The interior-disc and closed-disc configuration spaces are homotopy equivalent

The interior of $D^2$ is $\mathrm{int}{D}^2$. By [L4] the ball of radius $\epsilon$ about $z$ is $\{w:\mathrm{\mid }w-z\mathrm{\mid }<\epsilon \}$. If $\mathrm{\mid }z\mathrm{\mid }<1$, put $\epsilon :=1-\mathrm{\mid }z\mathrm{\mid }>0$; then every $w$ with $\mathrm{\mid }w-z\mathrm{\mid }<\epsilon$ has $\mathrm{\mid }w\mathrm{\mid }\le \mathrm{\mid }w-z\mathrm{\mid }+\mathrm{\mid }z\mathrm{\mid }<\epsilon +\mathrm{\mid }z\mathrm{\mid }=1$, so the ball lies in $D^2$, and by [L5] $z$ is an interior point. If $\mathrm{\mid }z\mathrm{\mid }=1$ and $\epsilon >0$, the point $w:=(1+\epsilon \mathrm{/}2)z$ has $\mathrm{\mid }w-z\mathrm{\mid }=(\epsilon \mathrm{/}2)\mathrm{\mid }z\mathrm{\mid }=\epsilon \mathrm{/}2<\epsilon$ but $\mathrm{\mid }w\mathrm{\mid }=1+\epsilon \mathrm{/}2>1$, so it lies outside $D^2$ and no ball about $z$ is contained in $D^2$. If $\mathrm{\mid }z\mathrm{\mid }>1$ then $z\mathrm{\notin }{D}^2$. Hence the interior of $D^2$ in $\mathbb{C}$ is exactly $\{z:\mathrm{\mid }z\mathrm{\mid }<1\}$.

Scaling by $\lambda \in (0,1]$ preserves configurations. Let $\lambda \in (0,1]$, let $m\in \mathbb{N}$ and let $x\in F_m(D^2)$ or $x\in F_m(\mathrm{int}{D}^2)$ according to the case, and put $\lambda x:=(\lambda x_1,\dots ,\lambda x_m)$. Then $\mathrm{\mid }\lambda x_i\mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }{x}_i\mathrm{\mid }=\lambda \mathrm{\mid }{x}_i\mathrm{\mid }\le \mathrm{\mid }{x}_i\mathrm{\mid }\le 1$, with $\mathrm{\mid }\lambda x_i\mathrm{\mid }\le \mathrm{\mid }{x}_i\mathrm{\mid }<1$ when all $\mathrm{\mid }{x}_i\mathrm{\mid }<1$; and $\lambda x_i=\lambda x_j$ implies $x_i={\lambda }^{-1}\lambda x_i={\lambda }^{-1}\lambda x_j=x_j$ because $\lambda \mathrm{\ne }0$. So $\lambda x\in F_m(D^2)$, and $\lambda x\in F_m(\mathrm{int}{D}^2)$ whenever $x\in F_m(\mathrm{int}{D}^2)$.

The quotient projection is open. Let $V\subseteq F_m(X)$ be open. A tuple $x$ lies in $p_X^{-1}(p_X(V))$ exactly when $x=\sigma \cdot v$ for some $\sigma \in S_m$ and $v\in V$, so $p_X^{-1}(p_X(V))={\bigcup }_{\sigma \in S_m}\sigma (V)$. Each $\sigma (V)$ is open because $\sigma$ acts by a homeomorphism of $F_m(X)$ ([L2]), so this finite union is open by [L5]; hence $p_X(V)$ is open in the quotient topology (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection). Thus $p_X$ is an open map.

