Satellite operators and concordance filtrations

The supplied collars give an open cover of $Y$ by a neighborhood of $W$ and a neighborhood of $X$, retracting onto $W,X$, with intersection retracting onto $A$. In collar coordinates this is obtained by adding a short collar segment across the seam to each side; shrinking its height linearly gives the three retractions. Hence [F1] gives the exact window $H_2(A;R)\stackrel{(i_W,-i_X)}{\to }{H}_2(W;R)\oplus H_2(X;R)\stackrel{j_W+j_X}{\to }{H}_2(Y;R)\stackrel{\delta }{\to }{H}_1(A;R)\stackrel{(i_W,-i_X)}{\to }{H}_1(W;R)\oplus H_1(X;R)\mathrm{.}$ The last arrow is injective, since its $X$ coordinate is injective by hypothesis. Exactness therefore says that $j_W+j_X$ is surjective. This statement is elementwise and does not choose a simultaneous family of preimages.

Write $Q_Y=H_2(Y;R)\mathrm{/}\mathrm{im}{H}_2(D\bigsqcup B;R)$ and let $\pi$ be its quotient map. Every $x\in H_2(X;R)$ is the image of a class in $H_2(B;R)$, by hypothesis. Its image in $H_2(Y;R)$ therefore lies in the boundary image, so $\pi j_X=0$. For any $y\in H_2(Y;R)$ choose, for this one $y$, a preimage $(w,x)$ under the surjection of step 1.1. Then $\pi (y)=\pi j_W(w)$. Thus $\pi j_W:H_2(W;R)\to Q_Y$ is surjective.

A class from $H_2(D;R)$ maps to zero under $\pi j_W$ because $D$ is a component union of $\mathrm{\partial }Y$. A class $a\in H_2(A;R)$ has $j_W{i}_W(a)=j_X{i}_X(a)$ by the zero composite of the first two Mayer–Vietoris arrows. step 2.1 makes the latter class zero in $Q_Y$. Since $\mathrm{\partial }W=A\bigsqcup D$, every boundary class of $W$ is a sum of these two kinds, so $\pi j_W$ vanishes on the full boundary image. It therefore factors uniquely through the displayed source quotient, and the factored map is surjective by step 2.1. This proves the claimed canonical map, without any assumption on $H_2(A;R)\to H_2(X;R)$.

If the source quotient is a vector space with a finite basis of size $r$, its $r$ images span the target. Finite elimination deletes linearly dependent images until a basis remains, so the target has dimension at most $r$. Only finitely many operations and selections occur. The proof also covers an empty seam or empty outer boundary, zero homology groups and the zero coefficient ring. No step selects representatives for infinitely many classes; surjectivity is used for a single class at a time. Thus the result and the finite-dimensional consequence require no choice principle.

The finite inertia calculations used below require no spectral theorem for an infinite-dimensional space. A nonzero Hermitian form on ${\mathbb{C}}^n$ has a vector with nonzero real value: if its diagonal values on the coordinate vectors all vanish, a nonzero off-diagonal entry is detected by a sum $e_i+e_j$ or $e_i+i{e}_j$. Split off the line of such a vector by subtracting its orthogonal projection, and repeat in the orthogonal complement. Finite induction gives a diagonal form with positive, negative and zero entries. If there are $p$ positive entries, the positive coordinate subspace has dimension $p$, and any subspace of dimension greater than $p$ meets the span of the nonpositive coordinates, so cannot be positive definite. Thus $p$ is the maximal positive dimension; the same argument gives the maximal negative dimension. Congruence preserves these maxima by an invertible change of coordinates, and the diagonal description proves additivity under block sum and sign reversal under negation. For a Hermitian $2\times 2$ matrix, its real eigenvalues are $(\mathrm{tr}C\pm \sqrt{(C_{11}-C_{22}{)}^2+4\mathrm{\mid }{C}_{12}{\mathrm{\mid }}^2})\mathrm{/}2$; their product is its determinant. Hence the trace/determinant tests used below follow explicitly. Every splitting is within a fixed finite-dimensional space.

For integers $b>0$ and $a$, put $w=a-b$, $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, $B=\left(\begin{array}{cc}1& w\\ w& w^2+b\end{array}\right)$ and $V(a,b)=\left(\begin{array}{cc}A& I_2\\ 0& B\end{array}\right)$. Both diagonal blocks are integral and symmetric, so $V(a,b)-V(a,b{)}^T=\left(\begin{array}{cc}0& I_2\\ -I_2& 0\end{array}\right)$ has determinant one. Also $B$ is positive definite: its real quadratic form is $(x+wy{)}^2+b{y}^2$, and the same expression with absolute squares works over $\mathbb{C}$. Its determinant is $b$ and its trace is positive. All these assertions follow directly from the displayed matrices.

