Thom Spectra and Unoriented Bordism Detection

The homotopy groups in the Eilenberg–Mac Lane definition give $(q-1)$-connectivity. Hurewicz gives $H_j(K_q;\mathbb{Z})=0$ for $0<j<q$ and $H_q(K_q;\mathbb{Z})={\mathbb{F}}_2$. The cohomological UCT then gives the claimed mod-two groups: in degree $q$ its Hom term is $\mathrm{Hom}({\mathbb{F}}_2,{\mathbb{F}}_2)={\mathbb{F}}_2$, its Ext term vanishes, and at degree one the possible $H_0$ Ext term also vanishes because $H_0=\mathbb{Z}$ is free. The normalized fundamental class is the identity evaluation class in the representability theorem. Apply the direct published mapping-path factorization to the marked inclusion $\ast \to K_q$. Its actual path-space total contracts by the supplier's explicit path reparametrization, and its strict fiber is $F=\mathrm{\Omega }{K}_q$. The fibration long exact sequence shows $F$ is path connected and its only nonzero positive homotopy group is $\mathbb{Z}\mathrm{/}2$ in degree $q-1$, marked by the connecting isomorphism. Apply the published CW approximation theorem to $F$, extending its marked basepoint as a one-point initial subcomplex. This gives a connected based CW complex $L$ and a based weak equivalence $\gamma :L\to F$. Transport the connecting marking to $L$; it is a marked CW $K(\mathbb{Z}\mathrm{/}2,q-1)$. The published marked uniqueness theorem supplies a homotopy equivalence $v:K_{q-1}\to L$ inducing that marking. If its marking uses basepoint transport, choose the corresponding path from $v$ of the marked point to the marked vertex of $L$, and use the published CW-point cofibration homotopy-extension property to homotope $v$ to a based map. The published moving-basepoint transport formula preserves the transported marking. (Choose the marked point of the CW source as a vertex.) Set $h=\gamma v$. It is a marked weak equivalence, and the published weak-equivalence homology lemma makes $h_{\ast }$ an isomorphism in integral homology in every degree. Naturality of coefficient UCT then makes $h_{\ast }$ an isomorphism in mod-two homology. Natural field evaluation duality makes $h^{\ast }$ an isomorphism in mod-two cohomology, and cup-product naturality makes it a unital graded-ring isomorphism. This chain of actual maps identifies the strict fiber's cohomology ring and fundamental class without an external CW-type-of-fibers theorem. The published multiplicative Serre theorem applies directly to the actual mapping-path fibration over the simply connected CW base; monodromy is trivial, and the just-proved ring isomorphism computes its strict fiber cohomology. Contractibility computes the positive-degree abutment as zero.

Important caveat. This lemma does not assert that every fiber class is transgressive or that taking its square commutes with a spectral-sequence differential in the manner required for the next induction. Those are additional claims, not consequences of ordinary Hurewicz or of spectral-sequence convergence alone.

In the published models, $S^{\mathrm{\infty }}=V_1({\mathbb{R}}^{\mathrm{\infty }})$ and ${\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}={\mathrm{Gr}}_1({\mathbb{R}}^{\mathrm{\infty }})$. Send a unit vector to its line. Over the open set of lines with nonzero coordinate $j$, choose the unique unit representative having positive $j$th coordinate; the other representative is its negative. These maps give two disjoint local sections and an evenly covered open set. Their formulas are continuous on every finite stage and therefore for the weak Grassmannian/Stiefel topologies. These open sets cover the base, so this is a two-sheeted covering with antipodal deck involution and discrete fiber $\{+1,-1\}$.

