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The ordered-to-unordered two-point quotient is not one-to-one

Statement refuted

Let $F_{2} (C)$ be the ordered configuration space of two points of the plane and let

$p : F_{2} (C) ⟶ C_{2} (C) = F_{2} (C) / S_{2}, p (x) : = S_{2} \cdot x = [x],$

be the quotient map onto the unordered configuration space (Unordered configuration spaces $C_{n} (X)$ , Ordered configuration spaces $F_{n} (X)$ ). Refuted claim: the natural quotient map $p$ is injective, hence a homeomorphism onto $C_{2} (C)$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological). It is not injective: for distinct points $z_{1} \neq z_{2}$ of $C$ the two ordered configurations $(z_{1}, z_{2})$ and $(z_{2}, z_{1})$ are distinct points of $F_{2} (C)$ with the same image under $p$ , because they lie in one $S_{2}$ -orbit. The claim refuted concerns the natural quotient map only: it is not asserted, and it does not follow, that $F_{2} (C)$ and $C_{2} (C)$ are never abstractly homeomorphic by some other map, and the example makes no statement about that question.

Facts & Assumptions

Given: The ordered configuration space $F_{2} (C)$ with $(0, 1), (1, 0) \in F_{2} (C)$ , the unordered configuration space $C_{2} (C)$ with its quotient map $p$ , and the nonidentity transposition $τ \in S_{2}$ of the coordinate permutation action.

[F1]

$F_{2} (C) = {(z_{1}, z_{2}) \in C^{2} : z_{1} \neq z_{2}}$ with the subspace topology, so $(0, 1)$ and $(1, 0)$ both lie in $F_{2} (C)$ ; two tuples in $C^{2}$ are equal exactly when they agree in every coordinate, so $(0, 1) \neq (1, 0)$ because $0 \neq 1$ in the field $C$ (Ordered configuration spaces $F_{n} (X)$ , The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , Field).

[F2]

The formula $(σ \cdot x)_{i} = x_{σ^{- 1} (i - 1) + 1}$ defines a continuous free left action of $S_{2}$ on $F_{2} (C)$ ; the nonidentity permutation $τ$ with $τ (0) = 1$ , $τ (1) = 0$ acts by $τ \cdot (z_{1}, z_{2}) = (z_{2}, z_{1})$ (The symmetric group acts continuously and freely on $F_{n} (X)$ by permuting labels, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

[F3]

$C_{2} (C) = F_{2} (C) / S_{2} = {S_{2} \cdot x : x \in F_{2} (C)}$ is the set of orbits with the quotient topology of $p (x) = S_{2} \cdot x = [x]$ ; $p$ is a quotient map, hence continuous and surjective, and $p (x) = p (y)$ exactly when $x$ and $y$ lie in the same orbit $S_{2} \cdot x = S_{2} \cdot y$ (Unordered configuration spaces $C_{n} (X)$ , Ordered configuration spaces $F_{n} (X)$ ).

[F4]

A function is injective when $f (x) = f (y)$ implies $x = y$ , and a homeomorphism is by definition a continuous bijection with continuous inverse; in particular a homeomorphism is injective, so a map that is not injective is not a bijection and not a homeomorphism (Injection, surjection, bijection, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Refutation

technique · direct

1.1

Two distinct ordered configurations. The tuples $x : = (0, 1)$ and $y : = (1, 0)$ lie in $F_{2} (C)$ , since $0 \neq 1$ ; they are distinct, because they differ in the first coordinate and coordinates determine an element of the product $C^{2}$ ; explicitly $x_{1} = 0 \neq 1 = y_{1}$ .

1.2

One orbit. By [F2] the transposition acts by $τ \cdot x = τ \cdot (0, 1) = (1, 0) = y$ , so $x$ and $y$ lie in the same $S_{2}$ -orbit $S_{2} \cdot x$ ; note $S_{2} = {id, τ}$ , so this is the whole orbit of $x$ .

2.1

Equal images, unequal points. By step 1.2 the two points $x, y$ lie in one orbit, so by [F3] their images agree: $p (x) = p (y)$ ; but $x \neq y$ by step 1.1. Hence $p$ is not injective, in the sense of [F4].

step 1.1step 1.2F3F4

3.1

The map is not a homeomorphism, and the scope of the refutation. A homeomorphism of $C_{2} (C)$ with domain $F_{2} (C)$ would be a bijection and hence injective by [F4]; since $p$ is not injective by step 2.1, the natural quotient map $p$ is not a homeomorphism. This refutes only the identification of the quotient map with a homeomorphism; the abstract question whether some other continuous bijection with continuous inverse exists between $F_{2} (C)$ and $C_{2} (C)$ is untouched by this witness, and no assertion about it is made here.

step 2.1F4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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

Sources

Juan González-Meneses, Basic results on braid groups, §§1.1–1.3 and 2.1, printed pp. 3–6, 11–13 (standard reference, not scraped)