For $0<\theta \le \pi$ write $c=1-e^{i\theta }$ and $h=2-2\cos \theta >0$, so $\mathrm{\mid }c{\mathrm{\mid }}^2=h$. The Hermitian matrix of $V(a,b)$ is $\left(\begin{array}{cc}hA& c{I}_2\\ \bar{c}{I}_2& hB\end{array}\right)$. Subtracting the appropriate multiple of the lower block from the upper block, and the conjugate multiple of its column, is an invertible block congruence to $(hA-B^{-1})\oplus hB$. Its second block contributes signature two. Put $C=hA-B^{-1}$. Since $\mathrm{tr}A=0$ and $B^{-1}$ is positive definite, $\mathrm{tr}C=-\mathrm{tr}{B}^{-1}<0$. Direct determinant expansion, using $\mathrm{tr}(BA)=2w$ and $\det (BA)=-b$, gives $\det C=b^{-1}\det (hBA-I_2)=b^{-1}(1-2wh-b{h}^2)\mathrm{.}$ The polynomial has exactly one positive root $s(a,b)=\frac{-w+\sqrt{w^2+b}}{b}\mathrm{.}$ For $0<h<s(a,b)$ its determinant is positive and its trace negative, so its two real eigenvalues are negative. At $h=s(a,b)$ it has one negative and one zero eigenvalue. For $h>s(a,b)$ its determinant is negative, so it has one positive and one negative eigenvalue. The signature of $H_{\theta }(V(a,b))$ is consequently $0,1,2$ in those three cases, respectively. At $\theta =0$ the Hermitian matrix is zero and its signature is zero.

The positive thresholds $s(a,b)$ are dense in $(0,4)$. Indeed fix $r\in (-1,1)$ and set $a_b=\lfloor br\rfloor$ for positive integers $b$. For all sufficiently large $b$, $a_b\mathrm{/}b<1$; put $d_b=1-a_b\mathrm{/}b>0$. Then $s(a_b,b)=d_b+\sqrt{d_b^2+1\mathrm{/}b},\mathrm{\mid }s(a_b,b)-2(1-r)\mathrm{\mid }\le 2\mathrm{/}b+1\mathrm{/}\sqrt{b}\mathrm{.}$ Here $\mathrm{\mid }{a}_b\mathrm{/}b-r\mathrm{\mid }\le 1\mathrm{/}b$ and $0\le \sqrt{d_b^2+1\mathrm{/}b}-d_b\le 1\mathrm{/}\sqrt{b}$, which prove the bound. For any desired positive tolerance a sufficiently large integer makes its right side smaller than that tolerance. Given any target in $(0,4)$, take $r$ to be one minus half the target; the explicit floor formula then supplies thresholds arbitrarily close to it. This is a sequence defined by a formula, without selecting any infinite family of witnesses.

The function $h(\theta )=2-2\cos \theta$ is continuous and strictly increasing from $0$ to $4$ on $[0,\pi ]$: its derivative is $2\sin \theta >0$ in the interior. Choose two real angles $\alpha <{\theta }_0<\beta$ in $I$ and small disjoint open intervals around them whose closures are contained respectively in $I\cap (0,{\theta }_0)$ and $I\cap ({\theta }_0,\pi )$. Their $h$ images are nonempty intervals in $(0,4)$. Step 3.1 supplies a pair $(a_{-},b_{-})$ with threshold in the first image and a pair $(a_{+},b_{+})$ with threshold in the second. If desired these finite witnesses can be specified by enumerating integer pairs by increasing $\mathrm{\mid }a\mathrm{\mid }+b$ and then increasing $a$ and taking the first pair whose explicit threshold is in the relevant interval. Thus the corresponding unique crossing angles satisfy ${\theta }_{-}<{\theta }_0<{\theta }_{+}$ and lie in $I$.

Define $V=V(a_{-},b_{-})\oplus (-V(a_{+},b_{+}))$. Each skew block has determinant one, also after negation, so $\det (V-V^T)=1$. Its Hermitian signature is the difference of the two signatures in step 2.1. Since the thresholds are ordered, it is zero below ${\theta }_{-}$, one at ${\theta }_{-}$, two on $({\theta }_{-},{\theta }_{+})$, one at ${\theta }_{+}$ and zero above ${\theta }_{+}$. It is also zero at zero. This proves all claims on $[0,\pi ]$, including nonzero value at ${\theta }_0$ and vanishing off $I$. Finally $V$ is real, so $H_{-\theta }(V)=\overline{H_{\theta }(V)}$; conjugation preserves the real eigenvalues and their multiplicities, proving symmetry. The only constructions were finite matrices, explicit integer formulas and two individual integer witnesses. Thus no choice or realization theorem has been used.