The published finite-local-strip covering-homotopy lemma makes this a Hurewicz fibration. Its total space is contractible by stable Stiefel contractibility; it is in particular connected and simply connected. The fibration long exact sequence gives ${\pi }_j({\mathbb{R}\mathbb{P}}^{\mathrm{\infty }})=0$ for $j\ge 2$, since positive homotopy groups of the fiber vanish. Path lifting associates to a based loop its endpoint deck transformation. This is a group homomorphism: lifting successive loops composes their endpoint deck transformations. It is surjective since a path from a unit vector to its negative exists in the connected sphere, and injective since a loop whose lift closes is nullhomotopic in the contractible total space and its contraction projects to the base. Thus ${\pi }_1=\mathbb{Z}\mathrm{/}2={\mathbb{F}}_2$, with the stated marking. The published Schubert theorem gives the CW base. This proves the marked Eilenberg–Mac Lane claim.

The published projective-space cohomology lemma gives a unique nonzero degree-one class $x$ and the ring ${\mathbb{F}}_2[x]$. The normalized fundamental class is nonzero by the published Eilenberg–Mac Lane representability theorem; uniqueness in degree one therefore identifies it with $x$. Marked Eilenberg–Mac Lane uniqueness transfers this ring to any chosen marked $K_1$.

Represent $x$ by a cocycle $u$ and extend $u$ by zero on singular simplices of $E$ not in $F$, obtaining an absolute cochain $b$. Then $c=db$ restricts to zero on $F$ and represents $\delta x$. If $0\le a\le m$, set $j=m-a$ and $b^{\mathrm{\prime }}=b{⌣}_{j+1}db+b{⌣}_{j}b\mathrm{.}$

Its restriction to $F$ is $u{⌣}_{j}u$, which represents $S{q}^{a}x$. The published cup-$i$ coboundary identity, $d^2=0$, and characteristic two give $d{b}^{\mathrm{\prime }}=db{⌣}_{j+1}db+b{⌣}_{j}db+db{⌣}_{j}b+db{⌣}_{j}b+b{⌣}_{j}db+b{⌣}_{j-1}b+b{⌣}_{j-1}b=c{⌣}_{j+1}c\mathrm{.}$

This is the relative representative of $S{q}^a[c]$, since $\mathrm{\mid }c\mathrm{\mid }=m+1$. Thus the two connector classes agree. For $a=m+1$, $S{q}^{m+1}x=0$ by instability, and $c{⌣}_{0}c=d(b{⌣}_{0}db)$; the primitive restricts to zero on $F$, proving that the other side also vanishes relatively. For $a>m+1$ both sides vanish by instability. The relative-carrier property guarantees all relative cochains used above vanish on $F$; no representative-selection family or choice axiom is needed.

Induct on $k$, assuming the base-axis maps are isomorphisms through column $k$. Column zero starts this induction. The tensor condition implies isomorphism on all $E_2^{p,q}$ with $p\le k$.

Induction on $r\ge 2$ gives simultaneously: $(A_r){\mathrm{\Phi }}_r^{p,q}\text{ is an isomorphism if }p\le k-r+1,(B_r){\mathrm{\Phi }}_r^{p,q}\text{ is injective if }p\le k\mathrm{.}$ For $r=2$ these follow from the preceding $E_2$ isomorphisms. Write $Z_r=\ker d_r$ and $B_r=\mathrm{im}{d}_r$, distinguishing the boundary group $B_r$ from the assertion $(B_r)$ by context. If $p\le k-r$, the domain map is an isomorphism and the outgoing-target map in column $p+r\le k$ is injective, so the map on cycles $Z_r^{p,q}$ is an isomorphism. If $p\le k-r+1$, both the domain and incoming-source maps are isomorphisms by $(A_r)$, so the maps on boundary groups and on the quotients by those boundaries are isomorphisms. Quotienting cycles by boundaries gives $(A_{r+1})$. For $p\le k$, the cycle map is injective by $(B_r)$, and the incoming-source map is surjective since $p-r\le k-r+1$. Thus every target boundary in the image of a source cycle lifts to a source boundary; the map on cycle/boundary quotients is injective. This gives $(B_{r+1})$.

