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✓ 7 results · all verified · 1 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 6 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Artin Presentation Completeness and Braid Combing

1 · Prerequisites

2 · Summary

This page proves that the Artin presentation is not merely a presentation that surjects onto the geometric braid group: the surjection $φ_{n} ⁣ : B_{n}^{A r t i n} \to B_{n}^{g e o m}$ that sends each abstract generator to the class of the elementary geometric half twist is an isomorphism. The proof is the braid-combing argument. The page first fixes the Zariski combing words $α_{i} = σ_{i} σ_{i + 1} \dots σ_{n - 1}$ and $x_{i} = α_{i + 1}^{- 1} σ_{i}^{2} α_{i + 1}$ , then shows that any word tracing the trivial braid can be rewritten, by free insertion of the cancelling blocks $α_{j} α_{j}^{- 1}$ alone, as a product of combing factors, one per letter, each factor being the record of where the last strand enters and leaves that letter. The six possible shapes of a combing factor then reduce, using only the two Artin relations and free cancellation, to either a lower-rank letter $σ_{1}, \dots, σ_{n - 2}$ or one of the words $x_{i}^{\pm 1}$ , and the conjugation table for the lower-rank letters moving past an $x$ -letter collects the whole word into the standard form $W \equiv W_{1} W_{2}$ with $W_{1}$ in the $x$ -letters and $W_{2}$ in the first $n - 2$ generators.

The second half of the page identifies the geometric content of that normal form. The words $x_{i}$ are conjugate, inside the free kernel of the forgetting map $P B_{n} \to P B_{n - 1}$ , to the standard pure generators $A_{i n}$ , and a left-inverse computation in the free group shows that the classes $φ_{n} (x_{1}), \dots, φ_{n} (x_{n - 1})$ form a free basis of that kernel; consequently a combed trivial word has both factors trivial, the left one by free cancellation of $x$ -pairs and the right one by the induction hypothesis on the number of strands. Induction on $n$ then proves completeness of the Artin presentation. Because the free-kernel input is the Axiom-of-Choice-dependent Fadell--Neuwirth sequence of the companion pure-braid page, the completeness theorem carries the Axiom of Choice, declared explicitly where it is used. The closing corollary composes the completeness theorem with the published configuration-space isomorphism and the boundary-fixed mapping-class isomorphism, tracking the generator through all four classical models of the braid group.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

The Zariski combing words alpha_i and x_i in the Artin presentation

Definition

Fix $n \geq 2$ and let $B_{n} = ⟨ σ_{1}, \dots, σ_{n - 1} ⟩$ be the abstract braid group of The braid group by Artin presentation, the group presented by the generators $σ_{1}, \dots, σ_{n - 1}$ and the relations

$σ_{i} σ_{i + 1} σ_{i} = σ_{i + 1} σ_{i} σ_{i + 1} (1 \leq i \leq n - 2),$

$σ_{i} σ_{j} = σ_{j} σ_{i} (1 \leq i, j \leq n - 1, ∣ i - j ∣ > 1) .$

Reading convention for words. A word $w_{1} w_{2} \dots w_{m}$ in this alphabet and its inverses is read first letter first: it denotes the product $[w_{1}] [w_{2}] \dots [w_{m}]$ in $B_{n}$ , and under the published stacking convention $[γ] [β] : = [γ ⋆ β]$ of Stacking of geometric braids is a well-defined associative operation on isotopy classes and The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, in which the first factor of a stacking is the upper one, its geometric image is the stacking $w_{1} ⋆ w_{2} ⋆ \dots ⋆ w_{m}$ whose first factor is the topmost layer. Here the geometric image is taken through the published surjection $φ ⁣ : B_{n} \to G_{n}$ , $φ (σ_{i}) = [σ_{i}]$ of The Artin presentation surjects onto the geometric braid group. In particular the empty word is the identity of $B_{n}$ . No injectivity or completeness of $φ$ is asserted, and nothing below depends on how many factors the stacking has.

The words $α_{i}$ . For $1 \leq i \leq n - 1$ put

$α_{i} : = σ_{i} σ_{i + 1} \dots σ_{n - 1} \in B_{n}, and α_{n} : = 1,$

the last being the empty word. Each $α_{i}$ is a word of length $n - i$ in the generators, and $α_{n - 1} = σ_{n - 1}$ .

The words $x_{i}$ . For $1 \leq i \leq n - 1$ put

$x_{i} : = σ_{n - 1}^{- 1} \dots σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} \dots σ_{n - 1} = α_{i + 1}^{- 1} σ_{i}^{2} α_{i + 1} \in B_{n} .$

The two displayed words for $x_{i}$ are literally the same word written in two ways: expanding $α_{i + 1}^{- 1}$ gives $σ_{n - 1}^{- 1} \dots σ_{i + 1}^{- 1}$ and expanding $α_{i + 1}$ gives $σ_{i + 1} \dots σ_{n - 1}$ , so the middle factor $σ_{i}^{2}$ sits in the same position in both readings. In particular $x_{n - 1} = σ_{n - 1}^{2}$ , and $x_{i}$ is a word of length $2 (n - i)$ in the generators and their inverses.

These words are the Zariski combing words of the Artin presentation. The definition is choice-free, it uses no relation of the presentation, and it asserts no property of $α_{i}$ or $x_{i}$ inside any geometric braid model; all of that is established, when needed, by the items that cite this definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Prefix insertion rewrites a trivial braid word into combing factors

Statement

Assume $n \geq 2$ and let $W = σ_{i_{1}}^{ε_{1}} \dots σ_{i_{m}}^{ε_{m}}$ be a word in the Artin letters and their inverses whose image under the published surjection $φ$ of The Artin presentation surjects onto the geometric braid group is the trivial geometric braid. Write $W_{k} : = σ_{i_{1}}^{ε_{1}} \dots σ_{i_{k}}^{ε_{k}}$ for $k = 0, \dots, m$ , and for $k = 0, \dots, m$ let $j_{k} : = π (φ (W_{k}))^{- 1} (n) \in {1, \dots, n}$ be the position at the bottom of the sub-braid $φ (W_{k})$ of the point that sits at position $n$ at its top, where $π ⁣ : G_{n} \to S_{n}$ is the endpoint permutation homomorphism; here $s_{r}$ denotes the transposition of $r$ and $r + 1$ . Then $j_{0} = n$ , the recursion $j_{k} = s_{i_{k}} (j_{k - 1}) (k = 1, \dots, m)$ holds, and $j_{m} = n$ because $φ (W) = 1$ . Equivalently, $j_{k}$ is the position of the point that starts at position $n$ at the top of $W$ after it has passed the first $k$ letters (the library's stacking puts the first letter of a word on top, so this point meets the letters in word order).

Using only free insertions of the words $α_{j} α_{j}^{- 1}$ and free deletions of cancelling pairs (no braid relation), $W$ is equivalent in the group $B_{n}$ of The braid group by Artin presentation to the product of combing factors $W \equiv \prod_{k = 1}^{m} (α_{j_{k - 1}}^{- 1} σ_{i_{k}}^{ε_{k}} α_{j_{k}}),$ where the words $α_{1}, \dots, α_{n}$ are those of The Zariski combing words alpha_i and x_i in the Artin presentation. The $k$ -th factor is the $k$ -th letter decorated by its two connectors: read bottom to top (the library's stacking puts the first letter of a word on top, so the last block of the factor is met first), the tracked point that starts at position $n$ travels through $α_{j_{k}}$ to position $j_{k}$ , is exchanged by the letter $σ_{i_{k}}^{ε_{k}}$ to position $j_{k - 1}$ when $j_{k} \in {i_{k}, i_{k} + 1}$ (and is fixed otherwise), and is carried back to position $n$ by $α_{j_{k - 1}}^{- 1}$ ; equivalently, in the source's bottom-up reading the point has position $j_{k}$ below the letter and $j_{k - 1}$ above it. It lies in the letter's support precisely when $j_{k} \in {i_{k}, i_{k} + 1}$ ; otherwise $j_{k - 1} = j_{k}$ and it remains fixed outside that support during the letter. In every case the recursion above is exactly the interchange rule $j_{k - 1} = s_{i_{k}} (j_{k})$ used by the six-case analysis, the empty word is allowed ( $m = 0$ , where $W$ is empty and $j_{0} = j_{m} = n$ ), and nothing but $φ (W) = 1$ is assumed about the geometric braid.

Facts & Assumptions

Given: An integer $n \geq 2$ , a word $W = σ_{i_{1}}^{ε_{1}} \dots σ_{i_{m}}^{ε_{m}}$ with $φ (W) = 1$ , its prefixes $W_{k}$ , and the words $α_{j}$ of The Zariski combing words alpha_i and x_i in the Artin presentation.

[F1]

$α_{j} = σ_{j} σ_{j + 1} \dots σ_{n - 1}$ for $1 \leq j \leq n - 1$ and $α_{n} = 1$ is the empty word; all these are words in the generators and their inverses (The Zariski combing words alpha_i and x_i in the Artin presentation).

[F2]

$B_{n}$ is the quotient of the free group on $σ_{1}, \dots, σ_{n - 1}$ by the normal closure of the two relation families; consequently words differing by insertions or deletions of adjacent inverse pairs $w^{\pm 1} w^{\mp 1}$ represent the same element, and a product of words telescopes whenever adjacent connector words cancel (The braid group by Artin presentation, The Zariski combing words alpha_i and x_i in the Artin presentation).

[F3]

The assignment $φ (σ_{i}) = [σ_{i}]$ extends to a surjective homomorphism $φ ⁣ : B_{n} \to G_{n}$ (The Artin presentation surjects onto the geometric braid group).

[F4]

The endpoint permutation $π ⁣ : G_{n} \to S_{n}$ is a homomorphism, the class of the half twist $σ_{r}$ has $π ([σ_{r}]) = s_{r}$ , and under the stacking convention of Stacking of geometric braids is a well-defined associative operation on isotopy classes the class of a word $u_{1} \dots u_{l}$ satisfies $π (φ (u_{1} \dots u_{l})) = π (φ (u_{1})) \circ \dots \circ π (φ (u_{l}))$ as functions, the first factor applied last (The elementary geometric half twist, its support disc, and its opposite, 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).

Proof

technique · direct

1.1F1F3F4

The connectors move the $n$ -th point down to position $j$ . By [F3] and [F1], $φ (α_{j}) = φ (σ_{j}) \dots φ (σ_{n - 1}) = [σ_{j}] \dots [σ_{n - 1}]$ , so by [F4] the endpoint permutation is $π (φ (α_{j})) = s_{j} \circ s_{j + 1} \circ \dots \circ s_{n - 1}$ as a function. Evaluating on positions, this function sends $n \mapsto n - 1 \mapsto n - 2 \mapsto \dots \mapsto j$ and fixes every $x < j$ , so it is the cycle $t_{j} : = (j j + 1 \dots n)$ with $t_{j} (n) = j$ ; in particular a point starting at position $n$ ends at position $j$ , and $t_{j}^{- 1} (j) = n$ . For $j = n$ the word $α_{n}$ is empty and $π (φ (α_{n})) = id$ with $id (n) = n$ .

1.2F3F4

The recursion and its endpoints. For each $k$ , [F3] and [F4] give $π (φ (W_{k})) = π (φ (σ_{i_{1}}^{ε_{1}})) \circ \dots \circ π (φ (σ_{i_{k}}^{ε_{k}})) = s_{i_{1}} \circ \dots \circ s_{i_{k}}$ , because $s_{r}^{\pm 1} = s_{r}$ . Taking inverses, $π (φ (W_{k}))^{- 1} = s_{i_{k}} \circ \dots \circ s_{i_{1}}$ , so $j_{k} : = π (φ (W_{k}))^{- 1} (n)$ satisfies $j_{0} = n$ (the empty product) and $j_{k} = s_{i_{k}} (j_{k - 1})$ for $1 \leq k \leq m$ . Since $W_{m} = W$ and $φ (W) = 1$ , $π (φ (W_{m})) = id$ and therefore $j_{m} = id (n) = n$ .

2.1F1F2step 1.2

The telescoping insertion. For $k = 1, \dots, m - 1$ insert the word $α_{j_{k}} α_{j_{k}}^{- 1}$ between the $k$ -th and $(k + 1)$ -st letter of $W$ and bracket the result as $W = α_{j_{0}}^{- 1} σ_{i_{1}}^{ε_{1}} α_{j_{1}} \cdot α_{j_{1}}^{- 1} σ_{i_{2}}^{ε_{2}} α_{j_{2}} \dots α_{j_{m - 1}}^{- 1} σ_{i_{m}}^{ε_{m}} α_{j_{m}} .$ Each interior position contributes $α_{j_{k}} α_{j_{k}}^{- 1} = 1$ , which is a free cancellation by [F2], and by [F1] and step 1.2 the two end connectors are $α_{j_{0}} = α_{n} = 1$ and $α_{j_{m}} = α_{n} = 1$ ; expanding the displayed product therefore returns the original word $W$ by free cancellations alone, and conversely $W$ is obtained from the displayed product by the inverse free moves. No defining relation of the Artin presentation is used.