On the peripheral torus $T$, the supplied meridian and longitude give $H_1(T;\mathbb{Z})=\mathbb{Z}[\mu ]\oplus \mathbb{Z}[\lambda ]$, as its finite square cell model shows. The tube retracts onto its core and has $H_1(N(K);\mathbb{Z})=\mathbb{Z}$. The longitude bounds the trimmed parallel Seifert surface in $X_K$, so its class there is zero, while it maps to the core generator in $N(K)$; the meridian maps to zero in that tube. Apply [F1] to $S^3=X_K\cup N(K)$ with slightly thickened collars. The segment $0=H_2(S^3)\to H_1(T)\to H_1(X_K)\oplus H_1(N(K))\to H_1(S^3)=0$ makes its middle map an isomorphism. Its values on the two displayed generators therefore give $H_1(X_K)=\mathbb{Z}[\mu ]$ exactly, including the marked generator.

For normal generation apply [F3] to $S^3=X_K\cup N(K)$. In the tube group the meridian becomes $1$ and the longitude becomes its core $c$, so the pushout is $({\pi }_1(X_K)\ast ⟨c⟩)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨\mu ,\lambda c^{-1}⟩\text{\hspace{0.17em}⁣}⟩$, naturally ${\pi }_1(X_K)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨\mu ⟩\text{\hspace{0.17em}⁣}⟩$. This group is trivial: cover $S^3$ by the complements of its north and south poles; each is ${\mathbb{R}}^3$ and contracts, and their overlap is path-connected, so [F3] gives ${\pi }_1(S^3)=1$. This proves normal generation of the exterior group. In the zero filling the meridian becomes the filling core and the longitude becomes $1$; the identical pushout calculation gives ${\pi }_1(M(K))={\pi }_1(X_K)\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨\lambda ⟩\text{\hspace{0.17em}⁣}⟩$. It too is normally generated by the image of $\mu$. No incompressibility or injection assertion about a torus group is used.

Signed intersection with the properly trimmed Seifert surface defines a homomorphism $H_1(X_K;\mathbb{Z})\to \mathbb{Z}$ taking $\mu$ to one. In this finite model its chain-level justification is explicit. Undo the supplied finite local roundings, work with its finite polygonal faces, and move each fixed finite loop or chain into general position with these faces by finite subdivision and small piecewise-linear perturbations. Each perturbation excludes only finitely many improper affine incidences, so finite rational choices suffice. These operations preserve the homology class by the straight simplex prisms in sufficiently small convex coordinate boxes. For a two-chain in the interior of $X_K$, its transverse intersection with the oriented surface is a finite oriented one-chain. Its boundary is the intersection with the two-chain boundary: no boundary term comes from the surface boundary, which lies on $T$ outside that interior. The signed sum of this one-chain's endpoints is zero, so intersection counts agree on homologous one-cycles. Boundary collars move any fixed finite chain into the interior before this calculation. The meridian meets the surface once with positive sign by the supplied peripheral orientation. Thus step 1.1 makes this intersection homomorphism an isomorphism. All loops in a small positive parallel copy $F^{+}$ of the trimmed surface miss the original surface, so their images in $H_1(X_K)$ are zero. The parallel copy and its boundary longitude are supplied by the finite oriented ribbon charts and tube.

By [F2], step 2.1 implies that the entire image of ${\pi }_1(F^{+})$ in ${\pi }_1(X_K)$ lies in its commutator subgroup. The oriented boundary loop of $F^{+}$ represents zero in $H_1(F^{+};\mathbb{Z})$: a finite triangulation of the supplied disk-band surface gives an oriented two-chain whose boundary is exactly that boundary loop, with all internal edges cancelling. Apply [F2] to $F^{+}$ to conclude that its boundary element belongs to $[{\pi }_1(F^{+}),{\pi }_1(F^{+})]$. Its image in ${\pi }_1(X_K)$ therefore belongs to $[{\pi }_1(X_K{)}^{(1)},{\pi }_1(X_K{)}^{(1)}]={\pi }_1(X_K{)}^{(2)}$. The compatible tube identifies this boundary with the preferred longitude up to conjugation. Derived subgroups are normal, as conjugation sends each finite commutator product to another, so the marked $\lambda$ has the same conclusion.

