==== TABLE OF CONTENTS ====
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//algebra.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//conclusion.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//cones.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//figures/proof-route.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//introduction.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//paper.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//powers.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//preamble.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//weights.tex
The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//references.bib
===========================

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//algebra.tex ====
\section{An obstruction in Steenrod modules}\label{sec:algebra}

The mapping-cone construction in Section~\ref{sec:cones} supplies
a nonzero tensor on a weight summand. In our application, its detecting
functional also lifts from an annihilated functional at the preceding level.
The obstruction below uses both inputs. Cartier halving of the lifted
functional gives a lower bound on the $d_0$-exponent. An
even-degree Tor injection then lets us halve the tensor without losing
its nonvanishing. The exponent bound controls the change of ideal at
each step.

Throughout this section $k\geq1$,
\[
 P_k=\F_2[x_1,\ldots,x_k],\qquad
 D_k=P_k^{\GL_k(\F_2)}=\F_2[d_0,\ldots,d_{k-1}],\qquad
 |d_j|=2^k-2^j,
\]
and $d_k=1$.  The invariants are normalized by
\begin{equation}\label{eq:algebra-dickson-polynomial}
 P(U)=\prod_{a\in\langle x_1,\ldots,x_k\rangle_{\F_2}}(U+a)
     =\sum_{j=0}^k d_jU^{2^j}.
\end{equation}
For $n\geq0$ we use the module
$M_k(n)=\Sigma^n d_0^nD_k$ of
Lemma~\ref{lem:weight-modules}.  A monomial in $M_k(n)$ means its
underlying polynomial monomial, followed by the grading shift.
In particular,
\begin{equation}\label{eq:algebra-degree-parity}
 \left|\Sigma^n d_0^c\prod_{j=1}^{k-1}d_j^{b_j}\right|
 =n+c(2^k-1)+\sum_{j=1}^{k-1}b_j(2^k-2^j)
 \equiv n+c\pmod2.
\end{equation}
An annihilated functional on a graded $A$-module $M$ is a homogeneous
linear functional $\lambda$ such that $\lambda(IM)=0$.

The Steenrod indecomposables of the ideals \(d_0^pD_k\) have been
studied by Giambalvo and Peterson \cite{GiambalvoPeterson2001}.
Here we first take coinvariants for the exterior algebra of Milnor
primitives, and retain the least allowed \(d_0\)-exponent throughout
the subsequent degree-halving argument. Both the lifted functional
and the tensor nonvanishing in the next statement will be needed.

The tensor product \(I\otimes_A M\) uses the right \(A\)-module
structure on \(I\); its balancing relation is
\((\alpha\beta)\otimes z=\alpha\otimes(\beta z)\), for
\(\alpha\in I\) and \(\beta\in A\).

\begin{lemma}[Algebra obstruction]\label{lem:algebra-obstruction}
Let $k\geq1$, $n=e+1$ with $e\geq0$, and $m\geq2$.  Let $v$ be
a degree-$m$ monomial of $M_k(n)$.  Suppose that its underlying
monomial is detected by an annihilated functional on $M_k(e)$ in
degree $m-1$; equivalently, $v$ is detected by the shifted restriction
of that functional along the ideal inclusion.
If
\begin{equation}\label{eq:obstruction-tensor}
 \Sq^m\otimes v\neq0
       \qquad\text{in } I\otimes_AM_k(n),
\end{equation}
then $m$ is a power of two and $n=k=1$.

Without the functional hypothesis, the tensor in
\eqref{eq:obstruction-tensor} is still zero whenever $n>0$ and
$m\geq3$ is odd.
\end{lemma}

The two halving arguments start at different exponent floors: the
functional is defined on $M_k(n-1)$, whereas the tensor lies over
$M_k(n)$. We first compute the Milnor coinvariants and Cartier map,
then obtain the functional's exponent bound. The Tor injection and
the remaining square calculations will turn that bound into the
asserted tensor obstruction.

\subsection{Milnor primitives and coinvariants}

Let $E\subset A$ be the exterior sub-Hopf algebra on the Milnor
primitives $q_i$, indexed by
$|q_i|=2^{i+1}-1$ for $i\geq0$.  Put $I_E=E^{>0}$.

\begin{lemma}\label{lem:algebra-milnor-quotient}
There is a quotient homomorphism
\[
 V:A\longrightarrow A,\qquad
 V(\Sq^{2a})=\Sq^a,\qquad V(\Sq^{2a+1})=0,
\]
where the target is regarded as having doubled degrees.  Its kernel is
\[
 K=I_EA=AI_E.
\]
Consequently $H_0(E,M)=M/I_EM$ is naturally a module over
$A/K$, for every $A$-module $M$.
\end{lemma}

\begin{proof}
Milnor's polynomial dual and coproduct description
\cite[\S5, Theorem~3, and Appendix~1]{Milnor1958} gives
$A_*=\F_2[\xi_1,\xi_2,\ldots]$. It implies that restriction to
$E_*$ is the quotient by the squares of these generators: their
images are the exterior generators dual to the Milnor primitives.
Squaring is an injective Hopf algebra
homomorphism on $A_*$; its degreewise dual is the indicated map $V$.
The formulas on the squares follow by pairing with the powers of
$\xi_1$.

For completeness, the kernel can be read directly from the Milnor
coproduct.  After restricting its first factor to $E_*$, that coproduct
becomes
\[
 \xi_i\longmapsto \bar\xi_i\otimes1+1\otimes\xi_i.
\]
The other terms have an even positive exponent in the first factor and
therefore vanish.  A polynomial has trivial coaction under this map
exactly when it is a square: the coefficients of the single generators
$\bar\xi_i$ are its ordinary partial derivatives, and their simultaneous
vanishing in characteristic two is equivalent to every exponent being
even.  Conversely, a square has trivial coaction since
$\bar\xi_i^2=0$.  Thus the annihilator of $I_EA$ is exactly the subalgebra
of squares, which also annihilates $\ker V$.  Degreewise duality proves
$I_EA=\ker V$.  Conjugation preserves $E$ and the subalgebra of squares
in $A_*$, so it also gives $AI_E=\ker V$.  Finally
$KM=I_EM$, using $K=I_EA$ and $AM=M$, and hence the stated quotient
module is well defined.
\end{proof}

\begin{lemma}\label{lem:milnor-coinvariants}
The first $k$ Milnor primitives act on $D_k$ by
\begin{equation}\label{eq:milnor-action}
 q_i=d_0\frac{\partial}{\partial d_{i+1}}\quad(0\leq i<k-1),
 \qquad
 q_{k-1}=d_0\sum_{j=0}^{k-1}d_j\frac{\partial}{\partial d_j}.
\end{equation}
Every later primitive is a linear combination of these with coefficients
that are squares in $D_k$.  All ideals $d_0^pD_k$, $p\geq0$, are $A$-submodules.

The quotient $H_0(E,d_0^pD_k)$ has a monomial basis consisting of all
monomials of $d_0$-exponent $p$, together with
\begin{equation}\label{eq:coinvariants}
 d_0^c\prod_{j=1}^{k-1}d_j^{b_j},\qquad
 c>p,\quad c\equiv k\pmod2,\quad b_j\text{ odd for every }j\geq1.
\end{equation}
\end{lemma}

\begin{proof}
The $q_i$ are derivations and $q_i x_s=x_s^{2^{i+1}}$.
The initial formulas on Dickson generators appear in
\cite[Corollary~2.3(b)]{Wilkerson1983} and are also recorded, with
indexing shifted by one, in \cite[Proposition~2.3]{Sum2020}; we give the
calculation.  Apply $q_i$ to $P(x_s)=0$.  Only the linear power of $x_s$
contributes a derivative, so
\[
 \sum_{j=0}^{k-1}q_i(d_j)x_s^{2^j}=d_0x_s^{2^{i+1}}.
\]
This identity holds on the whole binary span of the $x_s$, by additivity.
A linearized polynomial of degree at most $2^{k-1}$ is uniquely
determined by its values on that span: the difference of two such
polynomials cannot have $2^k$ distinct roots unless it is zero.
For $i<k-1$ this gives the partial-derivative formula.  For $i=k-1$,
substituting $x_s^{2^k}=\sum_{j<k}d_jx_s^{2^j}$ gives the Euler formula.

For $t\geq0$, raising the same root equation to the power $2^{t+1}$
gives, as an equality of derivations on $P_k$,
\[
 q_{k+t}=\sum_{j=0}^{k-1}d_j^{2^{t+1}}q_{j+t}.
\]
Induction on $t$ expresses every later primitive in the first $k$ with
square coefficients.  These coefficients belong to $D_k$.

Since $d_0$ is the product of the nonzero binary linear forms and the
total square of such a form $a$ is $a+a^2$, the total square of $d_0$
is divisible by $d_0$.  Its quotient is invariant and therefore belongs
to $D_k$.  The Cartan formula now proves the stability of every
$d_0^pD_k$.

Each nonzero term in \eqref{eq:milnor-action} raises the $d_0$-exponent
by exactly one.  Thus no monomial at exponent $p$ is hit.  Above this
floor, a monomial with some even $b_j$ is the $q_{j-1}$-image of the
monomial obtained by decreasing $c$ by one and increasing $b_j$ by one.
This preimage is still in $d_0^pD_k$.  If all $b_j$ are odd, the Euler
primitive on the monomial at exponent $c-1$ has coefficient
\[
 (c-1)+\sum_{j=1}^{k-1}b_j\equiv c+k\pmod2.
\]
It hits the given monomial exactly when $c\not\equiv k\pmod2$.
The images of the first $k$ primitives are spanned by these individual
monomials.  Later primitives add no new images: a square coefficient
passes inside any of the derivations.  This proves the basis assertion,
including the case $k=1$, when the list of $b_j$ is empty.
\end{proof}

\subsection{Cartier halving, with its exponent threshold}

We call the span of the monomials in \eqref{eq:coinvariants} the
\emph{above-floor part} of the coinvariants.  The following lemma keeps
track both of the action and of the first exponent in this part.

\begin{lemma}\label{lem:cartier}
The above-floor part of $H_0(E,d_0^pD_k)$ is an $A/K$-submodule.
After identifying $A/K$ with the Steenrod algebra by halving degrees,
the map
\begin{equation}\label{eq:cartier}
 C:\quad d_0^c\prod_{j=1}^{k-1}d_j^{b_j}
 \longmapsto
 d_0^{(c+k-2)/2}\prod_{j=1}^{k-1}d_j^{(b_j-1)/2}
\end{equation}
identifies it with an ideal in $D_k$.  More precisely, if
$c_0$ is the least integer greater than $p$ with $c_0\equiv k\pmod2$,
the target ideal is $d_0^{(c_0+k-2)/2}D_k$.
Internal polynomial degrees transform by $d\mapsto(d-k)/2$.
\end{lemma}

\begin{proof}
On $P_k$ define $C_x$ to be zero on a monomial unless all its exponents
are odd, and in the latter case set
\[
 C_x(x_1^{2a_1+1}\cdots x_k^{2a_k+1})
   =x_1^{a_1}\cdots x_k^{a_k}.
\]
This is the dual Kameko squaring map
\cite{Kameko1990}; see \cite[\S2, p.~4069]{Hung2005} for the
dual formula. It is also the polynomial top-form version of
Cartier's differential-form construction \cite{Cartier1957};
see also \cite[Chapter~II, \S6, pp.~200--203]{Cartier1958}.
We include the polynomial identities and naturality needed for its
restriction to the present ideals.
The binomial formula for squares on a one-variable polynomial, together
with the Cartan formula, gives
\begin{equation}\label{eq:algebra-cartier-action}
 C_x\Sq^{2i}=\Sq^iC_x,\qquad C_x\Sq^{2i+1}=0,
 \qquad C_x(a^2f)=aC_x(f).
\end{equation}
Indeed, on an odd exponent the parity of
$\binom{2a+1}{2i}$ is the parity of $\binom ai$; an odd square makes
that exponent even.  An even input exponent can never contribute an
odd output exponent, since the corresponding binomial coefficient then
vanishes.  Applying these facts in all variables proves the identities.

The operator $C_x$ commutes with $\GL_k(\F_2)$.  To see this, write a
polynomial uniquely as a sum of square-free monomials with square
coefficients.  The coefficient of $x_1\cdots x_k$ after a linear change
of variables is multiplied by its determinant; no square-free monomial
of smaller degree can contribute to that coefficient.  Over $\F_2$
the determinant is one, and taking the square root of the coefficient
commutes with the linear change of variables.  Hence $C_x$ preserves
$D_k$.

We compute this restriction by an elementary form of Cartier
naturality.  If $y_1,\ldots,y_k$ are algebraically independent
polynomials and $J=\det(\partial y_j/\partial x_i)$, then
\begin{equation}\label{eq:algebra-cartier-jacobian}
 C_x\bigl(Jf(y_1,\ldots,y_k)\bigr)
       =J\,C_y(f)(y_1,\ldots,y_k).
\end{equation}
Here $C_y$ is the same odd-exponent extractor in formal variables $y_j$.
For a proof, the polynomial differential-form complex in one variable
has cohomology generated by $1$ and $[x\,dx]$ over $\F_2[x^2]$.
Tensoring these complexes gives the corresponding assertion in $k$
variables.  Thus there is an isomorphism from the algebra of forms,
regarded without differential, to its cohomology, given by
\[
 f\longmapsto f^2,\qquad dx_i\longmapsto[x_i\,dx_i].
\]
This isomorphism is natural under polynomial substitution.  On
differentials the required fact is that $f\mapsto[f\,df]$ is additive
and satisfies the derivation rule with coefficients squared: its
additivity follows because the cross term is $d(fg)$, and its product
rule follows by expanding $(fg)d(fg)$.  In top degree the inverse
isomorphism sends $[h\,dx_1\cdots dx_k]$ to
$C_x(h)\,dx_1\cdots dx_k$.  Pulling back a top form in the $y$
variables proves \eqref{eq:algebra-cartier-jacobian}.

For Dickson coordinates $y_j=d_{j-1}$ the Jacobian is
\begin{equation}\label{eq:algebra-dickson-jacobian}
 J=d_0^{k-1}.
\end{equation}
In fact, differentiating $P(x_s)=0$ with respect to each $x_i$ gives
the matrix identity whose two factors have entries $x_s^{2^j}$ and
$\partial d_j/\partial x_i$, and whose product is $d_0$ times the
identity matrix.  The determinant of the first factor is $d_0$: it is
nonzero, is divisible by every nonzero binary linear form, and has
degree $2^k-1$.  Taking determinants gives
\eqref{eq:algebra-dickson-jacobian}.