Now consider the unknown bottom position $(k+1,0)$. It has no outgoing differential, and for each $r$ there is an exact sequence $Z_r^{k-r+1,r-1}⟶E_r^{k-r+1,r-1}\stackrel{d_r}{\to }{E}_r^{k+1,0}⟶E_{r+1}^{k+1,0}⟶0.$ The second term is mapped isomorphically by $(A_r)$. We claim the cycle term is mapped surjectively. Put $(u,v)=(k-r+1,r-1)$. For $s\ge r$, use $E_s^{u-s,v+s-1}⟶Z_s^{u,v}⟶E_{s+1}^{u,v}⟶0.$ Here the first map is the incoming differential restricted to cycles; this is exact because $d_s^2=0$. The map on the first term is an isomorphism by $(A_s)$, since $u-s=k-r-s+1\le k-s+1$. At all pages later than $s$, outgoing differentials from $(u,v)$ have negative fiber target, so $E_{s+1}^{u,v}=Z_{s+1}^{u,v}$. Start from stationarity, where its map is an isomorphism by the assumed limiting isomorphism, and descend on $s$; lifting an element of the third term and correcting by an element of the first proves surjectivity on $Z_s^{u,v}$, including $s=r$. The first quadrant gives a finite stationarity bound at every position, so this is finite downward induction.

Finally the bottom position itself is stationary for $r>k+1$. Starting from its limiting isomorphism, descend on $r$ in the displayed five-term exact sequence. Surjectivity on the first term and isomorphisms on the second and fourth show the third map is an isomorphism: for surjectivity lift its quotient in the fourth term and correct the difference by the second; for injectivity first lift a kernel element to the second, then lift its image in the first and subtract, using injectivity of the second. Thus ${\mathrm{\Phi }}_2^{k+1,0}$ is an isomorphism. Induction on $k$ proves the claim.

Grading and the counit give, for homogeneous m∈M_d with d>0, Δ_M(m)=u⊗m+m⊗u+R, where R lies in ⊕{0<i<d}M_i⊗M{d-i}. Indeed the degree-(0,d) component is u⊗m by (ε_M⊗id)Δ_M=id, since ε_M vanishes on positive degrees; the other counit identity gives the degree-(d,0) component. Tensor bidegrees are direct by [L5]. In degree zero, Δ_M(λu)=λu⊗u, as already justified. The same argument gives the two endpoints for Δ_A(a) in positive degree.

A⁺M is the span of a·m with a of positive degree. It is graded: decomposing a,m into homogeneous components expresses every element as a finite sum of homogeneous such products. It is an A-submodule, since b(a·m)=(ba)·m and each nonzero homogeneous product ba has positive degree. It has no degree-zero component. By [L2], Q is the direct sum of Q_d=M_d/(A⁺M)_d, π is graded and surjective, its kernel is A⁺M, and Q_0=M_0. Moreover π(a·m)=ε_A(a)π(m) for every a,m: positive-degree a is killed, and degree-zero a acts as a scalar.

Choose a k-basis of each Q_d and lift each basis vector to M_d; this is allowed by [L3] and AC. In degree zero use π(u), with lift u. Extend the lifts linearly on each degree and then on the direct sum to obtain a graded section f. Conversely any graded section has f(π(u))=u, because π:M_0→Q_0 is an isomorphism. The formula for Φ is balanced and bilinear over k, so [L4] makes it a well-defined map. It is graded, and A-linearity follows from (ba)f(q)=b(af(q)).

We prove surjectivity by induction on d≥0. In degree zero Φ is the scalar isomorphism k⊗k→k·u. Suppose every M_e with e<d is in its image, and take m∈M_d. The vector m-f(π(m)) lies in (A⁺M)_d, so it is a finite sum Σ a_t m_t with homogeneous a_t of positive degree and m_t of degree d-|a_t|<d. Such an expression is obtained by projecting any finite expression in A⁺M to degree d. By induction choose y_t∈A⊗Q with Φ(y_t)=m_t. Then m=Φ(1_A⊗π(m)+Σ a_t y_t). Thus Φ is surjective in each degree, and finite degree support proves surjectivity on M.

