inclusions, give $ι_{*}^{F} [z] = τ (ι_{*}^{F} [z_{σ_{1}^{2}}^{(2)}])$ . Since $τ$ preserves inverses, $φ (a) = τ (A_{12}^{(2)}) = a^{'}$ . By [F7], $a^{'}$ generates this quotient $P B_{2} ≅ Z$ .

2.1step 1.2

The conjugation by $a$ is conjugation by $w^{- 1}$ . Put $w : = b c \in \ker φ$ . By step 1.2, $a b c = z$ with $z$ central, so $a = z (b c)^{- 1} = z w^{- 1}$ , and for every $x \in P B_{3}$ , in particular for every $x \in \ker φ$ , $a x a^{- 1} = z w^{- 1} x w z^{- 1} = w^{- 1} x w .$

2.2F5F8F9F10step 1.3

The far-right section sends $a^{'}$ to $a$ . Write $z_{2} (t) = (z_{1} (t), z_{2} (t))$ for the coordinate path of $σ_{1}^{2}$ in $F_{2} (int D^{2})$ , a loop at $Q^{'}$ whose two entries lie in $U_{1}$ for every $t$ by [F10], and put $u_{i} (t) : = h^{- 1} (z_{i} (t))$ , so that the far-right lift of [F9] is $ς (t) : = (z_{1} (t), z_{2} (t), h (1 + ∣ u_{1} (t) ∣ + ∣ u_{2} (t) ∣)),$ a loop at $s^{'} (Q^{'})$ in $F_{3} (int D^{2})$ . Since $∣ z_{i} (t) ∣ \leq \frac{5}{32}$ by [F10], we have $∣ u_{i} (t) ∣ \leq \frac{5}{27}$ and, because $r \mapsto \frac{r}{1 + r}$ is increasing on $[0, \infty)$ , $h (1 + ∣ u_{1} (t) ∣ + ∣ u_{2} (t) ∣) \in [\frac{1}{2}, \frac{37}{64}],$ an interval in the positive real axis. The value of the far-right section at $Q^{'}$ is $s^{'} (Q^{'}) = (q_{1}, q_{2}, \frac{8}{15})$ : here $h^{- 1} (q_{1}) = (- 1 / 8) / (7 / 8) = - \frac{1}{7}$ and $h^{- 1} (q_{2}) = 0$ , so the third coordinate is $h (1 + \frac{1}{7}) = h (\frac{8}{7}) = \frac{8}{15}$ . Let $α (t) : = (q_{1}, q_{2}, \frac{1}{8} + t (\frac{8}{15} - \frac{1}{8}))$ be the path in the fibre from $Q$ to $s^{'} (Q^{'})$ , whose third coordinate runs along the real interval $[\frac{1}{8}, \frac{8}{15}]$ ; this interval, and likewise $[\frac{1}{2}, \frac{37}{64}]$ , is contained in $[\frac{1}{8}, \frac{37}{64}]$ , which is disjoint from $U_{1} \cup {q_{1}, q_{2}}$ by [F10]. By [F9] the section of [F8] is built from $σ = φ_{\overset{ˉ}{α}} \circ s_{*}^{'}$ with $σ ([z_{2}]) = [(α * ς) * \overset{ˉ}{α}]$ , and we claim this class is the class of the third-strand-fixed loop $ω_{0} (t) : = (z_{1} (t), z_{2} (t), q_{3})$ in $π_{1} (F_{3} (int D^{2}), Q)$ . To see this, let $v : I \to R$ denote the third coordinate path of $γ : = (α * ς) * \overset{ˉ}{α}$ , so that $v (0) = v (1) = q_{3}$ , $v$ takes values in $[\frac{1}{8}, \frac{37}{64}]$ , and $v$ coincides with the third coordinate of $α$ on the first quarter, with $h (1 + ∣ u_{1} ∣ + ∣ u_{2} ∣)$ on the second quarter and with the reversed third coordinate of $α$ on the last half. Let $ψ_{0}$ be the map that is $0$ on $[0, \frac{1}{4}]$ , $4 t - 1$ on $[\frac{1}{4}, \frac{1}{2}]$ and $1$ on $[\frac{1}{2}, 1]$ , and put $ψ_{r} (t) : = (1 - r) ψ_{0} (t) + r t$ for $(r, t) \in I \times I$ . Then $G (r, t) : = (z_{1} (ψ_{r} (t)), z_{2} (ψ_{r} (t)), (1 - r) v (t) + r q_{3})$ is continuous, lies in $F_{3} (int D^{2})$ pointwise: each $ψ_{r} (t)$ lies in $[0, 1]$ , so both $z_{i} (ψ_{r} (t))$ lie in $U_{1}$ and are distinct, while the third coordinate is a convex combination of two points of $[\frac{1}{8}, \frac{37}{64}]$ and therefore also lies in that interval, which is disjoint from $U_{1} \cup {q_{1}, q_{2}}$ by [F10]; moreover $ψ_{r} (0) = 0$ , $ψ_{r} (1) = 1$ and $v (0) = v (1) = q_{3}$ , so $G (r, 0) = G (r, 1) = Q$ for every $r$ . Finally $G (0, t) = γ (t)$ and $G (1, t) = ω_{0} (t)$ . Hence $G$ is a path homotopy relative to ${0, 1}$ from $γ$ to $ω_{0}$ and $σ ([z_{2}]) = [ω_{0}]$ . Now $(ι_{*}^{F})^{- 1} (a^{'}) = [z_{2}]^{- 1}$ in $π_{1} (F_{2} (int D^{2}), Q^{'})$ : by [F5] the element $a^{'} = φ (a) \in P B_{2}$ is $(ι_{*}^{F} [z_{2}])^{- 1}$ and $ι_{*}^{F}$ is an isomorphism. Since $σ$ is a homomorphism, $σ ([z_{2}]^{- 1}) = [ω_{0}]^{- 1}$ , and since the element $a = A_{12}$ of $P B_{3}$ is by [F5] the class $(ι_{*}^{F} [ω_{0}])^{- 1}$ (the coordinate path of the braid $σ_{1}^{2}$ in $G_{3}$ is $ω_{0}$ ), the section $s$ of [F9] satisfies $s (a^{'}) = ι_{*}^{F} (σ ((ι_{*}^{F})^{- 1} (a^{'}))) = ι_{*}^{F} ([ω_{0}]^{- 1}) = a .$

