Milnor invariants and link peripheral nilpotent quotients

Take CW structures with the basepoints as vertices and choose a maximal tree $T$ in the connected one-skeleton of $BG$, using [A1]. Send all vertices and $T$ to the basepoint of $BH$. For each oriented edge $e$ outside $T$, the tree paths to its endpoints turn $e$ into a based loop; send $e$ to a loop representing its image under $\varphi$. These loops generate ${\pi }_1(BG)$, and each two-cell attaching loop maps to the identity because $\varphi$ is a homomorphism on $G$. Thus the map extends over the two-cells. For a cell of dimension $k\ge 3$, its attaching sphere extends because ${\pi }_{k-1}(BH)=0$ by [F1]. AC chooses the tree, loop representatives, and cell extensions. The weak topology of the CW structure makes these compatible cellwise maps continuous, giving the required based map $B\varphi$.

Write $a=B\varphi \circ c_X$ and $b=c_Y\circ f$, so $a_{\ast }=b_{\ast }=\varphi$. Choose a maximal tree in $X^{(1)}$ rooted at $x_0$, and let ${\alpha }_v$ be its path from $x_0$ to each vertex $v$. Choose the vertex track from $a(v)$ to $b(v)$ to be $a({\alpha }_v{)}^{-1}b({\alpha }_v)$, with the track at $x_0$ constant. For an edge $e:v\to w$, the boundary of its homotopy square, conjugated by $a({\alpha }_v)$, represents $a_{\ast }[{\alpha }_{v}e{\alpha }_w^{-1}]b_{\ast }[{\alpha }_{v}e{\alpha }_w^{-1}{]}^{-1}=1$. Hence every such square fills. For each $k$-cell with $k\ge 2$, the boundary of its homotopy prism is a sphere $S^k\to BH$, which fills because ${\pi }_k(BH)=0$ by [F1]. AC chooses the fillings and the CW weak topology assembles them into a continuous homotopy fixed at $x_0$. Thus $a\simeq b$ by a based homotopy.

By [F3], the homotopy of step 2.1 gives $(B\varphi {)}_{\ast }\circ (c_X{)}_{\ast }=(c_Y{)}_{\ast }\circ f_{\ast }$. This is the asserted naturality, including degree two.

Fix the arbitrary map $c$ in the statement, rather than selecting a special classifying map. By [F1], its target is a connected CW $K(G,1)$, so ${\pi }_2(BG)=0$ and ${\pi }_1(c)$ is an isomorphism. For the derived-homology comparison in [F5], supply resolutions explicitly: over $\mathbb{Z}G$, take the free module on the underlying set of a module $M$, map its basis element $[m]$ to $m$, and repeat the same free-module construction on each successive kernel. The augmentations are onto and each image is the preceding kernel, so this is an exact free, hence projective, resolution for every $M$. This specified construction supplies the resolution convention; [A1] and [L1] supply DC.

Apply the external comparison [F6] to this CW model $BG$. With $T$ its universal cover, put $P_{\ast }=C_{\ast }^{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}(T)$. Choosing an oriented lift of each cell, permitted by AC, exhibits $P_{\ast }$ as free over $\mathbb{Z}G$. Contractibility makes its augmentation exact. Thus $P_{\ast }$ is a supplied resolution of the trivial left module; its comparison with the resolution in step 1.1 identifies the same derived group homology. Projection to $BG$ identifies $\mathbb{Z}{\otimes }_{\mathbb{Z}G}{P}_{\ast }$ with $C_{\ast }^{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}(BG)$, including their differentials. Taking homology, and using cellular-to-singular homology comparison as in [F6], gives $H_n(G;\mathbb{Z})\cong H_n(BG;\mathbb{Z})$.

Replace this prescribed $c$ by its mapping-path Serre fibration $p:E\to BG$, with $E\simeq X$ and $p$ corresponding to $c$. In [F2], the action on fiber components is transitive because $E$ is connected, and the component of the based fiber point has stabilizer all of ${\pi }_1(BG)$ because ${\pi }_1(c)$ is onto. Thus the fiber $F$ is path-connected. Since ${\pi }_2(BG)=0$ and ${\pi }_1(c)$ is injective, the same exact sequence gives ${\pi }_1(F)=1$. Hence $H_1(F;\mathbb{Z})=0$ by [F4].