For $c\geq k-1$, apply
\eqref{eq:algebra-cartier-jacobian} to
$f=d_0^{c-k+1}\prod_{j\geq1}d_j^{b_j}$.  It is zero unless
$c\equiv k\pmod2$ and every $b_j$ is odd; otherwise it is exactly
\eqref{eq:cartier}.  The same assertion holds for smaller nonnegative
$c$.  Multiply first by a sufficiently large even power $d_0^{2N}$,
use $C_x(d_0^{2N}f)=d_0^NC_x(f)$, and cancel $d_0^N$ in the polynomial
domain.  The output exponents in the nonzero cases are nonnegative.

The description of the hits in
Lemma~\ref{lem:milnor-coinvariants} now shows that $C_x$ descends to
coinvariants and is a bijection on the stated above-floor monomials.
Squares do not decrease the $d_0$-exponent: this follows from the
divisibility of the total square of $d_0$ used above and the Cartan
formula.  Consequently the above-floor span is a submodule.
Equation~\eqref{eq:algebra-cartier-action} gives the claimed action on
its image.  The least possible target exponent is the one displayed
in the statement, and a direct degree calculation in
\eqref{eq:cartier} gives the change $d\mapsto(d-k)/2$.
\end{proof}

\begin{remark}
On its image, the inverse monomial transformation in
\eqref{eq:cartier} is
\(f\mapsto f^2d_0^{2-k}d_1\cdots d_{k-1}\), with nonnegative
resulting exponents. Giambalvo and Peterson use this transformation
in their study of Dickson ideals \cite[\S4]{GiambalvoPeterson2001}.
Lemma~\ref{lem:cartier} specifies the simultaneous Milnor-coinvariant
quotient and the exact exponent floor used in the argument here.
\end{remark}

\subsection{The lifted functional and its exponent bound}

We first use Cartier halving on the functional from the preceding
level. Its ideal has exponent floor $n-1$, while the monomial it
detects already belongs to the next ideal, with floor $n$. The
resulting bound will control which ideal occurs when we halve the
tensor itself, including whether the floor can contribute.

\begin{lemma}\label{lem:algebra-functional-halving}
Let $n=e+1>0$, let $m$ be even, and let $v$ be a degree-$m$ monomial
in $M_k(n)$ with $d_0$-exponent $c$.  Suppose that its underlying
monomial is detected by an annihilated functional on $M_k(e)$ in
degree $m-1$.  If $b=2^{\nu_2(m)}$, then
\begin{equation}\label{eq:halving-bounds}
 n\equiv k\pmod b,\qquad c\geq n+b-2.
\end{equation}
If $m$ is a power of two, this functional hypothesis alone forces
$n=k=1$.
\end{lemma}

\begin{proof}
The monomial lies above the exponent floor $e$ in $M_k(e)$.
Since the functional factors through exterior coinvariants,
Lemma~\ref{lem:milnor-coinvariants} forces
$c\equiv k\pmod2$ and all its other exponents to be odd.
Its total degree there is $m-1$, so
\eqref{eq:algebra-degree-parity} gives $n\equiv k\pmod2$.

Restrict the induced functional to the above-floor submodule and use
Lemma~\ref{lem:cartier}.  The first allowed exponent in this submodule
is $c=n$, so its image has first exponent
\[
 e'=\frac{n+k-2}{2}.
\]
Give that image the grading shift $e'$, and put
\begin{equation}\label{eq:algebra-functional-recursion}
 m'=\frac m2,\qquad n'=\frac{n+k}{2},\qquad
 e'=n'-1,\qquad c'=\frac{c+k-2}{2}.
\end{equation}
The image is exactly $M_k(e')$, and its paired monomial has degree
$m'-1$: if the old internal degree is $d$, then
$e'+(d-k)/2=(m-2)/2$.
The new functional is again annihilated, because the Cartier map
intertwines the halved action.  If $m'$ is even, its paired monomial
cannot be at the floor $c'=e'$; a monomial at that floor has even
total degree, whereas $m'-1$ is odd.  It therefore lies in the next
ideal and satisfies the same hypothesis, allowing the step to repeat.

Iterate while the paired parameter $m$ is even, for
$t=\nu_2(m)$ steps.  With $b=2^t$, the resulting parameters obey
\[
 n_t=\frac{n+(b-1)k}{b},\qquad
 c_t-n_t+2=\frac{c-n+2}{b},\qquad c_t\geq n_t-1.
\]
The integrality of $n_t$ gives the congruence in
\eqref{eq:halving-bounds}, and the last two relations give its
inequality.  All shifts $e_t=n_t-1$ are nonnegative.

If $m=b$, the final paired monomial has degree zero in $M_k(e_t)$.
The smallest degree in that module is $2^ke_t$, so $e_t=0$.
Thus
\[
 n+(m-1)k=m.
\]
Since $n,k\geq1$, this equality forces $n=k=1$.
\end{proof}

\subsection{Injectivity for degree-one Tor}

The exact sequence $0\to I\to A\to\F_2\to0$ identifies
$\Tor_1^A(\F_2,M)$ with the kernel of the action map
$I\otimes_A M\to M$.  We refer to elements of this kernel as
\emph{cycles}.  A cycle represents zero precisely when it is zero in
the balanced tensor product.

\begin{lemma}\label{lem:tor-injection}
Let $n\geq0$ and $n\equiv k\pmod2$.  On cycles of even total degree,
the projection
\begin{equation}\label{eq:tor-injection}
 I\otimes_A M_k(n)\longrightarrow
 I_{A/K}\otimes_{A/K}H_0(E,M_k(n))
\end{equation}
is injective.
\end{lemma}

\begin{proof}
Put $M=M_k(n)$ and $B=A/K$.
The exact sequence of right $A$-modules
$K\to I\to I_B\to0$, followed by tensoring with $M$, shows that the
kernel of \eqref{eq:tor-injection} is the image of $K\otimes_A M$.
Here we have used
\[
 I_B\otimes_A M
 =I_B\otimes_B(B\otimes_A M)
 =I_B\otimes_BH_0(E,M).
\]
Multiplication gives a surjection $I_E\otimes_EA\to K=I_EA$;
tensoring again gives a surjection
$I_E\otimes_EM\to K\otimes_AM$.
Thus every element in this kernel has a preimage in
$I_E\otimes_EM$.  The action maps from both tensor products to $M$
commute with this map.  A preimage of a cycle is therefore a cycle
itself.

We prove that every even-total-degree cycle in $I_E\otimes_EM$ is
zero.  Write a homogeneous such cycle as
\[
 z=\sum_i q_i\otimes z_i.
\]
Each $z_i$ has odd total degree in $M$.  By
\eqref{eq:algebra-degree-parity}, its monomials have $d_0$-exponent
$c\geq n+1$ and parity opposite to $n$, hence opposite to $k$.
Lemma~\ref{lem:milnor-coinvariants} gives $z_i\in I_EM$.

This last property allows us to replace later primitives by the first
$k$ inside the tensor product.  If
$q_i=\sum_{j<k}a_j^2q_j$ on $M$, then
\begin{equation}\label{eq:algebra-hit-balancing}
 q_i\otimes z_i=\sum_{j<k}q_j\otimes a_j^2z_i
 \qquad(z_i\in I_EM).
\end{equation}
It suffices to check this for $z_i=q_sy$.  Balancing, commutativity of
the exterior generators, and their vanishing on squares give
\[
 \begin{aligned}
 q_i\otimes q_sy
 &=q_s\otimes q_iy
 =\sum_{j<k}q_s\otimes a_j^2q_jy\\
 &=\sum_{j<k}q_s\otimes q_j(a_j^2y)
 =\sum_{j<k}q_j\otimes a_j^2q_sy.
 \end{aligned}
\]
Thus $z$ has a cycle representative using only $q_j$, $j<k$, whose
coefficients still have exponents at least $n+1$.

Let $E_{[k]}=\Lambda(q_0,\ldots,q_{k-1})$ and
$L=D_k[d_0^{-1}]$.  We first show that $L$ is projective over
$E_{[k]}$.  For $T\subseteq\{0,\ldots,k-1\}$ write
$q_T=\prod_{i\in T}q_i$, and put
\[
 v_0=d_0^{-k}\prod_{j=1}^{k-1}d_j.
\]
The first $k-1$ primitives take $v_0$ to $d_0^{-1}$, and the last,
Euler primitive takes $d_0^{-1}$ to $1$.  Hence
$q_{\{0,\ldots,k-1\}}v_0=1$.
The action map from the free module $E_{[k]}\otimes_{\F_2}L$ to $L$
has the following section:
\begin{equation}\label{eq:algebra-projective-section}
 s(x)=\sum_{T\subseteq\{0,\ldots,k-1\}}
        q_T\otimes v_0\bigl(q_{T^c}x\bigr).
\end{equation}
It is $E_{[k]}$-linear: multiplying its first factor by $q_i$ and
reindexing the subsets gives $s(q_ix)$, since terms with repeated
$q_i$ vanish.  To compute the composite with the action map, expand
each $q_T(v_0q_{T^c}x)$ by the Cartan formula.  For a fixed subset $U$,
the term $(q_Uv_0)(q_{U^c}x)$ occurs once for every $T$ with
$U\subseteq T\subseteq\{0,\ldots,k-1\}$.  Its coefficient is
$2^{k-|U|}$, so only the full subset remains.  The composite is
therefore $x$.  The formula preserves degrees, since $v_0$ has the
negative of the degree of the product of all the $q_i$.
This proves the asserted projectivity.

Consider now the tensor product of the usual periodic free resolutions
for the one-generator exterior algebras.  After tensoring with an
$E_{[k]}$-module, its first differential is
\[
 (u_0,\ldots,u_{k-1})\longmapsto\sum_{i<k}q_iu_i.
\]
The degree-two terms have one generator for each repeated pair $i,i$
and each distinct pair $i,j$. Writing $e_i$ for the degree-one
generators, their images are
\[
 e_i\otimes q_i y,\qquad
 e_i\otimes q_jy+e_j\otimes q_iy\quad(i<j).
\]
The cokernel of this differential is
$I_{E_{[k]}}\otimes_{E_{[k]}}M$: the displayed relations become
$q_i^2\otimes y=0$ and
$(q_iq_j+q_jq_i)\otimes y=0$ under balancing.
Projectivity of $L$ implies that every degree-one
cycle in this complex with coefficients in $L$ has a degree-two
filling.

Apply this to the cycle representative obtained above, ignoring its
harmless shift by $n$.  Choose a filling with Laurent-polynomial
coefficients.  In each of the first two differentials, every nonzero
term raises the $d_0$-exponent of its coefficient by exactly one, by
\eqref{eq:milnor-action}.  Delete from the filling all monomials with
exponent less than $n$.  Their differential has exponents at most $n$,
whereas the retained part has exponents at least $n+1$.
Since the given cycle has only exponents at least $n+1$, the retained
part is still a filling.  Its coefficients belong to $d_0^nD_k$.
All Laurent polynomials and sums used here are finite, so this deletion
requires no completion.

We have proved that the representative is a boundary already over
$E_{[k]}$ with coefficients in $M$.  Equivalently, it is zero in
$I_{E_{[k]}}\otimes_{E_{[k]}}M$, and hence also in
$I_E\otimes_EM$.  Its image in $I\otimes_AM$ is zero, which proves
the lemma.
\end{proof}

\subsection{Proof of the obstruction lemma}

\begin{lemma}\label{lem:algebra-odd-tensor}
If $n>0$, $m\geq3$ is odd, and $v\in M_k(n)$ is homogeneous of
degree $m$, then $\Sq^m\otimes v=0$ in $I\otimes_AM_k(n)$.
\end{lemma}

\begin{proof}
The Adem relation $\Sq^m=\Sq^1\Sq^{m-1}$ gives
$\Sq^m\otimes v=\Sq^1\otimes\Sq^{m-1}v$.
For $n\geq2$ the second operation exceeds the internal polynomial
degree $m-n$, so this is zero.  For $n=1$ it is the internal square
$v^2$.  By linearity we may take
$v=d_0^c\prod_{j\geq1}d_j^{b_j}$, where $c\geq1$.
For $k\geq2$, formula~\eqref{eq:milnor-action} gives
\[
 v^2=\Sq^1\left(
 d_0^{2c-1}d_1^{2b_1+1}\prod_{j=2}^{k-1}d_j^{2b_j}
 \right).
\]
The preimage belongs to $d_0D_k$.  For $k=1$ use instead
$\Sq^1(d_0^{2c-1})=d_0^{2c}$.  Balancing once more and using
$(\Sq^1)^2=0$ proves the assertion.
\end{proof}

The even-degree argument below uses the following square formulas
only in its final case. These are the square-action formulas of Hai and Hung,
recorded in \cite[Theorem~B]{Hung1991}; the proof below uses the
root-polynomial method of \cite[\S II]{Wilkerson1983}.

\begin{lemma}\label{lem:algebra-near-top-square}
The total Steenrod square on a Dickson generator is
\begin{equation}\label{eq:algebra-total-dickson-square}
 \Sq(d_j)=d_j^2+
       \left(\sum_{i=0}^jd_i\right)
       \left(\sum_{a=j+1}^kd_a\right).
\end{equation}
In particular the operation one below the top on $d_j$ is
$d_0d_{j+1}$.  For a positive-degree monomial
$v=\prod_{j=0}^{k-1}d_j^{b_j}$ of internal degree $d$,
\begin{equation}\label{eq:algebra-one-below-top}
 \Sq^{d-1}v
 =v^2\sum_{j:\ b_j\ \text{odd}}\frac{d_0d_{j+1}}{d_j^2}.
\end{equation}
Every displayed summand is a polynomial.
\end{lemma}

\begin{proof}
Apply the total square simultaneously to the variables $x_s$ and $U$
in \eqref{eq:algebra-dickson-polynomial}.  Since
$\Sq(U+a)=(U+a)(U+a+1)$ and $P$ is additive,
\[
 \sum_{j=0}^k\Sq(d_j)(U+U^2)^{2^j}
 =P(U)P(U+1)=P(U)^2+P(1)P(U).
\]
Comparison of the coefficients of $U^{2^j}$ gives
$\Sq(d_0)=P(1)d_0$ and, for $1\leq j<k$,
\[
 \Sq(d_j)+\Sq(d_{j-1})=d_{j-1}^2+P(1)d_j.
\]
Summing these identities yields
\[
 \Sq(d_j)=P(1)\sum_{i=0}^jd_i+\sum_{i=0}^{j-1}d_i^2,
\]
which is \eqref{eq:algebra-total-dickson-square}.
The term of degree $2|d_j|-1$ in that formula is exactly
$d_0d_{j+1}$.  In the Cartan expansion of an operation one below the
top on a product, exactly one factor receives its operation one below
the top and every other factor receives its top operation.  Equal
choices cancel in pairs, leaving precisely
\eqref{eq:algebra-one-below-top}.
\end{proof}

\begin{lemma}\label{lem:algebra-even-vanishing}
Let $m$ be even and not a power of two, let $b=2^{\nu_2(m)}$, and
let $v\in M_k(n)$ be a degree-$m$ monomial with $n>0$ and
$d_0$-exponent $c$.  If \eqref{eq:halving-bounds} holds, then
$\Sq^m\otimes v=0$ in $I\otimes_AM_k(n)$.
\end{lemma}

\begin{proof}
The tensor is a cycle, because $\Sq^m$ exceeds the internal degree
$m-n$ of $v$.  We induct on $\nu_2(m)$, using
Lemma~\ref{lem:tor-injection} throughout; its parity assumption follows
from \eqref{eq:halving-bounds}. Here we halve coinvariants of the
current ideal, with floor $n$. This differs from the floor $n-1$
used for the functional in Lemma~\ref{lem:algebra-functional-halving}.