2.2F3F4step 1.1step 1.2

The bookkeeping inside a factor. Fix $k$ and read the factor $F_{k} = α_{j_{k - 1}}^{- 1} σ_{i_{k}}^{ε_{k}} α_{j_{k}}$ from the bottom upward, that is, starting from its last and lowest block $α_{j_{k}}$ and ending with its first and topmost block $α_{j_{k - 1}}^{- 1}$ (the word's first letter is the topmost layer in the stacking of [F4]). A point starting at position $n$ at the bottom of the factor is carried by the block $α_{j_{k}}$ to position $t_{j_{k}} (n) = j_{k}$ by step 1.1, then by the letter $σ_{i_{k}}^{ε_{k}}$ to position $s_{i_{k}}^{ε_{k}} (j_{k}) = s_{i_{k}} (j_{k}) = j_{k - 1}$ by [F4] and step 1.2 (if $j_{k} = j_{k - 1}$ the letter fixes it), and then by the block $α_{j_{k - 1}}^{- 1}$ to position $t_{j_{k - 1}}^{- 1} (j_{k - 1}) = n$ by step 1.1. Hence inside the $k$ -th factor the letter acts on the tracked point exactly through the interchange rule connecting the two connector positions $j_{k}$ (below the letter) and $j_{k - 1}$ (above it), and the tracked point returns to position $n$ at the top of every factor, as it must because at the top of $W$ it is again at position $n$ by $φ (W) = 1$ .

3.1F4step 1.2step 2.1step 2.2

Conclusion. Steps 1.2 and 2.1 establish the recursion, its endpoints, and the telescoping product. Step 2.2 gives the positions $j_{k}$ below and $j_{k - 1}$ above each letter. By the half-twist definition in [F4], the tracked point lies in the letter's support if $j_{k} \in {i_{k}, i_{k} + 1}$ ; otherwise it is a fixed base point outside that support. For $m = 0$ the product is empty. ∎

Remarks

The lemma is a pure bookkeeping statement: the group element is unchanged because each inserted connector is immediately cancelled, and the geometric input is only the published endpoint-permutation homomorphism, which fixes the positions $j_{k}$ by the triviality of $φ (W)$ .
In the source the same product is displayed with $α_{j_{0}} = α_{j_{m}} = 1$ , reading words bottom-up; the recursion $j_{k} = s_{i_{k}} (j_{k - 1})$ is identical in both conventions, and the six-case reduction of the next item depends only on this recursion and on the displayed shape of the factors.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Lower-rank Artin letters conjugate x-letters

Statement

Assume $n \geq 3$ , let $1 \leq i \leq n - 2$ and $1 \leq j \leq n - 1$ , and work in the group $B_{n}$ of The braid group by Artin presentation with the words $σ_{1}, \dots, σ_{n - 1}$ and $x_{1}, \dots, x_{n - 1}$ of The Zariski combing words alpha_i and x_i in the Artin presentation. Using only the two families of defining relations and free insertions and deletions of adjacent inverse letters:

(i) $σ_{i}^{- 1} x_{j} σ_{i}$ equals $x_{j}$ when $j < i$ or $j > i + 1$ , equals $x_{i + 1} x_{i} x_{i + 1}^{- 1}$ when $j = i + 1$ , and equals $x_{i + 1}$ when $j = i$ ;

(ii) $σ_{i} x_{j} σ_{i}^{- 1}$ equals $x_{j}$ when $j < i$ or $j > i + 1$ , equals $x_{i}$ when $j = i + 1$ , and equals $x_{i}^{- 1} x_{i + 1} x_{i}$ when $j = i$ .

Consequently, for every pair of signs $ε, δ \in {\pm 1}$ there is a word $w$ in the letters $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ with $σ_{i}^{ε} x_{j}^{δ} = w σ_{i}^{ε},$ so that in any word over the mixed alphabet $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}, x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ each occurrence of a lower-rank $σ$ -letter can be moved to the right of every $x$ -letter, the $x$ -letters changing only by further $x$ -letters and their inverses. All identities also hold in the geometric braid group under the published surjection $φ$ of The Artin presentation surjects onto the geometric braid group.

Facts & Assumptions

Given: Integers $n \geq 3$ , $1 \leq i \leq n - 2$ , $1 \leq j \leq n - 1$ , the group $B_{n}$ of The braid group by Artin presentation, and the elements $x_{1}, \dots, x_{n - 1} \in B_{n}$ of The Zariski combing words alpha_i and x_i in the Artin presentation.

[F1]

In $B_{n}$ the two defining families of relations hold: $σ_{r} σ_{r + 1} σ_{r} = σ_{r + 1} σ_{r} σ_{r + 1}$ for $1 \leq r \leq n - 2$ , and $σ_{r} σ_{s} = σ_{s} σ_{r}$ whenever $∣ r - s ∣ > 1$ ; words equal in the free group on $σ_{1}, \dots, σ_{n - 1}$ and their inverses represent the same element of $B_{n}$ , so adjacent inverse letters may be freely inserted and deleted (The braid group by Artin presentation, The Zariski combing words alpha_i and x_i in the Artin presentation).

[F2]

For every $r$ with $1 \leq r \leq n - 1$ one has the two displayed words $x_{r} = σ_{n - 1}^{- 1} \dots σ_{r + 1}^{- 1} σ_{r}^{2} σ_{r + 1} \dots σ_{n - 1} = α_{r + 1}^{- 1} σ_{r}^{2} α_{r + 1}$ , and $x_{r}^{- 1} = σ_{n - 1}^{- 1} \dots σ_{r + 1}^{- 1} σ_{r}^{- 2} σ_{r + 1} \dots σ_{n - 1}$ ; also $α_{s} = σ_{s} σ_{s + 1} \dots σ_{n - 1}$ for $1 \leq s \leq n - 1$ and $α_{n} = 1$ (The Zariski combing words alpha_i and x_i in the Artin presentation).

[F3]

The map $φ ⁣ : B_{n} \to G_{n}$ of The Artin presentation surjects onto the geometric braid group is a homomorphism with $φ (σ_{r}) = [σ_{r}]$ , and in the geometric braid group $G_{n}$ the two families of relations of [F1] hold: $[σ_{r}] [σ_{r + 1}] [σ_{r}] = [σ_{r + 1}] [σ_{r}] [σ_{r + 1}]$ and $[σ_{r}] [σ_{s}] = [σ_{s}] [σ_{r}]$ for $∣ r - s ∣ > 1$ (The geometric three strand braid relation, Far commutativity of elementary geometric half twists).

Proof

technique · direct

1.1F1

Two mixed forms of the braid relation. Let $1 \leq r \leq n - 2$ . Multiplying $σ_{r} σ_{r + 1} σ_{r} = σ_{r + 1} σ_{r} σ_{r + 1}$ on the left by $σ_{r}^{- 1}$ and on the right by $σ_{r + 1}^{- 1}$ gives $σ_{r}^{- 1} σ_{r + 1} σ_{r} = σ_{r + 1} σ_{r} σ_{r + 1}^{- 1},$ and multiplying the braid relation on the left by $σ_{r + 1}^{- 1}$ and on the right by $σ_{r}^{- 1}$ gives $σ_{r + 1}^{- 1} σ_{r} σ_{r + 1} = σ_{r} σ_{r + 1} σ_{r}^{- 1}$ , whose inverse is $σ_{r} σ_{r + 1}^{- 1} σ_{r}^{- 1} = σ_{r + 1}^{- 1} σ_{r}^{- 1} σ_{r + 1} .$ Both are consequences of the defining relations of [F1] alone.

1.2F1F2

The case $j > i + 1$ . Every letter $σ_{r}^{\pm 1}$ occurring in the displayed word for $x_{j}$ of [F2] has $r \geq j \geq i + 2$ , so $∣ i - r ∣ \geq 2$ and $σ_{i}$ commutes with that letter by the far-commutation relation of [F1]; repeating this letter by letter, $σ_{i}$ commutes with the whole word, so $σ_{i}^{\pm 1} x_{j} σ_{i}^{\mp 1} = x_{j}$ . In particular both assertions (i) and (ii) hold for $j > i + 1$ .

2.1F1F2step 1.1

The case $j < i$ : sliding $σ_{i}$ past $x_{j}$ . Decompose the word of [F2] for $x_{j}$ at the index $i$ , which satisfies $j < i$ , as $x_{j} = A σ_{i}^{- 1} K σ_{i} B, A : = σ_{n - 1}^{- 1} \dots σ_{i + 1}^{- 1}, K : = σ_{i - 1}^{- 1} \dots σ_{j + 1}^{- 1} σ_{j}^{2} σ_{j + 1} \dots σ_{i - 1}, B : = σ_{i + 1} \dots σ_{n - 1},$ where each of $A$ , $K$ , $B$ may be empty and the displayed equality is free cancellation in the two words of [F2]. Since $σ_{i}$ commutes with every letter of $A$ except $σ_{i + 1}^{- 1}$ , and $σ_{i} σ_{i + 1}^{- 1} σ_{i}^{- 1} = σ_{i + 1}^{- 1} σ_{i}^{- 1} σ_{i + 1}$ by step 1.1, one has $σ_{i} A σ_{i}^{- 1} = A σ_{i}^{- 1} σ_{i + 1}$ ; since $σ_{i + 1}$ commutes with every letter of $K$ (all its indices are at most $i - 1$ ), and since $B = σ_{i + 1} B^{'}$ with $B^{'} : = σ_{i + 2} \dots σ_{n - 1}$ , the computation $σ_{i} x_{j} = σ_{i} A σ_{i}^{- 1} K σ_{i} B = A σ_{i}^{- 1} σ_{i + 1} K σ_{i} B = A σ_{i}^{- 1} K σ_{i + 1} σ_{i} σ_{i + 1} B^{'} = A σ_{i}^{- 1} K σ_{i} σ_{i + 1} σ_{i} B^{'} = A σ_{i}^{- 1} K σ_{i} σ_{i + 1} B^{'} σ_{i} = x_{j} σ_{i}$ uses only far commutation, the braid relation, and the fact that $σ_{i}$ commutes with every letter of $B^{'}$ . Left-multiplying by $σ_{i}^{- 1}$ gives $σ_{i}^{- 1} x_{j} σ_{i} = x_{j}$ , and right-multiplying by $σ_{i}^{- 1}$ gives $σ_{i} x_{j} σ_{i}^{- 1} = x_{j}$ ; both assertions hold for $j < i$ .

2.2F1F2step 1.1

The case $j = i + 1$ . Put $C : = σ_{n - 1}^{- 1} \dots σ_{i + 2}^{- 1}$ and $D : = σ_{i + 2} \dots σ_{n - 1}$ , so that $C = D^{- 1}$ as words and, by [F2], $x_{i + 1} = C σ_{i + 1}^{2} D, x_{i} = C σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} D, x_{i + 1}^{- 1} = D^{- 1} σ_{i + 1}^{- 2} C^{- 1} .$ Since $σ_{i}$ commutes with every letter of $C$ and of $D$ , and using step 1.1, $σ_{i}^{- 1} x_{i + 1} σ_{i} = C σ_{i}^{- 1} σ_{i + 1}^{2} σ_{i} D = C (σ_{i}^{- 1} σ_{i + 1} σ_{i})^{2} D = C (σ_{i + 1} σ_{i} σ_{i + 1}^{- 1})^{2} D = C σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} D .$ Expanding the product $x_{i + 1} x_{i} x_{i + 1}^{- 1}$ with the three displayed words and using $D C = D D^{- 1} = 1$ and $C^{- 1} = D$ gives $x_{i + 1} x_{i} x_{i + 1}^{- 1} = C σ_{i + 1}^{2} (D C) σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} (D D^{- 1}) σ_{i + 1}^{- 2} C^{- 1} = C σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} D,$ so $σ_{i}^{- 1} x_{i + 1} σ_{i} = x_{i + 1} x_{i} x_{i + 1}^{- 1}$ .