Moving-basepoint lemma. Let $Y$ be a space, let $c:I\to Y$ be continuous and let $G:I\times I\to Y$ be continuous with $G(0,u)=G(1,u)=c(u)$ for every $u\in I$; write $G_0(s):=G(s,0)$ and $G_1(s):=G(s,1)$, loops at $c(0)$ and $c(1)$. Then $G_0\ast c\simeq c\ast G_1$ rel endpoints. Indeed, put $A(s):=(2s,0)$ for $s\le \frac{1}{2}$ and $A(s):=(1,2s-1)$ for $s\ge \frac{1}{2}$, and $B(s):=(0,2s)$ for $s\le \frac{1}{2}$ and $B(s):=(2s-1,1)$ for $s\ge \frac{1}{2}$; on each of the two closed halves a single continuous formula is given, and at $s=\frac{1}{2}$ both formulas for $A$ give $(1,0)$ and both formulas for $B$ give $(0,1)$, so by the finite closed pasting of [L5] $A$ and $B$ are continuous paths in $I\times I$ with $A(0)=B(0)=(0,0)$ and $A(1)=B(1)=(1,1)$. Put $P_t(s):=(1-t)A(s)+tB(s)$ for $(s,t)\in I\times I$, a continuous map into $I\times I$. Then $(s,t)\mapsto G(P_t(s))$ is continuous, for each $t$ it is a path from $G(0,0)=c(0)$ to $G(1,1)=c(1)$, and $G\circ P_0=G\circ A=G_0\ast c$, $G\circ P_1=G\circ B=c\ast G_1$, because for $s\le \frac{1}{2}$ one has $G(A(s))=G(2s,0)=G_0(2s)$ and $G(B(s))=G(0,2s)=c(2s)$, while for $s\ge \frac{1}{2}$ one has $G(A(s))=G(1,2s-1)=c(2s-1)$ and $G(B(s))=G(2s-1,1)=G_1(2s-1)$. So $G\circ P$ is a path homotopy rel endpoints from $G_0\ast c$ to $c\ast G_1$.

The scaling map and the homotopy are well defined. Put $\lambda (t):=1-t\mathrm{/}2$ for $t\in I$; then $1\mathrm{/}2\le \lambda (t)\le 1$, so $\lambda (t)\in (0,1]$, and $\lambda (0)=1$, $\lambda (1)=1\mathrm{/}2$. Define $H_t(x):=(\lambda (t)x_1,\dots ,\lambda (t)x_n)$ and ${\rho }^F:=H_1$. By step 1.2, $H_t$ maps $F_n(D^2)$ into itself and $F_n(\mathrm{int}{D}^2)$ into itself, so $H_1={\rho }^F$ is a well-defined map $F_n(D^2)\to F_n(\mathrm{int}{D}^2)$ and $H_0={\mathrm{id}}_{F_n(D^2)}$. Consequently $H_1={\iota }^F\circ {\rho }^F$ as maps $F_n(D^2)\to F_n(D^2)$, and the restriction of $H$ to $F_n(\mathrm{int}{D}^2)\times I$ is a homotopy within $F_n(\mathrm{int}{D}^2)$ from ${\mathrm{id}}_{F_n(\mathrm{int}{D}^2)}$ to ${\rho }^F\circ {\iota }^F$.

Joint continuity of $H$. The map $(x,t)\mapsto (\lambda (t),x_i)$ from $F_n(D^2)\times I$ to $\mathbb{C}\times \mathbb{C}$ is continuous: its first component is the composite of the projection to $I$ with the affine map $t\mapsto 1-t\mathrm{/}2$, its second the composite of the projection to $F_n(D^2)$ with the $i$-th coordinate projection of the product ${\mathbb{C}}^n$. Hence $(x,t)\mapsto \lambda (t)x_i$ is continuous as a composite with scalar multiplication [L5], so $(x,t)\mapsto H_t(x)$ is continuous into the product ${\mathbb{C}}^n$ by the component criterion and, since its values lie in the subspace, into $F_n(D^2)$; the restricted map $F_n(\mathrm{int}{D}^2)\times I\to F_n(\mathrm{int}{D}^2)$ is continuous for the same reason. Thus $H$ and its restriction are continuous homotopies.