3.1F6F8step 1.3step 2.1step 2.2∎

The action and the semidirect product. By step 2.2 the section of [F8] satisfies $s (a^{'}) = a$ , and by step 2.1 $a x a^{- 1} = w^{- 1} x w$ for every $x$ in the free kernel $\ker φ = ⟨ b, c ⟩$ of [F6]. Since the action of the section is $h \cdot x = s (h) x s (h)^{- 1}$ by [F8], the positive generator $a^{'}$ of $P B_{2} = ⟨ a^{'} ⟩ ≅ Z$ of step 1.3 acts by $x \mapsto w^{- 1} x w$ . The splitting of [F8] therefore exhibits $P B_{3} ≅ ⟨ b, c ⟩ ⋊ ⟨ a ⟩ ≅ F_{2} ⋊ Z$ with that action. The action is conjugation by the element $w^{- 1}$ of the free kernel, so it depends on the normalised section, no triviality of the action is claimed, and no direct-product decomposition is asserted.

Remarks

The computation of step 1.2 is a direct rank-three calculation inside $G_{3}$ : the full twist $Δ^{2} = (s t)^{3}$ is written as the product of the three standard generators $a, b, c$ , and centrality of that product is read off the single braid relation. The later centre theorem for $P B_{n}$ is not used, and neither is any Artin-presentation injectivity: only the surjectivity of The Artin presentation surjects onto the geometric braid group enters, through the generation of $G_{3}$ by $s$ and $t$ .
The section used above is the normalised far-right section: the fibre path is the straight segment from $q_{3}$ to $s^{'} (Q^{'})_{3}$ on the positive real axis, and on the small representative $σ_{1}^{2}$ of $a$ the far-right lift is homotopic to the loop with constant third coordinate, which is why $s (a^{'}) = a$ . Any other section $s^{'}$ of $φ$ has $s^{'} (a^{'}) = k a$ for some $k \in \ker φ$ , so its action is $x \mapsto k (w^{- 1} x w) k^{- 1}$ , an inner automorphism of the free kernel; the displayed formula is the one for this section, and clearing it of the normalisation would require a separate conjugation bookkeeping.
The action is by an inner automorphism of the free kernel, because $w \in ⟨ b, c ⟩$ itself. This example nevertheless asserts only the semidirect-product decomposition with the action of the chosen section; the classical direct-product decomposition $P B_{3} ≅ F_{2} \times Z$ is not derived here.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedaudited 2026-10-02Open item page →

Standard $A_{i j}$ as point pushes after relabeling

Example

Assume the Axiom of Choice, let $n \geq 2$ and $1 \leq i < j \leq n$ , and use the base configuration $Q = (q_{1}, \dots, q_{n})$ and positive half twists of The elementary geometric half twist, its support disc, and its opposite. The standard pure braid $A_{i j} \in P B_{n}$ is the image of the geometric word $W_{i j}$ of Standard geometric pure braid generators A_ij. Put $Y^{(j)} : = int D^{2} ∖ {q_{k} : k \neq j} .$ For a based loop $γ$ of $Y^{(j)}$ at $q_{j}$ , let $M_{j} (γ) (t) : = (q_{1}, \dots, q_{j - 1}, γ (t), q_{j + 1}, \dots, q_{n}) \in F_{n} (int D^{2}) .$ This is the ordered motion in which only the $j$ -th point moves.