Suppose first that $b\geq4$.  The image of $v$ in exterior
coinvariants is either zero, in which case the injectivity lemma gives
the result immediately, or is an above-floor survivor.  Indeed
$c\geq n+b-2\geq n+2$.  In this case the first allowed exponent in
the above-floor part of $M_k(n)$ is $n+2$, and its Cartier image has
first exponent
\[
 n'=\frac{n+k}{2}.
\]
With this shift the image is $M_k(n')$, with total degrees halved.
The image of our cycle in \eqref{eq:tor-injection} consequently comes
from
$\Sq^{m/2}\otimes v'$ in $I\otimes_AM_k(n')$, where $v'=C(v)$.
Put $c'=(c+k-2)/2$ and $b'=b/2$.  Then
\[
 n'\equiv k\pmod{b'},\qquad
 c'-n'+2=\frac{c-n+2}{2}\geq b'.
\]
Thus the same bounds hold for $m/2,n',c'$.  Since $m/2$ is still
not a power of two, induction makes this latter tensor zero, and
injectivity gives the required vanishing of the original cycle.

It remains to treat $b=2$.  Write $m=2L$, with $L\geq3$ odd.
All coinvariants of $M_k(n)$ have even total degree: this is clear
at the floor from \eqref{eq:algebra-degree-parity}, and above it from
$c\equiv k\equiv n\pmod2$.  Regrade them by half and call the
resulting $A$-module $C_n$.  The projected tensor is
\[
 \Sq^L\otimes[v]
   =\Sq^1\otimes\Sq^{L-1}[v].
\]
The second factor is the class of the original operation
$\Sq^{m-2}v$.  It suffices to show that this class is zero or an
image of $\Sq^1$ in $C_n$, since balancing then introduces
$(\Sq^1)^2=0$.

If $n\geq3$, the operation already exceeds the internal degree of
$v$.  If $n=2$, then $k$ is even and at least two, and the operation
is the internal square $v^2$.  This lies above the floor and has all
nonzero-index exponents even, so it is zero in coinvariants by
Lemma~\ref{lem:milnor-coinvariants}.

Suppose $n=1$, so $k$ is odd.  For $k=1$ the coinvariant basis,
including the floor, consists of all odd monomials in $d_0$.
The map $C_x$ identifies $C_1$ with $\Sigma D_1$, and the image of
$v=d_0^{2L-1}$ is $d_0^{L-1}$.  Its image under $\Sq^{L-1}$ is
\[
 d_0^{2L-2}=\Sq^1(d_0^{2L-3}),
\]
as required.

Finally, suppose $n=1$ and $k\geq3$ is odd.  The internal degree of
$v$ is $m-1$, so Lemma~\ref{lem:algebra-near-top-square} applies to
$\Sq^{m-2}v$.  In each summand of
\eqref{eq:algebra-one-below-top}, at most one exponent of
$d_j$ with $j\geq1$ is odd.  Since there are at least two such
indices, every term above the floor dies in exterior coinvariants.
The only possible floor term comes from the index $j=0$ when
$v=d_0w$, where $w$ involves only $d_j$, $j\geq1$.  It is
$d_0d_1w^2$.

This remaining term is an image for the halved first square.  Formula
\eqref{eq:algebra-total-dickson-square} gives
$\Sq^2d_2=d_1$ and $\Sq^1d_0=\Sq^2d_0=0$ for $k\geq3$.
Together with \eqref{eq:milnor-action} and the Cartan formula, it
therefore gives the explicit identity
\begin{equation}\label{eq:algebra-floor-witness}
 \Sq^2(d_0d_2w^2)
 =d_0d_1w^2+
   d_0^3d_2\left(\frac{\partial w}{\partial d_1}\right)^2.
\end{equation}
The extra term is an exterior hit in $d_0D_k$, with preimage
\[
 q_0\left(
 d_0^2d_1d_2\left(\frac{\partial w}{\partial d_1}\right)^2
 \right)
 =d_0^3d_2\left(\frac{\partial w}{\partial d_1}\right)^2.
\]
Hence the class of $d_0d_1w^2$ is the original $\Sq^2$-image of
$[d_0d_2w^2]$, or equivalently a $\Sq^1$-image after halving.
This completes the final case, and the injectivity lemma finishes
the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:algebra-obstruction}]
The odd assertion is Lemma~\ref{lem:algebra-odd-tensor}.
For even $m$, the functional hypothesis gives
\eqref{eq:halving-bounds} by
Lemma~\ref{lem:algebra-functional-halving}.  If $m$ is not a power
of two, Lemma~\ref{lem:algebra-even-vanishing} contradicts
\eqref{eq:obstruction-tensor}.  If $m$ is a power of two, the last
assertion of Lemma~\ref{lem:algebra-functional-halving} gives
$n=k=1$.
\end{proof}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//algebra.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//conclusion.tex ====
\section{The last nonzero suspension and the sphere}\label{sec:conclusion}

Let \(\alpha\in\pi_d^S\), \(d>0\), have nonzero mod-2 Hurewicz
image. Represent it by \(f_0:S^d\to Q_0S^0\), and write
\[
f_j:S^{d+j}\longrightarrow QS^j,\qquad
z_j=h(f_j)=\sigma_*^j h(f_0)
\]
for the successive adjoints and their Hurewicz images. All the \(z_j\)
are primitive and \(A\)-annihilated. There is a largest \(e\ge0\)
with \(z_e\ne0\). Indeed, when \(j>d\), the degree \(d+j\) lies
strictly between \(j\) and \(2j\); the free infinite-loop homology
description has no class in that interval.

\begin{lemma}\label{lem:last-root}
For some \(m\ge2\),
\begin{equation}\label{eq:last-square}
z_e=w^2,\qquad |w|=m-1,\qquad d+e=2m-2,
\end{equation}
where \(w\) is primitive and \(A\)-annihilated, and
\(\sigma_*w\ne0\). If \(e\ge1\), then \(|w|\) is odd.
\end{lemma}

\begin{proof}
Since \(\sigma_*z_e=0\), Lemmas~\ref{lem:suspension-kernel}
and~\ref{lem:primitive-roots} give \eqref{eq:last-square} with a
primitive \(A\)-annihilated root. Its degree is positive, so \(m\ge2\).

If \(e=0\) and \(\sigma_*w=0\), the same lemmas make \(w\) a
square. Then \(z_0\) is a fourth power, contradicting
Proposition~\ref{prop:no-fourth-powers}. Thus the root suspends
nontrivially in this case.

For \(e\ge1\), we instead use the preceding spherical class
\(z_{e-1}\) to prove that \(|w|\) is odd. First suppose that
\(e\ge2\), so both levels have the direct weight decomposition
of Lemma~\ref{lem:weight-modules}. Choose a nonzero weight \(2^l\)
component of \(z_e\). Here \(l\ge1\), since the degree \(d+e\)
is greater than the bottom degree \(e\). Its functional on
\(M_l(e)\) comes, after the grading shift, from the functional of
\(z_{e-1}\) on \(M_l(e-1)\). Vanishing of \(z_{e+1}\) says that
it annihilates \(d_0^{e+1}D_l\), by
\eqref{eq:module-suspension}. It must therefore pair nontrivially
with some floor monomial
\[
d_0^e\prod_{j=1}^{l-1}d_j^{b_j}.
\]
In \(M_l(e-1)\) this monomial is above the floor. Since the lifted
functional is \(A\)-annihilated, Lemma~\ref{lem:milnor-coinvariants}
gives
\[
e\equiv l\pmod2,\qquad b_j\text{ odd for every }j\ge1.
\]
The degree formula and \eqref{eq:last-square} yield
\begin{equation}\label{eq:root-parity}
m-1=2^{l-1}e+
\sum_{j=1}^{l-1}(2^{l-1}-2^{j-1})b_j.
\end{equation}
For \(l\ge2\), only the term with \(j=1\) is odd. For \(l=1\),
the right side is \(e\), which is odd by the congruence. Thus
\(m-1\) is odd.

If \(e=1\), \eqref{eq:last-square} shows that \(d\) is odd. Choose
the leading weight \(2^l\) of \(z_0\). Lemma~\ref{lem:zero-weights}
says that it survives to \(z_1\), with a functional lifted from the
annihilated functional on \(M_l(0)\). Since \(z_2=0\), the same
floor argument applies to this component of \(z_1\), and again
\eqref{eq:root-parity} proves that \(m-1\) is odd.

An odd-degree class cannot be a square. Therefore the suspension-kernel
and primitive-root lemmas imply \(\sigma_*w\ne0\) whenever \(e\ge1\).
\end{proof}

Set \(n=e+1\). The map
\[
f_n:S^{2m-1}\longrightarrow T=QS^n
\]
has zero Hurewicz image, and its previous adjoint has image \(w^2\).
The loop product under \(QS^{n-1}\simeq\Omega QS^n\) agrees with
the additive Pontryagin product. Thus Section~\ref{sec:cones} applies
with \(x=w\). The nonzero functional
\begin{equation}\label{eq:last-functional}
\lambda=\langle -,\sigma_*w\rangle
\quad\text{on }\bigl(Q\widetilde H^*(QS^n)\bigr)^m
\end{equation}
gives \(\Sq^m\otimes v\ne0\) for every monomial \(v\) on which
it has value one.

We first deduce that \(m\) is even. If \(m\) were odd, then
\(m\ge3\). Lemma~\ref{lem:algebra-obstruction} excludes a nonzero
pairing on every \(M_k(n)\) with \(k\ge1\). A bottom pairing
requires \(m=n\); on \(M_0(n)=\Sigma^n\F_2\), the balanced tensor
product is the shift of \(I/I^2\). But
\(\Sq^m=\Sq^1\Sq^{m-1}\in I^2\), contradicting
Corollary~\ref{cor:tensor-detection}. Since \(\lambda\ne0\), this
rules out odd \(m\).

\begin{proposition}[The positive terminal stage]\label{prop:hopf-case}
If \(e\ge1\), then \(\alpha\) has nonzero mod-2 stable Hopf
invariant and \(d\in\{1,3,7\}\).
\end{proposition}

\begin{proof}
Suppose that \(\sigma_*w\) has a nonzero component of weight
\(2^k\) with \(k\ge1\). Its degree-\(m\) functional on
\(M_k(n)\) is lifted from the \(A\)-annihilated functional of
\(w\) on \(M_k(e)\). The nonzero tensor supplied by
\eqref{eq:last-functional} therefore satisfies all the hypotheses of
Lemma~\ref{lem:algebra-obstruction}. That lemma forces \(n=1\),
contrary to \(n=e+1\ge2\). Hence \(\lambda\) is nonzero on the
bottom summand, giving
\[
m=n=e+1,\qquad d=e.
\]
Lemma~\ref{lem:cone-square} now says that the extension of the
weight-one bottom class in \(C_{f_n}\) has nonzero \(\Sq^n\), equal
to the top relative class.

Let \(\epsilon:\Sigma^\infty QS^n\to\Sigma^nS\) be the stable
counit and define
\[
\alpha_n=\epsilon\circ\Sigma^\infty f_n:
\Sigma^{2n-1}S\longrightarrow\Sigma^nS.
\]
This is the stable adjoint representing \(\alpha\) in stem
\(d=n-1\). The counit is the weight-one projection, by
Proposition~\ref{prop:free-ring}. The commutative square with identity
on the attaching sphere induces a map
\[
\Sigma^\infty C_{f_n}\longrightarrow C_{\alpha_n}
\]
to the stable two-cell cofiber. Its pullback carries the bottom class
to the weight-one bottom class and the top relative class to the top
relative class. Both top relative classes inject in mod-2 cohomology.
Naturality therefore makes \(\Sq^n\) nonzero in \(C_{\alpha_n}\).
This is nonzero mod-2 stable Hopf detection of \(\alpha\).

Adams's Hopf invariant one theorem, in its stable two-cell form
\cite[Theorem~1.1.1(d)]{Adams1960}, permits \(n=1,2,4,8\).
Since \(d=n-1>0\), we obtain \(d=1,3,7\).
\end{proof}

\begin{proposition}[The terminal stage zero]\label{prop:kervaire-case}
If \(e=0\), then for some \(j\ge1\),
\[
d=2^{j+1}-2,
\]
and \(\alpha\) has Kervaire invariant one.
\end{proposition}

\begin{proof}
\emph{Leading weight.}
Here \(n=1\), and we have proved that \(m\) is even. Thus
\(|w|=m-1\) is odd. Choose the leading weight \(2^k\) of \(w\)
in the level-zero expansion. It has \(k\ge1\), because \(w\) has
positive degree. By Lemma~\ref{lem:zero-weights}, this weight survives
in \(\sigma_*w\), with functional lifted from an annihilated
functional on \(D_k=M_k(0)\). Combining
Corollary~\ref{cor:tensor-detection} with
Lemma~\ref{lem:algebra-obstruction} gives
\[
k=1,\qquad m=2^j\quad\text{for some }j\ge1.
\]
Consequently \(z_0=w^2\) has leading weight four: its leading term
is the square of the nonzero weight-two term, and Frobenius is
injective in the associated free polynomial homology algebra. Also
\(d=2m-2=2^{j+1}-2\).

\emph{Adams filtration.}
We bound the Adams filtration of the given element.
Let \(\mathcal I\) be the augmentation ideal of
\(\Sigma^\infty_+QS^0\). Kuhn's lifting theorem
\cite[Theorems~1.2--1.3]{Kuhn2018}, applied to
\(X=S\) and the coefficient ring \(H\F_2\), says that an element
of mod-2 Adams filtration at least \(s_0\) has Hurewicz image lifted
from \(H\F_2\wedge\mathcal I^{2^{s_0}}\), where
\(\mathcal I^{2^{s_0}}\) is the corresponding derived augmentation-ideal power of
\(\Sigma^\infty_+QS^0\). Its hypotheses hold: both spectra are
connective and \(\pi_0S\to\pi_0H\F_2\) is onto. At the prime two
the localization away from \((p-1)!\) in that theorem is vacuous.
The theorem does not require \(QS^0\) to be connected.