The proof used the actual supplied finite surface model only to define and justify intersection counts and its finite fundamental two-chain. The marked tube, trimming and parallel copy were input or finite product constructions. All perturbations concerned one finite chain at a time; all homology comparisons and pushout maps were natural or uniquely defined. Consequently no simultaneous choice of representatives, triangulation of an arbitrary topological manifold, or locally flat four-dimensional normal-bundle assertion has been imported. This proves all stated peripheral and group conclusions in their full supplied-data scope.

A cochain $f:C_1\to N$ is a cocycle precisely when $f(B_1)=0$. Restricting to $Z_1$ therefore induces a map $H_1=Z_1\mathrm{/}{B}_1\to N$. A coboundary $g{d}_1$ vanishes on $Z_1$, so this induces the stated well-defined map $\kappa$. It is onto if every map $Z_1\to N$ vanishing on $B_1$ extends to $C_1$. It is injective if every map $B_0\to N$ extends to $C_0$: indeed a cocycle restricting to zero on $Z_1$ factors uniquely through $d_1:C_1\to B_0$, and such an extension expresses it as a coboundary.

For the principal-right-ideal alternative, $N$ is right divisible: given $n=q+R\in N$ and $0\mathrm{\ne }a\in R$, the element $q{a}^{-1}+R$ multiplied on the right by $a$ equals $n$. We prove by induction on $m$ that any right-linear map $f:S\to N$ from any right submodule $S\subseteq R^m$ extends to $R^m$. For $m=0$ this is immediate. Let the projection of $S$ to the last coordinate be the right ideal $J$. If $J=0$, apply induction on the first $m-1$ coordinates and give the last basis vector value zero. Otherwise write $J=aR$ with $a\mathrm{\ne }0$ and take $z=(z^{\mathrm{\prime }},a)\in S$. Then $S=S^{\mathrm{\prime }}\oplus zR$, where $S^{\mathrm{\prime }}=S\cap (R^{m-1}\times 0)$: subtract $zr$ from an element with last coordinate $ar$, and uniqueness follows from the absence of zero divisors.

If $H_0=0$, then $d_1:C_1\to C_0$ is onto. Lift the finitely many given basis vectors of $C_0$ and extend linearly to a section $s:C_0\to C_1$. This gives $C_1=Z_1\oplus s(C_0)$. Extend a map from $Z_1$ by zero on the second summand to prove surjectivity in step 1.1. Here $B_0=C_0$, so the extension required for injectivity is automatic. The section uses only finitely many selections.

Extend $f{\mathrm{\mid }}_{S^{\mathrm{\prime }}}$ to $g:R^{m-1}\to N$ by induction. Right divisibility supplies $v\in N$ with $va=f(z)-g(z^{\mathrm{\prime }})$. Define the extension on $R^m$ by $(u,r)\mapsto g(u)+vr$. It agrees with $f$ on $S^{\mathrm{\prime }}$ and on $z$, hence on all of $S$. This completes the finite induction, using one ideal generator, one lift, and one division at each of finitely many stages. It does not use a maximal extension argument or an infinite family of choices.

Apply the extension result of step 2.2 first to $Z_1\subseteq C_1$ and then to $B_0\subseteq C_0$, using the supplied finite bases. These are exactly the surjectivity and injectivity requirements of step 1.1. The first alternative was established in step 2.1, so $\kappa$ is an isomorphism in either case. Restriction and passage to cycles commute with every compatible left coefficient action; thus the inverse, being unique, also respects that action.

We first describe the finite linking calculation used in the construction. For disjoint oriented polygonal circles with a specified regular plane projection, the linking number is half the signed mixed-crossing sum. This can be proved with finite chains as follows. For the second curve $D$, extend each projected edge vertically downward from its actual height to a plane lying below both curves, and fill its resulting planar projected cycle by the finitely many oriented cone triangles from a plane point. The resulting finite two-chain has boundary $D$. Its intersections with the first curve $C$ occur exactly at the mixed crossings where $C$ is below $D$, with the usual oriented crossing signs. It therefore computes linking as the sum of that subset of signs. The full mixed sum is twice this sum: the difference between the sums where $C$ is below and above $D$ is the algebraic plane intersection of the two projected cycles, which is zero. To check the latter directly, fill one projected cycle by its finite planar cone chain and intersect the other with that chain; the signed endpoints of the resulting finite one-chain cancel. Independence of a finite spanning chain is equally finite: the difference of two such chains is a two-cycle in ${\mathbb{R}}^3$, bounded by its cone three-chain. Intersecting that three-chain with the closed curve gives a finite one-chain whose signed boundary sum is zero, so the two intersection numbers agree. Finite subdivisions and perturbations needed to avoid vertices or nontransverse affine intersections avoid only finitely many bad affine conditions; a rational point outside their union exists, for example by trying rational points in a fixed enumeration. Thus no transversality selection on an infinite family is involved. The same computation applies after small local smoothing or to parallel pushed curves, because these changes take place in disjoint boxes without introducing intersections between the two curves.

