Ordered configuration spaces cover the unordered ones regularly with deck group $S_n$

Punctured relative balls are path-connected. Let $d\ge 2$, $c\in {\mathbb{H}}^d$, $\epsilon >0$ and $R:=(B(c,\epsilon )\cap {\mathbb{H}}^d)\setminus \{c\}$. The set $B(c,\epsilon )\cap {\mathbb{H}}^d$ is convex by [L6], so segments between its points stay in it and are continuous paths; write $e_1:=(1,0,\dots ,0)$ and $e_d:=(0,\dots ,0,1)$, whose last coordinates are $0$ and $1$ respectively, so that $c+\lambda e_i\in R$ for $0<\lambda <\epsilon$ and $i\in \{1,d\}$, and $R\mathrm{\ne }\mathrm{\varnothing }$. Let $u\in R$, $\rho :=\mathrm{\mid }u-c\mathrm{\mid }\in (0,\epsilon )$ and $q_i:=c+\rho e_i\in R$. The segment $[u,q_i]$ contains $c$ only if $u-c$ is a negative multiple of $e_i$, which can happen for at most one of $i\in \{1,d\}$ because $e_1$ and $e_d$ are not parallel; choose $i$ with $c\mathrm{\notin }[u,q_i]$, so $[u,q_i]\subseteq R$ joins $u$ to $q_i$. The segment $[q_i,c+(\epsilon \mathrm{/}2)e_i]$ lies in $R$, since all its points are of the form $c+\lambda e_i$ with $\lambda >0$. The segment $[c+(\epsilon \mathrm{/}2)e_1,c+(\epsilon \mathrm{/}2)e_d]$ lies in $R$, since a point of it equals $c$ only if $(1-t)(\epsilon \mathrm{/}2)e_1+t(\epsilon \mathrm{/}2)e_d=0$, which forces $t=0$ and $t=1$ simultaneously as $e_1,e_d$ are linearly independent. Hence any two points of $R$ are joined by a polygonal path in $R$, so $R$ is path-connected.

Local form of $M$. Let $s\in M$ and let $O\subseteq M$ be open with $s\in O$. By [L5] there are an open $U\subseteq M$ containing $s$ and a homeomorphism $\psi$ of $U$ onto a relatively open $V\subseteq {\mathbb{H}}^d$. Replacing $U$ by $U\cap O$, which still contains $s$, we may suppose $U\subseteq O$. By [L6] there is $\epsilon >0$ with $B(\psi (s),\epsilon )\cap {\mathbb{H}}^d\subseteq V$; set $W:={\psi }^{-1}(B(\psi (s),\epsilon )\cap {\mathbb{H}}^d)$, an open neighbourhood of $s$ with $W\subseteq O$, homeomorphic to the relative ball $B(\psi (s),\epsilon )\cap {\mathbb{H}}^d$.

$p$ is an $n!$-sheeted covering. Let $y\in C_n(M)$; by [L2] there is $q\in F_n(M)$ with $p(q)=y$, and [L4] makes $p$ evenly covered at $y$ with $n!$ sheets. Hence $p$ is a covering map, and the fibre $p^{-1}(y)$ meets each of the $n!$ sheets in exactly one point, because on each sheet $p$ restricts to a homeomorphism; so every fibre has exactly $n!$ elements.

M is infinite. Apply step 1.2 with $O:=M$ and some $s\in M$, obtaining $W\cong B(\psi (s),\epsilon )\cap {\mathbb{H}}^d$. By step 1.1 the set $W\setminus \{s\}$, which corresponds to the punctured relative ball at $\psi (s)$, is nonempty and path-connected, so $M$ has at least two points. If $M$ were finite, then for $y\mathrm{\ne }s$ the sets $\{y\}$ are closed by [L8], so $\{s\}$ and $M\setminus \{s\}$ would be disjoint nonempty open sets covering $M$, a separation of the connected space $M$ by [L5]; hence $M$ is infinite.

$M$ is locally path-connected. Let $s\in M$ and let $O$ be a neighbourhood of $s$. Step 1.2 gives a neighbourhood $W$ of $s$ with $W\subseteq O$ homeomorphic to a relative ball $B(\psi (s),\epsilon )\cap {\mathbb{H}}^d$, which is path-connected by [L6]. A homeomorphism carries paths to paths, so $W$ is path-connected: the path-connected open sets form a neighbourhood basis of $s$.

$M$ is path-connected. It is connected and locally path-connected by [L5] and step 2.2, so [L7] makes it path-connected.