To use this with the leading weight, take the augmented algebra map
\[
\Sigma^\infty_+QS^0\longrightarrow\mathbb P_{\le7}(S\{u\})
\]
from \eqref{eq:weight-projections}, and let \(J\) denote the
positive ideal in its target. Regard \(J\) as a nonunital algebra
in the full category of nonnegatively graded spectra, with ordinary
Day convolution and with \(J_r=0\) for \(r=0\) and \(r\ge8\).
Using the reduced commutative operad \(\mathrm{Com}\), whose arities
are positive, its derived eighth power is the bar construction
\[
J^8=B(\mathrm{Com}_{\ge8},\mathrm{Com},J);
\]
see \cite[Definition~2.7]{Kuhn2018} and
\cite[Definition~3.1 and Example~3.2]{KuhnPereira2017}. Compute this
derived bar construction with a free resolution in
strictly positive weights. Every simplicial summand then has at least
eight inputs of weight at least one; no nullary operation can remove
such an input.
Its total weight is therefore at least eight.
The canonical map \(J^8\to J\) preserves weight and is zero, since
\(J\) has only weights one through seven. The direct-sum functor
from the full graded category is strong symmetric monoidal and
preserves realizations and homotopy orbits. It therefore carries the
graded free resolution and bar construction to the corresponding
ungraded derived construction, with the same zero canonical map.
Extension of coefficients to \(H\F_2\) also preserves this
construction \cite[Theorem~3.4(d)]{KuhnPereira2017}; in particular,
the canonical map remains zero after smashing with \(H\F_2\).

By naturality, the Hurewicz image of an element of Adams filtration
at least three would vanish under every positive-weight projection
below eight. The weight-four
image just found is nonzero. Thus \(\alpha\) has Adams filtration
at most two. This uses only a lower bound on possible weights; it does
not identify Adams filtration with an exact weight.

\emph{Kervaire detection.}
Positive stable stems are finite by Serre's finiteness theorem
\cite[Chapter~V, \S3, Proposition~3]{Serre1951}: choose an odd
sphere dimension \(N>d+1\), so that \(\pi_{N+d}(S^N)\) is both
stable and finite. Their odd-primary parts have zero
mod-2 Hurewicz image and zero mod-2 detecting invariants, so we may
discard them and use the convergent mod-2 Adams spectral sequence for
the remaining two-primary group. More precisely, Adams's convergence
theorem identifies its successive filtration quotients with
\(E_\infty\) and its filtration intersection with the prime-to-two
torsion, which has been removed
\cite[Theorem~2.1(i),(v),(vi)]{Adams1958}.
Write \(F^s\pi_d^S\) for the subgroup of Adams filtration at least
\(s\) in this two-primary group. Filtration zero has no positive-stem
sphere class. Filtration one has internal degrees \(2^a\), hence
stems \(2^a-1\); none is the positive even stem in question. In
filtration two, the classical calculation
\cite[Theorem~2.5.1(1)--(2)]{Adams1960} gives basis elements \(h_bh_a\),
where \(b\ge a\ge0\) and \(b\ne a+1\). The identity
\(2^a+2^b=2^{j+1}\) forces \(a=b=j\). Therefore
\[
\Ext_A^{2,2^{j+1}}(\F_2,\F_2)=\F_2\{h_j^2\},
\]
and the given element is detected by the surviving class \(h_j^2\).

Browder's detection theorem identifies the framed Kervaire invariant
of an individual stable sphere element with its detection by
\(h_j^2\). The original framed Arf identification and secondary
pairing occur in \cite[Theorem~3.2, Proposition~7.6 and
Theorems~7.7--7.8]{Browder1969}; the precise elementwise statement is
also recorded in \cite[Theorem~1.1]{BHHM2020}.
This statement includes the low framed cases \(j=1,2,3\).
It applies to the particular \(\alpha\) above, whose nonzero class
in \(F^2\pi_d^S/F^3\pi_d^S\) is represented by the surviving
\(h_j^2\), and gives Kervaire invariant one.

In particular the detector is unchanged by adding an element of
\(F^3\pi_d^S\). Such an element cannot be detected by a
filtration-two class, so the same elementwise theorem gives it
Kervaire invariant zero; additivity of the framed invariant gives
the assertion. This use of Browder's theorem concerns the invariant
of each element, in addition to the existence of a permanent cycle.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:curtis}]
Every positive-degree stable class with nonzero Hurewicz image has a
last nonzero suspension. Propositions~\ref{prop:hopf-case}
and~\ref{prop:kervaire-case} show that it has the stated nonzero
Hopf or Kervaire invariant. These two degree cases are disjoint.

In a Hopf degree, choose the corresponding class among
\(\eta,\nu,\sigma\). In a Kervaire degree, choose any
Kervaire-invariant-one representative \(\theta_j\); its existence
in this case is already supplied by the class under discussion. Each
detecting invariant is additive with values in \(\F_2\).
For the stable Hopf invariant, attach two top cells to the same bottom
sphere by representatives \(\alpha\) and \(\beta\). If \(u\)
is the bottom class and \(a_\alpha,a_\beta\) are the relative top
classes, naturality on the two one-cell cofibers gives
\(\Sq^{d+1}u=H(\alpha)a_\alpha+H(\beta)a_\beta\).
Pinching the attaching sphere induces a map from the cofiber of
\(\alpha+\beta\) to this two-attachment cofiber; it pulls both
relative classes to the single top class. Naturality therefore gives
\(H(\alpha+\beta)=H(\alpha)+H(\beta)\).
For the Kervaire invariant, disjoint union of framed manifolds takes
their quadratic forms to orthogonal sums and adds their Arf
invariants. Subtracting
the chosen representative therefore gives an invariant-zero class.
If this difference had nonzero Hurewicz image, the proved necessity
would force its invariant to be one, a contradiction. Thus its
Hurewicz image is zero.

Every nonzero image is consequently the image of the chosen Hopf or
Kervaire representative in that degree. These representative images
themselves belong to the image, so they span it. No existence assertion
for an additional \(\theta_j\) is used or obtained from this last
step. This proves Curtis's conjecture in all positive degrees.
\end{proof}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//conclusion.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//cones.tex ====
\section{Mapping-cone squares and a module extension}\label{sec:cones}

The mapping cone of an actual sphere map supplies the realization
information needed by the algebraic obstruction. We first recover a
cone square from the square Hurewicz image of the loop adjoint.
For targets \(QS^n\) with \(n>0\), we then retain the top cone class
while passing to indecomposable cohomology, and obtain a nonzero tensor
in \(I\otimes_A M_k(n)\). The first step is valid for every connected
target, including those that are not simply connected.

\begin{lemma}[The cone-square detector]\label{lem:cone-square}
Let \(T\) be a connected based space and \(m\ge2\). Suppose that
\(g:S^{2m-1}\to T\) has zero mod-2 Hurewicz image and that its adjoint
\(g^\flat:S^{2m-2}\to\Omega T\) has Hurewicz image \(x^2\), where
\(x\in H_{m-1}(\Omega_0T;\F_2)\) and the square uses the loop product.
If \(u\in H^m(T;\F_2)\) satisfies \(u^2=0\), then it has a unique
extension \(\widetilde u\in H^m(C_g;\F_2)\), and
\begin{equation}\label{eq:cone-square}
\widetilde u^2=\langle u,\sigma_*x\rangle a,
\end{equation}
where \(a\in H^{2m}(C_g;\F_2)\) is the image of the top relative class.
This relative class is nonzero.
\end{lemma}

\begin{proof}
The cohomology sequence of the attachment shows that restriction in
degree \(m\) is an isomorphism: \(m<2m-1\). It also shows that the
top relative class injects, since \(g^*:H^{2m-1}(T)\to H^{2m-1}(S^{2m-1})\)
is zero. Thus \(\widetilde u^2\) is a scalar multiple of \(a\).

To determine the scalar, we use the universal space for a cohomology
class equipped with a nullhomotopy of its square. Write
\(K(j)=K(\F_2,j)\) and set
\[
F=\hofib\bigl(\Sq^m:K(m)\longrightarrow K(2m)\bigr).
\]
A map representing \(u\) lifts to \(F\) because its square is zero.
Looping the defining map represents \(\Sq^m\) on the fundamental class
of degree \(m-1\), which is zero by instability. Consequently the
fibration
\[
K(2m-2)\longrightarrow\Omega F\longrightarrow K(m-1)
\]
is trivial up to fibre homotopy. This is a statement about spaces over
\(K(m-1)\); its splitting need not preserve loop products.

We need the additive square-zero property of Eilenberg--MacLane spaces:
every positive-degree mod-two homology class of \(K(j)\), \(j\ge1\),
has zero square for its additive Pontryagin product. This follows from
Serre's cohomology calculation \cite{Serre1953}; the following chain
argument gives the precise property used here. Use the simplicial
abelian group \(G_r=Z^j(\Delta[r];\F_2)\) of normalized simplicial
\(j\)-cocycles, with faces given by restriction. Its Moore complex is
\(\F_2\) in degree \(j\) and zero elsewhere: below that degree there
are no \(j\)-cochains; in higher degrees the vanishing faces and the
cocycle equation force the remaining face to vanish. Thus its realization
is an additive model for \(K(j)\). The normalized \emph{linearized}
complex \(N(\F_2[G])\), rather than the Moore complex of \(G\),
computes its homology. Addition induces the chain product through the
shuffle map and \([a][b]=[a+b]\). For a chain \(z\) of degree \(q>0\),
the terms in \(z*z\) pair under exchange of the two complementary
\((q,q)\)-shuffle blocks and of the two chosen simplices of \(z\).
This involution has no fixed point, and the paired summands agree because
addition is commutative. They cancel over \(\F_2\), even before
normalization. Hence \(z*z=0\), proving the property.

Let \(c\in H_{m-1}(\Omega F)\) be the bottom class. Its image in
\(H_{m-1}K(m-1)\) has square zero for the additive multiplication.
This is the additive square-zero property just proved. It follows that
\(c^2\) lies in
the kernel of the projection to \(K(m-1)\). In degree \(2m-2\) this
kernel is one-dimensional, generated by the bottom class of the fibre.
The homotopy long exact sequence gives
\(\pi_{2m-2}\Omega F=\F_2\), and the space splitting shows that its
Hurewicz map identifies its nonzero element with that fibre class.

It remains to show that \(c^2\) is the nonzero fibre class. We do this
by showing that the product of bottom classes survives the cobar
spectral sequence. The space \(F\) is
\((m-1)\)-connected, so it is simply connected even when \(m=2\).
Adams's cobar construction \cite{Adams1956} has a multiplicative
comparison with chains on its based loop space; we use the modern
comparison in \cite[Theorem~1]{Rivera2019}. Choose a 1-reduced Kan
model for \(F\); the comparison for its cobar algebra follows from
\cite[Corollary~8 and Theorem~9]{Rivera2019}. Each desuspended
reduced chain has positive degree, so
the tensor-length filtration is finite in each total degree. The
resulting spectral sequence therefore converges to actual loop
homology, without a completion ambiguity. Write its first page as
\[
 E_1^{s,t}=(\widetilde H_*(F)^{\otimes s})_t,
 \qquad E_1^{s,t}\Longrightarrow H_{t-s}(\Omega F),
 \qquad d_r:(s,t)\longmapsto(s+r,t+r-1).
\]
Let \(c_0\in H_m(F)\) be the bottom class. In length two and
internal degree \(2m\), the term
\([c_0\mid c_0]\) is not a boundary for the coproduct differential:
the relevant coefficient is dual to the square of the bottom
cohomology class of \(F\), and that square vanishes by the defining
fibre construction. The term is a cycle since the reduced coproduct
of \(c_0\) is zero.

There is no higher incoming differential: its source would be
\((2-r,2m-r+1)\), hence have negative length or length zero and
positive internal degree. A higher outgoing differential
would land in length \(2+r\) and internal degree \(2m+r-1\), where
\(r\ge2\). Such a term vanishes, since each reduced homology factor
has degree at least \(m\) and
\[
(2+r)m>2m+r-1.
\]
This argument includes \(m=2\): the term is then in \((s,t)=(2,4)\),
and a higher outgoing target has internal degree \(r+3<2(r+2)\).
The same connectivity bound shows that \([c_0]\) is permanent and
detects the unique bottom class \(c\). Cobar concatenation represents
the loop product on the associated graded. The nonzero surviving
product \([c_0\mid c_0]\) consequently implies \(c^2\ne0\) in
actual loop homology. It is therefore the nonzero fibre class.

Under any lift \(T\to F\), the loop image of \(x\) is
\(\langle u,\sigma_*x\rangle c\), by homology suspension and the
one-dimensional bottom homology. Thus the adjoint of the composite
\(S^{2m-1}\to T\to F\) has Hurewicz image
\(\langle u,\sigma_*x\rangle c^2\). The injectivity just proved
shows that this composite is null precisely when the coefficient is zero.

Finally, \(\widetilde u^2=0\) is equivalent to the existence of a lift
of its representing map over \(C_g\) to \(F\). Such a lift restricts
to a lift on \(T\) whose composite with \(g\) is null; conversely,
any such lift extends over the attachment. The calculation applies to
every choice of lift. One can also see the independence directly:
changes of lift are measured by \(H^{2m-1}(T)\), whose pullback along
\(g\) is zero. This proves \eqref{eq:cone-square}.
\end{proof}

Retain the maps and classes of Lemma~\ref{lem:cone-square},
specialize to \(T=QS^n\) with \(n>0\), and suppose in addition
that \(x\) is primitive and \(A\)-annihilated.
Put \(B=\widetilde H^*(T;\F_2)\).
The homology Hopf algebra is polynomial on primitives, so its graded
dual cohomology algebra is exterior: each divided-power factor is
exterior on the generators at powers of two. Degreewise finite type
justifies the duality. In particular every positive cohomology class
has square zero. Define
\[
\lambda:B^m\longrightarrow\F_2,
\qquad \lambda(u)=\langle u,\sigma_*x\rangle.
\]
This functional vanishes on decomposables and on positive Steenrod
images, since \(\sigma_*x\) is primitive and \(A\)-annihilated.

\begin{proposition}\label{prop:cone-extension}
Under these assumptions, the cohomology extension of the cone induces
an extension of graded \(A\)-modules
\begin{equation}\label{eq:indecomposable-cone-extension}
0\longrightarrow\F_2\{a\}\longrightarrow\mathcal E
\longrightarrow Q B\longrightarrow0,
\end{equation}
where \(|a|=2m\). On any lift of a degree-\(m\) class \(v\in Q B\),
the operation \(\Sq^m\) has value \(\lambda(v)a\).
\end{proposition}

\begin{proof}
The hypothesis \(h(g)=0\) gives an exact sequence of positive
cohomology \(A\)-modules
\[
0\longrightarrow\F_2\{a\}\longrightarrow\widetilde H^*(C_g)
\longrightarrow B\longrightarrow0.
\]
The relative module structure implies that \(a\) times any
positive-degree class is zero. Choose exterior algebra generators of
\(B\) lifting a homogeneous basis of \(Q B\), choosing the basis in
degree \(m\) so that at most one generator has nonzero \(\lambda\).
Choose lifts of these generators to \(\widetilde H^*(C_g)\).
Their squares are zero except possibly for that one distinguished
degree-\(m\) generator, whose square is \(a\) by
Lemma~\ref{lem:cone-square}.