Apply [F3] to $F\to E\to BG$. Since $F$ is path-connected, its $H_0$ local system is canonically the constant system $\mathbb{Z}$. The five-term sequence therefore has consecutive terms $H_2(E;\mathbb{Z})⟶H_2(BG;\mathbb{Z})⟶H_0(BG;{\mathcal{H}}_1(F;\mathbb{Z}))\mathrm{.}$ The last group is zero by step 2.2. Exactness makes the first map, the edge homomorphism $p_{\ast }$, surjective. Under $E\simeq X$ this map is $c_{\ast }$, and step 2.1 identifies its target with $H_2(G;\mathbb{Z})$, proving the assertion.

Parameterize component $i$ by $S^1=\mathbb{R}\mathrm{/}\mathbb{Z}$, and let $U_i=\{u_{i,1},\dots ,u_{i,m_i}\}$ be its nonempty finite set of underpassing parameters. Define $p_i(s)={\prod }_{a=1}^{m_i}(1-\cos (2\pi (s-u_{i,a})))\mathrm{.}$ Every factor is smooth and nonnegative, and it vanishes exactly when $s=u_{i,a}$ in $S^1$. Thus $p_i$ is smooth, vanishes exactly on $U_i$, and is positive somewhere because a finite subset does not exhaust $S^1$. Its maximum $P_i$ is positive by compactness. Set ${\widehat{\eta }}_i(s)=-1+p_i(s)\mathrm{/}{P}_i$. Then $-1\le {\widehat{\eta }}_i\le 0$, equality at $-1$ holds exactly on $U_i$, and all other heights are greater than $-1$.

At every projected double point, the overpassing parameter is not an underpassing parameter: regularity excludes triple points, so the two source parameters at that double point have distinct over/under roles. Hence the new height difference between the over and under branches is strictly positive. Let ${\eta }_i$ be the original height function and choose a smooth $\phi :[0,1]\to [0,1]$ equal to $0$ near $0$ and $1$ near $1$. The family $F_t(i,s)=({\gamma }_i(s),(1-\phi (t)){\eta }_i(s)+\phi (t){\widehat{\eta }}_i(s))$ is smooth and stationary near both endpoints. At each crossing its over-minus- under height is a convex combination of two positive numbers, so it remains positive. Since [F1] lists all projected multiple points, each $F_t$ is injective; its projection remains an immersion, so each compact-source map is a smooth embedding.

The track of the isotopy in step 2.1 is compact, so choose an open ball containing it and apply [F2] with that ball as the support neighborhood. The resulting ambient isotopy of ${\mathbb{R}}^3$ has compact support and therefore extends by the identity at $\mathrm{\infty }$ to an ambient isotopy of $S^3$. It carries the original link to $\widehat{L}$ and fixes the planar projection and over/under data throughout, because every slice of $F$ has those same data. This proves the stated normalized realization.

Put $U=M\setminus S$ and $V=N\cup T$. These are path-connected open sets covering $M$. Their intersection consists of $N_{-}$, $N_{+}$, and the tube joining their end balls. Cover the intersection by the two side components with the respective half of the tube. Each set is path-connected and simply connected, and their overlap is a contractible middle tube. Hence [F1] gives ${\pi }_1(U\cap V)=1$.

Join the centers of the two end balls by an embedded arc $\alpha$ in the connected smooth $3$-manifold $N$, and let $Q$ be a thin open tubular neighborhood of $\alpha$ in $N$, enlarged at its ends to meet those balls. The open sets $N$ and $W=T\cup Q$ cover $V$. The set $W$ is a thickened loop formed by the two tubes $T$ and $Q$, so ${\pi }_1(W)\cong \mathbb{Z}$; $N\cap W$ is $Q$ together with the two end balls and is contractible. Since $N$ is simply connected, [F1] gives ${\pi }_1(V)\cong \mathbb{Z}$.

Apply [F1] to the open cover $M=U\cup V$. Step 1.1 gives ${\pi }_1(U\cap V)=1$, while ${\pi }_1(U)={\pi }_1(M\setminus S)$ and step 2.1 gives ${\pi }_1(V)=\mathbb{Z}$. The pushout is therefore the free product ${\pi }_1(M\setminus S)\ast \mathbb{Z}$. A generator of the second factor is represented by $\alpha$ followed by ${\nu }^{-1}$, proving the stated description.

Choose connected CW models from [L1]. Build a based map $BG\to BQ$ inducing $G\to Q$ by mapping the one-skeleton to loops representing the quotient images, filling each two-cell because its attaching word maps to the identity in $Q$, and extending over each higher cell because ${\pi }_k(BQ)=0$ for $k>1$. The finite cell extensions are simultaneous over a set of cells; [A1] supplies those choices. Replace this map by its mapping-path fibration $p:E\to BQ$; the total space $E$ is homotopy equivalent to $BG$, and the fiber $F$ is its homotopy fiber.