Complements of finite sets are path-connected. Let $S\subseteq M$ be finite. Step 2.1 makes $M$ infinite, so $M\setminus S\mathrm{\ne }\mathrm{\varnothing }$; indeed $M\setminus S$ cannot be finite, for then $M=(M\setminus S)\cup S$ would be a union of two finite sets. For each $x\in M\setminus S$, steps 1.2 and 2.2 give a path-connected open neighbourhood $W_x\subseteq M\setminus S$ of $x$, since $M\setminus S$ is open by [L8]. Thus each path component $P$ of $M\setminus S$ is open in $M$: every $x\in P$ has such a $W_x\subseteq P$. No simultaneous choice of the neighbourhoods is required. For each of the finitely many $s\in S$ apply steps 1.1 and 1.2 with $O:=M\setminus (S\setminus \{s\})$, which is open by [L8]: this gives an open neighbourhood $W_s$ of $s$ with $W_s\subseteq O$ and $W_s\setminus \{s\}$ path-connected, hence contained in a single path component $P(s)$ of $M\setminus S$. For each path component $P$ of $M\setminus S$ define $A_P:=P\cup \{s\in S:P(s)=P\}$. It is open in $M$: $P$ is open, and for each added point $s$ the open set $W_s$ lies in $A_P$, since $W_s\setminus \{s\}\subseteq P$. The sets $A_P$ are pairwise disjoint and cover $M$, because each point of $M\setminus S$ belongs to exactly one path component and each $s\in S$ has exactly one assigned component $P(s)$. If there were two or more components, choose one $P$; then $A_P$ and the union of all $A_Q$ for $Q\mathrm{\ne }P$ would be disjoint nonempty open sets covering $M$, contradicting connectedness by [L5]. Hence $M\setminus S$ is path-connected.

$F_n(M)$ is path-connected. Let $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)$ be points of $F_n(M)$ with $n\ge 1$, and let $S:=\{x_1,\dots ,x_n,y_1,\dots ,y_n\}$, a finite set; by steps 2.1 and 4.1 the complement $M\setminus S$ is infinite, so choose distinct points $z_1,\dots ,z_n\in M\setminus S$ and put $z:=(z_1,\dots ,z_n)\in F_n(M)$. For $k=1,\dots ,n$ the set $M\setminus \{x_{k+1},\dots ,x_n,z_1,\dots ,z_{k-1}\}$ is path-connected by step 4.1, and both $x_k$ and $z_k$ lie in it, so there is a path in it from $x_k$ to $z_k$; replacing the $k$-th coordinate by that path while keeping the other coordinates fixed gives a path in $F_n(M)$ from $(z_1,\dots ,z_{k-1},x_k,\dots ,x_n)$ to $(z_1,\dots ,z_k,x_{k+1},\dots ,x_n)$, because every value of the moving coordinate avoids the finitely many fixed coordinates and the fixed coordinates are pairwise distinct. Concatenating these $n$ paths yields a path from $x$ to $z$, and the same construction with the roles of $x$ and $y$ exchanged yields a path from $y$ to $z$; reversing the latter and concatenating gives a path in $F_n(M)$ from $x$ to $y$. For $n=0$, $F_0(M)$ is a one-point space by [F1].

$C_n(M)$ is path-connected. $p$ is continuous and surjective, so for points $p(x),p(y)\in C_n(M)$ a path in $F_n(M)$ from $x$ to $y$, which exists by step 5.1, composes with $p$ to a path in $C_n(M)$ joining them.

Deck group. For $\sigma \in S_n$ the map ${\tau }_{\sigma }(x):=\sigma \cdot x$ is a homeomorphism of $F_n(M)$ by [L3], and $p\circ {\tau }_{\sigma }=p$ because $\sigma \cdot x$ lies in the orbit of $x$; hence ${\tau }_{\sigma }\in \mathrm{Deck}(p)$ by [L9]. The assignment $\sigma \mapsto {\tau }_{\sigma }$ is a group homomorphism, since ${\tau }_{\sigma \rho }(x)=(\sigma \rho )\cdot x=\sigma \cdot (\rho \cdot x)={\tau }_{\sigma }({\tau }_{\rho }(x))$ by the left-action law [L3], and it is injective: if ${\tau }_{\sigma }={\tau }_{\rho }$ then $\sigma \cdot x=\rho \cdot x$ for every $x$, so ${\rho }^{-1}\sigma$ fixes a point of $F_n(M)$, which is nonempty by step 5.1 and [F1], and freeness gives ${\rho }^{-1}\sigma =\mathrm{id}$. By step 5.1 the total space is connected, so by [L9] a deck transformation is determined by its value at a point; since every deck transformation permutes the fibre over $p(x)$, evaluation at any $x\in F_n(M)$ injects $\mathrm{Deck}(p)$ into that fibre, so $\mathrm{\mid }\mathrm{Deck}(p)\mathrm{\mid }\le n!$ by step 1.3, while the injective homomorphism exhibits $n!=\mathrm{\mid }{S}_n\mathrm{\mid }$ deck transformations. Therefore $\sigma \mapsto {\tau }_{\sigma }$ is a bijective homomorphism, hence by [L11] an isomorphism $S_n\cong \mathrm{Deck}(p)$.

Regularity. Let $y\in C_n(M)$ and let $x,x^{\mathrm{\prime }}\in p^{-1}(y)$. By [L2] there is $\sigma \in S_n$ with $x^{\mathrm{\prime }}=\sigma \cdot x={\tau }_{\sigma }(x)$, so the deck group acts transitively on the fibre; since $F_n(M)$ is path-connected by step 5.1, [L10] makes $p$ a regular covering.

Conclusion. Claim 1 is step 1.3, claim 2 is steps 5.1 and 6.1 together with [L7], claim 3 is step 6.2 and claim 4 is step 7.1; the case $n=0$ is [L2] and [F1].