The nonempty square-free monomials in these lifts, together with \(a\), form a
graded vector-space basis of \(\widetilde H^*(C_g)\). Quotienting by
all products could kill \(a\), since it may itself be a square.
Instead we remove only products of distinct generators.
Let \(L\) be the span of the products of at least two distinct lifted
generators, with no repetitions. Restriction maps \(L\) isomorphically
onto \(B^2\). We claim that \(L\) is an \(A\)-submodule. Apply
Cartan to such a product and expand each factor in the lifted basis.
A term involving \(a\) vanishes against any remaining positive factor.
A repeated generator also gives zero, except that two occurrences of
the distinguished generator give \(a\); if another positive factor
remains, this term again vanishes.

The only possible contribution outside \(L\) is therefore a product
of exactly two linear occurrences of the distinguished generator.
A positive Steenrod operation on a factor cannot supply such a linear
coefficient, because \(\lambda\) annihilates positive Steenrod
images. If both operations are zero, the original two distinct
generators cannot both be the distinguished generator. This proves
the claim.

Taking the quotient by \(L\) gives
\eqref{eq:indecomposable-cone-extension}. In degree \(m\), the
square detector is unaffected, and its value is still \(\lambda\).
\end{proof}

\begin{corollary}\label{cor:tensor-detection}
If \(\lambda(v)=1\) for a degree-\(m\) monomial \(v\) in a
summand \(M_k(n)\) of \(Q B\), then
\begin{equation}\label{eq:tensor-detection}
\Sq^m\otimes v\ne0\quad\text{in }I\otimes_A M_k(n).
\end{equation}
The same conclusion holds for the bottom summand
\(M_0(n)=\Sigma^n\F_2\) when its degree is \(m\).
\end{corollary}

\begin{proof}
Pull back \eqref{eq:indecomposable-cone-extension} to the summand
\(M=M_k(n)\) and choose a graded additive section \(s\). In total
degree \(2m\), define the scalar \(\varphi(\alpha\otimes z)\) by
\[
\alpha s(z)-s(\alpha z)=\varphi(\alpha\otimes z)a,
\qquad \alpha\in I.
\]
It respects balancing. Indeed, in comparing
\(\alpha\beta\otimes z\) and \(\alpha\otimes\beta z\), the
intermediate degree \(|\beta|+|z|\) is strictly less than \(2m\)
because \(|\alpha|>0\). The possible discrepancy in the section
therefore vanishes, and associativity gives equality. Thus
\(\varphi\) descends to \((I\otimes_A M)_{2m}\).

Since \(n>0\), the module \(M_k(n)\) is a positive suspension of
an unstable polynomial module; hence \(\Sq^m v=0\). This is also
true in the one-dimensional bottom summand. Proposition~\ref{prop:cone-extension}
therefore gives \(\varphi(\Sq^m\otimes v)=\lambda(v)=1\), proving
\eqref{eq:tensor-detection}.
\end{proof}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//cones.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//figures/proof-route.tex ====
\begin{figure}[t]
\centering
\small
\fbox{\begin{minipage}{0.88\linewidth}
\centering
An actual sphere map \(f_0:S^d\to Q_0S^0\), with \(h(f_0)\ne0\)\\[3pt]
\(z_j=\sigma_*^jh(f_0)\), with \(e=\max\{j:z_j\ne0\}\)
\end{minipage}}
\par\smallskip
\(\Downarrow\)\quad suspension kernel; the root survives by\\[2pt]
parity when \(e\ge1\), or exclusion of fourth powers when \(e=0\)
\par\smallskip
\fbox{\begin{minipage}{0.88\linewidth}
\centering
\(z_e=w^2\), with \(w\) primitive and Steenrod-annihilated,
\(\sigma_*w\ne0\)\\[3pt]
The cone of \(f_{e+1}\) supplies a nonzero square operation.
\end{minipage}}
\par\smallskip
\(\Downarrow\)\quad the algebra obstruction with the lifted functional
\par\smallskip
\begin{tabular}{@{}c@{\hspace{0.04\linewidth}}c@{}}
\fbox{\begin{minipage}{0.40\linewidth}\centering
\(e\ge1\)\\[3pt]
Only the bottom weight remains.\\
Stable Hopf detection\\
\(d=1,3,7\)
\end{minipage}}
&
\fbox{\begin{minipage}{0.40\linewidth}\centering
\(e=0\)\\[3pt]
The leading weight of \(z_0\) is four.\\
Adams filtration two\\
Kervaire detection
\end{minipage}}
\end{tabular}
\caption{The proof follows the successive adjoints of one stable
sphere class. The root \(w\) need not itself be spherical: the
next adjoint and its cone provide the realization information used
by the Steenrod-module obstruction.}
\label{fig:proof-route}
\end{figure}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//figures/proof-route.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//introduction.tex ====
\section{Introduction}

For \(n\ge0\), let
\(QS^n=\operatorname*{colim}_r\Omega^rS^{n+r}\), and let \(Q_0S^0\)
be its basepoint component. For \(d>0\), the identification
\(\pi_d(Q_0S^0)=\pi_d^S\) gives the mod-2 Hurewicz homomorphism
\[
h_d:\pi_d^S\longrightarrow H_d(Q_0S^0;\F_2).
\]
It sends a representative \(S^d\to Q_0S^0\) to the image of the
mod-2 fundamental class. Curtis's conjecture describes the positive-degree
image of these maps \cite{Curtis1975}.
The target has an explicit homology algebra, while the stable homotopy
groups in the source contain many families whose images are difficult
to determine. The image problem asks which of the readily described
homology classes are represented by maps from spheres.

Curtis proposed the Hopf--Kervaire description in his study of the
Dyer--Lashof and lambda algebras \cite{Curtis1975}. May's subsequent
account records a gap found by Wellington in the proposed proof
\cite[p.~484, footnote~3]{May1977}. We prove the following form of
the conjecture.

\begin{theorem}\label{thm:curtis}
The image of \(\bigoplus_{d>0}h_d\) is spanned over \(\F_2\) by
the Hurewicz images of the Hopf-invariant-one classes
\(\eta,\nu,\sigma\), in degrees \(1,3,7\), and the
Kervaire-invariant-one classes \(\theta_j\), in degrees
\(2^{j+1}-2\), \(j\ge1\), whenever they exist.
In particular, Curtis's conjecture is true in all positive degrees.
\end{theorem}

Here Kervaire invariant one means that the corresponding framed manifold,
under the Pontryagin--Thom identification, has Arf invariant one. The
notation \(\theta_j\) is used only when such a class exists. The theorem
does not posit existence in any additional degree.

Eccles's conjecture concerns the analogous realization problem for
\(QX\). In the formulation recorded in
\cite[pp.~14--15]{Zare2009Thesis}, for a path-connected CW complex
\(X\) of finite type, a positive-dimensional sphere
map to \(QX\) with nonzero mod-2 Hurewicz image has stable adjoint
detected either by homology or by a primary cohomology operation in
its mapping cone. Theorem~\ref{thm:curtis} gives this conclusion
when \(X\) is a positive-dimensional sphere.

For \(n>0\), write \(e_n\in H_n(QS^n;\F_2)\) for the bottom spherical
class, and write \(\sigma_*^n\) for \(n\)-fold homology suspension from
\(Q_0S^0\).

\begin{corollary}[Eccles's conjecture for spheres]\label{cor:eccles-spheres}
For every integer \(n>0\), each nonzero mod-2 spherical class in
\(H_*(QS^n;\F_2)\) is either \(e_n\), or a nonzero class of the form
\[
 \sigma_*^n h_d(\alpha),\qquad
 (\alpha,d)\in\{(\eta,1),(\nu,3),(\sigma,7)\}.
\]
Consequently Eccles's conjecture holds for every sphere \(S^n\), \(n>0\):
the alternatives are the bottom stably spherical class and
Hopf-invariant-one classes.
\end{corollary}
\begin{proof}
For a spherical representative \(f:S^{n+d}\to QS^n\) with \(d>0\),
adjunction gives \(f_0:S^d\to Q_0S^0\), and naturality gives
\(h(f)=\sigma_*^n h_d([f_0])\). By Theorem~\ref{thm:curtis}, the
level-zero image is spanned by the Hopf and Kervaire images. The latter
are squares killed by the first homology suspension, leaving the stated
Hopf possibilities; this is the Curtis-to-Eccles sphere implication
recorded in \cite[Chapter~5, pp.~81--82]{Zare2009Thesis}. In degree
\(n\), \(\pi_n(QS^n)=\pi_0^S=\Z\) gives the bottom image represented by
the unit \(S^n\to QS^n\); connectivity excludes lower degrees.
\end{proof}

\begin{corollary}[Finite positive stable Hurewicz image]\label{cor:finite-image}
The image of \(\bigoplus_{d>0}h_d\) is finite, and
\[
 h_d=0\qquad\text{if }d>0\text{ and }
 d\notin\{1,2,3,6,7,14,30,62,126\}.
\]
In particular, the positive stable mod-2 Hurewicz image vanishes above
degree \(126\).
\end{corollary}
\begin{proof}
Hill, Hopkins, and Ravenel prove that \(\theta_j\) can exist only for
\(j\le6\) \cite[Theorem~1.1]{HillHopkinsRavenel2016}. By Theorem~\ref{thm:curtis},
the positive image is therefore spanned by the named images in Hopf degrees
\(1,3,7\) and Kervaire degrees \(2,6,14,30,62,126\). Finitely many
generators over \(\F_2\) give a finite image and the stated support.
\end{proof}

Corollary~\ref{cor:eccles-spheres} concerns only positive-dimensional
spheres, not general spaces. The support in Corollary~\ref{cor:finite-image}
is an upper bound; it does not assert nonvanishing in every listed degree
or an exact dimension of the image.

Adams's Hopf invariant one theorem identifies the Hopf cases
\cite{Adams1960}. Browder's work relates the framed Kervaire invariant
to the classes \(h_j^2\) in the mod-2 Adams spectral sequence
\cite{Browder1969}. Hill--Hopkins--Ravenel's nonexistence theorem
leaves dimension \(126\) as the last possible Kervaire dimension.
In their preprint, Lin, Wang, and Xu prove existence in that dimension
\cite[Theorems~1.1 and~1.4]{LinWangXu2025}. This endpoint result is
separate from the upper bound used in Corollary~\ref{cor:finite-image}.

On the homology side, the free infinite-loop-space
calculation describes a large algebra generated by Dyer--Lashof
operations \cite{CLM1976}. A spherical class must be primitive and
annihilated by positive Steenrod operations, but these necessary
conditions alone do not establish its realization. The theorem concerns
the image of actual stable homotopy classes.

Eccles and Zare give restrictions on spherical squares and their roots,
including the exclusion of higher iterated squares
\cite[Theorem~3.10 and Lemma~3.12]{EcclesZare2009}.
Section~\ref{sec:powers} proves the exclusion of fourth powers by
spectrum pairings and integral characteristic classes; the conclusion
itself is already present in that prior work. The Dickson-algebra
approach to spherical classes also appears in Hung and Peterson's
study of the Lannes--Zarati homomorphism and products in the unstable
Adams \(E_2\)-term \cite{HungPeterson1998}.
Hunter's cohomological construction of the Curtis--Wellington spectral
sequence exhibits Dickson-algebra quotients in a width filtration
\cite[arXiv version~2, Theorem~1.2]{Hunter2025}.

The argument here combines classical detection by mapping-cone
operations with an obstruction for the specific Steenrod modules
dual to primitives. The cone of an \emph{actual} sphere map supplies
a nonzero extension. A detecting functional inherited from the
preceding suspension level then restricts that extension through the
algebraic obstruction proved in Section~\ref{sec:algebra}.

The free-algebra model constructed below is compatible with Kuhn's
structured splitting and completion maps \cite{Kuhn2006}; we prove
the required finite weight projections and their suspension
compatibility directly. For the final passage from leading
weight to Adams filtration, we use Kuhn's lifting theorem
\cite{Kuhn2018} and the derived ideal-power constructions of
\cite{KuhnPereira2017}.

\paragraph{Outline of the proof.}
The proof is organized around the last nonzero homology suspension of a
spherical class. For a representative \(f_0:S^d\to Q_0S^0\), let
\(f_j:S^{d+j}\to QS^j\) be its successive adjoints and put
\(z_j=h(f_j)\). Suspension gives \(z_{j+1}=\sigma_*z_j\), and the
homology degree bounds imply that \(z_j=0\) for \(j>d\).
Thus a nonzero initial image has a last nonzero term \(z_e\).
We show that this terminal class is a square \(z_e=w^2\) whose
primitive, Steenrod-annihilated root still has nonzero suspension.
At a positive terminal level, the preceding spherical suspension forces
the root to have odd degree. At level zero, we instead exclude fourth powers
using spectrum pairings and integral characteristic classes
(Section~\ref{sec:powers}). The mapping cone of the next adjoint then
converts the square into a nonzero Steenrod-module extension
(Section~\ref{sec:cones}).

At positive level \(n\), weight counts the inputs of an extended power.
The primitives of weight \(2^k\), \(k\ge1\),
are dual to \(M_k(n)=\Sigma^n d_0^nD_k\), where \(D_k\) is a
binary Dickson algebra and \(d_0\) is its generator of degree
\(2^k-1\); the bottom summand is \(M_0(n)=\Sigma^n\F_2\).
Section~\ref{sec:weights} constructs these modules, identifies
the dual of homology suspension with inclusion of the next \(d_0\)-ideal,
and supplies compatible finite weight projections at level zero.
Section~\ref{sec:algebra} shows that the cone extension on a nonbottom
weight, when its detecting functional lifts from the preceding
suspension level, is possible only for \(n=k=1\) and in a
power-of-two degree. Cartier halving and an injection on even-degree
Tor cycles retain the exponent-floor information needed for this
restriction.