Define T=(id_M⊗π)Δ_M:M→M⊗Q. Give M⊗Q the A-action on the first factor only. Then T is A-linear. Indeed apply id⊗π to the diagonal compatibility formula. Every summand with a₂ of positive degree vanishes by step 1.2. The surviving terms have a₂ in degree zero, so their Koszul signs are 1; the counit identity (id⊗ε_A)Δ_A(a)=a combines these terms to give T(a·m)=a·T(m). This calculation also treats a of degree zero and all inhomogeneous inputs by linearity.

For homogeneous q∈Q_d, the component of T(f(q)) with second degree d is exactly u⊗q. Every other component has second degree strictly less than d, by step 1.1. When d=0 there are no other components. Therefore TΦ(a⊗q)=ν(a)⊗q + terms of second degree less than d. This statement concerns second-factor degree, not total degree; multiplication on the first factor preserves that comparison.

Suppose z∈ker Φ. By [L5], using a homogeneous k-basis (q_j) of Q, write z uniquely as a finite sum Σ_j a_j⊗q_j with a_j∈A. If z≠0, take the largest degree d of a q_j with a_j≠0. Since TΦ(z)=0, its component of second degree d gives Σ_{|q_j|=d} ν(a_j)⊗q_j=0. The q_j in this equation are distinct basis vectors. Their coordinate functionals, tensored with id_M via [L4], give ν(a_j)=0 individually. Injectivity of ν gives a_j=0, contradicting the definition of d. Hence z=0 and Φ is injective. This finite maximum argument requires no finite-dimensionality of Q or its degree pieces.

By steps 2.1 and 4.1, Φ is a bijective graded A-linear map. Its inverse is A-linear and graded by uniqueness of preimages. Every element of A⊗Q has a unique finite expression Σ a_j⊗q_j, by [L5]; the first-factor action turns this into a free A-module with basis 1_A⊗q_j, in the sense of [L6]. Transporting that basis by Φ proves the theorem, with each generator in degree |q_j|. AC was used only for the homogeneous basis and its lifts; no further infinite selection occurs in the degree induction or finite-maximum argument.

On integral singular chains define τ(σ) as the sum of all d lifts of σ. The contractible simplex admits a lift for each point of the fiber over one vertex, and uniqueness of lifts shows these are all the lifts. Restriction to a face bijects these lift sets, so ∂τ=τ∂. This is precisely the chain construction proved in the published rational-transfer supplier; it precedes and is independent of its choice of coefficients. Precomposing R-valued cochains with τ gives T with Tp*=d id and pT=Σ_{g∈G}g. Thus p* is injective. Every pullback is invariant, and for invariant y, p*(d⁻¹Ty)=y. The ring identification follows from multiplicativity of pullback; transfer itself need not be multiplicative. This proves the upgrade over R, including R=Q and R=F_p with p∤d.

For a double cover with deck involution σ and 2 invertible, the sign local system downstairs corresponds to the anti-invariant cochain subcomplex upstairs. To verify this without invoking an unproved transfer with local coefficients, choose a lift of each singular simplex: the local-coefficient cochain assigns a coefficient in its orientation line; changing that lift changes its signed coordinate. Hence it corresponds exactly to an ordinary cochain c satisfying σc=−c. Transport along faces gives the ordinary cochain differential in these coordinates. The idempotents (1±σ)/2 split the entire cochain complex into its two eigenspaces; therefore cohomology of the anti-invariant subcomplex is the anti-invariant part of cohomology. This gives H*(X;O_R) ≅ H*(Y;R)^−.