3.1F1F2step 1.1

The case $j = i$ . With the same words $C, D$ of step 2.2, [F2] gives $x_{i} = C σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} D$ , and $σ_{i}$ commutes with every letter of $C$ and of $D$ , so $σ_{i}^{- 1} x_{i} σ_{i} = C σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} σ_{i} D = C (σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i}) (σ_{i} σ_{i + 1} σ_{i}) D .$ Inverting the first identity of step 1.1 gives $σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i} = σ_{i + 1} σ_{i}^{- 1} σ_{i + 1}^{- 1}$ , and the defining braid relation gives $σ_{i} σ_{i + 1} σ_{i} = σ_{i + 1} σ_{i} σ_{i + 1}$ ; substituting, $σ_{i}^{- 1} x_{i} σ_{i} = C σ_{i + 1} σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i + 1} σ_{i} σ_{i + 1} D = C σ_{i + 1} σ_{i}^{- 1} σ_{i} σ_{i + 1} D = C σ_{i + 1}^{2} D = x_{i + 1},$ the penultimate equality deleting the adjacent inverse pairs and the last equality being [F2] again.

4.1step 2.2step 3.1

The second orientation for $j = i$ and $j = i + 1$ . The map $c_{i} ⁣ : B_{n} \to B_{n}$ , $c_{i} (g) : = σ_{i}^{- 1} g σ_{i}$ , is the conjugation automorphism by $σ_{i}$ , with inverse $c_{i}^{- 1} (g) = σ_{i} g σ_{i}^{- 1}$ . Steps 2.2 and 3.1 give $c_{i} (x_{i}) = x_{i + 1}$ and $c_{i} (x_{i + 1}) = x_{i + 1} x_{i} x_{i + 1}^{- 1}$ , hence $c_{i} (x_{i}^{- 1} x_{i + 1} x_{i}) = x_{i + 1}^{- 1} (x_{i + 1} x_{i} x_{i + 1}^{- 1}) x_{i + 1} = x_{i}$ ; therefore $c_{i}^{- 1} (x_{i}) = x_{i}^{- 1} x_{i + 1} x_{i}$ , that is $σ_{i} x_{i} σ_{i}^{- 1} = x_{i}^{- 1} x_{i + 1} x_{i}$ , and $c_{i}^{- 1} (x_{i + 1}) = x_{i}$ , that is $σ_{i} x_{i + 1} σ_{i}^{- 1} = x_{i}$ . This is assertion (ii) in the two remaining cases.

5.1F2step 1.2step 2.1step 2.2step 3.1step 4.1

All four signs. Let $ε, δ \in {\pm 1}$ . If $δ = 1$ , then $σ_{i}^{ε} x_{j} = (σ_{i}^{ε} x_{j} σ_{i}^{- ε}) σ_{i}^{ε}$ ; by steps 1.2, 2.1, 2.2, 3.1 and 4.1 the middle factor equals $x_{j}$ (for $j < i$ or $j > i + 1$ ), or $x_{i + 1}$ , $x_{i + 1} x_{i} x_{i + 1}^{- 1}$ , $x_{i}$ , $x_{i}^{- 1} x_{i + 1} x_{i}$ (for $j = i$ or $j = i + 1$ , according to the sign of $ε$ ), so in every case it is a word in $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ . If $δ = - 1$ , then $σ_{i}^{ε} x_{j}^{- 1} = (σ_{i}^{ε} x_{j} σ_{i}^{- ε})^{- 1} σ_{i}^{ε}$ , and the same case list applies with the inverse word. Hence for all signs $σ_{i}^{ε} x_{j}^{δ} = w σ_{i}^{ε}$ with $w$ a word in the $x$ -letters and their inverses, and a leftmost occurrence of $σ_{i}^{\pm 1}$ in any mixed word can therefore be moved one $x$ -letter at a time to the right of all $x$ -letters, only $x$ -letters changing.

6.1F3step 1.1step 5.1

Transfer to the geometric braid group. Since $φ$ is a homomorphism with $φ (σ_{r}) = [σ_{r}]$ by [F3], and since the relations used in steps 1.1-5.1 are exactly the two families of [F1], applying $φ$ to each identity yields the corresponding identity in $G_{n}$ between $φ (σ_{i})^{\pm 1}$ and $φ (x_{j})^{\pm 1}$ : the images satisfy the braid relation and far commutation by the published geometric lemmas, and the free cancellations map to cancellations in the group $G_{n}$ .

7.1step 1.2step 2.1step 2.2step 3.1step 4.1step 5.1step 6.1

Assertion (i) is steps 1.2, 2.1, 2.2 and 3.1, assertion (ii) is steps 1.2, 2.1 and 4.1, the collection statement is step 5.1, and the geometric transfer is step 6.1. ∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Each combing factor reduces to a lower-rank letter or an x-letter

Statement

Assume $n \geq 2$ , work in the group $B_{n} = ⟨ σ_{1}, \dots, σ_{n - 1} ⟩$ of The braid group by Artin presentation, and use the words $α_{1}, \dots, α_{n}$ and $x_{1}, \dots, x_{n - 1}$ of The Zariski combing words alpha_i and x_i in the Artin presentation. Let $1 \leq j \leq n$ be a position, $1 \leq k \leq n - 1$ an index and $ε \in {\pm 1}$ a sign, and let the combing factor be the word $F = α_{j}^{- 1} σ_{k}^{ε} α_{j^{'}}, j^{'} : = {\begin{cases} k + 1, & j = k, \\ k, & j = k + 1, \\ j, & otherwise, \end{cases}$ so that $j^{'} = s_{k} (j)$ ; in the bottom-to-top reading of this factor, $j^{'}$ is the position below the letter and $j$ is the position above it, as in Prefix insertion rewrites a trivial braid word into combing factors. Then, using only the two Artin relations and free insertions and deletions of adjacent inverse letters:

(a) $F \equiv 1$ (the empty word) if $j = k$ and $ε = 1$ ;

(b) $F \equiv x_{k}^{- 1}$ if $j = k$ and $ε = - 1$ ;

(c) $F \equiv x_{k}$ if $j = k + 1$ and $ε = 1$ ;

(d) $F \equiv 1$ if $j = k + 1$ and $ε = - 1$ ;

(e) $F \equiv σ_{k}^{ε}$ if $k < j - 1$ ;

(f) $F \equiv σ_{k - 1}^{ε}$ if $k > j$ .

The six cases are mutually exclusive and exhaustive, and in every one of them the reduced form is a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}, x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ alone. In particular the letter $σ_{n - 1}$ and its inverse never survive the reduction outside an $x$ -letter, and all identities also hold in the geometric braid group $G_{n}$ under the published surjection $φ$ of The Artin presentation surjects onto the geometric braid group.

Facts & Assumptions

Given: An integer $n \geq 2$ , the group $B_{n}$ of The braid group by Artin presentation, the words $α_{j}$ and $x_{k}$ of The Zariski combing words alpha_i and x_i in the Artin presentation, a position $1 \leq j \leq n$ , an index $1 \leq k \leq n - 1$ , a sign $ε \in {\pm 1}$ , and the word $F = α_{j}^{- 1} σ_{k}^{ε} α_{j^{'}}$ with $j^{'} = s_{k} (j)$ as in the statement.

[F1]

In $B_{n}$ the two defining families of relations hold, $σ_{r} σ_{r + 1} σ_{r} = σ_{r + 1} σ_{r} σ_{r + 1}$ for $1 \leq r \leq n - 2$ and $σ_{r} σ_{s} = σ_{s} σ_{r}$ for $∣ r - s ∣ > 1$ , and two words that differ by insertions or deletions of adjacent inverse pairs $w^{\pm 1} w^{\mp 1}$ represent the same element of $B_{n}$ (The braid group by Artin presentation). Below we write $u \equiv v$ when the words $u$ and $v$ can be connected by these two families of relations together with such free insertions and deletions.

[F2]

$α_{j} = σ_{j} σ_{j + 1} \dots σ_{n - 1}$ for $1 \leq j \leq n - 1$ , $α_{n}$ is the empty word, and consequently the word identity $α_{k} = σ_{k} α_{k + 1}$ and its inverse form $α_{k}^{- 1} = α_{k + 1}^{- 1} σ_{k}^{- 1}$ hold for every $1 \leq k \leq n - 1$ . For every $1 \leq k \leq n - 1$ one has $x_{k} = σ_{n - 1}^{- 1} \dots σ_{k + 1}^{- 1} σ_{k}^{2} σ_{k + 1} \dots σ_{n - 1} = α_{k + 1}^{- 1} σ_{k}^{2} α_{k + 1}$ and hence $x_{k}^{- 1} = α_{k + 1}^{- 1} σ_{k}^{- 2} α_{k + 1}$ (The Zariski combing words alpha_i and x_i in the Artin presentation).

[F3]

The factor $F$ is exactly the shape of a combing factor in Prefix insertion rewrites a trivial braid word into combing factors: read bottom to top, the tracked point starts at position $n$ at the bottom, is carried by $α_{j^{'}}$ to position $j^{'}$ below the letter, is exchanged by $σ_{k}^{ε}$ to $j = s_{k} (j^{'})$ when $j^{'} \in {k, k + 1}$ and is fixed otherwise, and is carried by $α_{j}^{- 1}$ back to position $n$ at the top. Since $s_{k}$ is an involution, this is the same relation $j^{'} = s_{k} (j)$ used in the statement.

[F4]

The map $φ ⁣ : B_{n} \to G_{n}$ of The Artin presentation surjects onto the geometric braid group is a surjective homomorphism with $φ (σ_{r}) = [σ_{r}]$ , and in $G_{n}$ the two families of relations of [F1] hold: $[σ_{r}] [σ_{r + 1}] [σ_{r}] = [σ_{r + 1}] [σ_{r}] [σ_{r + 1}]$ for $1 \leq r \leq n - 2$ and $[σ_{r}] [σ_{s}] = [σ_{s}] [σ_{r}]$ for $∣ r - s ∣ > 1$ (The geometric three strand braid relation, Far commutativity of elementary geometric half twists).

Proof

technique · direct

1.1F1F2F3

The four cases in which the letter moves the tracked point. Assume $j = k$ or $j = k + 1$ ; by [F2] we have the word identities $α_{k} = σ_{k} α_{k + 1}$ , $α_{k}^{- 1} = α_{k + 1}^{- 1} σ_{k}^{- 1}$ , $x_{k} = α_{k + 1}^{- 1} σ_{k}^{2} α_{k + 1}$ and $x_{k}^{- 1} = α_{k + 1}^{- 1} σ_{k}^{- 2} α_{k + 1}$ . Substituting $α_{k}$ or $α_{k + 1}$ for the two connectors and cancelling the adjacent inverse pair $σ_{k}^{- 1} σ_{k}$ , or its inverse pair, by [F1]: for $j = k$ and $ε = 1$ , $F = α_{k}^{- 1} σ_{k} α_{k + 1} = α_{k + 1}^{- 1} σ_{k}^{- 1} σ_{k} α_{k + 1} \equiv α_{k + 1}^{- 1} α_{k + 1} \equiv 1,$ the empty word; for $j = k$ and $ε = - 1$ , $F = α_{k}^{- 1} σ_{k}^{- 1} α_{k + 1} = α_{k + 1}^{- 1} σ_{k}^{- 2} α_{k + 1} = x_{k}^{- 1};$ for $j = k + 1$ and $ε = 1$ , $F = α_{k + 1}^{- 1} σ_{k} α_{k} = α_{k + 1}^{- 1} σ_{k} σ_{k} α_{k + 1} = α_{k + 1}^{- 1} σ_{k}^{2} α_{k + 1} = x_{k};$ and for $j = k + 1$ and $ε = - 1$ , $F = α_{k + 1}^{- 1} σ_{k}^{- 1} α_{k} = α_{k + 1}^{- 1} σ_{k}^{- 1} σ_{k} α_{k + 1} \equiv α_{k + 1}^{- 1} α_{k + 1} \equiv 1.$ This gives (a), (b), (c) and (d).

1.2F1F2

The case $k < j - 1$ : far commutation. Here $j \notin {k, k + 1}$ , so $j^{'} = j$ and $F = α_{j}^{- 1} σ_{k}^{ε} α_{j}$ . Every letter $σ_{r}^{\pm 1}$ occurring in the word $α_{j}$ of [F2] has index $r \geq j \geq k + 2$ , hence $∣ r - k ∣ \geq 2$ and $σ_{k}$ commutes with that letter by the far-commutation relation of [F1]; iterating over the letters of $α_{j}$ (whose length is $n - j$ , possibly $0$ when $j = n$ ), we get $σ_{k}^{ε} α_{j} \equiv α_{j} σ_{k}^{ε}$ , whence $F = α_{j}^{- 1} σ_{k}^{ε} α_{j} \equiv α_{j}^{- 1} α_{j} σ_{k}^{ε} \equiv σ_{k}^{ε}$ by free cancellation. This is (e); note $k \leq j - 2$ and $j \leq n$ , so $k \leq n - 2$ .

