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Setwise puncture preservation does not imply purity

Statement refuted

Refuted claim: for the punctured disc, preserving the marked set $Q_{n}$ setwise is the same as being pure. Precisely: every homeomorphism $f$ of $D^{2}$ that fixes $\partial D^{2}$ pointwise and satisfies $f ({q_{1}, \dots, q_{n}}) = {q_{1}, \dots, q_{n}}$ is isotopic rel $\partial D^{2}$ through homeomorphisms preserving $Q_{n}$ setwise at every time to a homeomorphism that fixes every $q_{i}$ individually, so that $PMod (D^{2}, Q_{n}; \partial D^{2}) = Mod (D^{2}, Q_{n}; \partial D^{2});$ equivalently, every class of the setwise stabiliser is a pure class.

The witness is the explicit supported positive half twist $H_{1}$ of A supported half-twist homeomorphism. For $n \geq 2$ and an adjacent index $i$ , the class of $H_{1}$ lies in $Mod (D^{2}, Q_{n}; \partial D^{2})$ but not in $PMod (D^{2}, Q_{n}; \partial D^{2})$ : the homeomorphism $H_{1}$ fixes $\partial D^{2}$ pointwise and preserves $Q_{n}$ setwise, yet it exchanges $q_{i}$ and $q_{i + 1}$ and fixes all other marked points, so it induces the transposition of $i$ and $i + 1$ rather than the identity permutation of the marked set.

What is and is not claimed. What fails is exactly the implication "setwise preservation $\Rightarrow$ purity" and the resulting equality of the two groups. Nothing here asserts that the transposition is the only permutation that can occur, and nothing here computes an isotopy invariant beyond the permutation of the marked set.

Facts & Assumptions

Given: The Axiom of Choice, the natural number $n \geq 2$ , the adjacent index $1 \leq i \leq n - 1$ , the base configuration $Q_{n} = (q_{1}, \dots, q_{n})$ , and the explicit homeomorphism $H_{1}$ of A supported half-twist homeomorphism with its collar function and support disc $U_{i}$ .

[F1]

The explicit half rotation $H$ of the example satisfies: every $H_{s}$ is a homeomorphism of $D^{2}$ fixing $\partial D^{2}$ pointwise and fixing every point outside the support disc $U_{i}$ ; on the two adjacent punctures $H_{s} (q_{i}) = m_{i} + h (- \cos π s, - \sin π s)$ and $H_{s} (q_{i + 1}) = m_{i} + h (\cos π s, \sin π s)$ ; every other base point is fixed throughout; and at $s = 1$ the two moving punctures are exchanged, so $H_{1}$ preserves $Q_{n}$ setwise and lies in ${Homeo}^{+} (D^{2}, \partial D^{2}; Q_{n})$ (A supported half-twist homeomorphism).

[F2]

${Homeo}^{+} (D^{2}, \partial D^{2}; {\hat{Q}}_{n})$ denotes the subgroup of homeomorphisms fixing $\partial D^{2}$ pointwise and every marked point, and $PMod (D^{2}, Q_{n}; \partial D^{2})$ is its set of path components; the inclusion of the pointwise stabiliser into the setwise stabiliser induces an injection, so $PMod (D^{2}, Q_{n}; \partial D^{2})$ is identified with the subgroup of $Mod (D^{2}, Q_{n}; \partial D^{2})$ consisting of the classes whose permutation of $Q_{n}$ is trivial; the permutation induced by a representative is locally constant along a path in the setwise stabiliser, because each strand $s \mapsto f_{s} (q_{j})$ is continuous and lands in the finite discrete set ${q_{1}, \dots, q_{n}}$ (Pure boundary-fixed mapping classes).

[F3]

${Homeo}^{+} (D^{2}, \partial D^{2}; Q_{n})$ is the setwise stabiliser of $Q_{n}$ in the boundary-fixing group, on $\partial D^{2}$ each element fixes the circle pointwise, each $f$ in it has a unique permutation $π (f) \in S_{n}$ with $f (q_{j}) = q_{π (f) (j - 1) + 1}$ for $1 \leq j \leq n$ , using the identification $κ (j) = j - 1$ of point labels with ${0, \dots, n - 1}$ , and $Mod (D^{2}, Q_{n}; \partial D^{2})$ is its set of isotopy classes, two elements being isotopic exactly when they are joined by a path in this stabiliser (Boundary-fixed mapping class group of a punctured disk).