Equivariance and the induced map ${\iota }^C$. For $\sigma \in S_n$, $x\in F_n(D^2)$, $t\in I$ and every label $i$ one has $(H_t(\sigma \cdot x){)}_i=\lambda (t)(\sigma \cdot x{)}_i=\lambda (t)x_{{\sigma }^{-1}(i-1)+1}=(\sigma \cdot H_t(x){)}_i$, so $H_t(\sigma \cdot x)=\sigma \cdot H_t(x)$; with $t=1$ this gives ${\rho }^F(\sigma \cdot x)=\sigma \cdot {\rho }^F(x)$. Hence $p_{D^2}\circ {\iota }^F$ is constant on the fibres of $p_{\mathrm{int}{D}^2}$, and since it is continuous, [L6] factors it uniquely through a continuous ${\iota }^C$ with ${\iota }^C\circ p_{\mathrm{int}{D}^2}=p_{D^2}\circ {\iota }^F$, sending the orbit of $x$ to the same orbit viewed in $F_n(D^2)$; ${\iota }^C$ is injective because two orbits of $F_n(\mathrm{int}{D}^2)$ that coincide as subsets of $F_n(D^2)$ are equal. It is also open onto its image: for $V\subseteq C_n(\mathrm{int}{D}^2)$ open, $W:=p_{\mathrm{int}{D}^2}^{-1}(V)$ is open, $F_n(\mathrm{int}{D}^2)=(\mathrm{int}{D}^2{)}^n\cap F_n(D^2)$ is open in $F_n(D^2)$, and the saturated set ${\bigcup }_{\sigma \in S_n}\sigma (W)=p_{D^2}^{-1}({\iota }^C(V))$ is open in $F_n(D^2)$, so ${\iota }^C(V)$ is open in $C_n(D^2)$; a continuous injective open map is a homeomorphism onto its image.

Injectivity for the ordered spaces. Let $\delta$ be a loop in $F_n(\mathrm{int}{D}^2)$ at $q$ with $[{\iota }^F\circ \delta ]$ trivial, and let $F:I\times I\to F_n(D^2)$ with $F(s,0)={\iota }^F(\delta (s))$, $F(s,1)=q$ and $F(0,u)=F(1,u)=q$ be the nullhomotopy rel endpoints. Then ${\rho }^F\circ F$ is a nullhomotopy rel endpoints of ${\rho }^F\circ \delta$ in $F_n(\mathrm{int}{D}^2)$, so $[{\rho }^F\circ \delta ]=1$. Apply step 1.4 to $G(s,u):=H_u(\delta (s))$, which by step 2.1 takes values in $F_n(\mathrm{int}{D}^2)$ and satisfies $G(0,u)=G(1,u)=c(u)$ for $c(u)=H_u(q)$: it gives $\delta \ast c\simeq c\ast ({\rho }^F\circ \delta )$ rel endpoints. Since ${\rho }^F\circ \delta$ is nullhomotopic, [L7] gives $c\ast ({\rho }^F\circ \delta )\simeq c$ and hence $\delta \ast c\simeq c$; right-concatenating with $\bar{c}$ and using that $c\ast \bar{c}$ is nullhomotopic with constants absorbed, $\delta \simeq \delta \ast (c\ast \bar{c})\simeq c\ast \bar{c}\simeq c_q$. So $[\delta ]=1$ and ${\iota }_{\ast }^F$ is injective.

Claim 1. By steps 2.1 and 3.1, $H$ is a homotopy from ${\mathrm{id}}_{F_n(D^2)}=H_0$ to $H_1={\iota }^F\circ {\rho }^F$, and its restriction to $F_n(\mathrm{int}{D}^2)\times I$ is a homotopy from ${\mathrm{id}}_{F_n(\mathrm{int}{D}^2)}$ to ${\rho }^F\circ {\iota }^F$; equivalently ${\iota }^F\circ {\rho }^F\simeq {\mathrm{id}}_{F_n(D^2)}$ and ${\rho }^F\circ {\iota }^F\simeq {\mathrm{id}}_{F_n(\mathrm{int}{D}^2)}$. By [L3], ${\iota }^F$ is a homotopy equivalence with homotopy inverse ${\rho }^F$.