This applies also when the spaces or cochain groups are infinite: the idempotents, not a finite-dimensional averaging argument, give the splitting.

Use cellular approximation to replace $f$ by a cellular map $g$ through a homotopy. If the homotopy moves the basepoint, its track gives the canonical target basepoint-transport isomorphism, so both the hypotheses and conclusions transfer between $f$ and $g$. Form the CW mapping cylinder $M_g$, with source inclusion $j:X\hookrightarrow M_g$ and target deformation retraction $r:M_g\to Y$, so $rj=g$. The homology exact sequence contains $H_i(X)\stackrel{j_{\ast }}{\to }{H}_i(M_g)⟶H_i(M_g,X)⟶H_{i-1}(X)\stackrel{j_{\ast }}{\to }{H}_{i-1}(M_g)\mathrm{.}$ For $1\le i\le N$, the first arrow is surjective and the last injective. Exactness therefore forces $H_i(M_g,X)=0$. The component condition gives $H_0(M_g,X)=0$ as well.

Both $X$ and $M_g$ are path connected and simply connected. The relative homotopy exact sequence consequently gives ${\pi }_1(M_g,X)=\ast$: every relative path has its initial endpoint connected to the basepoint inside $X$, and the resulting based loop is null in $M_g$. Induct on $m=2,\dots ,N$. If the relative groups below $m$ vanish, the CW pair $(M_g,X)$ is $(m-1)$-connected, and $X$ is nonempty simply connected. Relative Hurewicz identifies ${\pi }_m(M_g,X)$ with $H_m(M_g,X)=0$. This proves all relative groups through $N$ vanish.

For $2\le i<N$, the two adjacent relative groups in ${\pi }_{i+1}(M_g,X)⟶{\pi }_i(X)\stackrel{j_{\ast }}{\to }{\pi }_i(M_g)⟶{\pi }_i(M_g,X)$ vanish, so $j_{\ast }$ is injective and surjective. At $i=N$, vanishing of the last term gives surjectivity. At $i=1$ both absolute groups are trivial. Composition with $r_{\ast }$ and the basepoint-transport isomorphism proves the claims for $f$. Every Hurewicz invocation is within its stated simple-connectivity range.

Endpoint justification. Homology isomorphisms through degree $L$ imply homotopy isomorphisms only through degree $L-1$ by this argument: apply the lemma with $N=L$. To deduce an isomorphism at $L$, one also needs homology surjectivity at $L+1$. This loss of one degree is essential to this proof, since injectivity at $L$ uses ${\pi }_{L+1}(M_g,X)$.

Apply the published localization/tensor theorem with Z and the nonzero positive integers, whose ring localization is Q. In the fraction definition equality of a/s with 0/1 means u a=0 for some positive u. A finite sum of tensors a_t⊗(r_t/s_t) has a common denominator, and hence is a single fraction, so this criterion applies to every element. Exactness and direct sums are precisely the published localization statements.

For a chain complex C, apply exact rationalization to 0→Z_j(C)→C_j→B_{j−1}(C)→0 and 0→B_j(C)→Z_j(C)→H_j(C)→0. It identifies the cycles and boundaries in C⊗Q with the rationalizations of the original cycles and boundaries, hence identifies homology. The basis of singular simplices gives the literal chain isomorphism C_(Y;Z)⊗Q=C_(Y;Q), compatible with every continuous map. Finally choose a basis of V under AC. Tensoring a rational complex with V is a direct sum of copies of that complex. Kernels and images of its coordinate differential are direct sums of kernels and images, since elements have finite support. This proves the V assertion; the natural tensor map, not the auxiliary basis, supplies the isomorphism.