Section~\ref{sec:conclusion} assembles the argument. At a positive
terminal level the obstruction leaves only the bottom weight, giving
stable Hopf detection. At terminal level zero it forces the root to
have leading weight two, so the original spherical class has leading
weight four. Kuhn's theorem bounds its Adams filtration by two, and
Browder's identification gives Kervaire detection. Additivity of the
detecting invariants then gives the asserted span.

Figure~\ref{fig:proof-route} summarizes the two terminal cases.
\input{figures/proof-route}

\paragraph{Conventions.}
Spaces are based and of CW homotopy type; for a based space \(X\),
write \(QX=\Omega^\infty\Sigma^\infty X\).
Homology and cohomology have coefficients in \(\F_2\) unless stated
otherwise. Let \(A\) be the mod-2 Steenrod algebra and \(I=A^{>0}\)
its augmentation ideal. The dual homology action is denoted
\(\Sq_*^i\). A homology class is \(A\)-annihilated if every
\(\Sq_*^i\), \(i>0\), kills it. Equivalently, its functional on
cohomology vanishes on the image of \(I\). For a positive graded
cohomology algebra \(B\), write \(Q B=B/B^2\) for its
indecomposables.

Primitivity refers to the ordinary space diagonal, with the basepoint
as coaugmentation, unless the linear coproduct is explicitly specified.
Products in homology of an infinite loop
space are its additive Pontryagin products, written \(*\) when a
distinction is needed. The block pairing introduced later is written
\(\circ\). Homology suspension between infinite loop levels is
\(\sigma_*\). We distinguish it from stable homology evaluation
\(H_*(\Omega^\infty E)\to H_*(E)\), induced by the counit.

\begin{samepage}
The symbol \(S\) denotes the sphere spectrum. We write
\(D_rT=(T^{\wedge r})_{h\Sigma_r}\) for an extended power and
\(\mathbb P(T)=\bigvee_{r\ge0}D_rT\) for the free commutative ring
spectrum on \(T\).
All constructions are derived and homotopy coherent, including free
algebras, pushouts, group completion, localization and parametrized
families of maps. We use the usual mod-2 Dyer--Lashof operations on
infinite loop spaces and commutative ring spectra
\cite{CLM1976,BMMS1986}. Whenever a module is written with a suspension
shift, polynomial exponents refer to the unshifted monomial; its total
degree also includes that shift.
\par
\end{samepage}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//introduction.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//paper.tex ====
\documentclass[11pt]{article}
\input{preamble}
\title{The Stable Hurewicz Image of the Sphere at Two}
\author{OpenAI}
\date{September 25, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove Curtis's conjecture: in every positive degree, the mod-two
stable Hurewicz image of the sphere is spanned by the images of
\(\eta,\nu,\sigma\) and the Kervaire-invariant-one classes that exist.
Consequently, Eccles's conjecture holds for every sphere \(S^n\) with
\(n>0\).
\end{abstract}
\setcounter{tocdepth}{1}
\tableofcontents
\input{introduction}
\input{weights}
\input{cones}
\input{powers}
\input{algebra}
\input{conclusion}
\clearpage
\begingroup
\small
\setlength{\parskip}{0pt}
\bibliographystyle{plain}
\bibliography{references}
\endgroup
\end{document}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//paper.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//powers.tex ====
\section{Excluding fourth and higher powers}
\label{sec:powers}

A spherical class whose homology suspension vanishes is a Pontryagin
square.  To apply the cone-square argument to such a class, we will need
its square root to have nonzero suspension.  At level zero this requires
a separate argument: a fourth power and its square root both disappear
under one homology suspension.  We exclude this possibility using the
weight projections and an integral characteristic class.

In this section, $*$ denotes the additive Pontryagin product.  A pairing
of spectra $E_1\wedge E_2\to E_3$ induces a based pairing of their zero
spaces, and we write $\circ$ for the corresponding pairing on homology.
Thus $*$ and $\circ$ are different operations.  We write $a^{*N}$ and
$a^{\circ N}$ for their respective powers whenever both are present.

\begin{proposition}
\label{prop:no-fourth-powers}
There is no positive-dimensional stable sphere class whose nonzero
mod-$2$ Hurewicz image is a fourth or higher iterated Pontryagin
square.  Equivalently, if $f:S^d\to Q_0S^0$, $d>0$, and
\[
 h(f)=z^{*2^a},\qquad a\geq2,
\]
then $h(f)=0$.
\end{proposition}

Eccles and Zare exclude higher iterated squares when the root survives
homology suspension \cite[Theorem~3.10]{EcclesZare2009}, and state that
the root of a spherical square is odd-dimensional
\cite[Lemma~3.12]{EcclesZare2009}. Here we prove the required exclusion
using spectrum pairings and integral characteristic classes.

The space maps adjoint to individual weight projections need not
preserve Pontryagin squares. We first compute their correction term,
which is expressed through the block pairings of extended powers.

For $v\geq1$, set
\[
 X_v=(B\Sigma_v)_+,\qquad E_v=D_vS\simeq\Sigma^\infty X_v.
\]
Block inclusion gives pairings $E_v\wedge E_w\to E_{v+w}$, associative
and commutative coherently, and thus associative and commutative
pairings $\circ$ on their zero-space homology.  Write
\[
 \varepsilon_v:\Sigma^\infty QX_v\longrightarrow E_v
\]
for the stable counit; its induced map on positive-dimensional homology
is stable homology evaluation.

Fix an integer $N>0$.  Let
\[
 \tau:\Sigma^\infty_+QS^0\longrightarrow\mathbb P_{\leq N}(S\{u\})
\]
be the augmented commutative ring projection of
\eqref{eq:weight-projections}.  For $1\leq i\leq N$, let
\[
 j_i:QS^0\longrightarrow\Omega^\infty E_i=QX_i
\]
be the based adjoint of its weight-$i$ projection, and write
$J_i=(j_i)_*$ on positive-dimensional homology.  These are generally
\emph{not} additive infinite loop maps.  Their multiplication formula
is the following.

\begin{lemma}[Multiplication under the weight projections]
\label{lem:weight-product}
For points $x,y\in QS^0$, there is a homotopy
\begin{equation}
\label{eq:weight-product}
 j_i(x+y)\simeq j_i(x)+j_i(y)
       +\sum_{0<t<i}j_t(x)\circ j_{i-t}(y),
\end{equation}
where addition in the target is its additive infinite loop structure.
For positive primitives $A_1,A_2\in H_*Q_0S^0$, this implies
\begin{equation}
\label{eq:primitive-weight-product}
 J_i(A_1*A_2)
 =(J_iA_1)*(J_iA_2)
   +\sum_{0<t<i}(J_tA_1)\circ(J_{i-t}A_2).
\end{equation}
Each $J_i$ preserves primitivity and $A$-annihilation.  In particular,
\begin{equation}
\label{eq:weight-square}
 J_i(A^{*2})=(J_iA)^{*2}
 +\begin{cases}(J_{i/2}A)^{\circ2},&i\text{ even},\\0,&i\text{ odd}.
 \end{cases}
\end{equation}
\end{lemma}

\begin{proof}
Under the adjunction between suspension ring spectra and
$E_\infty$ spaces, $\tau$ gives an $E_\infty$ map from the additive
structure of $QS^0$ to the multiplicative space of the truncated ring.
The source is grouplike, so this map lands in units.  Its weight-zero
coordinate is constantly $1$, since $\tau$ is augmented; the
positive coordinates at the source basepoint are zero.  Multiplication
in the truncated graded ring has weight-$i$ coordinate equal to the
right side of \eqref{eq:weight-product}.  This proves that formula,
including its compatibility with all lower coordinates in the single
chosen truncation.  Group completion causes no further terms:
$1+u$ is a unit in the target, so the map is precisely the extension
of the free-monoid map through the localization
\eqref{eq:zero-localization}.

To evaluate \eqref{eq:weight-product} on $A_1\times A_2$, distribute
the two primitive diagonals across its additive summands.  Either the
two positive inputs enter the two linear terms, giving their
Pontryagin product, or both enter one bilinear summand, giving its
block pairing.  Every other allocation pairs a positive class with
the zero point and vanishes.  This gives
\eqref{eq:primitive-weight-product}.  Although $j_i$ need not preserve
addition, it is a based space map, so it preserves ordinary
primitivity and commutes with the dual Steenrod operations.
When $A_1=A_2$, the terms with indices $t$ and $i-t$ cancel by block
commutativity; only the equal-block term can remain.  This proves
\eqref{eq:weight-square}.
\end{proof}

To remove the extra block-square term in \eqref{eq:weight-square},
we will use nilpotence for primitives whose stable homology evaluation
vanishes. The block pairings arise from spectrum maps, so they have
coherent additive structure in each variable. The following calculation
uses that structure; it does not require the maps $j_i$ to be additive.

\begin{lemma}[Dyer--Lashof operations and spectrum pairings]
\label{lem:pairing}
Let $E_1\wedge E_2\to E_3$ be a pairing of spectra.  If
$u\in H_a\Omega^\infty E_1$ is homogeneous and
$y\in H_b\Omega^\infty E_2$, where $b>0$, is $A$-annihilated, then,
for $i\geq a$,
\begin{equation}
\label{eq:pairing-rule}
 (Q^i u)\circ y=
 \begin{cases}
 Q^i(u\circ y),&i\geq a+b,\\
 0,&a\leq i<a+b.
 \end{cases}
\end{equation}
If, in addition, $y$ is primitive and belongs to the basepoint component,
then $(u_1*u_2)\circ y=0$ for positive-dimensional $u_1,u_2$.
Translating a positive-dimensional first factor by a component point
does not change its pairing with $y$.
\end{lemma}

\begin{proof}
Put $X=\Omega^\infty E_1$ and $Y=\Omega^\infty E_2$.
The pairing is a coherent family of additive infinite loop maps in its
first variable.  Indeed its adjoint is a spectrum map
$E_2\to F(E_1,E_3)$; passing to zero spaces gives a map from $Y$ to the
space of spectrum maps $E_1\to E_3$, and hence to the space of additive
infinite loop maps.  Connective covers may be taken without changing
the zero spaces.  In particular, the naturality homotopies for the
additive $E_\infty$ structure vary coherently over $Y$.

For an unbased space $U$, write
\[
 \mathcal E_2U=EC_2\times_{C_2}(U\times U),
\]
where $C_2$ interchanges the factors.  The naturality just described
factors the pairing of an additive square operation with $Y$ through
\[
 \delta:\mathcal E_2X\times Y\longrightarrow\mathcal E_2(X\times Y),
 \qquad [e;x_1,x_2],y\longmapsto[e;(x_1,y),(x_2,y)].
\]
Let $W$ be the free periodic $\F_2[C_2]$-resolution with generators
$e_j$ and differential $d e_j=(1+\tau)e_{j-1}$.
The chain model
\[
 W\otimes_{C_2}C_*(U)^{\otimes2}
\]
computes $H_*\mathcal E_2U$.  Choosing cycles for a homology basis of
$U$ gives an equivariant quasi-isomorphism before taking homotopy
orbits.  It follows that a basis of the resulting homology consists of
unordered unequal pairs in resolution degree zero and classes
$e_j\otimes z\otimes z$, $j\geq0$, on equal basis elements.

We claim the universal formula
\begin{equation}
\label{eq:external-pairing}
 \delta_*\bigl((e_j\otimes u\otimes u)\times y\bigr)
 =\sum_{t\geq0}e_{j-b+2t}\otimes
       (u\times\Sq_*^t y)\otimes(u\times\Sq_*^t y),
\end{equation}
with terms of negative resolution degree omitted.  This is the
extended-square calculation underlying May's formula
\cite[\S2, formula~(2.11), pp.~246--247]{Madsen1975}; we include its verification.

Transfer from $\mathcal E_2U$ to $U^2$ sends an unequal pair to the sum
of its two ordered versions and kills all equal-pair classes.  Thus it
detects the unequal-pair part injectively.  Transfer is natural for
$\delta$, since its pulled-back double cover is the double cover of
$\mathcal E_2X\times Y$ obtained from $\mathcal E_2X$.
The input of \eqref{eq:external-pairing} has transfer zero.  The output
therefore has no unequal-pair terms.

For $\beta\in H^qU$, let $P(\beta)\in H^{2q}\mathcal E_2U$ be its
external square, and let $h\in H^1BC_2$ be the characteristic class of
the double cover.  In the dual resolution model, a cocycle $f$
representing $\beta$ gives $P(\beta)$ by $f\otimes f$ in resolution
column zero.  This description is independent of $f$: if
$f'=f+dc$, then $P(f')-P(f)$ is the total coboundary of the cochain
whose columns zero and one are, respectively,
\[
 f\otimes c+c\otimes f+c\otimes dc,
 \qquad c\otimes c.
\]
Multiplication by $h^l$ is the resolution shift, so
$h^lP(\beta)$ is represented by $f\otimes f$ in column $l$.
It evaluates on $e_j\otimes z\otimes z$ as
$\delta_{lj}\langle\beta,z\rangle^2$.

External squares respect cross products over the common base $BC_2$.
Moreover, pulling $P(\beta)$ back along the Borel diagonal
$BC_2\times U\to\mathcal E_2U$ gives
\[
 \sum_{t=0}^{q}h^{q-t}\Sq^t\beta;
\]
this is the total-square definition of the Steenrod operations.
Consequently, for $\alpha\in H^*X$ and $\beta\in H^qY$,
\[
 \delta^*\bigl(h^lP(\alpha\times\beta)\bigr)
 =h^lP(\alpha)\sum_{t=0}^{q}h^{q-t}\Sq^t\beta.
\]
Its evaluation on the input in \eqref{eq:external-pairing} can be
nonzero only when
\[
 |\alpha|=a,\qquad q+t=b,\qquad j=l+q-t=l+b-2t.
\]
The coefficient is then
$\langle\alpha,u\rangle\langle\beta,\Sq_*^t y\rangle$.
Tensor cohomology classes detect all equal-pair coefficients; in each
evaluation only finitely many homology basis elements are involved.
Together with the transfer calculation, these evaluations prove
\eqref{eq:external-pairing}.

Now take $j=i-a$ and apply the label pairing, followed by the additive
square operation.  The term with index $t$ becomes
$Q^{i+t}(u\circ\Sq_*^t y)$.  When $y$ is $A$-annihilated, only $t=0$
remains.  Its resolution index is $i-a-b$, which proves both cases of
\eqref{eq:pairing-rule}.