Choose the compatible family of local meridian circles and stems constructed by the puncture-avoiding fan argument in the proof of The $A_{i n}$ are meridian generators of the forgetful free kernel. Thus, for each $r < n$ , take $0 < ϵ_{r} \leq h_{n} / 10$ , let $C_{r} = {z : ∣ z - q_{r} ∣ = ϵ_{r}}$ and $c_{r} = q_{r} + (ϵ_{r}, 0)$ , and use the local stem $τ_{r}$ from $q_{r + 1}$ to $c_{r}$ selected in that compatible family, inside $U_{r} ∖ {q_{r}}$ . For $i < j$ , let $h_{i, j} : = H_{j - 1} \circ \dots \circ H_{i + 1},$ where $H_{s}$ represents the positive half twist $σ_{s}$ and the empty composition for $j = i + 1$ is the identity. The associated meridian stem from $q_{j}$ to $C_{i}$ is $h_{i, j} \circ τ_{i}$ ; let $γ_{i, j}^{c w}$ be the based loop that follows this stem, traverses $C_{i}$ clockwise once, and returns along the reverse stem. These are the compatible standard meridian stems obtained by transporting the adjacent local stem through the successive half twists. Then:

Point push. For every $1 \leq i < j \leq n$ , $A_{i j} = ι_{*}^{F} [M_{j} (γ_{i, j}^{c w})] \in P B_{n} .$ Thus the positive standard generator is the class of the motion that holds the other $n - 1$ labelled points fixed and moves the $j$ -th point clockwise once around $q_{i}$ along the stated stem.
Relabeled form. Let $ρ \in S_{n}$ satisfy $ρ (j) = n$ , $ρ (k) = k - 1$ for $j < k \leq n$ , and $ρ (k) = k$ for $k < j$ . The coordinate permutation $(R x)_{k} : = x_{ρ^{- 1} (k)}$ gives homeomorphisms $R^{\circ}$ and $R^{D}$ on the open and closed ordered configuration spaces, respectively. It takes $Q$ to $Q^{ρ} = (q_{1}, \dots, q_{j - 1}, q_{j + 1}, \dots, q_{n}, q_{j}) .$ Put $Y^{ρ} : = int D^{2} ∖ {q_{k} : k \neq j}$ , and define the open terminal-coordinate inclusion ${\tilde{κ}}^{ρ} : π_{1} (Y^{ρ}, q_{j}) ⟶ π_{1} (F_{n} (int D^{2}), Q^{ρ}), [γ] ⟼ [(q_{1}, \dots, q_{j - 1}, q_{j + 1}, \dots, q_{n}, γ)] .$ With $ι_{*}^{F, ρ}$ the open-to-closed map at basepoint $Q^{ρ}$ , set $κ^{ρ} : = ι_{*}^{F, ρ} \circ {\tilde{κ}}^{ρ}$ . Then $R_{*}^{D} (A_{i j}) = κ^{ρ} ([γ_{i, j}^{c w}]) \in π_{1} (F_{n} (D^{2}), Q^{ρ}),$ the loop class in which the last coordinate moves clockwise around $q_{i}$ and all other coordinates remain fixed.
Terminal mapping-class sign. For $i < n$ , $Θ_{n} (A_{i n}) = {Push}_{n} ([γ_{i, n}^{c w}]), Θ_{n} : = Ψ_{n}^{m c} \circ (Ψ_{n}^{c o n f})^{- 1},$ where $Θ_{n}$ is the isomorphism of Point pushing is the kernel of forgetting the last disk puncture. The inverse-endpoint convention in Point pushing the last puncture makes the clockwise fibre meridian correspond to the positive generator. The raw ordered slice of the positive standard word runs counterclockwise; the configuration identification inverts that slice.