For n≥1, forgetting orientation is genuinely a two-sheeted cover BSO(n)=Gr⁺_n(R∞)→BO(n)=Gr_n(R∞) in the chosen model. On a base graph chart, Gram–Schmidt supplies a continuous orthonormal frame of the tautological bundle. Its two possible orientations give two disjoint open subsets of the oriented chart, each mapped homeomorphically to the base chart. These charts cover BO(n). This includes n=1: SO(1) is trivial, Gr⁺_1(R∞)=S∞, and forgetting orientation is v↦Rv from the unit sphere to RP∞; over a chart with a chosen unit representative it has the two sections v and −v. Rank zero is excluded here and treated as a point below.

Lift each Schubert characteristic disk D^d→BO(n) to its two sheets; this is possible because the disk is contractible, and the two lifts restrict to the two lifts over every boundary cell. Choose a label for the two lifts of each cell using AC. Their open interiors map homeomorphically to the underlying Schubert open cell, and the lifted boundary maps into the inverse image of the lower Schubert skeleton. Thus these disks supply attaching maps with finite boundary support. Each finite Schubert skeleton is compact Hausdorff with finitely many cells; its two-sheeted preimage is compact Hausdorff, and the finite attaching quotient maps continuously and bijectively onto that preimage. It is therefore a homeomorphism.

The infinite topology is the weak topology of these lifted cells. To check the nontrivial direction, let C be a subset whose inverse image in every lifted characteristic disk is closed. Over any evenly covered open set U of BO(n), restrict to one sheet U⁺. On each base characteristic disk, the preimage of U splits into relatively open-and-closed pieces on which a chosen lift lies in U⁺; the lifted-cell test makes the preimage of C∩U⁺ closed on each such piece, hence on the preimage of U. Therefore the preimage of U⁺\C is open relative to the preimage of U. Since that preimage of U is open in each characteristic disk, this restricted preimage is open in the full disk. The base CW weak-topology test makes U⁺\C open in U⁺, hence C∩U⁺ closed there. The same holds on the other sheet, and these sheets cover BSO(n), so C is closed upstairs. Conversely a closed subset has closed inverse image under each characteristic map by continuity. The lift topology is therefore exactly the CW weak topology. Finite cell counts in each dimension and closure finiteness are inherited from the Schubert structure, with two lifts per cell. For n=0, BSO(0) is a point by the published definition. This is the lifted-cover CW prerequisite of the shared MSO prespectrum definition.

The unit is the only degree-zero word. For $N>0$, every normalized word has at most $N$ entries. Pad its sequence on the right with zeros to length $N$ and order these finite sequences lexicographically, reading from the left. There are finitely many such sequences of total sum $N$.

If a word is not admissible, choose a positive adjacent pair $(a,b)$ with $a<2b$. Each nonzero summand in the Adem replacement has the pair $(a+b-j,j)$, where $0\le j\le \lfloor a\mathrm{/}2\rfloor <b$. Its first changed entry is therefore $a+b-j>a$. If $j=0$, remove $s_0=1$ and normalize; this removal occurs after the strictly increased entry, so the normalized padded word remains lexicographically larger. Total degree is preserved, and normalized length still is at most $N$.

Descending induction on this finite ordered set proves that each word is a sum of admissible words: terminal words cannot have a replaceable pair, while each replacement uses only words already covered by the induction. Sums over words are finite. Applying the well-defined quotient-to-operation map proves the second assertion. This proves spanning only; independence follows below.

The published projective-space lemma gives the polynomial ring of ${\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}$ and skeletal restriction isomorphisms through degree $L$. Finite projective space has one mod-two cellular generator in every degree $0,\dots ,L$, zero cellular differential, and no cells above $L$; the cellular comparison gives degreewise finite-free homology, and field evaluation duality makes cohomology vanish above $L$. Thus restriction computes its ring as ${\mathbb{F}}_2[x]\mathrm{/}(x^{L+1})$, and iterative application of the published finite-free Künneth theorem gives the displayed product ring. Its distinct surviving monomials are linearly independent.