The descended scaling map ${\rho }^C$. Since ${\rho }^F$ is continuous and $S_n$-equivariant by step 3.2, the continuous composite $p_{\mathrm{int}{D}^2}\circ {\rho }^F$ is constant on the fibres of $p_{D^2}$: equivariance makes the images of orbit representatives belong to the same target orbit. By [L6] this composite factors uniquely through a continuous map ${\rho }^C:C_n(D^2)\to C_n(\mathrm{int}{D}^2)$ with $p_{\mathrm{int}{D}^2}\circ {\rho }^F={\rho }^C\circ p_{D^2}$.

Surjectivity for the ordered spaces. Let $\gamma$ be a loop in $F_n(D^2)$ at $q\in F_n(\mathrm{int}{D}^2)$ and put $c(u):=H_u(q)$ for $u\in I$. By steps 2.1 and 3.1, $c$ is a path in $F_n(\mathrm{int}{D}^2)$ from $q$ to ${\rho }^F(q)$ and $G(s,u):=H_u(\gamma (s))$ is a continuous map $I\times I\to F_n(D^2)$ with $G(0,u)=G(1,u)=c(u)$, $G_0=\gamma$ and $G_1={\rho }^F\circ \gamma$, so step 1.4 gives $\gamma \ast c\simeq c\ast ({\rho }^F\circ \gamma )$ rel endpoints. Then $\eta :=c\ast ({\rho }^F\circ \gamma )\ast \bar{c}$ is a loop in $F_n(\mathrm{int}{D}^2)$ at $q$, and right-concatenating that homotopy with $\bar{c}$, using [L7] that concatenation respects path homotopy and that $c\ast \bar{c}$ is nullhomotopic with constants absorbed, gives $\gamma \simeq c\ast ({\rho }^F\circ \gamma )\ast \bar{c}={\iota }^F\circ \eta$ rel endpoints. Hence $[\gamma ]={\iota }_{\ast }^F([\eta ])$ by [L8] and ${\iota }_{\ast }^F$ is surjective.

The descended homotopy $H^C$. Put $q:=p_{D^2}\times {\mathrm{id}}_I:F_n(D^2)\times I\to C_n(D^2)\times I$. This $q$ is continuous and surjective, and it is a quotient map: if $O\subseteq F_n(D^2)\times I$ is open and $(x,t)\in O$, the box basis [L5] gives a box $U\times J\subseteq O$ with $x\in U$ and $t\in J$, and by step 1.3 $p_{D^2}(U)$ is open, so the box $p_{D^2}(U)\times J$ is an open subset of $q(O)$ containing $q(x,t)$, since any $(y,t^{\mathrm{\prime }})$ in it equals $q(x^{\mathrm{\prime }},t^{\mathrm{\prime }})$ for some $x^{\mathrm{\prime }}\in U$. Hence $q(O)$ is open. The formula $H^C(y,t):=p_{D^2}(H_t(x))$ for $x\in p_{D^2}^{-1}(y)$ is well defined by the equivariance of $H$ (step 3.2), and $H^C\circ q=p_{D^2}\circ H$ is continuous, so [L6] makes $H^C:C_n(D^2)\times I\to C_n(D^2)$ continuous. It satisfies $H_0^C={\mathrm{id}}_{C_n(D^2)}$, $p_{D^2}\circ H_t=H_t^C\circ p_{D^2}$, and $H_1^C={\iota }^C\circ {\rho }^C$, the last because on classes $H_1^C([x])=[H_1(x)]=[{\rho }^F(x)]={\iota }^C({\rho }^C([x]))$ by steps 3.2 and 4.2.

Claim 3 for the ordered spaces. Let $q\in F_n(\mathrm{int}{D}^2)$ and let ${\iota }_{\ast }^F$ be the induced homomorphism of [L8]. If $n=0$ then $F_0(\mathrm{int}{D}^2)$ and $F_0(D^2)$ are one-point spaces, all their loops are constant, so both fundamental groups are one-element groups and ${\iota }_{\ast }^F$ is a bijection. If $n\ge 1$, steps 4.3 and 3.3 exhibit ${\iota }_{\ast }^F$ as surjective and injective. In both cases [L8] makes ${\iota }_{\ast }^F$ a group isomorphism.