1.3F1F2

The case $k > j$ : sliding the letter to the right. Here again $j \notin {k, k + 1}$ , so $j^{'} = j$ and $F = α_{j}^{- 1} σ_{k}^{ε} α_{j}$ . First take $ε = 1$ . Since $j < k \leq n - 1$ , the word $α_{j} = σ_{j} \dots σ_{n - 1}$ splits, with the three groups possibly empty, as the word $α_{j} = (σ_{j} \dots σ_{k - 2}) σ_{k - 1} σ_{k} (σ_{k + 1} \dots σ_{n - 1}),$ where the first group contains exactly the letters with indices $r \leq k - 2$ and the last exactly those with indices $r \geq k + 1$ . Every letter of the first group has $k - r \geq 2$ , so $σ_{k}$ commutes with each of them and moves right past them; every letter of the last group has $r - (k - 1) \geq 2$ , so $σ_{k - 1}$ commutes with each of them and moves right past them; and the three middle letters satisfy the braid relation $σ_{k} σ_{k - 1} σ_{k} \equiv σ_{k - 1} σ_{k} σ_{k - 1}$ by [F1]. Combining the three moves gives the chain $σ_{k} α_{j} \equiv (σ_{j} \dots σ_{k - 2}) σ_{k} σ_{k - 1} σ_{k} (σ_{k + 1} \dots σ_{n - 1}) \equiv (σ_{j} \dots σ_{k - 2}) σ_{k - 1} σ_{k} σ_{k - 1} (σ_{k + 1} \dots σ_{n - 1}) \equiv (σ_{j} \dots σ_{k - 2}) σ_{k - 1} σ_{k} (σ_{k + 1} \dots σ_{n - 1}) σ_{k - 1} = α_{j} σ_{k - 1} .$ For $ε = - 1$ , left-multiply this identity in the group $B_{n}$ by $σ_{k}^{- 1}$ : it becomes $α_{j} = σ_{k}^{- 1} α_{j} σ_{k - 1}$ , hence $σ_{k}^{- 1} α_{j} = α_{j} σ_{k - 1}^{- 1}$ . In both signs, therefore, $σ_{k}^{ε} α_{j} \equiv α_{j} σ_{k - 1}^{ε}$ and $F = α_{j}^{- 1} σ_{k}^{ε} α_{j} \equiv α_{j}^{- 1} α_{j} σ_{k - 1}^{ε} \equiv σ_{k - 1}^{ε}$ by free cancellation. This is (f); here $k > j \geq 1$ gives $k - 1 \geq 1$ and $k \leq n - 1$ gives $k - 1 \leq n - 2$ .

2.1F1F4step 1.1step 1.2step 1.3

Transfer to the geometric braid group. Since $φ$ is a homomorphism with $φ (σ_{r}) = [σ_{r}]$ by [F4], applying $φ$ to each of the reductions of steps 1.1, 1.2 and 1.3 turns it into the corresponding identity in $G_{n}$ : the free cancellations become $[γ] [γ]^{- 1} = 1$ , and the two Artin relations used are the geometric relations supplied by the published The geometric three strand braid relation and Far commutativity of elementary geometric half twists.

3.1step 1.1step 1.2step 1.3step 2.1

Conclusion. The conditions of the six cases are exactly: $j = k$ (with either sign), $j = k + 1$ (with either sign), $j \notin {k, k + 1}$ with $k \leq j - 2$ , and $j \notin {k, k + 1}$ with $k \geq j + 1$ ; if $j \notin {k, k + 1}$ then either $j > k + 1$ , that is $k < j - 1$ , or $j < k$ , that is $k > j$ , so the list is exhaustive, and the conditions are visibly mutually exclusive. Steps 1.1, 1.2 and 1.3 establish the reductions (a)-(f), and the reduced forms are the empty word, $x_{k}^{\pm 1}$ , or $σ_{k}^{ε}$ with $k \leq n - 2$ , or $σ_{k - 1}^{ε}$ with $k - 1 \leq n - 2$ , so all of them are words in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}, x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ ; in particular no copy of $σ_{n - 1}$ survives the reduction outside an $x$ -letter. Step 2.1 transfers each reduction to $G_{n}$ . ∎

Remarks

The case list is exactly the source's list for the factors $(σ_{n - 1}^{- 1} \dots σ_{i}^{- 1}) σ_{k}^{\pm 1} (σ_{i} \dots σ_{n - 1})$ , written with the library's first-letter-first convention; the slide identity $σ_{k} α_{j} \equiv α_{j} σ_{k - 1}$ is the source's displayed relation (3.2), and it is the only place where the braid relation is used in cases (e) and (f).
Cases (a)-(f) are the mechanism by which a combing factor that meets the trivial point either disappears, becomes an $x$ -letter, or degenerates to a letter of rank at most $n - 2$ ; the surviving lower-rank letters are collected to the right of the $x$ -letters by Lower-rank Artin letters conjugate x-letters.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Every trivial braid word combs as W_1W_2

Statement

Assume $n \geq 2$ , and let $W$ be a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 1}^{\pm 1}$ whose geometric image under the surjection $φ$ of The Artin presentation surjects onto the geometric braid group is the trivial geometric braid. Then $W$ is equivalent to a product $W_{1} W_{2}$ , using only the two Artin relations and free insertions and deletions of adjacent inverse pairs $σ^{\pm 1} σ^{\mp 1}$ , in which

$W_{1}$ is a word in the $x$ -letters $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ of The Zariski combing words alpha_i and x_i in the Artin presentation, and
$W_{2}$ is a word in the lower-rank letters $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ .

Both $W_{1}$ and $W_{2}$ may be empty, and no letter $σ_{n - 1}$ or $σ_{n - 1}^{- 1}$ occurs in either of the two words.

Facts & Assumptions

Given: An integer $n \geq 2$ , the group $B_{n} = ⟨ σ_{1}, \dots, σ_{n - 1} ⟩$ of The braid group by Artin presentation, a word $W$ in the Artin letters and their inverses with $φ (W) = 1$ , and the words $α_{1}, \dots, α_{n}$ and $x_{1}, \dots, x_{n - 1}$ of The Zariski combing words alpha_i and x_i in the Artin presentation.

[F1]

In $B_{n}$ the two families of defining relations hold, and two words differing by insertions or deletions of adjacent inverse pairs $w^{\pm 1} w^{\mp 1}$ represent the same element; write $u \equiv v$ when the words $u, v$ can be connected by these moves and the two relation families (The braid group by Artin presentation). The moves are symmetric, so $\equiv$ is an equivalence relation, and it is compatible with concatenation in the sense that $u \equiv v$ implies $p u t \equiv p v t$ for words $p, t$ .

[F2]

$φ ⁣ : B_{n} \to G_{n}$ is the surjective homomorphism of The Artin presentation surjects onto the geometric braid group with $φ (σ_{i}) = [σ_{i}]$ .

[F3]

Prefix insertion (Prefix insertion rewrites a trivial braid word into combing factors): if $W = σ_{i_{1}}^{ε_{1}} \dots σ_{i_{m}}^{ε_{m}}$ and $φ (W) = 1$ , then there are positions $j_{0}, j_{1}, \dots, j_{m} \in {1, \dots, n}$ with $j_{0} = j_{m} = n$ and $j_{k} = s_{i_{k}} (j_{k - 1})$ such that $W \equiv \prod_{k = 1}^{m} (α_{j_{k - 1}}^{- 1} σ_{i_{k}}^{ε_{k}} α_{j_{k}})$ , where each factor is a combing factor in the sense of Each combing factor reduces to a lower-rank letter or an x-letter with $j = j_{k - 1}$ , $k \mapsto i_{k}$ , $ε \mapsto ε_{k}$ and $j^{'} = j_{k}$ .

[F4]

Each combing factor reduces to a word of at most one letter in the mixed alphabet $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}, x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ : the empty word, or $x_{i_{k}}^{\pm 1}$ , or $σ_{i_{k}}^{ε_{k}}$ with $i_{k} \leq n - 2$ , or $σ_{i_{k} - 1}^{ε_{k}}$ with $i_{k} - 1 \leq n - 2$ (Each combing factor reduces to a lower-rank letter or an x-letter).

[F5]

Conjugation table (Lower-rank Artin letters conjugate x-letters): for every $1 \leq i \leq n - 2$ , $1 \leq j \leq n - 1$ and signs $ε, δ \in {\pm 1}$ there is a word $w$ in the letters $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ with $σ_{i}^{ε} x_{j}^{δ} = w σ_{i}^{ε}$ in $B_{n}$ . In particular a contiguous pair consisting of a lower-rank $σ$ -letter followed immediately by an $x$ -letter can be replaced by a word of $x$ -letters followed by that same $σ$ -letter.

Proof

technique · direct

1.1F2F3

Prefix insertion. By [F3] and $φ (W) = 1$ , there are positions $j_{0} = j_{m} = n$ with $j_{k} = s_{i_{k}} (j_{k - 1})$ and $W \equiv \prod_{k = 1}^{m} F_{k}$ , $F_{k} : = α_{j_{k - 1}}^{- 1} σ_{i_{k}}^{ε_{k}} α_{j_{k}}$ .

1.2F1F4

Reducing the factors. Fix $k$ and apply [F4] to $F_{k}$ , whose letter has index $i_{k}$ and whose connector position is $j_{k - 1}$ with $j_{k} = s_{i_{k}} (j_{k - 1})$ ; the factor is equivalent to the empty word, or to a single letter $x_{i_{k}}$ or $x_{i_{k}}^{- 1}$ , or to a single letter $σ_{i_{k}}^{ε_{k}}$ with $i_{k} \leq n - 2$ , or to a single letter $σ_{i_{k} - 1}^{ε_{k}}$ with $i_{k} - 1 \leq n - 2$ . Deleting the factors that reduce to the empty word and choosing one such reduced word in each remaining factor, we obtain a word $U$ in the mixed alphabet $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}, x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ with $W \equiv U$ , because $\equiv$ is compatible with concatenation by [F1].

1.3F1F5

Collection of the lower-rank letters. We show: for every word $U$ in the mixed alphabet $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}, x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ there are a word $u$ in the $x$ -letters and a word $v$ in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ with $U \equiv u v$ . Proceed by induction on the number $r$ of $σ$ -letters occurring in $U$ . If $r = 0$ , take $u : = U$ and $v$ empty. If $r \geq 1$ , let $s$ be the last (rightmost) $σ$ -letter of $U$ and write $U = P s A$ , where $A$ is the (possibly empty) $x$ -word following $s$ , so that no $σ$ -letter occurs in $A$ . While $A$ is nonempty, let $x$ be its first letter and replace the adjacent pair $s x$ by $w s$ , where $s = σ_{i}^{ε}$ with $i \leq n - 2$ , $x = x_{j}^{δ}$ and $σ_{i}^{ε} x_{j}^{δ} = w σ_{i}^{ε}$ is the identity of [F5]; this is a permitted rewrite, and the new word again has $s$ as its rightmost $σ$ -letter, now followed by $A$ with its first letter deleted, because the $x$ -word $w$ stands immediately to the left of $s$ . Hence the number of letters strictly to the right of $s$ decreases by exactly one at each rewrite, so after finitely many steps we obtain a word $U^{'} \equiv U$ whose letters strictly to the right of the rightmost $σ$ -letter $s$ are none, that is, $U^{'} = P^{'} s$ where $P^{'}$ is a word in the mixed alphabet with exactly $r - 1$ $σ$ -letters. By the induction hypothesis applied to $P^{'}$ , there are an $x$ -word $u^{'}$ and a $σ$ -word $v^{'}$ with $P^{'} \equiv u^{'} v^{'}$ ; then $U \equiv U^{'} \equiv u^{'} (v^{'} s)$ by [F1], where $u^{'}$ is an $x$ -word and $v^{'} s$ is a word in the $σ$ -letters of rank at most $n - 2$ , so the induction is complete.

2.1F1step 1.2step 1.3

Assembly. By step 1.2 there is a mixed word $U$ with $W \equiv U$ , and by step 1.3 there are an $x$ -word $W_{1}$ and a lower-rank $σ$ -word $W_{2}$ with $U \equiv W_{1} W_{2}$ ; since $\equiv$ is transitive by [F1], $W \equiv W_{1} W_{2}$ , which is the required product.