Facts & Assumptions

Given: AC, $n \geq 2$ , $1 \leq i < j \leq n$ , the base configuration $Q$ , the half twists $σ_{1}, \dots, σ_{n - 1}$ and their supports $U_{1}, \dots, U_{n - 1}$ , the words $W_{i j}$ , and the maps in the statement.

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

In ZF, AC implies DC and DC implies countable choice (AC implies DC implies countable choice).

[F2]

The Statement of The $A_{i n}$ are meridian generators of the forgetful free kernel says that, under AC, the last-coordinate fibre inclusion $κ$ identifies $F_{m - 1}$ with the kernel of forgetting $P B_{m} \to P B_{m - 1}$ , and the $m - 1$ configuration-group images $Ψ ([A_{1 m}]), \dots, Ψ ([A_{m - 1, m}])$ of the standard geometric classes form a free basis. For the compatible stem family constructed in its proof, if $[λ_{r}]$ is the counterclockwise meridian class in the fibre, then $Ψ ([A_{r m}]) = (κ_{*} [λ_{r}])^{- 1} .$ Here $[A_{r m}] = [W_{r m}]$ is the geometric braid class and, by [F3], its configuration-group image is the element denoted $A_{r m}$ in this example; $κ_{*} [λ]$ is the closed-disc image of the open fibre loop. The supplier Statement makes this last-column assertion; its proof also supplies the local winding and conjugation arguments used below for arbitrary $j$ . It does not assert that the whole word motion is braid-isotopic to a one-coordinate motion.

[F3]

The standard generators are $A_{r s} = Ψ_{n}^{c o n f} ([W_{r s}])$ , where $W_{r s} = σ_{s - 1} \dots σ_{r + 1} σ_{r}^{2} σ_{r + 1}^{- 1} \dots σ_{s - 1}^{- 1} .$ The rightmost factor is the bottom one, $[γ ⋆ β] = [γ] [β]$ , and $Ψ_{n}^{c o n f} ([β]) = (ι_{*}^{F} [z_{β}])^{- 1}$ for the ordered coordinate loop $z_{β}$ (Standard geometric pure braid generators A_ij, Pure geometric braids and ordered configuration loops).

[F4]

The positive half twist $σ_{r}$ exchanges $q_{r}, q_{r + 1}$ , is supported in $U_{r} = B (m_{r}, 3 h_{n} / 2)$ , and fixes every other base point; $U_{r} \cap U_{s} = \emptyset$ when $∣ r - s ∣ > 1$ . Here the base points are equally spaced by $2 h_{n}$ , so the center of $U_{r + 1}$ is distance $3 h_{n}$ from $q_{r}$ (The elementary geometric half twist, its support disc, and its opposite).

[F5]

Geometric braids form a group under stacking, with the right factor running first; coordinate paths of pure braids are loops in $F_{n} (int D^{2})$ , and reversed braid paths represent inverse classes (The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[F6]

Coordinate permutations act by homeomorphisms on both $F_{n} (int D^{2})$ and $F_{n} (D^{2})$ , and the open-to-closed inclusions commute with these permutations (The symmetric group acts continuously and freely on $F_{n} (X)$ by permuting labels, The interior-disc and closed-disc configuration spaces are homotopy equivalent).

[F7]

The maps $Ψ_{n}^{c o n f} : G_{n}^{p u r e} \to P B_{n}$ and $Ψ_{n}^{m c} : G_{n} \to Mod (D^{2}, Q; \partial D^{2})$ are group isomorphisms, with $Ψ_{n}^{c o n f} ([β]) = (ι_{*}^{F} [z_{β}])^{- 1}$ and, for the raw unordered slice $S (β)$ , $Ψ_{n}^{m c} ([β]) = δ ([S (β)]^{- 1})$ ; on a half twist, $Ψ_{n}^{m c} ([σ_{r}]) = [H_{r}]$ , where $H_{r}$ is orientation-preserving, supported in $U_{r}$ , and exchanges $q_{r}, q_{r + 1}$ (Pure geometric braids and ordered configuration loops, Braid group as boundary-fixed punctured-disk mapping classes).

[F8]

The point-pushing map for the last puncture is ${Push}_{n} ([γ]) = δ ([\overset{ˉ}{γ}])$ , where $\overset{ˉ}{γ}$ is the unordered loop of the ordered motion that moves only $q_{n}$ and $δ$ is the inverse-endpoint boundary map. The braid-to-mapping-class map sends that raw geometric motion to $δ ([\overset{ˉ}{γ}]^{- 1}) = {Push}_{n} ([γ])^{- 1}$ (Point pushing the last puncture, Boundary map from point motions).

