A choice-free continuous section of planar coordinate forgetting

The plane section. Define $s(z_1,\dots ,z_{n-1}):=(z_1,\dots ,z_{n-1},1+{\sum }_{i<n}\mathrm{\mid }{z}_i\mathrm{\mid })$ on $F_{n-1}(\mathbb{C})$. The last coordinate is a positive real number and $1+{\sum }_i\mathrm{\mid }{z}_i\mathrm{\mid }>\mathrm{\mid }{z}_i\mathrm{\mid }$ for every $i$, so it differs from each of $z_1,\dots ,z_{n-1}$; the first $n-1$ coordinates are pairwise distinct because $(z_1,\dots ,z_{n-1})\in F_{n-1}(\mathbb{C})$ by [F1]. Hence $s$ takes values in $F_n(\mathbb{C})$. The absolute-value and sum operations are continuous, and a tuple of continuous coordinate maps is continuous, so $s$ is continuous; forgetting the last coordinate returns the given tuple, that is $p\circ s=\mathrm{id}$.

Complements of finite sets in the disc are path-connected. Let $Q\subseteq \mathrm{int}{D}^2$ be finite and let $x,y\in \mathrm{int}{D}^2\setminus Q$. If $x=y$ the constant path joins them, so assume $x\mathrm{\ne }y$ and choose $r$ with $\max \{\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\}<r<1$ and $r\mathrm{\ne }\mathrm{\mid }w\mathrm{\mid }$ for every $w\in Q$; only finitely many radii are forbidden, so such an $r$ exists, and then the circle $C_r=\{z:\mathrm{\mid }z\mathrm{\mid }=r\}$ is disjoint from $Q$ and contains $x,y$ in its interior. For each $w\in Q$ let $R_x(w):=\{x+t(w-x):t\ge 1\}$ and $R_y(w):=\{y+t(w-y):t\ge 1\}$ be the rays from $x$ and from $y$ through $w$ extended beyond $w$; each meets $C_r$ in at most one point, so only finitely many points of $C_r$ are excluded. Choose $z\in C_r$ outside this finite excluded set. If some $w\in Q$ lay on the segment $[x,z]$, then $z=x+t(w-x)$ with $t\ge 1$, contradicting the exclusion of $z$; thus $[x,z]\cap Q=\mathrm{\varnothing }$, and likewise $[z,y]\cap Q=\mathrm{\varnothing }$. Both segments lie in $\mathrm{int}{D}^2$ because that disc is convex and all three endpoints do, so the concatenation $[x,z]\cup [z,y]$ is a path in $\mathrm{int}{D}^2\setminus Q$ from $x$ to $y$.

Conjugation and the constant loop. By [F3] every path $c$ from $x_0$ to $x_1$ gives an isomorphism ${\phi }_c:{\pi }_1(X,x_0)\to {\pi }_1(X,x_1)$ with inverse ${\phi }_{\bar{c}}$; by [F4] the constant loop at a point represents the identity class, so if $c$ is the constant path at $x_0$ then ${\phi }_c$ is the identity map of ${\pi }_1(X,x_0)$, since $(\bar{c}\ast \alpha )\ast c$ differs from $\alpha$ only by insertions of constant loops at the endpoints.

