String-link end inclusions identify finite nilpotent peripheral quotients

By [F1], each $f_{\pm }$ induces an isomorphism on topological $H_1$ and an epimorphism on topological $H_2$. Naturality of [F2] identifies the $H_1$ map with the map between the abelianizations of ${\pi }_1(P_{\pm })$ and ${\pi }_1(X_{\tau })$. Thus each group homomorphism $(f_{\pm }{)}_{\ast }$ is an isomorphism on group $H_1$.

By [F6], $P_{\pm }$ and $X_{\tau }$ have CW structures, so choose classifying maps $c_{\pm }:P_{\pm }\to B{\pi }_1(P_{\pm })$ and $c_X:X_{\tau }\to B{\pi }_1(X_{\tau })$. For either sign and any $y\in H_2({\pi }_1(X_{\tau });\mathbb{Z})$, choose $z\in H_2(X_{\tau };\mathbb{Z})$ with $(c_X{)}_{\ast }(z)=y$ using [F3]. By [F1] choose $w\in H_2(P_{\pm };\mathbb{Z})$ with $(f_{\pm }{)}_{\ast }(w)=z$. Write ${\varphi }_{\pm }=(f_{\pm }{)}_{\ast }$. Naturality [F4] gives $(B{\varphi }_{\pm }{)}_{\ast }((c_{\pm }{)}_{\ast }(w))=(c_X{)}_{\ast }((f_{\pm }{)}_{\ast }(w))=y\mathrm{.}$ Thus every class $y$ in $H_2({\pi }_1(X_{\tau });\mathbb{Z})$ is in the image of the induced group-homology map $H_2({\pi }_1(P_{\pm });\mathbb{Z})\to H_2({\pi }_1(X_{\tau });\mathbb{Z})$.

By [F5], each $(f_{\pm }{)}_{\ast }$ now satisfies both Stallings hypotheses. Applying it separately to $f_{-}$ and $f_{+}$ proves that each induced map on $F_r\mathrm{/}{\mathrm{\Gamma }}_q$ is an isomorphism for every finite $q\ge 1$.

Since both end maps are isomorphisms at every such $q$, their inverse/composite is defined and is an isomorphism. The meridian labels and basing path are carried through the same maps, so this is exactly the stated peripheral comparison. The proof uses full AC only through the classifying-space and Stallings suppliers; no additional choice is used in the elementwise quotient argument.

Choose arbitrary independently based positive endpoint meridians. Their map $a:F_r\to {\pi }_1(P_{\pm })$ is the identity on the ordered $H_1$ basis by [F1] and [F2]. Also $H_2({\pi }_1(P_{\pm });\mathbb{Z})=0$: [F1] gives $H_2(P_{\pm })=0$, and [F3] maps that group onto the group-homology group. Thus $a$ is onto on $H_2$, independently of its source's $H_2$. Apply [F5] to $a$. It induces an isomorphism at every finite lower-central quotient, even though $a$ need not be an isomorphism on the full groups. Compose it with the already proved end isomorphism of step 3.1. This proves the arbitrary-basing branch and its stated inverse/composite, using exactly the selected meridian conjugations rather than an incorrect free-basis assertion.