[F9]

Under AC, $Θ_{n} = Ψ_{n}^{m c} ∣_{G_{n}^{p u r e}} \circ (Ψ_{n}^{c o n f})^{- 1}$ is an isomorphism and $Θ_{n} \circ κ = {Push}_{n}$ for the terminal-coordinate fibre inclusion (Point pushing is the kernel of forgetting the last disk puncture).

[F10]

$F_{n} (X)$ is the space of pairwise distinct tuples in $X^{n}$ , $P B_{n} = π_{1} (F_{n} (D^{2}), Q)$ , and $ι_{*}^{F} : π_{1} (F_{n} (int D^{2}), Q) \to P B_{n}$ is an isomorphism (Ordered configuration spaces $F_{n} (X)$ , The pure braid group $P B_{n}$ as the fundamental group of an ordered configuration space, The interior-disc and closed-disc configuration spaces are homotopy equivalent).

[F11]

The local two-point winding computation in the proof of The $A_{i n}$ are meridian generators of the forgetful free kernel identifies $F_{2} (U_{r}) ≅ S^{1} \times C$ with $C$ convex and shows that the raw coordinate loop of $σ_{r}^{2}$ has relative winding $+ 1$ , equal to a counterclockwise local meridian motion of $q_{r + 1}$ around $q_{r}$ . By the inverse-endpoint convention, $Ψ_{n}^{m c} ([σ_{r}^{2}]) = {Push}_{q_{r + 1}} ([λ_{r}^{c c w}])^{- 1} = {Push}_{q_{r + 1}} ([λ_{r}^{c w}]) .$ The proof's conjugation argument establishes, for any mapping class $h$ taking a marked point $p$ to $p^{'}$ , the typed naturality $h {Push}_{p} ([α]) h^{- 1} = {Push}_{p^{'}} ([h \circ α]),$ where $α$ is a loop in the complement of $Q ∖ {p}$ based at $p$ and $h \circ α$ is based at $p^{'}$ in the complement of $Q ∖ {p^{'}}$ . There ${Push}_{p}$ is the inverse endpoint of an ambient lift of this single-point motion; for $p = q_{n}$ it agrees with [F8].

Verification

technique · direct

Choice bookkeeping. AC supplies DC and countable choice, so the cited fibre exact sequence, braid identifications, and point-pushing identifications are available. [A1, F1, F2, F7, F9]

1.1A1F1F2F7F9

The choice hypothesis discharges the cited fibration and mapping-class identifications.

Terminal fibre case. Fix $i < n$ and let $[λ_{i}^{c c w}]$ be the counterclockwise based meridian class in the last-coordinate fibre at $q_{n}$ from [F2]. Write $[λ_{i}^{c w}] = [λ_{i}^{c c w}]^{- 1}$ . The standard generator in $P B_{n}$ is $A_{i n} = Ψ_{n}^{c o n f} ([W_{i n}])$ ; applying $Ψ_{n}^{c o n f}$ to $A_{i n}$ again would be ill-typed. The source formula and the fact that the fibre inclusion is a homomorphism give $A_{i n} = (κ_{*} [λ_{i}^{c c w}])^{- 1} = κ_{*} [λ_{i}^{c w}] = ι_{*}^{F} [M_{n} (λ_{i}^{c w})] .$ This proves the terminal instance of claim 1. [F2, F3, F10]

2.1step 1.1F2F3F10

Thus the terminal generator is the closed-disc image of the clockwise last-coordinate meridian.

A point-motion loop maps to its point push. Let $p = q_{j}$ and let $γ$ be any based loop in $Y^{(j)}$ at $q_{j}$ . By the geometric-braid/configuration identification [F3] and the open-to-closed map [F10], the ordered loop $M_{j} (γ)$ defines a pure geometric braid $β_{γ}$ whose unordered slice is $\overset{ˉ}{γ}$ . Its inverse braid $β_{γ}^{- 1}$ has coordinate loop class $[M_{j} (γ)]^{- 1}$ and unordered slice class $[\overset{ˉ}{γ}]^{- 1}$ . The braid/configuration and braid/mapping-class maps [F7] therefore give $Ψ_{n}^{c o n f} (β_{γ}^{- 1}) = (ι_{*}^{F} [M_{j} (γ)]^{- 1})^{- 1} = ι_{*}^{F} [M_{j} (γ)], Ψ_{n}^{m c} (β_{γ}^{- 1}) = δ (([\overset{ˉ}{γ}]^{- 1})^{- 1}) = δ ([\overset{ˉ}{γ}]) .$ For any marked point $q_{j}$ , the inverse-endpoint point-motion map constructed in the kernel lemma's proof is ${Push}_{q_{j}} ([γ]) = δ ([\overset{ˉ}{γ}])$ ; for $j = n$ this agrees with [F8]. Thus $Θ_{n} (ι_{*}^{F} [M_{j} (γ)]) = Ψ_{n}^{m c} (β_{γ}^{- 1}) = {Push}_{q_{j}} ([γ]) .$ This identity will compare classes by the isomorphism $Θ_{n}$ and uses no fibre-inclusion injectivity. [F3, F7, F8, F10, F11]