If $g=0$, use an oriented disk. Otherwise start with an oriented disk and $2g$ labeled orientable rectangular bands. In the cyclic order of their attachment intervals on the disk, put the endpoints of each pair of bands $(2j-1,2j)$ in the alternating order $2j-1,2j,2j-1,2j$, and put the pairs in disjoint successive sectors. Attaching the first band of a pair splits the current boundary into two components; attaching the second joins those two, as direct traversal of the four alternating attachment intervals shows. Each pair therefore preserves one boundary component and raises genus by one. The finite relative cell calculation for the disk and bands has one degree-one generator per band and no relative degree-two cells: collapse the disk, and collapse each band across its width. The relative degree-one boundary is zero because both band ends lie in the connected disk. The pair sequence, with the disk contractible and the degree-zero inclusion an isomorphism, therefore identifies $H_1(F;\mathbb{Z})$ with ${\mathbb{Z}}^{2g}$. The band cores completed by disk arcs map to its coordinate basis. Choose the disk arcs so that only paired arcs meet, once in their sector. The embedded bands can be routed through a finite regular ribbon diagram: give each band a private twist box and, for every unordered pair, give those two bands one private common box with two parallel strands. Connect these finitely many specified visits by polygonal plane arcs, avoiding the finite existing vertices and arranging transverse double crossings without triple crossings by finitely many rational perturbations. At every other diagram crossing assign one strand above the other, then lift it in a small crossing box. Distinct crossing boxes are disjoint; sufficiently narrow ribbons following these finitely many arcs, and the specified orientable attachments, give an embedded disk-band surface. Arbitrary incidental diagram crossings merely determine its initial matrix and cause no restriction.

Let $W_0$ be the positive-push matrix of this initial surface, computed by step 1.1. Its skew part has zero entries for core pairs whose disk arcs are disjoint. Indeed the annulus sweeping a core from its negative to its positive push misses the other core in that case. In the explicit normal ribbons, sliding both members of the pair $(a_i^{-},a_j)$ a small equal distance in the positive normal direction identifies its linking with that of $(a_i,a_j^{+})$: their normal heights stay distinct throughout, including over a disk crossing. Linking is symmetric by step 1.1, so the signed-intersection calculation gives equality of the two ordered entries. For a paired pair the sweep meets the other core once in the disk, with sign $+1$ or $-1$, so the difference of its two ordered entries is that sign. In local disk coordinates the sweep is the product of the core arc with a normal interval; its single transverse intersection has multiplicity one. Orient the second core of each pair so that this difference is $+1$. Thus $W_0-W_0^T$ is exactly the displayed standard skew matrix. This calculation is entirely within the given disk and finite ribbon chart; it does not assume a surface-classification or duality theorem. Consequently $W-W_0$ is symmetric.

In the private box of band $i$, insert one full ribbon twist with sign chosen so that the two mixed crossings between its core and its positive parallel have signed sum $+2$. Step 1.1 says this increases $(W_0{)}_{ii}$ by one. The opposite twist decreases it by one. Every other core and its push miss that box, so its pairings with this modification are unchanged, as are all other matrix entries. Insert the finite integer number of twists required for each $i$ to make the diagonal equal that of $W$. Full twists preserve orientability and the attachment intervals, hence the abstract disk-band surface and its one boundary component.

In the private common box of bands $i<j$, replace the two parallel ribbon strands by the two-strand full braid whose two mixed crossings have sign $\epsilon \in \{+1,-1\}$. Give each ribbon its diagram framing, with no individual full twist. The two crossings between the positive parallel of the first core and the second core have signed sum $2\epsilon$, so step 1.1 changes entry $(i,j)$ by $\epsilon$. The same two crossings between the positive parallel of the second and the first give the same change in entry $(j,i)$. Each ribbon has no self crossing or extra rotation of its framing in this box, so its diagonal entry is unchanged. All other cores and parallels miss the box. Thus this modification changes exactly those two symmetric entries. Since the remaining difference from $W$ is symmetric with zero diagonal by steps 3.1–4.1, insert the prescribed finite number of positive or negative full braids in each pair box. None affects another pair box or previously corrected entries. The resulting surface has matrix exactly $W$.