3.1step 1.1step 1.2step 1.3step 2.1

Conclusion. Given $W$ with $φ (W) = 1$ , steps 1.1-1.3 rewrite it, using only the two Artin relations and free insertions and deletions of adjacent inverse pairs, first into the product of its combing factors, then into a word $U$ in the mixed alphabet, and finally into a product $W_{1} W_{2}$ with $W_{1}$ a word in $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ and $W_{2}$ a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ ; the name $n - 2$ in the statement is justified because [F4] bounds every surviving $σ$ -index by $n - 2$ . ∎

Remarks

The collection step is the source's "we can collect all the $σ_{i}^{\pm 1}$ on the right". The naive measure "number of pairs (an $x$ -letter left of a $σ$ -letter)" is not monotone, because a pair $s x$ can be replaced by a word $w s$ in which $w$ has up to three letters, for instance $σ_{i} x_{i} = x_{i}^{- 1} x_{i + 1} x_{i} σ_{i}$ ; the proof above instead processes the $σ$ -letters from right to left, and each swap strictly shortens the segment to the right of the processed letter.
Only the two Artin relations, the conjugation table of Lower-rank Artin letters conjugate x-letters, and the tracking of Prefix insertion rewrites a trivial braid word into combing factors are used; in particular the geometric input is only $φ (W) = 1$ .
For $n = 2$ the mixed alphabet contains no $σ$ -letters of rank at most $0$ , so the conclusion reads $W \equiv W_{1}$ with $W_{1}$ a word in $x_{1}^{\pm 1}$ ; the lemma is not used to determine how many $x_{1}$ -factors occur, and the induction of the completeness theorem below supplies that in the trivial case.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

The combed geometric decomposition is unique

Statement

Assume AC and $n \geq 2$ , and work with the abstract Artin group $B_{n}$ of The braid group by Artin presentation, the words $x_{1}, \dots, x_{n - 1}$ of The Zariski combing words alpha_i and x_i in the Artin presentation and the published surjection $φ ⁣ : B_{n} \to G_{n}$ of The Artin presentation surjects onto the geometric braid group. Let $W_{1}$ be a word in $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ and $W_{2}$ a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ such that $φ (W_{1} W_{2})$ is the trivial geometric braid; such a pair exists for every combed word by Every trivial braid word combs as W_1W_2. Write $Ψ_{n} ⁣ : G_{n}^{p u r e} \to P B_{n}$ for the canonical isomorphism of Pure geometric braids and ordered configuration loops, and $ρ ⁣ : P B_{n} \to P B_{n - 1}$ for the forgetting map of The Fadell-Neuwirth short exact sequence for pure braids. Write $A_{i j} \in G_{n}$ for the geometric generators of Standard geometric pure braid generators A_ij, and write ${\tilde{A}}_{i j} : = σ_{j - 1} \dots σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} \dots σ_{j - 1}^{- 1} \in B_{n}$ for their Artin-word lifts, so $φ ({\tilde{A}}_{i j}) = A_{i j}$ . Put ${\tilde{P}}_{i} : = {\tilde{A}}_{i + 1, n} \dots {\tilde{A}}_{n - 1, n}$ in $B_{n}$ and $P_{i} : = A_{i + 1, n} \dots A_{n - 1, n}$ in $G_{n}$ , with both empty when $i = n - 1$ . Then:

(i) every $φ (x_{i})$ is a pure braid and $φ (x_{i}) = P_{i}^{- 1} A_{i n} P_{i},$ so that the conjugating element lies in the subgroup generated by the standard generators with larger first index; at the word level this is $x_{i} \equiv {\tilde{P}}_{i}^{- 1} {\tilde{A}}_{i n} {\tilde{P}}_{i}$ in $B_{n}$ . Moreover $Ψ_{n} (φ (x_{1})), \dots, Ψ_{n} (φ (x_{n - 1}))$ form a free basis of the free kernel $\ker ρ$ of the forgetting map of The Fadell-Neuwirth short exact sequence for pure braids;

(ii) $φ (W_{1}) = 1$ , and $W_{1}$ , as a word in the alphabet $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ , reduces to the empty word by free cancellations of adjacent inverse pairs $x_{i}^{\pm 1} x_{i}^{\mp 1}$ (each of which, after expanding both $x_{i}$ 's, is a permitted deletion of $σ$ -pairs);

(iii) $φ (W_{2}) = 1$ , and, viewing the same letter word as an element of $B_{n - 1}$ , its geometric image under the rank- $(n - 1)$ surjection $φ_{n - 1} ⁣ : B_{n - 1} \to G_{n - 1}$ is trivial: $φ_{n - 1} (W_{2}) = 1 \in G_{n - 1}$ . This asserts triviality of the geometric image, not that $W_{2} = 1$ in $B_{n - 1}$ .

Facts & Assumptions

Given: An integer $n \geq 2$ , the groups $B_{n}$ and $G_{n}$ of The braid group by Artin presentation and The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, the words $α_{j}, x_{j}$ of The Zariski combing words alpha_i and x_i in the Artin presentation, words $W_{1}$ in $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ and $W_{2}$ in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ with $φ (W_{1} W_{2}) = 1$ , the standard pure braid generators $A_{i j}$ of Standard geometric pure braid generators A_ij, and the identifications of The Fadell-Neuwirth short exact sequence for pure braids and Pure geometric braids and ordered configuration loops.

[F1]

In $B_{n}$ the two Artin relations hold and adjacent inverse $σ$ -pairs may be freely inserted and deleted. For $1 \leq i < j \leq n$ , set ${\tilde{A}}_{i j} : = σ_{j - 1} \dots σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} \dots σ_{j - 1}^{- 1} \in B_{n}$ ; its image under $φ$ is the geometric standard generator $A_{i j}$ of Standard geometric pure braid generators A_ij, since $φ (σ_{r}) = [σ_{r}]$ and $φ$ preserves the displayed stacking product (The braid group by Artin presentation, The Artin presentation surjects onto the geometric braid group, Standard geometric pure braid generators A_ij). The words $α_{j}$ satisfy $α_{j} = σ_{j} \dots σ_{n - 1}$ and $α_{n} = 1$ , and $x_{i} = α_{i + 1}^{- 1} σ_{i}^{2} α_{i + 1}$ (The Zariski combing words alpha_i and x_i in the Artin presentation).

[F2]

Assumption AC, and the published consequences that are used to identify the free kernel: AC implies dependent choice and countable choice (The Axiom of Choice, AC implies DC implies countable choice), so that the Fadell--Neuwirth fibrations of The Fadell-Neuwirth short exact sequence for pure braids are available: forgetting the last strand is a homomorphism $ρ ⁣ : P B_{n} \to P B_{n - 1}$ with kernel a free group of rank $n - 1$ , and under the fiber-inclusion identification the elements $Ψ_{n} (A_{1 n}), \dots, Ψ_{n} (A_{n - 1, n})$ are a free basis of that kernel (The $A_{i n}$ are meridian generators of the forgetful free kernel). Forgetting a strand is realized by the coordinate projection: under the identifications [F3] at ranks $n$ and $n - 1$ , a pure braid whose coordinate loop is $(Z_{1}, \dots, Z_{n})$ satisfies $ρ (Ψ_{n} ([β])) = (ι_{*}^{F} [(Z_{1}, \dots, Z_{n - 1})])^{- 1}$ , with target basepoint $(q_{1}, \dots, q_{n - 1})$ . This follows because coordinate projection commutes with the open-to-closed inclusion and its induced homomorphism preserves inverses.

[F3]

For every rank $m \geq 0$ , $Ψ_{m} ⁣ : G_{m}^{p u r e} = \ker π_{g e o} \to P B_{m}$ is an isomorphism, and for a pure braid $[β]$ with coordinate path $z_{β}$ it is $Ψ_{m} ([β]) = (ι_{*}^{F} [z_{β}])^{- 1}$ , the inverse of the class of the coordinate loop carried to the closed-disc configuration space; the same sign convention is used at every rank, so the inverse cancels in the comparisons between ranks below. At every interior base configuration the open-to-closed inclusion induces an isomorphism on fundamental groups (The interior-disc and closed-disc configuration spaces are homotopy equivalent). Consequently $[β] = 1$ if and only if $[z_{β}] = 1$ (Pure geometric braids and ordered configuration loops, Ordered configuration spaces $F_{n} (X)$ , The homomorphism on fundamental groups induced by a pointed continuous map). Moreover $π_{c o n f} (Φ ([β])) = π_{g e o} ([β])$ , so a geometric braid is pure exactly when its class lies in $P B_{n}$ under these identifications (The geometric endpoint permutation matches covering monodromy).

[F4]

Geometric conventions. The base configuration is $Q = (q_{1}, \dots, q_{n})$ with $q_{j} = ((2 j - n - 1) h, 0)$ , $h = \frac{1}{4 (n + 1)}$ , and the (n−1)-strand base configuration is $Q^{(n - 1)} = (q_{1}^{'}, \dots, q_{n - 1}^{'})$ with $q_{j}^{'} = ((2 j - n) h^{'}, 0)$ , $h^{'} = \frac{1}{4 n}$ (Geometric braids in the disc with setwise endpoints). The elementary half twist $σ_{i}$ acts as $m_{i} + ρ (t), m_{i} - ρ (t)$ on the two points $q_{i}, q_{i + 1}$ and fixes all others, where $m_{i} = q_{i} + (h, 0)$ and $ρ$ is the diamond path of size $h$ (The elementary geometric half twist, its support disc, and its opposite); stacking is first-under-second with $π (γ ⋆ β) = π (γ) \circ π (β)$ (Stacking of geometric braids is a well-defined associative operation on isotopy classes, The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism). The two configurations are compared through the similarity $A (w) : = a w - \frac{1}{4 (n + 1)}$ with $a : = \frac{n}{n + 1}$ . Since $\frac{1}{4 (n + 1)} = \frac{1 - a}{4}$ , this is the dilation about $c = - \frac{1}{4}$ , namely $A (w) = c + a (w - c)$ . It satisfies $A (Q^{(n - 1)}) = (q_{1}, \dots, q_{n - 1})$ ; its image $A (D^{\circ})$ is the open disc of radius $a$ centered at $A (0) = - \frac{1}{4 (n + 1)}$ , whose closure lies in $D^{\circ}$ because $a + \frac{1}{4 (n + 1)} < 1$ .

[F5]

Moving-homotopy transport: for a path $γ ⁣ : x_{0} \to x_{1}$ in a space $X$ and each $n \geq 1$ there is a transport isomorphism $β_{γ} ⁣ : π_{n} (X, x_{1}) \to π_{n} (X, x_{0})$ depending only on the endpoint-fixed class of $γ$ ; if $H ⁣ : f ≃ g$ is a homotopy with basepoint track $γ$ , then $f_{*} = β_{γ} \circ g_{*}$ . In degree one $β_{γ} [a] = [γ * a * \overset{ˉ}{γ}]$ (Higher homotopy basepoint transport and moving homotopies).