2.2step 1.1F3F7F8F10F11

For every marked point, the image under $Θ_{n}$ of its one-coordinate motion is the corresponding point push.

Transport the adjacent point push. Fix $i < j$ . Choose the local circle $C_{i}$ and stem $τ_{i}$ from the statement, and let $λ_{i}^{c w}$ be the loop following $τ_{i}$ , once clockwise around $C_{i}$ , and back. Put $g_{i, j} : = [σ_{j - 1}] \dots [σ_{i + 1}] \in G_{n}$ and let $h_{i, j} = H_{j - 1} \circ \dots \circ H_{i + 1}$ be its mapping-class representative, with the rightmost map acting first. Hence $h_{i, j} (q_{i + 1}) = q_{j}$ and $h_{i, j} (q_{i}) = q_{i}$ . It fixes $C_{i}$ pointwise: each support $U_{r}$ for $r \geq i + 2$ is disjoint from $U_{i}$ , while every point of $C_{i}$ is at distance at least $3 h_{n} - ϵ_{i} \geq 2.9 h_{n} > 3 h_{n} / 2$ from the center of $U_{i + 1}$ . Thus $h_{i, j} \circ τ_{i}$ is a stem from $q_{j}$ to $C_{i}$ , and $h_{i, j} \circ λ_{i}^{c w} = γ_{i, j}^{c w}$ . The path avoids all punctures other than its basepoint because $h_{i, j}$ permutes $Q$ and sends the omitted point $q_{i + 1}$ to $q_{j}$ . The word identity [F3], first-under-second product, and the local winding and typed naturality in [F11] now give $Θ_{n} (A_{i j}) = Ψ_{n}^{m c} ([W_{i j}]) = h_{i, j} Ψ_{n}^{m c} ([σ_{i}^{2}]) h_{i, j}^{- 1} = h_{i, j} {Push}_{q_{i + 1}} ([λ_{i}^{c w}]) h_{i, j}^{- 1} = {Push}_{q_{j}} ([γ_{i, j}^{c w}]) .$ For $j = i + 1$ , $h_{i, j}$ is the identity and this is exactly the local winding case. [F3, F7, F11]

2.3step 1.1F3F4F5F7F11

The conjugated geometric generator has the point-push image along the transported standard stem.

Point-push claim for every pair. By step 2.2, the image under $Θ_{n}$ of the one-coordinate motion along $γ_{i, j}^{c w}$ is ${Push}_{q_{j}} ([γ_{i, j}^{c w}])$ . By step 2.3, this equals $Θ_{n} (A_{i j})$ . Since $Θ_{n}$ is an isomorphism by [F9], it is injective, and therefore $A_{i j} = ι_{*}^{F} [M_{j} (γ_{i, j}^{c w})] .$ This proves claim 1 without an isotopy assertion about the full word motion. [F9, step 2.2, step 2.3]

3.1F9step 2.2step 2.3

Equality under $Θ_{n}$ proves the point-push statement for every pair.

Relabeled form and open-to-closed maps. Let $M_{n}^{ρ} (γ) = (q_{1}, \dots, q_{j - 1}, q_{j + 1}, \dots, q_{n}, γ)$ be the open ordered loop based at $Q^{ρ}$ . Pointwise, $R^{\circ} \circ M_{j} (γ) = M_{n}^{ρ} (γ)$ . The coordinate-permutation square commutes with the open-to-closed inclusions, so $R_{*}^{D} (ι_{*}^{F} [M_{j} (γ)]) = ι_{*}^{F, ρ} [M_{n}^{ρ} (γ)] = κ^{ρ} ([γ]) .$ Apply this identity to the loop from the point-push claim to obtain $R_{*}^{D} (A_{i j}) = κ^{ρ} ([γ_{i, j}^{c w}]) .$ This is the relabeled terminal-coordinate form. The argument uses functoriality only and asserts no injectivity of ${\tilde{κ}}^{ρ}$ or $κ^{ρ}$ . [F6, F10, step 3.1]

4.1F6F10step 3.1

The coordinate permutation carries the proven open motion to the stated closed-disc terminal-coordinate class.