All polygonal bends, band attachments and diagram height changes in this description lie in finitely many disjoint small coordinate boxes. Replace each polygonal bend by a smooth arc in its plane, agreeing with its incident straight pieces outside the box; a fixed smooth cutoff flat at the endpoints supplies this replacement. Along a band use the same interpolation for its two edges and framing. At a disk attachment the construction is the product of its attachment interval with a rounded planar corner, so it too has an explicit local smooth replacement. Choose the smoothing boxes and ribbon widths smaller than the minimum of the finitely many positive separations between nonincident pieces. The replacements remain embedded, retain the band parametrizations and preserve every chosen crossing sign and push-off separation. They therefore preserve the exact linking computations in steps 3.1–5.1. The resulting surface is smooth and oriented with one smooth boundary component, which is a knot in $S^3$. The band collapse still provides the displayed basis, and its positive normal is the actual oriented smooth normal.

Every list in the construction has length bounded in terms of the finite matrix entries: the bands, visits, vertices, crossings, twist counts and pair-clasp counts are finite. Rational perturbations exclude finitely many affine conditions at each of finitely many stages, and local smoothing uses an explicit cutoff and a finite minimum. No arbitrary sequence of embeddings, bases, surfaces, generic projections or normal fibers is selected. The final surface and ordered basis realize every entry of the supplied $W$ with the positive push convention in the Statement. This proves the claimed finite, choice-free realization.

Each $P^{i+1}L$ is a subgroup of $P^{i}L$. An automorphism of $L$ preserves $P^{i}L$ inductively, and induces an automorphism of its abelianization and of the displayed tensor group; their maps commute with $a\mapsto a\otimes 1$. Its kernel is therefore preserved, proving that all $P^{i}L$ are characteristic and in particular normal. Also $L^{(i)}\subseteq P^{i}L$ by induction: the assertion is equality at zero, and commutators of elements of $L^{(i)}$ lie in $[P^{i}L,P^{i}L]$, which maps to zero under abelianization and hence lies in $P^{i+1}L$. Derived subgroups are descending by their finite commutator definition. Consequently $N\subseteq G^{(r)}\subseteq G^{(i)}\subseteq P^{i}G$ for every $i\le r$.

We prove $P^{i}H=q(P^{i}G)$ through index $r$. It holds at zero. Suppose it holds at $i<r$ and put $A=P^{i}G$. The restriction $q:A\to P^{i}H$ is onto with kernel $N$. Moreover $N\subseteq G^{(r)}\subseteq G^{(i+1)}=[G^{(i)},G^{(i)}]\subseteq [A,A]\mathrm{.}$ The abelianization of $A\mathrm{/}N$ is $A\mathrm{/}[A,A]$: expanding commutators shows $[A\mathrm{/}N,A\mathrm{/}N]=[A,A]\mathrm{/}N$, since every finite commutator word in the quotient lifts by finitely many individual lifts. Thus the induced abelianization map is an isomorphism. Tensoring this isomorphism with $R_i$ is an isomorphism too, because tensoring its inverse gives an inverse on every generator $a\otimes u$; exactness of tensor is not being assumed. The two maps to these isomorphic tensor groups commute with $q$. Their kernels consequently satisfy $P^{i+1}H=q(P^{i+1}G)$: a fixed element of the former has a lift in $A$, and its tensor value is zero by the isomorphism, hence that lift is in $P^{i+1}G$. This proves the finite induction.

The kernel of each restricted $q:P^{i}G\to P^{i}H$ is $N$, already contained in $P^{i}G$ by step 1.1, proving the first displayed isomorphism. The preimage of $P^{i}H=q(P^{i}G)$ is $P^{i}G$, since it already contains $N$; this proves the second. Similarly the preimage in $P^{i}G$ of $P^{i+1}H$ is $P^{i+1}G$, giving the successive-quotient isomorphisms. For ordinary derived subgroups, expanding commutators and lifting each finite commutator word proves inductively $q(G^{(i)})=H^{(i)}$. Since $N\subseteq G^{(r)}$, the same preimage argument gives $G\mathrm{/}{G}^{(r)}\cong H\mathrm{/}{H}^{(r)}$. None of these conclusions follows merely from characteristicness: their actual kernels and preimages have been identified.

When all $R_i=\mathbb{Q}$, each quotient $P^{i}G\mathrm{/}{P}^{i+1}G$ embeds, by the kernel definition, in the additive group $(P^{i}G\mathrm{/}[P^{i}G,P^{i}G]){\otimes }_{\mathbb{Z}}\mathbb{Q}$. That group is a rational vector space by scalar multiplication on the second tensor coordinate, so its additive group is torsion-free abelian: multiplication by a nonzero integer has inverse multiplication by its reciprocal. The finite normal series $G\mathrm{/}{P}^{r}G⊵P^{1}G\mathrm{/}{P}^{r}G⊵\cdots ⊵P^{r}G\mathrm{/}{P}^{r}G=1$ therefore has torsion-free abelian factors and is precisely a PTFA series, with trivial factors permitted. For $r=0$ the quotient is trivial and the series is empty. For other ring lists steps 1.1–3.1 are unchanged, but a positive-characteristic tensor group can have additive torsion, so this last PTFA argument is intentionally not asserted. Every lift above concerned one finite word or one individual quotient element; no infinite set of representatives or vector-space basis was selected. This proves the full natural quotient assertions without a choice principle.