The map on fundamental groups is the quotient $G\to Q$. Exactness of [F1] gives ${\pi }_0(F)=\{\ast \}$, ${\pi }_1(F)\cong \ker (G\to Q)=N$, and ${\pi }_k(F)=0$ for every $k>1$, because both $BG$ and $BQ$ have zero higher homotopy groups by [L1]. By the external input [F5], $F$ has CW homotopy type. Choose a connected CW model $F^{\mathrm{\prime }}\simeq F$ and identify ${\pi }_1(F^{\mathrm{\prime }})$ with $N$. The uniqueness theorem [F4] gives $F^{\mathrm{\prime }}\simeq BN$, so $H_q(F;\mathbb{Z})\cong H_q(N;\mathbb{Z})$ for every $q$.

A lift $g\in G$ of a loop class $q\in Q={\pi }_1(BQ)$ transports the fiber fundamental group by $n\mapsto gn{g}^{-1}$. Choose a based homotopy equivalence $h:F^{\mathrm{\prime }}\to F$ as in step 2.1 and a based homotopy inverse $\bar{h}:F\to F^{\mathrm{\prime }}$, using the basepoint path supplied by the transport. The map $T_g^{\mathrm{\prime }}=\bar{h}\circ T_g\circ h$ is a based self-map of $F^{\mathrm{\prime }}$ inducing $c_g(n)=gn{g}^{-1}$ on ${\pi }_1(F^{\mathrm{\prime }})\cong N$. By [F6], the classifying map $c:F^{\mathrm{\prime }}\to BN$ satisfies $c\circ T_g^{\mathrm{\prime }}\simeq B(c_g)\circ c$, so the monodromy on $H_q(F^{\mathrm{\prime }})$ agrees under $h$ with the bar-classifying-space map induced by $c_g$. If a different lift is chosen, it differs by an element of $N$ and changes this map by an inner automorphism of $N$. Inner conjugation is homotopic to the identity on the bar classifying space: the conjugation functor on the one-object category of $N$ is naturally isomorphic to the identity by the conjugating element. Hence the induced action on $H_q(F;\mathbb{Z})\cong H_q(N;\mathbb{Z})$ depends only on $q$, and is exactly the stated $Q$-module action.

Apply [F2] to $p:E\to BQ$. By step 2.1 the fiber homology is $H_q(N;\mathbb{Z})$, and step 3.1 identifies its monodromy local system with the conjugation action. Therefore the $E^2$ page is $H_p(Q;H_q(N;\mathbb{Z}))$ and the spectral sequence strongly converges to $H_{p+q}(E;\mathbb{Z})$.

The total-space equivalence from step 1.1 identifies $H_{p+q}(E;\mathbb{Z})$ with $H_{p+q}(BG;\mathbb{Z})$. By [F4], this classifying CW model is homotopy equivalent to the bar model used in [F3], so its homology is $H_{p+q}(G;\mathbb{Z})$. The base and fiber are connected CW models, so the Serre filtration is first quadrant and finite on each total-degree diagonal. AC supplies the DC hypothesis of [F3] by [L2]. The fiber action and Serre filtration introduce no further choice.

Supply the resolution convention needed by [F3] as follows. For each relevant group $K$ and left $\mathbb{Z}K$-module $M$, take the free module on the underlying set of $M$, augment $[m]\mapsto m$, and repeat on each successive kernel. Each augmentation is onto its kernel target, so the resulting complex is an exact free, hence projective, resolution, specified uniformly in $M$. By [A1] and [L1], DC also holds. Apply [F2] to [F1]. Since $H_0(N;\mathbb{Z})=\mathbb{Z}$ with trivial $Q$-action, its terms are $E_{2,0}^2=H_2(Q;\mathbb{Z})$, $E_{0,1}^2=H_0(Q;H_1(N;\mathbb{Z}))$, $H_1=H_1(G;\mathbb{Z})$, and $E_{1,0}^2=H_1(Q;\mathbb{Z})$. This gives an exact sequence with the coinvariant group in the middle.