Terminal mapping-class sign. For $i < n$ , the fibre formula and [F9] give $Θ_{n} (A_{i n}) = Θ_{n} (κ_{*} [λ_{i}^{c w}]) = {Push}_{n} ([λ_{i}^{c w}]) .$ The source formula [F2] identifies the configuration-group image of $[W_{i n}]$ as $(κ_{*} [λ_{i}^{c c w}])^{- 1}$ , while [F3] identifies the same image as $(ι_{*}^{F} [z_{W_{i n}}])^{- 1}$ for the raw ordered coordinate loop. Equating and inverting gives $ι_{*}^{F} [z_{W_{i n}}] = κ_{*} [λ_{i}^{c c w}]$ : the raw ordered slice is counterclockwise in the fibre. The configuration identification inverts it, so $A_{i n} = κ_{*} [λ_{i}^{c w}]$ . The inverse-endpoint convention then gives exactly the clockwise point push. [F2, F3, F8, F9, step 2.1]

5.1F2F3F8F9step 2.1∎

This proves the terminal mapping-class sign in claim 3 with the positive generator clockwise.

Remarks

The stem for $A_{i j}$ is the image of the adjacent local stem under the actual mapping-class representative $H_{j - 1} \circ \dots \circ H_{i + 1}$ . This specifies the compatible meridian path and preserves its clockwise orientation.
The relabeling is a coordinate-permutation homeomorphism from basepoint $Q$ to $Q^{ρ}$ . The open terminal-coordinate map and its closed-disc composite have distinct codomains; only the latter is denoted $κ^{ρ}$ in the statement.
The example identifies standard generators and makes no new generation or presentation claim. The proof uses the local two-point winding and typed point-push conjugation already proved in The $A_{i n}$ are meridian generators of the forgetful free kernel.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

The short exact sequence to $S_{n}$ does not prove $B_{n}$ torsion-free

Statement refuted

Let $1 \to K \to G \to Q \to 1$ be a short exact sequence of groups with $K$ torsion-free and $Q$ finite. Then $G$ is torsion-free. In particular the configuration braid short exact sequence $1 \to P B_{n} \to B_{n}^{c o n f} \to S_{n} \to 1$ , whose kernel $P B_{n}$ is torsion-free and whose quotient $S_{n}$ is finite, would by itself show that $B_{n}^{c o n f}$ is torsion-free.

Facts & Assumptions

Given: the Axiom of Choice, an integer $n \geq 0$ , the groups $(Z, +, 0)$ and $(Z / 2, +)$ , and their external direct product $Z \times (Z / 2)$ .

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

Assume AC. Then $P B_{n}$ is torsion-free for every $n \geq 0$ : if $g \in P B_{n}$ and $g^{m} = e$ for some $m \geq 1$ , then $g = e$ (Pure braid groups are torsion-free). AC is used only for this citation (AC implies DC implies countable choice).

[F2]

For every $n \geq 0$ the sequence $1 \to P B_{n} \overset{p_{*}}{\to} B_{n}^{c o n f} \overset{π}{\to} S_{n} \to 1$ of the configuration groups is short exact: $p_{*}$ is injective, $π$ is surjective and $im p_{*} = \ker π$ (The configuration braid short exact sequence $1 \to P B_{n} \to B_{n}^{c o n f} \to S_{n} \to 1$ ); moreover $∣ S_{n} ∣ = n!$ , so $S_{n}$ is finite (The Lehmer code gives $∣ S_{n} ∣ = n!$ again).

[F3]

$(Z, +, \cdot, 0, 1)$ is a commutative ring and a totally ordered ring: the order is total and compatible with addition, so $x \leq y$ implies $x + z \leq y + z$ , and exactly one of $x < y$ , $x = y$ , $y < x$ holds for all integers (The integers form a commutative ring, The integers form a totally ordered ring).

[F4]

$(Z / 2, +, [0]_{2})$ is an abelian group with $[a]_{2} + [b]_{2} = [a + b]_{2}$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold). Its exactly two classes are $[0]_{2}$ and $[1]_{2}$ (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ , The congruence class $[a]_{n}$ and the quotient set $Z / n$ ), so $[1]_{2} \neq [0]_{2}$ and $[1]_{2} + [1]_{2} = [2]_{2} = [0]_{2}$ ; hence it has order $2$ and is generated by $[1]_{2}$ .