The supplied contractible basis consists of path-connected open sets, so connectedness implies path-connectedness: path components are open and their complements are open. Form the set of paths $\gamma :o\to x$, identifying two with the same endpoint when the loop $\gamma \ast ({\gamma }^{\mathrm{\prime }}{)}^{-1}$ belongs to $N$. The subgroup laws make this an equivalence relation. For a member $U$ of the supplied contractible basis and a path $\gamma$ ending in $U$, let $U_{[\gamma ]}$ consist of $[\gamma \ast \eta ]$ for paths $\eta$ lying in $U$ and starting at that endpoint. The projection to $U$ is bijective because any two such paths with the same endpoint are homotopic rel endpoints in $U$. If two such sets over $U$ meet they coincide; for two different basis neighborhoods, a smaller basis neighborhood of a common endpoint shows their intersections contain a basic set of this same form. They therefore define a topology for which $p^{-1}(U)$ is a disjoint union of open sets each homeomorphic to $U$. Surjectivity uses a path to each individual endpoint, rather than selecting all such paths. Thus this is an actual covering space constructed without a representative section.

For completeness the degree-one constant-coefficient calculation needs no universal-coefficient theorem. For any path-connected based space $Y$, concatenate paths by the first-loop-first rule. A constant edge is the boundary of a constant triangle; a path homotopy gives the boundary of its two-triangle prism up to constant endpoint edges; and $uv$ is homologous to $u+v$ by mapping a triangle to the concatenated path with vertex parameters $0,1\mathrm{/}2,1$. A reversed path is homologous to the negative path, since its concatenation with the original contracts by retracing. These finite relations show that $[u{]}_{\mathrm{a}\mathrm{b}}\otimes c\mapsto c[u]\in H_1(Y;k)$ is a well-defined additive balanced map $S:{\pi }_1(Y{)}_{\mathrm{a}\mathrm{b}}{\otimes }_{\mathbb{Z}}k\to H_1(Y;k)$. For a fixed finite cycle $z={\sum }_e{c}_{e}e$, choose paths ${\gamma }_v$ from the basepoint to the finitely many endpoint images and put $T_{\gamma }(z)={\sum }_e[{\gamma }_{e(0)}\ast e\ast {\gamma }_{e(1)}^{-1}{]}_{\mathrm{a}\mathrm{b}}\otimes c_e\mathrm{.}$ Finite choice follows by finite induction in ZF, so these paths exist without an axiom of choice.

A path $\gamma$ lifts from the constant-path point by $t\mapsto [\gamma {\mathrm{\mid }}_{[0,t]}]$, with the restricted path reparametrized and held stationary after its endpoint. This is continuous on the basic sets of step 1.1, and reaches $[\gamma ]$, so $W_N$ is path-connected. For any fixed starting lift, path lifting exists uniquely by finite subdivision into evenly covered basis neighborhoods. A homotopy lifts uniquely by the same finite-grid construction on its compact square: choose a finite fine grid subordinate to the inverse-image cover and extend sheet by sheet, with uniqueness on overlaps. The square is compact metric, so this finite construction uses no choice principle. Hence $p_{\ast }$ is injective on fundamental groups, and its image consists exactly of based loops whose lifted endpoint is the constant-path point, that is, exactly $N$. This identifies ${\pi }_1(W_N)$ with $N$ by the actual projection. The same lifting statement for a singular simplex follows directly by sending a point to the lift obtained along the straight segment from its first vertex; segments joined by a triangle are homotopic inside the simplex, so this gives the unique continuous lift with prescribed first vertex.