By [F3], $H_1(N;\mathbb{Z})=N\mathrm{/}[N,N]$ under the map sending $n$ to its class. The $Q$-coinvariants impose the relations $[gn{g}^{-1}]=[n]$ for every lift $g\in G$ and $n\in N$. Their subgroup in $N\mathrm{/}[N,N]$ is exactly $[G,N]\mathrm{/}[N,N]$, because $[N,N]\subseteq [G,N]$. Therefore $H_0(Q;H_1(N;\mathbb{Z}))\cong N\mathrm{/}[G,N]$.

Substituting the identification of step 1.2 into step 1.1 gives the displayed sequence. To check naturality, take a homomorphism of extensions. The bar classifying-space maps give a commuting square from $BG\to BQ$ to $B{G}^{\mathrm{\prime }}\to B{Q}^{\mathrm{\prime }}$; mapping-path replacement gives a map of their fibrations. Its fiber map induces the specified homomorphism $N\to N^{\mathrm{\prime }}$ on fundamental groups. Under the fiber identifications with $BN,B{N}^{\mathrm{\prime }}$ in [F1], [F5] identifies its homology map with that induced by $N\to N^{\mathrm{\prime }}$. By [F4], the map of spectral sequences commutes with $d^2$, all edge maps, and the abutment filtration. The five-term splice in [F2] uses only these maps and the natural subquotient identifications, so its squares commute. Finally, step 1.2 sends each $n$ to its class in the coinvariants, hence commutes with the homomorphism on $N$ and its conjugation action. This proves naturality of the displayed sequence, including the transgression.

Supply the resolution convention for [F2] by taking, for every module over each relevant group ring, the free module on its underlying set, augmented $[m]\mapsto m$, and repeating this construction on successive kernels. These surjections give an exact free resolution. By [A1] and [L1], DC also holds. The case $n=1$ is the map between two trivial quotients. For $n=2$, the bar degree-one calculation in [F2] identifies $A\mathrm{/}{\mathrm{\Gamma }}_{2}A\to B\mathrm{/}{\mathrm{\Gamma }}_{2}B$ with the assumed isomorphism on $H_1$. These are the base cases.

Fix $n\ge 2$ and assume $A\mathrm{/}{\mathrm{\Gamma }}_{n}A\to B\mathrm{/}{\mathrm{\Gamma }}_{n}B$ is an isomorphism. Apply [F1] to the extensions $1\to {\mathrm{\Gamma }}_{n}A\to A\to A\mathrm{/}{\mathrm{\Gamma }}_{n}A\to 1$ and $1\to {\mathrm{\Gamma }}_{n}B\to B\to B\mathrm{/}{\mathrm{\Gamma }}_{n}B\to 1$. Since ${\mathrm{\Gamma }}_{n}X\subseteq [X,X]$ for $n\ge 2$, the maps $H_1(X)\to H_1(X\mathrm{/}{\mathrm{\Gamma }}_{n}X)$ are isomorphisms. Exactness therefore makes the transgression $H_2(X\mathrm{/}{\mathrm{\Gamma }}_{n}X)\to {\mathrm{\Gamma }}_{n}X\mathrm{/}[X,{\mathrm{\Gamma }}_{n}X]$ onto, with kernel the image of $H_2(X)$. By definition $[X,{\mathrm{\Gamma }}_{n}X]={\mathrm{\Gamma }}_{n+1}X$.

Naturality of [F1] makes the map $H_2(A\mathrm{/}{\mathrm{\Gamma }}_{n}A)\to H_2(B\mathrm{/}{\mathrm{\Gamma }}_{n}B)$ compatible with the two transgressions. It is an isomorphism because the quotient-group map is an isomorphism by the induction hypothesis. The epimorphism $H_2(A)\to H_2(B)$ makes the image of $H_2(A)$ in $H_2(A\mathrm{/}{\mathrm{\Gamma }}_{n}A)$ map onto the image of $H_2(B)$ in $H_2(B\mathrm{/}{\mathrm{\Gamma }}_{n}B)$. The induced map on the two cokernels is consequently an isomorphism. By step 1.2 and exactness, those cokernels are ${\mathrm{\Gamma }}_{n}A\mathrm{/}{\mathrm{\Gamma }}_{n+1}A$ and ${\mathrm{\Gamma }}_{n}B\mathrm{/}{\mathrm{\Gamma }}_{n+1}B$.