Finally, for a primitive $y$ in the basepoint component,
$\Delta y=y\otimes[0]+[0]\otimes y$.  Distributivity of a spectrum
pairing over addition gives
\[
 (u_1*u_2)\circ y
 =(u_1\circ y)*(u_2\circ[0])
  +(u_1\circ[0])*(u_2\circ y)=0,
\]
because pairing a positive-dimensional class with the zero point is
zero.  If one first factor is a component point $[c]$, its pairing
with $[0]$ is the additive unit $[0]$ in the target; the same formula
proves the translation assertion.
\end{proof}

\begin{lemma}[Triple block nilpotence]
\label{lem:block-nilpotence}
Suppose $B\in H_i(Q_0X_v)$, $i>0$, is primitive and $A$-annihilated.
If $\varepsilon_{v*}B=0$, then
\[
 B^{\circ3}=0.
\]
\end{lemma}

\begin{proof}
The free infinite loop space homology theorem and group completion
describe $H_*Q_0X_v$ as a polynomial Pontryagin algebra
\cite[Part I]{CLM1976}.  Its generators are the positive-dimensional
particle classes from a basis of $H_*B\Sigma_v$, and the admissible
positive-excess Dyer--Lashof words of positive length on particle
classes, including the degree-zero particle.  Each is translated to
the basepoint component.  There is one degree-zero particle component
to invert in group completion; these translations do not change
stable homology evaluation in positive dimensions.

The counit takes the length-zero positive-dimensional generators
independently to their corresponding basis classes in $H_*E_v$.
Stable homology evaluation kills products of positive-dimensional classes.
It also kills positive-length words: upper-index operations commute with
homology suspension, and repeated suspension eventually makes their
outer operation unstable.  Therefore, modulo Pontryagin decomposables,
the hypothesis $\varepsilon_{v*}B=0$ expresses $B$ as a sum of
positive-length words only.

Pair this expression with $B$.  The preceding lemma discards the
decomposables and translations in the first factor.  A remaining term
has the form $Q^q c$ with $q+|c|=i$.  Its pairing can be nonzero only if
\[
 q\geq |c|+i=q+2|c|,
\]
so $|c|=0$.  In this case
\[
 (Q^i c)\circ B=Q^i(c\circ B)=(c\circ B)^{*2}.
\]
Thus $B\circ B$ is Pontryagin decomposable, and another pairing with
$B$ is zero by Lemma~\ref{lem:pairing}.

We also record a property used below: the block pairing of two
positive primitives in their basepoint components is primitive.
Indeed, naturality of the ordinary diagonal leaves the two end terms;
each crossed term pairs a positive-dimensional class with the zero
point and vanishes.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:no-fourth-powers}]
It suffices to exclude fourth powers, since every higher iterated square
is a fourth power. Suppose, therefore, that
\[
 h(f)=z^{*4}\ne0.
\]
Applying Lemma~\ref{lem:primitive-roots} twice shows that the
homogeneous positive-dimensional root \(z\) is primitive and
\(A\)-annihilated.

\emph{Choosing the detector.}
Let \(r=2^{k_0}\ge2\) be the leading weight of this root, as in
Lemma~\ref{lem:zero-weights}, and let \(z_r\in H_*D_rS\) be its
leading component. Put
\[
 x=z^{*2},\qquad b=|x|,\qquad s=2r=2^\ell.
\]
Here \(b\) is positive and even, \(\ell\ge2\), and \(d=2b\).
In the polynomial associated weight algebra the leading component of
\(x\) is
\[
 p=z_r^2\in H_bD_sS,
\]
and is nonzero. This square is taken in the graded homology algebra
\(H_*\mathbb P(S)\), rather than by the unstable pairing \(\circ\).

The class $p$ is primitive for the linear weight coproduct and is a
product of positive-dimensional factors.  It therefore dies under
the weight suspension \eqref{eq:weight-suspension}.
One may see this before group completion: homology suspension from
the free monoid kills positive-dimensional Pontryagin products, and
Lemma~\ref{lem:weight-suspension} identifies its weight components.
Thus, under the dual description in
Lemma~\ref{lem:weight-modules}, the functional defined by $p$ on the
Dickson algebra $D_\ell$ kills $d_0D_\ell$.
It is nonzero on some monomial in $d_1,\ldots,d_{\ell-1}$.
The restriction description of that lemma lifts $d_j$ to
$w_{s-2^j}(\rho_s)$, where $\rho_s$ is the permutation representation
of $\Sigma_s$.  Hence there is a class
\begin{equation}
\label{eq:even-characteristic-detector}
 v\in H^b(B\Sigma_s),\qquad \langle v,p\rangle=1,
\end{equation}
which is a polynomial in the \emph{even-index}
Stiefel--Whitney classes of $\rho_s$.

\emph{Removing the block correction.}
Use one truncation through weight \(s=2r\) to define all the maps
\(j_i\). We claim
\begin{equation}
\label{eq:power-projection}
 \begin{split}
 J_s h(f)
  &=(J_sx)^{*2}+(J_rx)^{\circ2}\\
  &=(J_sx)^{*2}+(J_{r/2}z)^{\circ4}\\
  &=(J_sx)^{*2}.
 \end{split}
\end{equation}
The first equality is \eqref{eq:weight-square}. For the second,
apply that formula to \(x=z^{*2}\):
\[
 J_rx=(J_rz)^{*2}+(J_{r/2}z)^{\circ2}.
\]
Each \(J_i z\) lies in the basepoint component and is primitive and
\(A\)-annihilated, by Lemma~\ref{lem:weight-product}. Thus
\((J_rz)^{*2}\) is a primitive Pontryagin decomposable, and its
block square is zero by Lemma~\ref{lem:pairing}. Block commutativity
over \(\F_2\) cancels the mixed terms. This gives
\((J_rx)^{\circ2}=(J_{r/2}z)^{\circ4}\).
The indices are integral because \(r\) is a power of two at least two,
and all block products have total weight at most \(s\).

For the last equality in \eqref{eq:power-projection}, observe that
stable counit followed by an adjoint recovers the original spectrum
map.  Thus, on positive-dimensional homology,
\begin{equation}
\label{eq:projected-counit}
 \varepsilon_{i*}J_i=(\operatorname{pr}_i\tau)_*.
\end{equation}
In particular \(B=J_{r/2}z\) has zero stable homology evaluation,
because \(r/2\) is below the leading weight of \(z\). It is a
primitive, \(A\)-annihilated class in \(Q_0X_{r/2}\).
Lemma~\ref{lem:block-nilpotence} gives \(B^{\circ3}=0\), hence
\(B^{\circ4}=0\). This proves the last equality in
\eqref{eq:power-projection}.
The same counit identity gives
\begin{equation}
\label{eq:root-stable-projection}
 \varepsilon_{s*}J_sx=p.
\end{equation}

\emph{Contradictory cone squares.}
Let
\[
 g:S^{2b+1}\longrightarrow T=Q\Sigma X_s
\]
be the adjoint of $j_s\circ f:S^{2b}\to Q_0X_s$, using the
equivalence $QX_s\simeq\Omega Q\Sigma X_s$.
The space $\Sigma X_s$, and hence $T$, is connected.
The loop product under this equivalence is the additive Pontryagin
product.  Equation~\eqref{eq:power-projection} therefore says that
the loop adjoint has Hurewicz class $(J_sx)^{*2}$, and gives
$h(g)=0$ after homology suspension.

Let $\varepsilon:\Sigma^\infty T\to\Sigma^\infty\Sigma X_s$ be
the stable counit and define
\[
 u=\varepsilon^*(\Sigma v)\in H^{b+1}T.
\]
Counit compatibility with homology suspension and
\eqref{eq:root-stable-projection} give
\[
 \langle u,\sigma_*J_sx\rangle
   =\langle v,\varepsilon_{s*}J_sx\rangle
   =\langle v,p\rangle=1.
\]
Stable Steenrod naturality gives
\[
 u^2=\Sq^{b+1}u
 =\varepsilon^*(\Sigma\Sq^{b+1}v)=0.
\]
With $m=b+1$, the hypotheses of Lemma~\ref{lem:cone-square} now hold.
The unique extension $\widetilde u\in H^mC_g$ consequently satisfies
\begin{equation}
\label{eq:nonzero-power-cone-square}
 \Sq^m\widetilde u\ne0.
\end{equation}

On the other hand, the restriction of
$\Sq^{m-1}\widetilde u$ to $T$ is
\[
 \Sq^b u=\varepsilon^*(\Sigma\Sq^b v)
        =\varepsilon^*(\Sigma(v^2)).
\]
This class has an integral torsion lift.  Indeed, for real vector
bundles the reduction of the integral Pontryagin class is
\[
 \rho_2(p_i)=w_{2i}^{\,2}.
\]
For completeness, this follows from
$p_i(\xi)=(-1)^ic_{2i}(\xi\otimes\C)$: the underlying real bundle
of $\xi\otimes\C$ is $\xi\oplus\xi$, and reduction of its Chern
classes gives its even Stiefel--Whitney classes.
Because $v$ is a polynomial in even-index Stiefel--Whitney classes,
its square is the reduction of an integral polynomial
$V\in H^{2b}(B\Sigma_s;\Z)$ in the Pontryagin classes of $\rho_s$.
The degree is positive.  Restriction to the trivial subgroup followed
by transfer shows that positive-degree integral cohomology of the
finite group $\Sigma_s$ is annihilated by $|\Sigma_s|$; thus $V$ is
torsion.  The integral class
\[
 U=\varepsilon^*(\Sigma V)\in H^{2b+1}(T;\Z)
\]
is torsion as well.

Since $H^{2b+1}(S^{2b+1};\Z)=\Z$ is torsion-free, $g^*U=0$.
The integral cohomology sequence of the mapping cone supplies an
extension $\widetilde U\in H^{2b+1}(C_g;\Z)$.
Restriction
\[
 H^{2b+1}(C_g;\F_2)\longrightarrow H^{2b+1}(T;\F_2)
\]
is injective, since the preceding group in the cone sequence is
$\widetilde H^{2b}(S^{2b+1};\F_2)=0$.
Consequently
\[
 \rho_2(\widetilde U)=\Sq^b\widetilde u.
\]
The mod-$2$ Bockstein $\Sq^1$ annihilates an integral reduction.
Finally $b$ is even, so the Adem relation gives
\[
 \Sq^{b+1}\widetilde u
 =\Sq^1\Sq^b\widetilde u
 =\Sq^1\rho_2(\widetilde U)=0,
\]
contradicting \eqref{eq:nonzero-power-cone-square}.
\end{proof}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//powers.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//preamble.tex ====
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\hypersetup{pdftitle={The Stable Hurewicz Image of the Sphere at Two},pdfauthor={OpenAI},pdfsubject={September 25, 2026}}
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==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//weights.tex ====
\section{Weights, primitives, and homology suspension}
\label{sec:weights}

Homology suspension supplies two inputs to the last nonzero-suspension
argument. Its kernel identifies a terminal spherical class as a
Pontryagin square. On primitive weight components, its dual identifies
a detecting functional with the restriction of a functional at the
preceding level. We establish these statements together with finite
weight projections at level zero, where group completion requires a
filtration in place of the positive-level splitting.

For a spectrum $T$, put
\[
 D_rT=(T^{\wedge r})_{h\Sigma_r},\qquad
 \mathbb P(T)=\bigvee_{r\geq0}D_rT,\qquad D_0T=S.
\]
Thus $\mathbb P(T)$ is the free commutative ring spectrum on $T$, and its
$r$th summand has \emph{weight} $r$. Throughout, extended powers, free
algebras, pushouts, and localizations are derived constructions. We work
with their natural maps and families in the relevant homotopy coherent
categories. In particular, naturality of a construction includes its
action on mapping spaces, not just on homotopy classes of individual maps.

\subsection{A structured free-algebra model}

Let $C(X)$ be the reduced free $E_\infty$ monoid on a based space $X$:
the basepoint of $X$ is identified with the monoid unit. The usual
approximation and homology theorems identify its group completion with
$QX$ and calculate its homology by Dyer--Lashof operations
\cite[Part~I, Theorems~4.1--4.2 and Corollary~4.5]{CLM1976}.
We first record the structured splitting that will specify
our weights, including their behavior in families.

\begin{proposition}\label{prop:free-ring}
There is a natural equivalence of augmented commutative ring spectra
\[
 \Sigma^\infty_+ C(X)\simeq\mathbb P(\Sigma^\infty X).
\]
The reduced suspension spectrum of $C(X)$, based at its unit, corresponds
to the sum of the positive weights. The adjoint of projection to weight
one is the canonical map
\[
 C(X)\longrightarrow QX.
\]
This map is group completion and is an equivalence when $X$ is connected.
In the connected case, the weight-one projection is consequently the
stable counit
$\Sigma^\infty QX\to\Sigma^\infty X$.
\end{proposition}

\begin{proof}
Let $F$ denote the free unbased commutative monoid in spaces. The reduced
free monoid is the homotopy pushout
\[
 C(X)\simeq F(X)\amalg_{F(*)}*.
\]
Indeed, mapping this pushout into a commutative monoid $Y$ gives the
homotopy fibre of
$\operatorname{Map}(X,Y)\to\operatorname{Map}(*,Y)=Y$
over the unit. This is exactly the mapping space of based maps from $X$
to $Y$, based at its unit. The pushout therefore has the required reduced
free universal property, including the basepoint homotopy.

The functor $\Sigma^\infty_+$ is a strong symmetric monoidal left adjoint
from cartesian spaces to spectra. On commutative algebra objects it is
again a left adjoint, and hence preserves this pushout. Write
$T=\Sigma^\infty X$. The basepoint section and augmentation give a natural
stable splitting
\[
 \Sigma^\infty_+X\simeq S\vee T.
\]
Under it, the spectrum map induced by the basepoint is the inclusion of
the first summand. Since the free commutative algebra functor sends a
direct sum to a coproduct of algebras, the pushout after suspension is
\[
 \begin{aligned}
 \Sigma^\infty_+C(X)
 &\simeq\mathbb P(S\vee T)\amalg_{\mathbb P(S)}S\\
 &\simeq
 \bigl(\mathbb P(S)\amalg\mathbb P(T)\bigr)
       \amalg_{\mathbb P(S)}S
 \simeq\mathbb P(T).
 \end{aligned}
\]
Here $\mathbb P(S)\to S$ sends its generator to the ring unit. The
augmentation of the resulting algebra sends $T$ to zero. Thus the
positive-weight summands are precisely the augmentation fibre, or
equivalently the reduced suspension spectrum at the unit. In this
description the generating spectrum map is the \emph{difference} between
the inclusion of $X$ and its basepoint section.