[F5]

For any groups $G, H$ the componentwise operation $(g, h) (g^{'}, h^{'}) = (g g^{'}, h h^{'})$ makes $G \times H$ a group with identity $(e_{G}, e_{H})$ , and the coordinate projections are group homomorphisms (The external direct product $G \times H$ with componentwise multiplication, $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

[F6]

In a group, $g^{0} = e$ and $g^{k + 1} = g^{k} \cdot g$ for natural powers, and $g$ has finite order $k \geq 1$ exactly when $k$ is the least positive integer with $g^{k} = e$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Counterexample

technique · direct

1.1F3F4

The two factors. By [F3] the additive group $(Z, +, 0)$ is a group, and by [F4] $(Z / 2, +)$ is an abelian group of order $2$ with $[1]_{2} \neq [0]_{2}$ and $[1]_{2} + [1]_{2} = [2]_{2} = [0]_{2}$ .

1.2F3F6

$Z$ is torsion-free. Let $a \in Z$ with $a \neq 0$ and let $m \geq 1$ . By trichotomy [F3] either $a > 0$ or $- a > 0$ ; assume $a > 0$ , the other case being symmetric with $m (- a) = - (m a)$ . Since $0 < a$ and the order is compatible with addition, $a < a + a = 2 a$ and inductively $k a < (k + 1) a$ for every $k \geq 1$ : if $k a > 0$ then $(k + 1) a = k a + a > k a > 0$ . Hence $m a > 0$ for every $m \geq 1$ , so $m a \neq 0$ ; an element $a$ of finite additive order would have $m a = 0$ for some $m \geq 1$ , so no nonzero element of $Z$ has finite order and $(Z, +, 0)$ is torsion-free.

2.1step 1.2F4F5

The sequence $1 \to Z \to Z \times (Z / 2) \to Z / 2 \to 1$ . By [F5] the set $G : = Z \times (Z / 2)$ with componentwise addition is a group with identity $e_{G} = (0, [0]_{2})$ , and the second-coordinate projection $π : G \to Z / 2$ is a group homomorphism; it is surjective because $π (a, [b]_{2}) = [b]_{2}$ for every $[b]_{2}$ . Its kernel is $K : = {(a, [0]_{2}) : a \in Z}$ , and the first-coordinate map $K \to Z$ , $(a, [0]_{2}) \mapsto a$ , is a group isomorphism; hence $K$ is torsion-free by step 1.2, and $G / K ≅ Z / 2$ has order $2$ by [F4], so the quotient is finite. Thus $1 \to K \to G \overset{π}{\to} Z / 2 \to 1$ is a short exact sequence with torsion-free kernel and finite quotient.

3.1step 2.1F4F5F6

The middle group has an element of order two. Put $g : = (0, [1]_{2}) \in G$ . Using the componentwise operation of [F5] and the addition of [F4], $g^{2} = (0 + 0, [1]_{2} + [1]_{2}) = (0, [2]_{2}) = (0, [0]_{2}) = e_{G}$ , while $g \neq e_{G}$ because $[1]_{2} \neq [0]_{2}$ . By [F6] the element $g$ has order $2$ , so $G$ is not torsion-free although $K$ is torsion-free and $G / K$ is finite. This refutes the general implication of the statement: a torsion-free kernel and a finite quotient do not force the middle group to be torsion-free.

4.1A1step 2.1step 3.1F1F2

The braid reading. Under the standing assumption [A1], which discharges the Axiom-of-Choice hypothesis that [F1] attaches to its own citation, the configuration braid sequence of [F2] has the two abstract features used above: its kernel $P B_{n}$ is torsion-free by [F1], and its quotient $S_{n}$ is finite by [F2]. Since the general implication fails already for the infinite cyclic kernel $Z$ and the two-element quotient $Z / 2$ of step 2.1, those two features alone cannot establish torsion-freeness of $B_{n}^{c o n f}$ ; only a finer argument about the specific groups can do that. This counterexample makes no claim that $B_{n}^{c o n f}$ itself has torsion.

∎

Pure Braids, Fadell–Neuwirth, and Asphericity — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The two-strand pure braid group is infinite cyclic

Example

Facts & Assumptions

Verification

PB_3 as F_2 by Z, with its section action

Example

Facts & Assumptions

Verification

Remarks

Standard AijAij​ as point pushes after relabeling

Example

Facts & Assumptions

Verification

Remarks

The short exact sequence to SnSn​ does not prove BnBn​ torsion-free

Statement refuted

Facts & Assumptions

Counterexample

Sources

Standard $A_{i j}$ as point pushes after relabeling

The short exact sequence to $S_{n}$ does not prove $B_{n}$ torsion-free