The inclusions and quotient maps give exact central extensions $1\to {\mathrm{\Gamma }}_{n}X\mathrm{/}{\mathrm{\Gamma }}_{n+1}X\to X\mathrm{/}{\mathrm{\Gamma }}_{n+1}X\to X\mathrm{/}{\mathrm{\Gamma }}_{n}X\to 1,X=A,B,$ because $[X,{\mathrm{\Gamma }}_{n}X]={\mathrm{\Gamma }}_{n+1}X$. The maps on the kernel and quotient terms are isomorphisms by step 2.1 and the induction hypothesis. The middle map is injective: an element in its kernel projects to the identity in the quotient, hence lies in the kernel term, where injectivity holds. It is surjective: for a target element, lift its quotient class through the quotient isomorphism and then correct the chosen middle lift by a kernel element using the kernel isomorphism. Thus $A\mathrm{/}{\mathrm{\Gamma }}_{n+1}A\to B\mathrm{/}{\mathrm{\Gamma }}_{n+1}B$ is an isomorphism.

Steps 1.1 and 3.1 prove by induction the isomorphism for every finite $n$. Apart from the inherited AC used to obtain the homology sequence [F1], the induction uses only exactness and a single element lift at a time; no additional choice principle is used.

Let $A_L^{\circ }=(({\mathbb{R}}^2\times (-1,\mathrm{\infty }))\cup \{\mathrm{\infty }\})\setminus L_D$. By [F1], ${\pi }_1(A_L^{\circ })\cong {\pi }_1(A_L)$. A small neighborhood $H$ of $\mathrm{\infty }$ in $A_L^{\circ }$ is the open upper half-ball together with its cone vertex $\mathrm{\infty }$, disjoint from $L_D$; other points of the level-sphere boundary have been deleted. The open cover $A_L^{\circ }=M\cup H$ has intersection $H\setminus \{\mathrm{\infty }\}$, which is homeomorphic to $(0,1)\times \mathrm{int}(S_{+}^2)$ and is contractible because the open hemisphere $\mathrm{int}(S_{+}^2)$ is an open disk. Since $H$ is also contractible, [F2] gives ${\pi }_1(M)\cong {\pi }_1(A_L^{\circ })$.

Each $D_j$ is the image of the open region $(s,t)\mapsto (\gamma (s),-1+t(\eta (s)+1))$ with $s\in I_j$ and $0<t<1$; this is an embedded open disk. At a projected double point, the underpassing branch has height $-1$ and is absent from $M$, while the overpassing branch has height greater than $-1$. Thus sheets over distinct intervals are disjoint. Each sheet is closed in $M$: its only limit points outside itself lie on the excluded plane $z=-1$ or on the excluded link. Hence it is a proper smooth embedded simply connected surface.

At a fixed vertical line, every point of $Z$ lies above all link points on that line. Since every link height is at most $0$, the straight homotopy $(x,y,z)\mapsto (x,y,(1-t)z+t)$ ($0\le t\le 1$) stays above those link points and contracts $Z$ onto the plane $z=1$. This plane is contained in $Z$, so $Z$ is contractible.

For $j=1,\dots ,n$, put $M_j=M\setminus {\bigcup }_{k>j}{D}_k$. Then $M_0=Z$ and $M_j=M_{j-1}\cup D_j$. At each stage $M_{j-1}$ is path-connected: it contains path-connected $Z$, and every point of an earlier sheet joins to $Z$ along a short normal segment. Orient $D_j$ by its interval and height coordinates. The ambient orientation orients its normal line, so that line bundle is trivial; a bundle metric gives the unique positive unit vector in each fibre. By [A1, F3], choose a tubular neighborhood disjoint from the other sheets. Radial contraction shows the whole neighborhood and each of its two side components are simply connected. Choose a small positive meridian circle around an interior point of the $j$th strand in a ball disjoint from all other sheets. It meets $D_j$ transversely once. Remove a short open arc about that intersection; the remaining compact arc lies in $M_{j-1}$ and joins the two sides of $D_j$. Shrink the tubular neighborhood so that it meets this arc only in two end balls. This gives the corridor required by [F4], and its complement hypothesis holds because $M_j\setminus D_j=M_{j-1}$ is path-connected.

The cut lemma [F4] gives ${\pi }_1(M_j)\cong {\pi }_1(M_{j-1})\ast \mathbb{Z}$. Its cyclic generator is represented by the short meridian arc crossing $D_j$ followed by the complementary arc from step 3.1. Joining this loop to the basepoint by its access path in $Z$ gives exactly $x_j$, with positive orientation fixed by [F5]. Since ${\pi }_1(M_0)={\pi }_1(Z)=1$, induction yields that ${\pi }_1(M)$ is free on $x_1,\dots ,x_n$. Step 1.1 identifies this with ${\pi }_1(A_L)$.