Proof

technique · direct

1.1F1

The combing identity, by free cancellation. For $1 \leq j \leq n - 1$ put $δ_{j} : = σ_{n - 1} σ_{n - 2} \dots σ_{j + 1}$ and $γ_{j} : = σ_{j + 1} σ_{j + 2} \dots σ_{n - 1}$ (empty words for $j = n - 1$ ); then $δ_{j} = δ_{j + 1} σ_{j + 1}$ and $γ_{j} = σ_{j + 1} γ_{j + 1}$ as words, and by [F1] the Artin word ${\tilde{A}}_{j n} = δ_{j} σ_{j}^{2} δ_{j}^{- 1}$ lifts the geometric generator $A_{j n}$ , while $x_{j} = γ_{j}^{- 1} σ_{j}^{2} γ_{j}$ is the combing word. First, for every $j$ the Artin word ${\tilde{Q}}_{j} : = {\tilde{A}}_{j, n} {\tilde{A}}_{j + 1, n} \dots {\tilde{A}}_{n - 1, n}$ freely reduces to $δ_{j} σ_{j}^{2} γ_{j}$ : this is a downward induction on $j$ , in which both words are $σ_{n - 1}^{2}$ for $j = n - 1$ , and in which the substitution of ${\tilde{A}}_{j, n} = δ_{j} σ_{j}^{2} δ_{j}^{- 1}$ and of the reduced form ${\tilde{Q}}_{j + 1} \equiv δ_{j + 1} σ_{j + 1}^{2} γ_{j + 1}$ gives ${\tilde{Q}}_{j} \equiv δ_{j} σ_{j}^{2} (δ_{j}^{- 1} δ_{j + 1}) σ_{j + 1}^{2} γ_{j + 1} = δ_{j} σ_{j}^{2} σ_{j + 1}^{- 1} (δ_{j + 1}^{- 1} δ_{j + 1}) σ_{j + 1}^{2} γ_{j + 1},$ where the block $δ_{j + 1}^{- 1} δ_{j + 1}$ cancels freely to the empty word -- its middle pair $σ_{n - 1}^{- 1} σ_{n - 1}$ cancels first, after which the next pair is adjacent, and so on outwards -- leaving $δ_{j} σ_{j}^{2} γ_{j}$ . For $1 \leq i \leq n - 2$ this gives ${\tilde{P}}_{i} : = {\tilde{Q}}_{i + 1} \equiv δ_{i + 1} σ_{i + 1}^{2} γ_{i + 1}$ , and the inverse word is ${\tilde{P}}_{i}^{- 1} \equiv γ_{i + 1}^{- 1} σ_{i + 1}^{- 2} δ_{i + 1}^{- 1}$ ; substituting these reduced forms into ${\tilde{P}}_{i}^{- 1} {\tilde{A}}_{i n} {\tilde{P}}_{i}$ , and using $δ_{i + 1}^{- 1} δ_{i} \equiv σ_{i + 1}$ and $δ_{i}^{- 1} δ_{i + 1} \equiv σ_{i + 1}^{- 1}$ (the same middle-outward cancellation applied to $δ_{i} = δ_{i + 1} σ_{i + 1}$ ), gives ${\tilde{P}}_{i}^{- 1} {\tilde{A}}_{i n} {\tilde{P}}_{i} \equiv γ_{i + 1}^{- 1} σ_{i + 1}^{- 2} (δ_{i + 1}^{- 1} δ_{i}) σ_{i}^{2} (δ_{i}^{- 1} δ_{i + 1}) σ_{i + 1}^{2} γ_{i + 1} \equiv γ_{i + 1}^{- 1} σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} γ_{i + 1} = γ_{i}^{- 1} σ_{i}^{2} γ_{i} = x_{i} .$ The remaining index $i = n - 1$ has the empty product ${\tilde{P}}_{n - 1}$ and $x_{n - 1} = {\tilde{A}}_{n - 1, n} = σ_{n - 1}^{2}$ , so $x_{n - 1} = {\tilde{P}}_{n - 1}^{- 1} {\tilde{A}}_{n - 1, n} {\tilde{P}}_{n - 1}$ holds trivially. No Artin relation is used in this step, only free insertions and deletions. Applying $φ$ gives the stated geometric identity.

1.2F4

The two models of a lower-rank word. Put $a : = n / (n + 1)$ and consider the similarity $A (w) = a w - h_{n}$ with $h_{n} = \frac{1}{4 (n + 1)}$ , so that $A (w) = a w - \frac{1}{4 (n + 1)}$ . One computes $A (q_{j}^{'}) = q_{j}$ for $1 \leq j \leq n - 1$ , $A (m_{i}^{'}) = m_{i}$ for $1 \leq i \leq n - 2$ where $m_{i}^{'} = q_{i}^{'} + (h^{'}, 0)$ is the (n−1)-strand midpoint, and $a ρ^{'} (t) = ρ (t)$ for the two displacement paths, since $a h^{'} = h$ and each diamond-path coordinate is linear in its size. The reflected paths for the negative letters satisfy the same scaling identity. Since $A$ is affine, $A (m_{i}^{'} \pm ρ^{'} (t)) = A (m_{i}^{'}) \pm a ρ^{'} (t) = m_{i} \pm ρ (t)$ . Thus for every $1 \leq i \leq n - 2$ and both signs, $A ((σ_{i}^{(n - 1)})_{j} (t)) = (σ_{i}^{(n)})_{j} (t) (1 \leq j \leq n - 1), A (q_{j}^{'}) = q_{j},$ where $σ_{i}^{(k)}$ is the elementary half twist of the $k$ -strand model of [F4]. Every lower-rank moving point stays in the support disc $U_{i}^{'}$ of the letter currently being run, or is one of the fixed base points. Relative to $c : = (- \frac{1}{4}, 0)$ , these base points have radii $∣ q_{j}^{'} - c ∣ = j / (2 n) \leq (n - 1) / (2 n)$ , and the support disc $U_{i}^{'}$ has center $m_{i}^{'}$ with $∣ m_{i}^{'} - c ∣ = (2 i + 1) / (4 n)$ and radius $3 / (8 n)$ ; hence every such path point has radius at most $(4 n - 3) / (8 n)$ . This local agreement is all that will be used: $A$ maps the lower standard-generator paths to the corresponding $n$ -strand paths, but it is not used to map arbitrary lower braids into the complement of $q_{n}$ . Consequently, for every word $V$ in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ , if $z = (z_{1}, \dots, z_{n - 1})$ is its (n−1)-strand coordinate motion and $Z = (Z_{1}, \dots, Z_{n})$ its n-strand coordinate motion, then $Z_{j} = A \circ z_{j} (1 \leq j \leq n - 1), Z_{n} \equiv q_{n},$ by induction on its letters: the generator paths agree by the displayed identity, and stacking has the same coupling permutation on ${1, \dots, n - 1}$ in both models and fixes $n$ . Thus $φ_{n} (V)$ is represented by $(A \circ z_{1}, \dots, A \circ z_{n - 1}, q_{n})$ .

2.1F2F3F4F5

The straight-strand extension is a well-defined injective homomorphism. Let $c : = (- \frac{1}{4}, 0)$ , $a : = n / (n + 1)$ , $R_{0} : = \frac{1}{2} - \frac{1}{4 n}$ and $r_{n} : = n / (2 (n + 1))$ . For $θ \in R$ let $u_{θ} : = (\cos θ, \sin θ)$ and let $R (θ) : = \frac{1}{4} \cos θ + \sqrt{1 - \frac{1}{16} \sin^{2} θ},$ the distance from $c$ to the boundary of $D^{\circ}$ along the ray $c + r u_{θ}$ ; in particular $R (θ) \geq \frac{3}{4} > R_{0}$ . Define the strictly increasing radial function $λ (r, θ) : = {\begin{cases} a r, & 0 \leq r \leq R_{0}, \\ a R_{0} + (r_{n} - a R_{0}) \frac{r - R_{0}}{R (θ) - R_{0}}, & R_{0} \leq r < R (θ), \end{cases}$ and define $E (c + r u_{θ}) : = c + λ (r, θ) u_{θ}$ , with $E (c) : = c$ . Since $a R_{0} < r_{n}$ , each ray is mapped increasingly onto the ray segment of length $r_{n}$ ; thus $E$ is a homeomorphism from $D^{\circ}$ onto the open disk $L_{n} : = B (c, r_{n})$ . The closure of $L_{n}$ lies in $D^{\circ}$ because $∥ c ∥_{2} + r_{n} < \frac{3}{4} < 1$ , and $q_{n}$ lies on $\partial L_{n}$ because $∣ q_{n} - c ∣ = n / (2 (n + 1)) = r_{n}$ . Moreover, $E = A$ whenever $∣ w - c ∣ \leq R_{0}$ . The basepoints satisfy $∣ q_{j}^{'} - c ∣ = j / (2 n) \leq (n - 1) / (2 n) < R_{0}$ , and for $n \geq 3$ every lower support disc satisfies $\max_{1 \leq i \leq n - 2} \sup_{w \in U_{i}^{'}} ∣ w - c ∣ \leq \frac{4 n - 3}{8 n} < R_{0}$ (for $n = 2$ there are no such support discs). Therefore $E (Q^{(n - 1)}) = (q_{1}, \dots, q_{n - 1})$ , and [F4] with step 1.2 shows that $E$ maps every lower-rank standard-generator path to its corresponding n-strand path. If $β \in G_{n - 1}^{p u r e}$ with representative coordinate loop $z = (z_{1}, \dots, z_{n - 1})$ , define $\overset{ˇ}{s} (β)$ to be the class of $(E \circ z_{1}, \dots, E \circ z_{n - 1}, q_{n})$ . The tuple is a pure n-strand braid: $E$ is injective, its image $L_{n}$ avoids $q_{n}$ , and the endpoints are $E (Q^{(n - 1)}) = (q_{1}, \dots, q_{n - 1})$ . Applying $E$ to a braid isotopy preserves pairwise distinctness and avoids $q_{n}$ , so $\overset{ˇ}{s}$ is well defined on isotopy classes; applying $E$ to the stacking formula shows it is a homomorphism. To prove injectivity, first note that the induced map $E_{*} : π_{1} (F_{n - 1} (D^{\circ}), Q^{(n - 1)}) \to π_{1} (F_{n - 1} (D^{\circ}), E (Q^{(n - 1)}))$ is injective. Indeed, for $0 \leq t \leq 1$ let $λ_{t} (r, θ) : = (1 - t) r + t λ (r, θ)$ and $E_{t} (c + r u_{θ}) : = c + λ_{t} (r, θ) u_{θ}$ . For each $t$ , $λ_{t}$ is strictly increasing along every ray and, when $r < R (θ)$ , $λ_{t} (r, θ) < (1 - t) R (θ) + t r_{n} \leq R (θ),$ so $E_{t}$ is an embedding $D^{\circ} \to D^{\circ}$ , with $E_{0} = id$ and $E_{1} = E$ . Hence $F_{n - 1} (E_{t})$ is a homotopy on the ordered configuration space from the identity to $F_{n - 1} (E)$ , whose basepoint follows $η (t) : = F_{n - 1} (E_{t}) (Q^{(n - 1)})$ ; by [F5], ${id}_{*} = β_{η} \circ E_{*}$ , and since $β_{η}$ is an isomorphism, $E_{*}$ is injective. If $\overset{ˇ}{s} (β) = 1$ , [F3] and the coordinate-projection description of [F2] imply $E_{*} [z] = 1$ ; injectivity gives $[z] = 1$ , and [F3] implies $β = 1$ .

2.2F2F3step 1.1

The conjugating identity in the geometric group. Applying $φ$ to the word identity of step 1.1 gives $φ (x_{i}) = P_{i}^{- 1} A_{i n} P_{i} in G_{n},$ where $P_{i} = A_{i + 1, n} \dots A_{n - 1, n}$ is the geometric product defined in the Statement. Each factor $A_{j, n}$ of $P_{i}$ and $A_{i n}$ itself lies in the pure subgroup and maps under $Ψ_{n}$ into $\ker ρ$ by [F2, F3]. Since $\ker ρ$ is a subgroup, $Ψ_{n} (φ (x_{i}))$ lies in $\ker ρ$ ; in particular $φ (x_{i})$ is pure.

3.1F2F3step 2.1

The straight-strand extensions meet the kernel trivially. By [F2] the forgetting map acts on coordinate loops by dropping the last coordinate. For the straight-strand extension of step 2.1 the coordinate loop is $(E \circ z_{1}, \dots, E \circ z_{n - 1}, q_{n})$ , whose dropped loop is $E \circ z$ . Thus $ρ (Ψ_{n} (\overset{ˇ}{s} (β))) = ι_{*}^{F} (E_{*} [z])^{- 1}$ at the basepoint $(q_{1}, \dots, q_{n - 1}) = E (Q^{(n - 1)})$ . If $γ = Ψ_{n} (\overset{ˇ}{s} (β))$ also lies in $\ker ρ$ , this class is trivial. The inclusion map $ι_{*}^{F}$ is injective at that basepoint by [F3], so $E_{*} [z] = 1$ . Injectivity of $E_{*}$ (step 2.1) gives $[z] = 1$ , and [F3] gives $β = 1$ and $γ = 1$ . In particular $Ψ_{n} (im \overset{ˇ}{s}) \cap \ker ρ = {1}$ , which is the uniqueness statement used below.