For a degree-one generator, normalization gives $Sq(x)=x+x^2$. Cartan on a product of $m$ copies gives, by the finite binomial expansion, $S{q}^a(x^m)=\left(\genfrac{}{}{0px}{}{m}{a}\right)x^{m+a}\mathrm{.}$ In particular, for $m=2^h$ the polynomial identity $(1+t{)}^{2^h}=1+t^{2^h}$ over ${\mathbb{F}}_2$ gives $S{q}^0(x^{2^h})=x^{2^h}$, $S{q}^{2^h}(x^{2^h})=x^{2^{h+1}}$, and $S{q}^a(x^{2^h})=0$ for all other nonnegative $a$.

Consequently every term in an iterated action on $P$ arises by a finite schedule of doublings. At the step labelled $i_j$, a set of variables is doubled whose current exponents sum to $i_j$. Different schedules may give the same final monomial, so their parity must be considered; their supports are not disjoint.

We prove the largest-monomial assertion by induction on $k$. For $k=1$, exactly $i_1$ of the variables are doubled. Since $r\ge e(I)=i_1$, the largest term doubles the first $i_1$ variables, uniquely. It has coefficient one.

For $k>1$, a variable reaches exponent $2^k$ only if it is doubled at every one of the $k$ steps. The first step to act is $i_k$, when all exponents equal one; hence at most $i_k=d_k$ variables can reach $2^k$. Achieving $i_k$ such variables exhausts that first-step budget, and their later costs are forced to be $2^{k-j}{i}_k$ at step $j$. Subtract those costs from the earlier budgets and discard the now exhausted final step. The remaining sequence is $I^{\mathrm{\prime }}=(i_1-2^{k-1}{i}_k,i_2-2^{k-2}{i}_k,\dots ,i_{k-1}-2i_k)\mathrm{.}$ All its entries are nonnegative by admissibility, and its successive admissibility differences are $d_1,\dots ,d_{k-1}$. Remove any terminal zeros. Its excess is $e(I)-i_k$, and $r-i_k\ge e(I^{\mathrm{\prime }})$. Thus the induction hypothesis constructs the largest remaining term on the remaining variables. Taking the first $i_k$ variables for the full doubling chains constructs a nonzero term with the stated exponent vector.

A monomial with fewer than $i_k$ variables of exponent $2^k$ is lexicographically smaller than that term after placing its exponents in decreasing order. Symmetry of $P$ and of its square action ensures that arranging exponents in decreasing order maximizes lexicographic order. If precisely $i_k$ variables attain exponent $2^k$, their forced all-step chains consume exactly the costs above, so comparison of the remaining exponents is the induction problem for $I^{\mathrm{\prime }}$. Therefore none is larger. For the specified leading monomial, its first $i_k$ variables have uniquely forced chains and the residual leading term has coefficient one by induction; hence the full coefficient is one, not an unproved parity assertion. No displayed term is truncated: a variable's exponent cannot exceed $1+\mathrm{\mid }I\mathrm{\mid }$, because every doubling cost contributes to the total increase $\mathrm{\mid }I\mathrm{\mid }$. Thus $L\ge \mathrm{\mid }I\mathrm{\mid }+1$ suffices. Finally the multiplicities $d_j$ recover the sequence by the backward recursion $i_j=d_j+2i_{j+1}$. Since a normalized sequence has $d_k=i_k>0$, the largest exponent also recovers its length. Distinct admissible sequences therefore have distinct leading monomials. Empty sequences and $r=0$ give the identity action on the unit.

On $P=xyz$, $S{q}^3(P)=x^2{y}^2{z}^2,S{q}^{2}S{q}^1(P)=x^{4}yz+x{y}^{4}z+xy{z}^4+x^2{y}^2{z}^2\mathrm{.}$ Both composites are admissible and their supports overlap. Their leading monomials differ, which is exactly the property needed for independence.