[F4]

The standard positive half twist $σ_{i}$ is a braid whose endpoint permutation exchanges the point labels $i$ and $i + 1$ , that is, it is the transposition $(i - 1 i)$ in $S_{n}$ ; this is not the identity permutation (The elementary geometric half twist, its support disc, and its opposite, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

[F5]

The Axiom of Choice is assumed (The Axiom of Choice).

Counterexample

1.1F1F2F3F4F5

The witness and the permutation it induces. By [F1], which is available under the standing Axiom of Choice [F5], the homeomorphism $H_{1}$ fixes $\partial D^{2}$ pointwise, fixes every point outside $U_{i}$ , and satisfies $H_{1} (q_{i}) = q_{i + 1}$ , $H_{1} (q_{i + 1}) = q_{i}$ , and $H_{1} (q_{k}) = q_{k}$ for $k \notin {i, i + 1}$ , so $H_{1}$ preserves the marked set $Q_{n}$ setwise and its unique permutation $π (H_{1}) \in S_{n}$ of [F3] is the transposition $(i - 1 i)$ on ${0, \dots, n - 1}$ (exchanging point labels $i$ and $i + 1$ ), which differs from the identity permutation by [F4]; in particular $H_{1}$ does not fix every marked point individually, and $H_{1}$ is a member of the setwise stabiliser but not of the pointwise stabiliser of [F2].

1.2F2F3

The permutation is an isotopy invariant. Let $s \mapsto f_{s}$ be any path in ${Homeo}^{+} (D^{2}, \partial D^{2}; Q_{n})$ , so that it is an isotopy rel $\partial D^{2}$ from $f_{0}$ to $f_{1}$ ; for each label $j$ the map $s \mapsto f_{s} (q_{j})$ is continuous and takes values in the finite set ${q_{1}, \dots, q_{n}}$ , which is discrete in the subspace topology, hence is constant on the connected interval $I$ ; therefore each $π (f_{s})$ is defined and independent of $s$ , and every representative of the isotopy class of $f_{0}$ induces the same permutation $π (f_{0})$ .

2.1F2F4step 1.1step 1.2

The class is not pure. Since $H_{1}$ lies in the setwise stabiliser, its class $[H_{1}]$ lies in $Mod (D^{2}, Q_{n}; \partial D^{2})$ by [F3], and by step 1.2 every representative of $[H_{1}]$ induces the transposition of $i$ and $i + 1$ on the marked set; this permutation is not trivial by [F4], so by the identification of [F2] the class $[H_{1}]$ is not an element of $PMod (D^{2}, Q_{n}; \partial D^{2})$ .

3.1step 2.1F2

Conclusion. Step 1.1 exhibits a homeomorphism fixing $\partial D^{2}$ pointwise that preserves $Q_{n}$ setwise without fixing its points, and step 2.1 shows that its isotopy class lies outside $PMod (D^{2}, Q_{n}; \partial D^{2})$ ; hence setwise preservation of the punctures does not imply purity, not even for a single representative class, and the two subgroups of $Mod (D^{2}, Q_{n}; \partial D^{2})$ are distinct whenever $n \geq 2$ , with $PMod$ the strict subgroup of classes of trivial permutation. ∎

Remarks

The witness is the geometric half twist: the isotopy class of the positive braid generator $σ_{i}$ acts on the marked set by a transposition, while the boundary-fixed pure subgroup is by definition the part of the mapping class group acting trivially. The distinction is visible already for $n = 2$ , where the only non-identity permutation is the transposition realised by the half twist.
The discreteness of the permutation is what makes the counterexample robust: no isotopy rel $\partial D^{2}$ can convert the transposition into the identity, because the strands would have to leave the marked set, which the setwise condition forbids.

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Sources

Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 1.3, author manuscript pp. 5-7 (standard reference, not scraped)
Juan Gonzalez-Meneses, Basic results on braid groups, section 1.4, printed pp. 6-7 (standard reference, not scraped)