The disc section. Put $s^{\mathrm{\prime }}:=h^{(n)}\circ s\circ (h^{-1}{)}^{(n-1)}$ on $F_{n-1}(\mathrm{int}{D}^2)$, where $h^{(k)}(u_1,\dots ,u_k):=(h(u_1),\dots ,h(u_k))$; explicitly $s^{\mathrm{\prime }}(v_1,\dots ,v_{n-1})=(v_1,\dots ,v_{n-1},h(1+{\sum }_{i<n}\mathrm{\mid }{u}_i\mathrm{\mid }))$ with $u_i=h^{-1}(v_i)$. This is continuous as a composite of continuous maps, and it takes values in $F_n(\mathrm{int}{D}^2)$: the last coordinate $h(1+{\sum }_i\mathrm{\mid }{u}_i\mathrm{\mid })$ lies in $\mathrm{int}{D}^2$, and it differs from $v_i=h(u_i)$ because $h$ is injective and $1+{\sum }_i\mathrm{\mid }{u}_i\mathrm{\mid }=u_i$ is impossible — taking moduli would give $1+{\sum }_i\mathrm{\mid }{u}_i\mathrm{\mid }=\mathrm{\mid }{u}_i\mathrm{\mid }\le {\sum }_i\mathrm{\mid }{u}_i\mathrm{\mid }$. Composing with $\tilde{p}$ returns the given tuple, so $\tilde{p}\circ s^{\mathrm{\prime }}=\mathrm{id}$; thus $s^{\mathrm{\prime }}$ is a continuous section of $\tilde{p}$, transported from $s$ as defined.

The based splitting. Let $F:={\tilde{p}}^{-1}(q^{\mathrm{\prime }})=\{(q_1,\dots ,q_{n-1},x):x\in \mathrm{int}{D}^2\setminus \{q_1,\dots ,q_{n-1}\}\}$ be the fibre over $q^{\mathrm{\prime }}$; it contains $q$, since $q_n\mathrm{\ne }{q}_i$ for $i<n$, and it contains $s^{\mathrm{\prime }}(q^{\mathrm{\prime }})$ by step 2.1. The fibre is homeomorphic to $\mathrm{int}{D}^2\setminus \{q_1,\dots ,q_{n-1}\}$ and hence path-connected by step 1.2 applied to the finite set $Q=\{q_1,\dots ,q_{n-1}\}$. Choose a path $\alpha :I\to F$ from $q$ to $s^{\mathrm{\prime }}(q^{\mathrm{\prime }})$; such a path exists, and choosing it is a single selection, not an instance of AC. Write $\iota :F\to F_n(\mathrm{int}{D}^2)$ for the inclusion and ${\phi }_{\alpha }:{\pi }_1(F_n(\mathrm{int}{D}^2),q)\to {\pi }_1(F_n(\mathrm{int}{D}^2),s^{\mathrm{\prime }}(q^{\mathrm{\prime }}))$ for the conjugation isomorphism of [F3], and define $\sigma :={\phi }_{\bar{\alpha }}\circ s_{\ast }^{\mathrm{\prime }}:{\pi }_1(F_{n-1}(\mathrm{int}{D}^2),q^{\mathrm{\prime }})\to {\pi }_1(F_n(\mathrm{int}{D}^2),q)$, where $s_{\ast }^{\mathrm{\prime }}$ is induced at the basepoint $q^{\mathrm{\prime }}$ and ${\phi }_{\bar{\alpha }}([\beta ])=[(\alpha \ast \beta )\ast \bar{\alpha }]$. For $[\gamma ]\in {\pi }_1(F_{n-1}(\mathrm{int}{D}^2),q^{\mathrm{\prime }})$, the path $\tilde{p}\circ ((\alpha \ast (s^{\mathrm{\prime }}\circ \gamma ))\ast \bar{\alpha })=(\tilde{p}\circ \alpha )\ast (\tilde{p}\circ s^{\mathrm{\prime }}\circ \gamma )\ast (\tilde{p}\circ \bar{\alpha })$ has the constant paths $\tilde{p}\circ \alpha$ and $\tilde{p}\circ \bar{\alpha }$ at $q^{\mathrm{\prime }}$ as outer factors, because $\alpha$ lies in the fibre over $q^{\mathrm{\prime }}$, and its middle factor is $\tilde{p}\circ s^{\mathrm{\prime }}\circ \gamma =\gamma$ by step 2.1; by [F4] and step 1.3 this class equals $[\gamma ]$ in ${\pi }_1(F_{n-1}(\mathrm{int}{D}^2),q^{\mathrm{\prime }})$, so ${\tilde{p}}_{\ast }\circ \sigma =\mathrm{id}$ and ${\tilde{p}}_{\ast }$ is split surjective.