The restricted homotopy on $C_n(\mathrm{int}{D}^2)$. Define $K:C_n(\mathrm{int}{D}^2)\times I\to C_n(\mathrm{int}{D}^2)$ by letting $K(y,t)$ be the orbit in $F_n(\mathrm{int}{D}^2)$ of $H_t(x)$ for any $x\in p_{\mathrm{int}{D}^2}^{-1}(y)$; this is well defined by step 3.2 and its values lie in $C_n(\mathrm{int}{D}^2)$ because $H_t$ maps $F_n(\mathrm{int}{D}^2)$ into itself (step 2.1). By the embedding property of step 3.2, $K$ is continuous if and only if ${\iota }^C\circ K$ is, and ${\iota }^C\circ K(y,t)=H_t^C({\iota }^C(y))$ is the composite of the continuous map ${\iota }^C\times {\mathrm{id}}_I$ with the continuous $H^C$ of step 5.1; so $K$ is continuous, with $K_0={\mathrm{id}}_{C_n(\mathrm{int}{D}^2)}$ and $K_1={\rho }^C\circ {\iota }^C$.

Claim 2. Steps 5.1 and 6.1 give ${\iota }^C\circ {\rho }^C=H_1^C\simeq H_0^C={\mathrm{id}}_{C_n(D^2)}$ and ${\rho }^C\circ {\iota }^C=K_1\simeq K_0={\mathrm{id}}_{C_n(\mathrm{int}{D}^2)}$, so by [L3] ${\iota }^C$ is a homotopy equivalence with homotopy inverse ${\rho }^C$, and the three compatibility identities of the statement hold by steps 3.2, 4.2 and 5.1.

Claim 3 for the unordered spaces. Let $y:=[q]\in C_n(\mathrm{int}{D}^2)$ and let $\gamma$ be a loop in $C_n(D^2)$ at $y$. Put $c(u):=H_u^C(y)$ and $G(s,u):=H_u^C(\gamma (s))$; by steps 5.1 and 6.1 these are continuous, $G(0,u)=G(1,u)=c(u)$, $G_0=\gamma$ and $G_1={\rho }^C\circ \gamma$ because $H_1^C={\iota }^C\circ {\rho }^C$, and $c$ is a path in $C_n(\mathrm{int}{D}^2)$ from $y$ to ${\rho }^C(y)$. Step 1.4 gives $\gamma \ast c\simeq c\ast ({\rho }^C\circ \gamma )$, and $\eta :=c\ast ({\rho }^C\circ \gamma )\ast \bar{c}$ is a loop in $C_n(\mathrm{int}{D}^2)$ at $y$ with ${\iota }^C\circ \eta \simeq \gamma$, so ${\iota }_{\ast }^C$ is surjective. For injectivity let $\delta$ be a loop in $C_n(\mathrm{int}{D}^2)$ at $y$ with ${\iota }_{\ast }^C([\delta ])=1$, witnessed by a nullhomotopy $F$ of ${\iota }^C\circ \delta$; then ${\rho }^C\circ F$ nullhomotopes ${\rho }^C\circ \delta$, and step 1.4 applied to $G(s,u):=K_u(\delta (s))$ of step 6.1 gives $\delta \ast c\simeq c\ast ({\rho }^C\circ \delta )\simeq c$, whence $\delta \simeq c_q$ by the same cancellation as in step 3.3. So ${\iota }_{\ast }^C$ is injective, and for $n=0$ both groups are one-element as in step 5.2. By [L8], ${\iota }_{\ast }^C$ is an isomorphism.

Conclusion. Claim 1 is step 4.1, claim 2 is step 7.1 and claim 3 is steps 5.2 and 7.2, so all the assertions of the statement hold.