Put $P=P{K}_{n+1}$ and $B=K_{n+1}$. Choose a based map $f:S^n\to F$ representing the marked nonzero element of ${\pi }_n(F)=\mathbb{Z}\mathrm{/}2$. Its absolute Hurewicz image $h_F[f]=f_{\ast }[S^n]$ is the nonzero element of $H_n(F;\mathbb{Z})$: for $n>1$, use the marked weak equivalence $h:K_n\to F$ of the path-loop input lemma. Absolute Hurewicz is an isomorphism on the CW source $K_n$; the weak-equivalence definition gives an isomorphism on homotopy, and the integral weak-equivalence homology theorem in [F1] gives the isomorphism on integral homology, and Hurewicz naturality therefore makes $h_F$ an isomorphism too. For $n=1$, absolute Hurewicz says abelianization for every path-connected space, and ${\pi }_1(F)=\mathbb{Z}\mathrm{/}2$ is already abelian.

Since $P$ is contractible, $f$, considered as a map into $P$, has a nullhomotopy. Cone that nullhomotopy to obtain a continuous map $A:D^{n+1}\to P$ restricting to $f$ on the boundary; identify the disk with the cone on the sphere and choose its boundary orientation accordingly. Let $c_D$ be a finite triangulated integral fundamental chain of the disk, whose boundary $c_S$ is the corresponding fundamental cycle of its sphere. The chain $A_{\ast }{c}_D$ is a relative cycle for $(P,F)$ since its boundary is $f_{\ast }{c}_S$ in $F$. Write $z=[A_{\ast }{c}_D]\in H_{n+1}(P,F;\mathbb{Z})\mathrm{.}$ The homology pair-LES computes its boundary directly as $\mathrm{\partial }z=[f_{\ast }{c}_S]=h_F[f]$. Because $P$ is contractible and $n\ge 1$, that boundary map is an isomorphism $H_{n+1}(P,F;\mathbb{Z})\to H_n(F;\mathbb{Z})$. Thus $z$ is its nonzero generator. No relative Hurewicz map or theorem has been invoked.

The composite $pA$ maps the whole boundary sphere to the marked basepoint. Hence it factors through the collapsed disk, giving $g:D^{n+1}\mathrm{/}\mathrm{\partial }{D}^{n+1}\cong S^{n+1}⟶B\mathrm{.}$ In the fibration connecting construction, $A$ is a lift of this disk representative and its boundary lift is $f$; therefore the connecting homomorphism sends $[g]$ to $[f]$. It is an isomorphism because the path-space total is contractible. The fiber marking was defined by precisely this connecting isomorphism, so $[g]$ is the marked nonzero base element of ${\pi }_{n+1}(B)$.

On chains, $p_{\ast }z$ is the relative class of $(pA{)}_{\ast }{c}_D$. Collapsing the disk boundary sends its oriented relative fundamental class to the fundamental sphere class. The sphere boundary is a nonempty closed subspace with a radial collar retracting onto it, so cor-homology-of-good-pairs-is-reduced-homology-of-the-quotient identifies relative disk homology with the reduced quotient homology. In the one-top-cell description of $D^{n+1}\mathrm{/}\mathrm{\partial }{D}^{n+1}$ the quotient sends the characteristic disk generator to the same top-cell generator; naturality of thm-cellular-homology-computes-singular-homology makes this the asserted comparison of actual singular classes. Choose the quotient sphere orientation accordingly. Thus, under $H_{n+1}(B,\ast ;\mathbb{Z})\cong H_{n+1}(B;\mathbb{Z})$, $p_{\ast }z=g_{\ast }[S^{n+1}]=h_B[g]\mathrm{.}$ Absolute Hurewicz of the $n$-connected base identifies this with the nonzero generator of $H_{n+1}(B;\mathbb{Z})=\mathbb{Z}\mathrm{/}2$, including $n=1$, where the base is simply connected and its first nonzero homotopy degree is two. This is the required generator comparison entirely through the pair-LES and the two absolute Hurewicz maps.