3.2F2step 2.2

The free basis by the left-inverse argument. Let $K : = \ker ρ \leq P B_{n}$ be the free kernel, and let $F : = F (a_{1}, \dots, a_{n - 1})$ be the abstract free group. The map $ȷ : F \to K$ given by $ȷ (a_{j}) : = Ψ_{n} (A_{j n})$ is an isomorphism by the free-basis clause of [F2]. Write $w_{i} : = a_{i + 1} a_{i + 2} \dots a_{n - 1} \in F$ and $t_{i} : = w_{i}^{- 1} a_{i} w_{i} \in F$ . By step 2.2 and the homomorphism property of $Ψ_{n}$ , $ȷ (t_{i}) = Ψ_{n} (φ (x_{i}))$ . First, the elements $t_{1}, \dots, t_{n - 1}$ generate $F$ : downward induction on $i$ shows $⟨ t_{i}, a_{i + 1}, \dots, a_{n - 1} ⟩ = ⟨ a_{i}, \dots, a_{n - 1} ⟩$ , because $a_{i} = w_{i} t_{i} w_{i}^{- 1}$ lies in the left side and $t_{i}$ lies in the right side; at $i = n - 1$ one has $w_{n - 1} = 1$ and $t_{n - 1} = a_{n - 1}$ . Second, they are independent: define an endomorphism $θ$ of $F$ by descending recursion on $i$ by $θ (a_{i}) : = θ (w_{i}) a_{i} θ (w_{i})^{- 1}$ , where $θ (w_{i})$ is the image of the word $w_{i}$ under the already defined $θ$ on the generators $a_{i + 1}, \dots, a_{n - 1}$ (at $i = n - 1$ this gives $θ (a_{n - 1}) = a_{n - 1}$ ); this prescription specifies an element $θ (a_{i}) \in F$ for every free generator, hence defines a unique endomorphism $θ ⁣ : F \to F$ . Then $θ (t_{i}) = θ (w_{i})^{- 1} θ (a_{i}) θ (w_{i}) = θ (w_{i})^{- 1} θ (w_{i}) a_{i} θ (w_{i})^{- 1} θ (w_{i}) = a_{i}$ for every $i$ . If $ρ^{'} ⁣ : F \to F$ denotes the homomorphism with $ρ^{'} (a_{i}) = t_{i}$ , then $θ (ρ^{'} (a_{i})) = θ (t_{i}) = a_{i}$ , so $θ \circ ρ^{'}$ is the identity on a free basis, hence $θ \circ ρ^{'} = {id}_{F}$ and $ρ^{'}$ is injective. Therefore $t_{1}, \dots, t_{n - 1}$ freely generate $F$ , and applying $Ψ_{n}^{- 1} \circ ȷ$ carries this free basis to the free basis $φ (x_{1}), \dots, φ (x_{n - 1})$ of $Ψ_{n}^{- 1} (K)$ .

4.1step 2.2step 3.2step 1.2step 2.1step 3.1

The three claims. (i) is step 2.2 together with step 3.2, the free-basis clause of [F2] being what turns $Ψ_{n} (A_{1 n}), \dots, Ψ_{n} (A_{n - 1, n})$ into a basis of the free kernel. For (ii) and (iii), note first that $φ (W_{1})$ is a product of the pure braids $φ (x_{i})$ of step 2.2 and that $φ (W_{2}) = φ (W_{1})^{- 1}$ because $φ (W_{1} W_{2}) = 1$ ; by (i) the elements $φ (x_{1}), \dots, φ (x_{n - 1})$ generate $\ker ρ$ , so both $φ (W_{1})$ and $φ (W_{2})$ lie in the free kernel. The lower-rank word $W_{2}$ has the same reading in the two models, so the endpoint permutation of its n-strand realization is the identity on the labels $1, \dots, n - 1$ (the last strand is fixed) and the permuted labels $1, \dots, n - 1$ agree with those of its (n-1)-strand realization; since the permutation of $φ (W_{2})$ is trivial, the (n-1)-strand realization is a pure braid and step 1.2 identifies $φ (W_{2}) = \overset{ˇ}{s} (φ_{n - 1} (W_{2}))$ in $G_{n}^{p u r e}$ . Hence $Ψ_{n} (φ (W_{2})) \in Ψ_{n} (im \overset{ˇ}{s}) \cap \ker ρ = {1}$ by step 3.1, so $φ (W_{2}) = 1$ ; then injectivity of $\overset{ˇ}{s}$ (step 2.1) gives $φ_{n - 1} (W_{2}) = 1$ , which is (iii), and $φ (W_{1}) = φ (W_{2})^{- 1} = 1$ . Finally, since $x_{1}, \dots, x_{n - 1}$ are carried to a free basis of $\ker ρ$ by (i), the homomorphism from the free group on the letters $x_{1}, \dots, x_{n - 1}$ to $G_{n}^{p u r e}$ sending $x_{i} \mapsto φ (x_{i})$ is injective; a word in this alphabet whose image is $φ (W_{1}) = 1$ is therefore freely trivial, and each cancellation of an adjacent pair $x_{i}^{\pm 1} x_{i}^{\mp 1}$ expands, after writing out both expanded words $x_{i}^{\pm 1}$ , into a sequence of permitted deletions of adjacent inverse $σ$ -pairs. This is (ii). ∎

Remarks

The load-bearing in-run inputs are the batch-21 items The Fadell-Neuwirth short exact sequence for pure braids, The $A_{i n}$ are meridian generators of the forgetful free kernel and Standard geometric pure braid generators A_ij; steps 2.1 and 3.1 in particular use the convention that the forgetting map is realized by the coordinate projection, which the current statement of The Fadell-Neuwirth short exact sequence for pure braids asserts (the map induced by $(x_{1}, \dots, x_{n}) \mapsto (x_{1}, \dots, x_{n - 1})$ ); the suppliers' certification remains the owner-held obligation before the item is accepted.
The combing identity of step 1.1 is the free-cancellation identity ${\tilde{Q}}_{j} \equiv δ_{j} σ_{j}^{2} γ_{j}$ , hence ${\tilde{P}}_{i} = {\tilde{Q}}_{i + 1} \equiv δ_{i + 1} σ_{i + 1}^{2} γ_{i + 1}$ , and no relation of the Artin presentation is used in it; applying $φ$ gives the geometric conjugation with $P_{i}$ in step 2.2.
The axiom of choice is used only through the batch-21 Fadell--Neuwirth inputs and the identification of the free kernel; the combing identity, the $θ$ argument and the two-model comparison of steps 1.1, 2.1 and 3.1 are choice-free.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

The Artin presentation is complete for geometric braids

Statement

Assume AC. For every $n \geq 1$ the published surjection $φ_{n} ⁣ : B_{n}^{A r t i n} \to B_{n}^{g e o m}$ of The Artin presentation surjects onto the geometric braid group is an isomorphism. Equivalently, every word in $σ_{1}^{\pm 1}, \dots, σ_{n - 1}^{\pm 1}$ whose geometric braid is trivial is equivalent to the empty word using only the two Artin relations and free insertions and deletions of adjacent inverse pairs, so that the Artin presentation of The braid group by Artin presentation is a presentation of the geometric braid group.

Facts & Assumptions

Given: A natural number $n \geq 1$ ; the abstract Artin group $B_{n}^{A r t i n} = ⟨ σ_{1}, \dots, σ_{n - 1} ∣ σ_{i} σ_{i + 1} σ_{i} = σ_{i + 1} σ_{i} σ_{i + 1} (1 \leq i \leq n - 2), σ_{i} σ_{j} = σ_{j} σ_{i} (∣ i - j ∣ > 1) ⟩$ of The braid group by Artin presentation, trivial for $n \leq 1$ ; the geometric braid group $B_{n}^{g e o m} = G_{n}$ of The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism; and the surjective homomorphism $φ_{n}$ of The Artin presentation surjects onto the geometric braid group, which sends the abstract letter $σ_{i}$ to the class of the elementary geometric half twist.

[F1]

Permitted moves. Two words in the letters $σ_{1}^{\pm 1}, \dots, σ_{n - 1}^{\pm 1}$ are called equivalent when one can be obtained from the other by a finite sequence of the following operations: replacing a subword $σ_{i} σ_{i + 1} σ_{i}$ by $σ_{i + 1} σ_{i} σ_{i + 1}$ or conversely; replacing a subword $σ_{i} σ_{j}$ by $σ_{j} σ_{i}$ or conversely when $∣ i - j ∣ > 1$ ; and inserting or deleting a subword $σ_{i}^{ϵ} σ_{i}^{- ϵ}$ . Equivalence is an equivalence relation compatible with concatenation, and equivalent words represent the same element of $B_{n}^{A r t i n}$ and, through $φ_{n}$ , the same geometric braid. (The braid group by Artin presentation, Group presentation by generators and relations.)

[F2]

Combing. Assume $n \geq 2$ and let $W$ be a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 1}^{\pm 1}$ whose image under $φ_{n}$ is the trivial geometric braid. Then $W$ is equivalent, by the permitted moves of [F1], to a product $W_{1} W_{2}$ in which $W_{1}$ is a word in $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ and $W_{2}$ is a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ , where $x_{j} = α_{j + 1}^{- 1} σ_{j}^{2} α_{j + 1}$ are the combing words of The Zariski combing words alpha_i and x_i in the Artin presentation (Every trivial braid word combs as W_1W_2).

[F3]

Uniqueness. Assume AC, $n \geq 2$ , and that $W_{1}$ is a word in $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ and $W_{2}$ a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ with $φ_{n} (W_{1} W_{2}) = 1$ . Then $φ_{n} (W_{1}) = 1$ , the word $W_{1}$ reduces to the empty word by free cancellations of adjacent inverse pairs $x_{j}^{\pm 1} x_{j}^{\mp 1}$ , each of which expands into permitted deletions of $σ$ -pairs; and $φ_{n - 1} (W_{2}) = 1$ , where $φ_{n - 1} ⁣ : B_{n - 1}^{A r t i n} \to B_{n - 1}^{g e o m}$ is the rank $n - 1$ surjection applied to the same word read on $n - 1$ strands (The combed geometric decomposition is unique).

[F4]

AC holds, and AC implies dependent choice and countable choice (The Axiom of Choice, AC implies DC implies countable choice); this is the hypothesis under which [F3] is available. The free kernel of the forgetting map $P B_{n} \to P B_{n - 1}$ is free with basis $A_{1 n}, \dots, A_{n - 1, n}$ for the standard pure braid generators $A_{i j}$ of Standard geometric pure braid generators A_ij (The $A_{i n}$ are meridian generators of the forgetful free kernel, The Fadell-Neuwirth short exact sequence for pure braids), and $Ψ_{m} ⁣ : G_{m}^{p u r e} \to P B_{m}$ is the canonical isomorphism at every rank $m$ (Pure geometric braids and ordered configuration loops).

Proof

technique · direct induction on $n$

1.1F1

Base case. For $n = 1$ there is no index $i$ with $1 \leq i \leq n - 1$ , so the only word in the displayed alphabet is the empty word, and it is equivalent to itself by the empty sequence of permitted moves; the claim holds at $n = 1$ .

2.1F2F3step 1.1

Induction step. Assume $n \geq 2$ , that the claim holds at rank $n - 1$ , and let $W$ be a word in $σ_{1}^{\pm 1}, \dots, σ_{n - 1}^{\pm 1}$ with $φ_{n} (W) = 1$ . By [F2] the word $W$ is equivalent to a product $W_{1} W_{2}$ with $W_{1}$ in the $x$ -letters $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ and $W_{2}$ in the lower-rank letters $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ ; since equivalence is compatible with concatenation and does not change the geometric braid, $φ_{n} (W_{1} W_{2}) = φ_{n} (W) = 1$ . By [F3] applied to the pair $(W_{1}, W_{2})$ , the word $W_{1}$ reduces to the empty word by free cancellations of adjacent inverse $x$ -pairs, each of which expands into permitted deletions of $σ$ -pairs, so $W_{1}$ is equivalent to the empty word by the moves of [F1]; and $φ_{n - 1} (W_{2}) = 1$ . The word $W_{2}$ lies in the alphabet $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}$ of the rank $n - 1$ presentation, so the induction hypothesis applies to it: $W_{2}$ is equivalent to the empty word using the rank $n - 1$ moves. Every rank $n - 1$ move is also a permitted rank $n$ move of [F1], because the generators $σ_{1}, \dots, σ_{n - 2}$ with the braid and far-commutation relations among them are part of the rank $n$ presentation, and the intermediate free insertions and deletions are the same operation. Hence $W \equiv W_{1} W_{2} \equiv W_{2} \equiv 1$ , so the claim holds at rank $n$ .

