The maximal abelian cover retracts onto the cover over the compact link exterior

For $x={\mathrm{\Phi }}_i(v)\in N(K_i)\setminus K_i$, write $\rho =\parallel v\parallel >0$ and define $F_s(x)={\mathrm{\Phi }}_i\text{\hspace{0.17em}⁣}\left(((1-s)\rho +s{ϵ}_i)\frac{v}{\rho }\right),0\le s\le 1.$ On $E_L$ set $F_s(x)=x$. The formula is continuous on each punctured tube and agrees with the identity on its boundary $\rho ={ϵ}_i$. The products with $I$ of the finitely many punctured closed tubes and $E_L$ form a finite closed cover of $X_L\times I$; [F5] therefore makes $F:X_L\times I\to X_L$ continuous. It starts at the identity, fixes $E_L$ pointwise, and at $s=1$ maps every punctured tube to its boundary, so it is a strong deformation retraction onto $E_L$.

The link complement is path-connected by [F6], so its deformation retract $E_L$ is path-connected. The tube construction makes $E_L$ a smooth manifold with boundary, hence locally path-connected and semilocally simply connected. By [F8] and [F9], inclusion induces an isomorphism $i_{\ast }:{\pi }_1(E_L,b)\to {\pi }_1(X_L,b)$. By [F8] and [F10], it also induces an isomorphism $i_{\ast }:H_1(E_L;\mathbb{Z})\to H_1(X_L;\mathbb{Z})$. The positive meridians lie in $E_L$ and map to the same classes, so the meridian basis is preserved. An $E_L$-loop closes in the restricted cover exactly when its image under $i_{\ast }$ lies in $\ker \varphi$, so the component containing $\tilde{b}$ induces the subgroup $i_{\ast }^{-1}(\ker \varphi )=\ker (\varphi \circ i_{\ast })\mathrm{.}$ This subgroup is exactly the kernel of the abelianization of ${\pi }_1(E_L,b)$ because $i_{\ast }$ is an isomorphism. To check connectedness, take $y\in E_L$, a path $\delta$ in $E_L$ from $b$ to $y$, and a lift $\tilde{\delta }$ from a chosen point $\tilde{b}$ over $b$. Any other point over $y$ is $h\tilde{\delta }(1)$ for some deck transformation $h$, by [F7]. Surjectivity of $i_{\ast }$ gives a loop $\alpha$ in $E_L$ whose lift from $\tilde{b}$ ends at $h\tilde{b}$. The reverse of $\tilde{\delta }$, this lifted loop, and $h\tilde{\delta }$ concatenate to a path in $p^{-1}(E_L)$ from $\tilde{\delta }(1)$ to $h\tilde{\delta }(1)$. Thus the restriction is connected. By [F4] it is the connected maximal abelian cover of $E_L$, with the same deck group $H_1(X_L;\mathbb{Z})\cong H_1(E_L;\mathbb{Z})$.

Apply [F2] to $F$ with initial lift the identity of $X_L^{\mathrm{a}\mathrm{b}}$. For a point over $E_L$, its constant path is a lift of its path under $F$; uniqueness makes the lifted homotopy fix every point of $p^{-1}(E_L)$. At time $1$ the image lies in $p^{-1}(E_L)$, so the lifted homotopy is a strong deformation retraction of $X_L^{\mathrm{a}\mathrm{b}}$ onto that restricted cover.

For every deck transformation $h$, the maps $(x,s)\mapsto h{\tilde{F}}_s(x)$ and $(x,s)\mapsto {\tilde{F}}_s(hx)$ lift the same homotopy $F(p(x),s)$ and agree at $s=0$. Uniqueness in [F2] makes them equal. The retraction is therefore deck-equivariant, and its induced isomorphism on first homology is ${\mathrm{\Lambda }}_r$-linear. By [F1] the target homology of the full cover is $A_L$, proving the asserted module isomorphism.