Continuous local sections for disk point evaluation

Small displacements give boundary-fixed homeomorphisms. Fix an ordered configuration $y=(y_1,\dots ,y_n)\in F_n(\mathrm{int}{D}^2)$. Put $d:=1-{\max }_i\parallel y_i{\parallel }_2>0$, let $\eta :={\min }_{i<j}\parallel y_i-y_j{\parallel }_2$ if $n\ge 2$ and $\eta :=1$ if $n=1$, and set $\rho :=\min (\eta ,d)\mathrm{/}3>0$. Choose Lipschitz functions ${\chi }_i:{\mathbb{R}}^2\to [0,1]$ with ${\chi }_i=1$ on a neighbourhood of $y_i$, $\mathrm{supp}{\chi }_i$ contained in the open ball of radius $\rho$ about $y_i$, and a common Lipschitz constant $C\mathrm{/}\rho$; the supports are pairwise disjoint and lie in $\mathrm{int}{D}^2$, the latter because $\rho \le d\mathrm{/}3$ leaves a positive margin to the boundary. For $z=(z_1,\dots ,z_n)$ with $v_i:=z_i-y_i$ satisfying ${\sum }_i\parallel v_i{\parallel }_2<\rho \mathrm{/}(2C)$, put $h_{y,z}(w):=w+{\sum }_i{\chi }_i(w)v_i$. Then the displacement $\phi :=h_{y,z}-\mathrm{id}$ is Lipschitz with constant at most ${\sum }_{i}C\parallel v_i{\parallel }_2\mathrm{/}\rho <1\mathrm{/}2<1$, so for every $w^{\mathrm{\prime }}\in {\mathbb{R}}^2$ the map $w\mapsto w^{\mathrm{\prime }}-\phi (w)$ is a contraction of the complete space ${\mathbb{R}}^2$ and [L1] and [L2] give it a unique fixed point; the resulting inverse is Lipschitz, since $\parallel h_{y,z}(u)-h_{y,z}(v){\parallel }_2\ge (1-\mathrm{Lip}(\phi ))\parallel u-v{\parallel }_2$, and is a two-sided inverse of $h_{y,z}$, and it shows simultaneously that $h_{y,z}$ is bijective with continuous inverse, hence a homeomorphism. Since $h_{y,z}$ is the identity outside $\mathrm{int}{D}^2$, bijectivity prevents an interior point from mapping outside the disc; thus it restricts to a homeomorphism of $D^2$ and fixes $\mathrm{\partial }{D}^2$ pointwise, and $h_{y,z}(y_i)=y_i+v_i=z_i$ because ${\chi }_i=1$ near $y_i$ and ${\chi }_j(y_i)=0$ for $j\mathrm{\ne }i$. Moreover, if $z^{(k)}\to z$ with all members admissible, then ${\sup }_w\parallel h_{y,z^{(k)}}(w)-h_{y,z}(w){\parallel }_2\le {\sum }_i\parallel v_i^{(k)}-v_i{\parallel }_2\to 0$, so $z\mapsto h_{y,z}$ is continuous for the topology of uniform convergence, which on $D^2$ is the compact-open topology by [L3].

Local order of an unordered configuration. By [L4] the quotient $p:F_n(\mathrm{int}{D}^2)\to C_n(\mathrm{int}{D}^2)$ is a covering map with path-connected total space and base. Given $\xi \in C_n(\mathrm{int}{D}^2)$ and a chosen preimage $x\in F_n(\mathrm{int}{D}^2)$ with $[x]=\xi$, [L5] supplies an evenly covered open $U\ni \xi$ and the sheet through $x$ gives a continuous local section $\sigma :U\to F_n(\mathrm{int}{D}^2)$ with $\sigma (\xi )=x$ and $p\circ \sigma ={\mathrm{id}}_U$; the passage from an unordered configuration to one of its enumerations is a single selection from a nonempty set and costs no choice.

Evaluation is surjective. Let $\xi \in C_n(\mathrm{int}{D}^2)$ and by [L7] choose $x=(x_1,\dots ,x_n)\in F_n(\mathrm{int}{D}^2)$ with $[x]=\xi$. By the path-connectedness in [L4] and the compactness of $[0,1]$ choose a path $\beta :[0,1]\to F_n(\mathrm{int}{D}^2)$ with $\beta (0)=Q_n$ and $\beta (1)=x$ together with a finite partition $0=t_0<t_1<\cdots <t_m=1$ such that for every $k$ the configurations $y:=\beta (t_{k-1})$, $z:=\beta (t_k)$ satisfy the admissibility bound of step 1.1; this partition exists because the configurations along a path stay at a positive distance from one another and from the boundary and both quantities are uniformly continuous on the compact interval. Put $g:=h_{\beta (t_{m-1}),\beta (t_m)}\circ \cdots \circ h_{\beta (t_0),\beta (t_1)}$; by step 1.1 each factor is a homeomorphism of $D^2$ fixing $\mathrm{\partial }{D}^2$, so $g$ is one too, and $g(q_i)=x_i$ for every $i$. Hence $\mathrm{ev}(g)=[x]=\xi$, which proves surjectivity.

Continuous local sections. Fix ${\xi }^0\in C_n(\mathrm{int}{D}^2)$ and, using step 2.1, choose $g^0\in {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ with $\mathrm{ev}(g^0)={\xi }^0$; write $x^0:=(g^0(q_1),\dots ,g^0(q_n))\in F_n(\mathrm{int}{D}^2)$, so $[x^0]={\xi }^0$. Let $U$ and $\sigma$ be the evenly covered neighbourhood and local section of step 1.2 through $x^0$, so that $\sigma ({\xi }^0)=x^0$, and set $s(\xi ):=h_{x^0,\sigma (\xi )}\circ g^0(\xi \in U)\mathrm{.}$ Shrink $U$ to the open neighbourhood on which the displacement bound of step 1.1 holds; this is possible because $\sigma$ is continuous and $\sigma ({\xi }^0)=x^0$. Then $s$ is well defined on all of this smaller $U$; it is continuous as a composite of the continuous maps $\sigma$, $z\mapsto h_{x^0,z}$ and right composition with the fixed homeomorphism $g^0$. Finally $\mathrm{ev}(s(\xi ))=[\sigma (\xi )]=\xi$ by step 1.1 and $p\circ \sigma ={\mathrm{id}}_U$, so $s$ is the required continuous local section.