3.1F4step 2.1

Conclusion. By steps 1.1 and 2.1, for every $n \geq 1$ each word in $σ_{1}^{\pm 1}, \dots, σ_{n - 1}^{\pm 1}$ whose image under $φ_{n}$ is trivial is equivalent to the empty word; since equivalent words represent the same element of $B_{n}^{A r t i n}$ , the kernel of $φ_{n}$ is trivial. The published proposition gives that $φ_{n}$ is surjective, so $φ_{n}$ is an isomorphism of groups. ∎

Remarks

The first nontrivial rank is $n = 2$ : there the free kernel of the forgetting map $P B_{2} \to P B_{1}$ is all of $P B_{2}$ , freely generated by the single standard generator $A_{12} = σ_{1}^{2}$ by The $A_{i n}$ are meridian generators of the forgetful free kernel, and $x_{1} = σ_{1}^{2}$ with $P_{1}$ the empty product; the uniqueness clause of [F3] therefore has content already at $n = 2$ , where it says that a word in $x_{1}^{\pm 1}$ with trivial geometric image is freely trivial.
No injectivity of any Artin presentation is assumed anywhere: the induction reduces words in the kernel to the empty word, and injectivity is a conclusion. The only use of AC is through [F3], whose suppliers invoke dependent and countable choice.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

All four classical braid models realize the Artin presentation

Statement

Assume AC and $n \geq 1$ . Write $B_{n}^{A r t i n}$ for the Artin-presentation group of The braid group by Artin presentation, $G_{n} = B_{n}^{g e o m}$ for the geometric braid group at the base tuple $Q_{n}$ of The elementary geometric half twist, its support disc, and its opposite, $B_{n}^{c o n f} : = π_{1} (C_{n} (D^{2}), [Q_{n}]), π_{1} (C_{n} (int D^{2}), [Q_{n}])$ for the unordered configuration-space fundamental groups of Unordered configuration spaces $C_{n} (X)$ at the same base configuration, the second identified with the first by the open-to-closed inclusion, and $Mod (D^{2}, Q_{n}; \partial D^{2})$ for the boundary-fixed punctured-disk mapping class group of Boundary-fixed mapping class group of a punctured disk. Then:

the four models $B_{n}^{A r t i n}$ , $G_{n}$ , $B_{n}^{c o n f} ≅ π_{1} (C_{n} (int D^{2}), [Q_{n}])$ and $Mod (D^{2}, Q_{n}; \partial D^{2})$ are pairwise connected by the canonical isomorphisms: the completeness isomorphism $φ_{n}$ of The Artin presentation is complete for geometric braids, the published inverse-loop isomorphism of The geometric and configuration braid models agree at the fixed base configuration, the open-to-closed identification, and the AC-dependent boundary-fixed mapping-class isomorphism of Braid group as boundary-fixed punctured-disk mapping classes (an in-run batch-20 scaffold, not a published supplier);
under these identifications, for every $1 \leq i \leq n - 1$ the Artin generator $σ_{i}$ corresponds to the class $[σ_{i}]$ of the elementary geometric half twist, to the configuration loop class $Φ ([σ_{i}])$ whose endpoint monodromy is the adjacent transposition $(i i + 1)$ , and to the class $[H_{i}]$ of the half twist supported near the $i$ -th and $(i + 1)$ -st punctures, that is, of the explicit boundary-fixed homeomorphism supported in the disc $U_{i}$ and exchanging $q_{i}$ and $q_{i + 1}$ ;
consequently each of the four models carries the Artin presentation with these corresponding generators: for each model the assignment $σ_{i} \mapsto$ its generator extends to a group isomorphism from $B_{n}^{A r t i n}$ onto the model, so the model is presented by the generators $σ_{1}, \dots, σ_{n - 1}$ subject to the two Artin relations and to no further relations.

For $n = 1$ there is no generator, all four groups are trivial, and clauses 2 and 3 are vacuous.

Facts & Assumptions

Given: AC, an integer $n \geq 1$ , the Artin-presentation group $B_{n}^{A r t i n} = ⟨ X ∣ R ⟩$ of The braid group by Artin presentation with generating set $X = {σ_{1}, \dots, σ_{n - 1}}$ and its two families of Artin relators interpreted in the sense of Group presentation by generators and relations, the four models of the statement, and an index $i$ with $1 \leq i \leq n - 1$ .

[F1]

For every $n \geq 1$ the surjection $φ_{n} ⁣ : B_{n}^{A r t i n} \to G_{n}$ of The Artin presentation surjects onto the geometric braid group is an isomorphism, and it carries each generator $σ_{i}$ to the class $[σ_{i}]$ of the elementary geometric half twist of The elementary geometric half twist, its support disc, and its opposite; the endpoint permutation of that class is the transposition of $i$ and $i + 1$ (The Artin presentation is complete for geometric braids, The Artin presentation surjects onto the geometric braid group, The elementary geometric half twist, its support disc, and its opposite).

[F2]

The inverse-loop slicing map $Φ ⁣ : G_{n} \to B_{n}^{c o n f} = π_{1} (C_{n} (D^{2}), [Q_{n}])$ , $Φ ([β]) = (ι_{*}^{C} [S (β)])^{- 1}$ , is a group isomorphism intertwining the endpoint maps, $π_{c o n f} (Φ ([β])) = π_{g e o} ([β])$ for every $[β] \in G_{n}$ ; and the open-to-closed inclusion induces an isomorphism $ι_{*}^{C} ⁣ : π_{1} (C_{n} (int D^{2}), [Q_{n}]) \to π_{1} (C_{n} (D^{2}), [Q_{n}])$ at the same basepoint (The geometric and configuration braid models agree at the fixed base configuration, The geometric endpoint permutation matches covering monodromy, The interior-disc and closed-disc configuration spaces are homotopy equivalent).

[F3]

The composite $Ψ : = δ \circ (ι_{*}^{C})^{- 1} \circ Φ ⁣ : G_{n} \to Mod (D^{2}, Q_{n}; \partial D^{2})$ is a group isomorphism, and for every $1 \leq i \leq n - 1$ it satisfies $Ψ ([σ_{i}]) = [H_{i}]$ , where $H_{i}$ is the explicit boundary-fixed homeomorphism supported in the support disc $U_{i}$ of The elementary geometric half twist, its support disc, and its opposite and exchanging $q_{i}$ and $q_{i + 1}$ . The theorem supplying $Ψ$ is an in-run batch-20 scaffold of this run, not a published supplier (Braid group as boundary-fixed punctured-disk mapping classes, Boundary-fixed mapping class group of a punctured disk).

[F4]

AC holds, and AC implies dependent choice and countable choice (The Axiom of Choice, AC implies DC implies countable choice); this is the hypothesis under which the mapping-class isomorphism of [F3] and the free-kernel suppliers of the completeness theorem of [F1] are available.

[F5]

Presentation transport along an isomorphism. If a group has a presentation $B = ⟨ X ∣ R ⟩ = F (X) / ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)}$ and $θ ⁣ : B \to M$ is a group isomorphism, then the composite $q_{θ} ⁣ : F (X) \to B \overset{θ}{\to} M$ of the quotient map with $θ$ is a surjective homomorphism with kernel $⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)}$ : surjectivity is clear, and $q_{θ} (w) = e_{M}$ holds exactly when $q (w) \in \ker θ = {e_{B}}$ , that is, exactly when $w \in ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)}$ . The first isomorphism theorem therefore gives $M ≅ F (X) / ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)} = ⟨ X ∣ R ⟩$ , the isomorphism carrying the class of each $x \in X$ to $θ ([x])$ (Group presentation by generators and relations, First isomorphism theorem for groups: $G / \ker f ≅ im f$ , The braid group by Artin presentation).

Proof

technique · direct

1.1F1

The abstract and geometric models. By [F1] the map $φ_{n} ⁣ : B_{n}^{A r t i n} \to G_{n}$ is a group isomorphism and $φ_{n} (σ_{i}) = [σ_{i}]$ for every $i$ , so the abstract model and the geometric model are identified generator by generator.

2.1F1F2step 1.1

The two configuration models. By [F2] the inverse-loop slicing map $Φ ⁣ : G_{n} \to B_{n}^{c o n f}$ is a group isomorphism with $π_{c o n f} \circ Φ = π_{g e o}$ , and $ι_{*}^{C}$ is an isomorphism $π_{1} (C_{n} (int D^{2}), [Q_{n}]) \to B_{n}^{c o n f}$ ; hence the composites $Φ \circ φ_{n}$ and $(ι_{*}^{C})^{- 1} \circ Φ \circ φ_{n}$ , being composites of group isomorphisms, are group isomorphisms from $B_{n}^{A r t i n}$ onto $B_{n}^{c o n f}$ and onto $π_{1} (C_{n} (int D^{2}), [Q_{n}])$ respectively. The generator $σ_{i}$ is carried to $Φ ([σ_{i}])$ , whose endpoint monodromy is $π_{c o n f} (Φ ([σ_{i}])) = π_{g e o} ([σ_{i}])$ , the transposition of $i$ and $i + 1$ by [F1] and [F2]; in the open-disc model it is carried to $(ι_{*}^{C})^{- 1} Φ ([σ_{i}])$ , the same configuration loop class read through the inclusion.

2.2F3step 1.1

The mapping-class model. By [F3] the composite $Ψ$ is a group isomorphism $G_{n} \to Mod (D^{2}, Q_{n}; \partial D^{2})$ with $Ψ ([σ_{i}]) = [H_{i}]$ , so $Ψ \circ φ_{n}$ is a group isomorphism $B_{n}^{A r t i n} \to Mod (D^{2}, Q_{n}; \partial D^{2})$ carrying $σ_{i}$ to $[H_{i}]$ , the class of the boundary-fixed half twist supported in $U_{i}$ that exchanges $q_{i}$ and $q_{i + 1}$ .

3.1F5step 1.1step 2.1step 2.2

Presentation transport to each model. Let $R$ be the set of the two families of Artin relators, so that $B_{n}^{A r t i n} = ⟨ X ∣ R ⟩ = F (X) / ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)}$ by The braid group by Artin presentation and Group presentation by generators and relations. Apply [F5] to the identity isomorphism of $B_{n}^{A r t i n}$ and to the isomorphisms of steps 1.1, 2.1 and 2.2: the identity on $B_{n}^{A r t i n}$ , $φ_{n}$ , $Φ \circ φ_{n}$ , $(ι_{*}^{C})^{- 1} \circ Φ \circ φ_{n}$ and $Ψ \circ φ_{n}$ . Each of the five models is therefore isomorphic to $F (X) / ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩$ through the composite of the quotient map with that isomorphism, with the class of $σ_{i}$ mapping to the corresponding generator displayed in steps 1.1, 2.1 and 2.2; in particular each model is generated by those $n - 1$ elements and satisfies no relation among them beyond the Artin relators.

4.1F1F3F4step 1.1step 2.1step 2.2step 3.1

Conclusion and the one-strand case. Steps 1.1, 2.1 and 2.2 identify the four models pairwise through the stated isomorphisms and track $σ_{i}$ to the half twist, to the loop of monodromy $(i i + 1)$ and to the supported half twist, and step 3.1 transports the presentation to each of them, proving all three clauses for $n \geq 2$ . For $n = 1$ the index range $1 \leq i \leq n - 1$ is empty, so the generator clauses are vacuous, and $B_{1}^{A r t i n}$ is the trivial group given by the empty presentation by The braid group by Artin presentation; the isomorphisms of [F1]–[F3] then identify the other three models with it, so each of the four models is trivial and carries the empty presentation of the trivial group, which is clause 3 at $n = 1$ . AC enters only through [F3] and through the free-kernel suppliers of the completeness theorem recorded in [F4]. ∎

Remarks

The corollary does not reprove the mapping-class or configuration identifications: it composes them with the completeness theorem and tracks the generator through the composite. Its only genuinely new input beyond the suppliers is the bookkeeping that the generator correspondence survives each composite, which is why the configuration and mapping-class models inherit the Artin presentation.
The mapping-class isomorphism is an in-run batch-20 draft (Braid group as boundary-fixed punctured-disk mapping classes, precheck PASS; not a published supplier), flagged in the dispatch report together with the consuming step 2.2 and the cross-batch edge recorded in frontier-37-owner-30-batch-22.cross-batch-dependencies.json; no published theorem supplies it.
The construction is choice-free apart from the AC hypothesis: the presentations are finite, the free group on $X$ is explicit, and no connecting path or lift is chosen in the composites, all of which use the fixed basepoint $Q_{n}$ of the suppliers.

5 · Examples, counterexamples and false statements

None yet.

Artin Presentation Completeness and Braid Combing

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The Zariski combing words alpha_i and x_i in the Artin presentation

Definition

Prefix insertion rewrites a trivial braid word into combing factors

Statement

Facts & Assumptions

Proof

Remarks

Lower-rank Artin letters conjugate x-letters

Statement

Facts & Assumptions

Proof

Each combing factor reduces to a lower-rank letter or an x-letter

Statement

Facts & Assumptions

Proof

Remarks

Every trivial braid word combs as W_1W_2

Statement

Facts & Assumptions

Proof

Remarks

The combed geometric decomposition is unique

Statement

Facts & Assumptions

Proof

Remarks

The Artin presentation is complete for geometric braids

Statement

Facts & Assumptions

Proof

Remarks

All four classical braid models realize the Artin presentation

Statement

Facts & Assumptions

Proof

Remarks

5 · Examples, counterexamples and false statements

Sources