Disjoint coordinate neighbourhoods evenly cover the unordered configuration space

Disjoint coordinate neighbourhoods exist. Since $X$ is Hausdorff and $q\in F_n(X)$ has pairwise distinct coordinates, [F1] supplies, for every label $i\in \{1,\dots ,n\}$, an open neighbourhood $U_i$ of $q_i$ with $U_i\cap U_j=\mathrm{\varnothing }$ for $i\mathrm{\ne }j$.

The case $n=0$. By [L3], $F_0(X)$ and $C_0(X)$ are one-point spaces and $p$ is their unique homeomorphism; its only fibre has $0!=1$ element and the single point of $C_0(X)$ is evenly covered by the single sheet $F_0(X)$.

The set $U$ and its translates. With $U:=\left({\prod }_{i=1}^n{U}_i\right)\cap F_n(X)$, the set ${\prod }_{i=1}^n{U}_i$ is open in $X^n$ and $U$ is open in $F_n(X)$ by [L5]; moreover $q\in U$, because $q_i\in U_i$ for every $i$ and $q\in F_n(X)$. For $\sigma \in S_n$ the translate $\sigma (U)=\{\sigma \cdot x:x\in U\}$ is open in $F_n(X)$, being the image of the open set $U$ under the homeomorphism $x\mapsto \sigma \cdot x$ of [L4].

The translates are disjoint and cover the preimage of $V$. Suppose $x\in \sigma (U)\cap \tau (U)$ for $\sigma ,\tau \in S_n$. Writing $x=\sigma \cdot u=\tau \cdot u^{\mathrm{\prime }}$ with $u,u^{\mathrm{\prime }}\in U$, the coordinate formula of [F1] gives, for every label $k$, the element $x_k=u_{{\sigma }^{-1}(k-1)+1}=u_{{\tau }^{-1}(k-1)+1}^{\mathrm{\prime }}$ of $U_{{\sigma }^{-1}(k-1)+1}\cap U_{{\tau }^{-1}(k-1)+1}$; by step 1.1 the sets $U_i$ are pairwise disjoint, so ${\sigma }^{-1}(k-1)+1={\tau }^{-1}(k-1)+1$ for every $k$, that is $\sigma =\tau$. Hence the translates $\sigma (U)$ are pairwise disjoint. A tuple $x\in F_n(X)$ lies in $p^{-1}(p(U))$ exactly when $p(x)=p(u)$ for some $u\in U$, that is, by [L3], exactly when $x=\sigma \cdot u$ for some $\sigma \in S_n$ and $u\in U$; therefore $p^{-1}(V)={\bigcup }_{\sigma \in S_n}\sigma (U)$, and this union is disjoint.

$V$ is an open neighbourhood of $[q]$. By step 2.1, $p^{-1}(V)$ is a finite union of open sets, hence open in $F_n(X)$ by [L5], so $V$ is open in $C_n(X)$ by [L6]. It contains $p(q)=[q]$ because $q\in U$ by step 1.3.

$p{\mathrm{\mid }}_U:U\to V$ is a homeomorphism. The restriction is continuous, and it is injective: if $p(u)=p(u^{\mathrm{\prime }})$ for $u,u^{\mathrm{\prime }}\in U$, then $u^{\mathrm{\prime }}\in U\cap \sigma (U)$ for some $\sigma$ by step 2.1, so $u^{\mathrm{\prime }}=u$ by the disjointness proved there. It is surjective onto $V=p(U)$ by definition. Finally it is open: for $A\subseteq U$ open, $p^{-1}(p(A))={\bigcup }_{\sigma \in S_n}\sigma (A)$ is a finite union of images of $A$ under the homeomorphisms of [L4], hence open in $F_n(X)$, so $p(A)$ is open in $C_n(X)$ by [L6] and therefore in $V$. By [L7], $p{\mathrm{\mid }}_U$ is a homeomorphism onto $V$.

Every translate maps homeomorphically onto $V$. Let $\sigma \in S_n$ and let $h_{\sigma }(x):=\sigma \cdot x$ be the homeomorphism of $F_n(X)$ given by [L4], which maps $U$ onto $\sigma (U)$. For $y=\sigma \cdot u\in \sigma (U)$ with $u\in U$ one has $p(y)=p(\sigma \cdot u)=p(u)$, since orbits are permuted by $\sigma$; hence $p{\mathrm{\mid }}_{\sigma (U)}=p{\mathrm{\mid }}_U\circ (h_{\sigma }{\mathrm{\mid }}_U{)}^{-1}$ is a composite of homeomorphisms and therefore a homeomorphism onto $V$ by [L7].

Conclusion. By steps 1.1, 1.3, 2.1 and 5.1, the open neighbourhood $V$ of $[q]$ has preimage $p^{-1}(V)$ equal to the disjoint union of the open sets $\sigma (U)$, $\sigma \in S_n$, each of which is carried homeomorphically onto $V$ by $p$; by [L8] and $\mathrm{\mid }{S}_n\mathrm{\mid }=n!$ these are exactly $n!$ sheets, so $V$ is evenly covered. The case $n=0$ is step 1.2.