Let $H=⟨m_1,\dots ,m_r⟩$. By [F1] and [F2], the images of these meridians generate $G\mathrm{/}{\mathrm{\Gamma }}_{2}G$, so $G=H{\mathrm{\Gamma }}_{2}G$.

For every $k\ge 2$, the quotient ${\mathrm{\Gamma }}_{k}G\mathrm{/}{\mathrm{\Gamma }}_{k+1}G$ is generated by iterated commutators of weight $k$ in the selected meridians. For $k=2$, [F4] makes the pairing $G\mathrm{/}{\mathrm{\Gamma }}_2\times G\mathrm{/}{\mathrm{\Gamma }}_2\to {\mathrm{\Gamma }}_2\mathrm{/}{\mathrm{\Gamma }}_3$ biadditive, so [F1] implies this layer is generated by brackets of the $m_i$. If the claim holds at weight $k$, then [F3] writes the next layer as generated by brackets of a degree-one class with a weight-$k$ class; [F4] makes that pairing biadditive, so [F1] again reduces the degree-one entries to the $m_i$, and the induction hypothesis reduces the other entries to brackets of those meridians. Each such iterated commutator is a word in $H$ lying in ${\mathrm{\Gamma }}_{k+1}G$, so its class is in the image of $H\cap {\mathrm{\Gamma }}_{k+1}G$ in the graded layer.

The conclusion of step 2.1 implies ${\mathrm{\Gamma }}_{k}G\subseteq H{\mathrm{\Gamma }}_{k+1}G$ for every $k\ge 2$, because each generator of the graded layer is represented by a word in the $m_i$ that lies in $H$. Starting from $G=H{\mathrm{\Gamma }}_{2}G$ in step 1.1 and applying these inclusions successively gives $G=H{\mathrm{\Gamma }}_{q}G$ for every finite $q$. Therefore the image of $H$, and hence of the free group map in the statement, is all of $G\mathrm{/}{\mathrm{\Gamma }}_{q}G$. For $q=1$ the quotient is trivial.

In any filtered associative ring, write $U=1+a$ and $V=1+b$ with $a\in {\mathfrak{m}}^p$ and $b\in {\mathfrak{m}}^q$. Since $U,V$ are units, $[U,V]-1=(UV-VU)U^{-1}{V}^{-1}=(ab-ba)U^{-1}{V}^{-1}\mathrm{.}$ By [F1], $ab-ba\in {\mathfrak{m}}^{p+q}$, and multiplication by units preserves this two-sided ideal. Therefore $[U,V]\in 1+{\mathfrak{m}}^{p+q}$. This is a ring calculation conditional on the two degree bounds; no lower-central inclusion is assumed here.

By [F1], $E({\mathrm{\Gamma }}_1{F}_r)=E(F_r)\subseteq 1+\mathfrak{m}$. Assume $E({\mathrm{\Gamma }}_q{F}_r)\subseteq 1+{\mathfrak{m}}^q$. For a generator $[u,v]$ of ${\mathrm{\Gamma }}_{q+1}{F}_r=[F_r,{\mathrm{\Gamma }}_q{F}_r]$, [F1] gives $E(u)-1\in \mathfrak{m}$ and the induction hypothesis gives $E(v)-1\in {\mathfrak{m}}^q$. The ring calculation in step 1.1 therefore puts its image in $1+{\mathfrak{m}}^{q+1}$. Products and inverses remain in that subgroup: their images modulo the two-sided ideal ${\mathfrak{m}}^{q+1}$ are all $1$. Thus $E({\mathrm{\Gamma }}_{q+1}{F}_r)\subseteq 1+{\mathfrak{m}}^{q+1}$, and induction proves the inclusion for every $q$. Now for arbitrary $u\in {\mathrm{\Gamma }}_p{F}_r$ and $v\in {\mathrm{\Gamma }}_q{F}_r$, the established inclusions give both hypotheses of step 1.1, which yields $E([u,v])\in 1+{\mathfrak{m}}^{p+q}$.

If two words represent the same element of $F_r\mathrm{/}{\mathrm{\Gamma }}_{d+1}{F}_r$, their quotient lies in ${\mathrm{\Gamma }}_{d+1}{F}_r$. By step 2.1 its expansion is $1$ modulo ${\mathfrak{m}}^{d+1}$. Multiplicativity of $E$ then makes the two truncated expansions equal, proving representative-independence of every coefficient of degree at most $d$.