Another finite path family changes each loop term by $d_{e(0)}-d_{e(1)}$ in the abelianization. After tensoring, its total coefficient at a vertex is that of the cycle boundary, hence zero. Thus $T_{\gamma }(z)$ has one uniquely defined value $T(z)$, independent of paths or enlargement of the finite endpoint list. On a singular triangle the loops for edges $01,12,02$ satisfy $L_{01}{L}_{12}=L_{02}$ because the triangle fills their concatenation, so $T$ kills its signed boundary. For any finite two-chain use the finite union of its vertex images to get the same conclusion termwise, with its $k$ coefficients. Thus $T$ descends to homology. The relations in step 1.2 say ${\gamma }_{e(0)}\ast e\ast {\gamma }_{e(1)}^{-1}$ is homologous to ${\gamma }_{e(0)}+e-{\gamma }_{e(1)}$; in the weighted sum the extra path terms cancel by the cycle condition. Hence $ST=\mathrm{i}\mathrm{d}$. On a based loop use the constant connector at its endpoint, giving $TS=\mathrm{i}\mathrm{d}$ on every generator $[u{]}_{\mathrm{a}\mathrm{b}}\otimes c$. This proves the natural isomorphism $H_1(Y;k)\cong {\pi }_1(Y{)}_{\mathrm{a}\mathrm{b}}\otimes k$. Naturality follows by mapping a finite connector family forward; independence then removes that family from the map. No paths to all points of $Y$ and no basis of an infinite vector space were chosen.

For $g\in G$ and any representative loop $a$, define the right action by $[\gamma ]\cdot g=[a^{-1}\ast \gamma ]$. Replacing $a$ by another representative gives the same path class, since $N$ is normal; replacing $\gamma$ by an equivalent path also does. The action law follows from $b^{-1}\ast a^{-1}=(a\ast b{)}^{-1}$, and its maps carry the basic sets in step 1.1 to basic sets. On each fiber it is free and transitive: for paths $\gamma ,\delta$ to the same endpoint take the quotient class of $\gamma \ast {\delta }^{-1}$, which sends $[\gamma ]$ to $[\delta ]$. The formula specifies the action uniquely without choosing representatives for all quotient elements. Free $k$-linearization of each fiber gives ${\mathcal{R}}_x$; path transport commutes with this right action because it appends a path at the end, while deck action prepends $a^{-1}$. Thus it is a right-$k[G]$ local system, each fiber individually having a torsor point as a free basis. At the basepoint the constant-path point is a specified basis, and transport along $a$ sends it to $[a]=[1]\cdot g^{-1}$, giving exactly the stated inverse convention and a right-linear map.

A basis generator in the local chain group consists of a base singular simplex $\sigma$ together with one point of the covering fiber over its first vertex and a scalar in $k$. Send it to the unique lift of $\sigma$ with that first vertex, given in step 2.1, with the same scalar. Pairs of this kind are in bijection with covering singular simplices: projection gives the inverse map. Both directions are uniquely specified on generators, hence define inverse maps on all finite chains without selecting a lift for each base simplex. For the face omitting the first vertex, the first point of the lifted face is the transport of the old first point along that edge, exactly the local coefficient boundary rule. The other faces keep the first point. Thus the bijection commutes with all signed boundary terms. It commutes with the right deck action on generators too, so gives the claimed actual chain isomorphism of right $k[G]$-module complexes.

Apply steps 1.2 and 2.2 to $Y=W_N$ and combine with steps 2.1 and 4.1. A based loop $u\in N$ lifts to the based loop whose projection is $u$, so its coefficient-one class maps precisely to $[u]\otimes 1$. For the right action, move the constant-path base lift by $g=[a]$. The lift of $a^{-1}$ joins the old base lift to that new one. Rebase the translated lift of $u$ by this path and its inverse; its projection is $a^{-1}ua$. Connector contributions cancel in first homology by step 1.2, so this gives the stated conjugation action on $N_{\mathrm{a}\mathrm{b}}\otimes k$. Different representatives of $g$ change it by an inner automorphism of $N$, trivial on $N_{\mathrm{a}\mathrm{b}}$. The zero criterion is now exactly the displayed isomorphism and its coefficient-one formula; for $N=P^s\pi$ in the finite mixed series defined in the Statement, it is the next kernel $P^{s+1}\pi$. For a based continuous map $f:(W,o)\to (W^{\mathrm{\prime }},o^{\mathrm{\prime }})$ between spaces with the stated connected, Hausdorff and contractible-neighborhood hypotheses, if $f_{\ast }(N)\subseteq N^{\mathrm{\prime }}◃{\pi }_1(W^{\mathrm{\prime }},o^{\mathrm{\prime }})$, the covering map is explicitly $[\gamma ]\mapsto [f\circ \gamma ]$; the equivalence relation makes it well-defined. It commutes with endpoint projection, the induced quotient-group action and simplex lifting, so step 4.1 is natural for its coefficient maps. Steps 1.2 and 2.2 already prove naturality of the constant-coefficient identification, yielding the stated commutative homology comparison. All covering, chain and homology maps were explicit quotient maps or uniquely defined finite-lift maps, so the full coefficient and choice scope in the Statement is proved.