We identify the weight-one adjoint using the algebra projection
\[
 \mathbb P(T)\longrightarrow S\oplus T,
\]
where the target is the split square-zero algebra. For any spectrum $M$,
the fibre over the unit point of
\[
 \mathrm{GL}_1(S\oplus M)\longrightarrow\mathrm{GL}_1S
\]
identifies with $\Omega^\infty M$, via $y\mapsto1+y$, as a grouplike
commutative monoid. To see the structure as well as the underlying space,
expand a multiproduct using distributivity in
$(S\oplus M)^{\wedge t}$. The term with no $M$ factors gives the base
unit. The terms with one $M$ factor give the sum of the inputs. Every term
with two or more $M$ factors vanishes by the square-zero structure. These
formulas respect permutations and iterated products: the structure maps
on the fibre are the additive sum-over-fibres maps. They therefore give
the stated identification with the additive infinite loop space of $M$.

The adjunction from the suspension ring now gives an $E_\infty$ map
$C(X)\to\Omega^\infty T$. On the generating based space it is the usual
map $X\to\Omega^\infty\Sigma^\infty X$. Reduced freeness identifies it
with the canonical stabilization map. Equivalently, maps from its group
completion into a grouplike commutative monoid $\Omega^\infty E$ are
based maps $X\to\Omega^\infty E$, or spectrum maps
$\Sigma^\infty X\to E$. This identifies the group completion with $QX$.
For connected $X$, the reduced free monoid has trivial $\pi_0$, so it is
already grouplike and no group completion is needed.
Under $C(X)\simeq QX$, the adjoint of the weight-one projection is then
the identity of $QX$. By the suspension-spectrum adjunction this says
exactly that the projection is the stable counit.
\end{proof}

\begin{remark}
For connected $X$, the underlying stable decomposition is the Snaith
splitting \cite[Theorem~1.1]{Snaith1974}. The argument above also
establishes the augmented commutative ring-spectrum structure and based
naturality needed for multiplication and suspension below. For a
free-algebra formulation of this structured splitting, see
\cite[Theorem~2.2 and Appendix~A]{Kuhn2006}.
\end{remark}

\subsection{Suspension and the linear coproduct}

\begin{lemma}\label{lem:weight-suspension}
Under the splitting of Proposition~\ref{prop:free-ring}, the stable
adjoint suspension on reduced pieces is weight preserving. At weight
$r$ it is the natural map
\begin{equation}\label{eq:weight-suspension}
 \Sigma D_rT\longrightarrow D_r(\Sigma T),
 \qquad T=\Sigma^\infty X,
\end{equation}
formed by using the same suspension coordinate in all $r$ factors. For
$T=\Sigma^nS$, this is the Thom spectrum map induced by the inclusion
of the diagonal trivial line into the permutation representation.

Under group completion the map is the ordinary homology suspension
$\sigma_*:H_*QX\to H_{*+1}Q\Sigma X$. It commutes with the upper-index
Dyer--Lashof operations on reduced classes and kills products of two
positive-dimensional classes. If $c$ is the class of a component point
and $z$ has positive degree, then
\[
 \sigma_*(c*z)=\sigma_*z.
\]
\end{lemma}

\begin{proof}
For $v\in S^1$, the based map $x\mapsto v\wedge x$ induces
$C(X)\to C(\Sigma X)$. These maps form a pointed family, since the
basepoint of $S^1$ induces the constant map to the monoid unit. Its
adjoint is
\[
 S^1\wedge C(X)\longrightarrow C(\Sigma X).
\]
Since $\Sigma X$ is connected, the target is $Q\Sigma X$. On the
generating space $X$, adjunction identifies this family with the usual
stabilization into $\Omega Q\Sigma X\simeq QX$. Reduced freeness and
group completion therefore identify its adjoint with the usual structure
map of the infinite loop spaces.

The positive-weight equivalence in Proposition~\ref{prop:free-ring} is
a natural equivalence between the reduced functors
\[
 X\longmapsto\operatorname{fib}(\Sigma^\infty_+C(X)\to S),
 \qquad
 X\longmapsto\bigvee_{r>0}D_r(\Sigma^\infty X).
\]
Both send the one-point based space to zero. Their action on mapping
spaces therefore respects the basepoint given by factorization through
that space, including its nullhomotopy. Apply this naturality to the
pointed mapping-space family
$v\mapsto(v\wedge-)$, and then take the stable adjunction. On the
$r$th extended power the same $v$ is used in every tensor factor. This
gives \eqref{eq:weight-suspension} with no mixing of weights. For a
sphere, $D_r(\Sigma^nS)$ is the Thom spectrum of $n\rho_r$ over
$B\Sigma_r$, where $\rho_r$ is the permutation representation. The
map in question is the Thom map associated to
\[
 n\rho_r\oplus\mathbb R\longrightarrow(n+1)\rho_r,
\]
using the diagonal line $\mathbb R\subset\rho_r$.

The commutation rule with upper-index Dyer--Lashof operations is the
usual suspension rule for free infinite loop space homology
\cite[\S20, p.~481]{May1977}. The product rules can also be read directly from the
adjunction of loop multiplication: the suspension of loop concatenation
factors through the pinch map and gives the sum of the two suspensions.
Thus a cross product of two positive-dimensional classes has zero image.
For a product with a component point, the surviving term is the
suspension of the positive-dimensional factor, multiplied by the
augmentation of the point class, which is one.
\end{proof}

For every $n\geq0$ we may equip $\mathbb P(\Sigma^nS)$ with the
\emph{linear coproduct}: it is the commutative ring map whose restriction
to the generating spectrum is the sum of its inclusions in the two
factors. At $n=0$ this is auxiliary structure and is not asserted to be
the space diagonal. At positive levels it is the space diagonal, as the
following observation shows.

\begin{lemma}\label{lem:linear-diagonal}
For a suspension $X$, the space diagonal on $C(X)$ corresponds under
Proposition~\ref{prop:free-ring} to the linear coproduct on
$\mathbb P(\Sigma^\infty X)$. Its component from weight $r$ to weights
$(i,r-i)$ is the stable transfer for
\[
 \Sigma_i\times\Sigma_{r-i}\subset\Sigma_r.
\]
\end{lemma}

\begin{proof}
In terms of the reduced generator, the diagonal is the sum of its two
linear inclusions and the term induced by the reduced diagonal
$X\to X\wedge X$. The latter is null for a suspension. Because the
source ring is free, a homotopy on the generating spectrum gives the
required homotopy of commutative ring maps.

Apply $D_r$ to the linear map $T\to T\vee T$. The terms choosing the
first summand $i$ times are indexed by the transitive $\Sigma_r$-set
$\Sigma_r/(\Sigma_i\times\Sigma_{r-i})$. Equivariantly, the map into
these terms is the diagonal into a finite sum indexed by that set.
On homotopy orbits it is the stable transfer to the indicated subgroup.
This describes every component, including the two end components.
\end{proof}

\subsection{The Steenrod modules in a primitive weight}

At weight $r>0$, let $P_r(n)$ denote the subspace of
$H_*D_r(\Sigma^nS)$ primitive for the linear coproduct: all components
with $0<i<r$ vanish. Its degreewise dual is the quotient of
$H^*D_r(\Sigma^nS)$ by the sum of the images of the proper two-block
cohomology transfers. At each fixed weight these groups are of finite
type, so this duality is exact degree by degree. For $n>0$, put
\[
 B_n=\widetilde H^*(QS^n),\qquad QB_n=B_n/B_n^2.
\]
Lemma~\ref{lem:linear-diagonal} identifies the transfer quotients with
the corresponding weight parts of $QB_n$.

At weight $2^k$, restriction to the regular elementary abelian subgroup
of $\Sigma_{2^k}$ will identify this quotient with an ideal in a
polynomial invariant ring. We first describe that ring.

For $k\geq1$, write
\[
 P_k=\F_2[x_1,\ldots,x_k],\qquad |x_i|=1,
 \qquad D_k=P_k^{\GL_k(\F_2)}
       =\F_2[d_0,\ldots,d_{k-1}],
\]
where $|d_j|=2^k-2^j$. Here $D_k$ is a Dickson algebra, in contrast to
the extended-power spectrum $D_rT$. We suppress the rank $k$ on the
generators $d_j$ and set $d_k=1$. The Dickson generators are the
coefficients of the additive polynomial
\begin{equation}\label{eq:dickson-polynomial}
 \prod_{a\in\operatorname{span}\{x_1,\ldots,x_k\}}(U+a)
   =\sum_{j=0}^k d_jU^{2^j}.
\end{equation}
The invariant-ring description is Dickson's theorem
\cite{Dickson1911}. We recall the root-polynomial proof of the
binary case, also developed in Wilkerson's account
\cite[\S I]{Wilkerson1983}.

\begin{lemma}\label{lem:dickson-ring}
The coefficients \(d_0,\ldots,d_{k-1}\) in
\eqref{eq:dickson-polynomial} are algebraically independent, have
degrees \(2^k-2^j\), and generate
\(P_k^{\GL_k(\F_2)}\).
\end{lemma}

\begin{proof}
The product over a finite binary vector space is additive in \(U\).
Indeed, if \(W'=W\oplus\langle a\rangle\) and its root polynomial
for \(W\) is \(P_W\), then
\[
 P_{W'}(U)=P_W(U)P_W(U+a)
          =P_W(U)^2+P_W(a)P_W(U).
\]
Induction from \(P_{\{0\}}(U)=U\) proves additivity and the displayed
powers of \(U\). Homogeneity gives the degrees of the coefficients.
They are invariant because an invertible binary linear change of
variables permutes the roots.

Put \(B=\F_2[d_0,\ldots,d_{k-1}]\subset P_k\) and
\(L=\F_2(x_1,\ldots,x_k)\). Each \(x_i\) satisfies the monic
root polynomial, so \(P_k\) is integral over \(B\). Therefore
\(B\) has transcendence degree \(k\); its \(k\) indicated
generators are algebraically independent and \(B\) is integrally
closed. The derivative of the root polynomial is the nonzero product
\(d_0\) of all nonzero binary linear forms. Its splitting field
over \(\operatorname{Frac}B\) is \(L\). Every automorphism of
this splitting field acts invertibly and linearly on the root space;
conversely every such linear transformation fixes the coefficients.
Consequently \(L^{\GL_k(\F_2)}=\operatorname{Frac}B\).
An invariant polynomial lies in this fraction field and is integral
over \(B\), hence belongs to \(B\). This proves the assertion.
\end{proof}

\begin{lemma}\label{lem:weight-modules}
The only nonzero positive-weight transfer quotients are at weights
$r=2^k$, $k\geq0$. As Steenrod modules they are
\begin{equation}\label{eq:weight-modules}
 M_0(n)=\Sigma^n\F_2,\qquad
 M_k(n)=\Sigma^n d_0^nD_k\quad(k\geq1),\qquad n\geq0.
\end{equation}
For $n>0$ these give the weight decomposition of $QB_n$. The action on
$d_0^nD_k$ is its ordinary action as an ideal in $P_k$.

The dual of \eqref{eq:weight-suspension} on these quotients is, after
the common grading suspension, the inclusion
\begin{equation}\label{eq:module-suspension}
 d_0^{n+1}D_k\hookrightarrow d_0^nD_k
 \qquad(k\geq1).
\end{equation}
In weight one it is the identity with the corresponding shift.
\end{lemma}

\begin{proof}
We first determine dimensions and the weights in which primitives can
occur. The free infinite loop space homology theorem gives, for $n>0$,
\begin{equation}\label{eq:free-homology-basis}
 H_*QS^n=
 \F_2\bigl[Q_{j_1}\cdots Q_{j_k}g_n:
          0<j_1\leq\cdots\leq j_k\bigr],
\end{equation}
including the empty word $g_n$, where
$Q_jz=Q^{j+|z|}z$ and $|g_n|=n$ \cite{CLM1976}.
The generator $g_n$ is primitive. Naturality of the operations for the
coproduct, together with Cartan and the vanishing of positive
operations on the unit, shows that every displayed polynomial
generator is primitive.

For $n=0$, the same polynomial description holds for
$H_*\mathbb P(S)$ on the free degree-zero generator, which we denote
here by $u$. One may see this from the free unbased monoid
$\coprod_rB\Sigma_r$, whose suspension ring is free on a degree-zero
spectrum generator. The Dyer--Lashof operations for commutative ring
spectra agree with the space operations on such suspension rings
\cite{BMMS1986}. Naturality for the \emph{linear} ring coproduct again
makes the polynomial generators primitive. This argument does not
identify that coproduct with the actual space diagonal at level zero.

In both calculations, the bottom generator has weight one, a
Dyer--Lashof operation doubles weight, and multiplication adds weights.
In a polynomial Hopf algebra on primitive generators, the primitives
have basis the iterated $2$-powers of those generators. For example,
this follows by writing the primitive identity as
$f(\mathbf x+\mathbf y)=f(\mathbf x)+f(\mathbf y)$: a monomial involving
distinct variables produces a nonzero cross term, and for a single
variable all interior binomial coefficients vanish exactly when the
exponent is a power of two. Every polynomial involves only finitely
many variables, so the argument also applies to the infinite
polynomial algebra.

It follows that a primitive basis is given by the words
\[
 Q_{j_1}\cdots Q_{j_k}g_n,
 \qquad 0\leq j_1\leq\cdots\leq j_k,
\]
with $u$ in place of $g_n$ at $n=0$. Initial zero indices record
iterated squares. A word of length $k$ has weight $2^k$ and degree
\[
 2^kn+j_1+2j_2+\cdots+2^{k-1}j_k.
\]
Thus primitives vanish at the other positive weights. Setting
$a_1=j_1$ and $a_i=j_i-j_{i-1}$ for $i>1$ rewrites this degree as
\begin{equation}\label{eq:primitive-degree-count}
 2^kn+\sum_{i=1}^k(2^k-2^{i-1})a_i,
 \qquad a_i\geq0.
\end{equation}
These are exactly the monomial degrees in
$\Sigma^nd_0^nD_k$. At $n=0$ we perform this count at fixed weight,
allowing degree zero; it remains finite in each degree at that weight.

We next identify the Steenrod module, rather than just its graded
dimension. Fix $r=2^k$ with $k\geq1$ and let
$V=(\Z/2)^k\subset\Sigma_r$ be the regular elementary abelian
subgroup. The extended power $D_r(\Sigma^nS)$ is the Thom spectrum
of $n\rho_r$ over $B\Sigma_r$. Restrict its cohomology to $BV$.
Every proper two-block transfer restricts to zero. Indeed, the double
coset formula expresses its restriction as transfers from subgroups
\[
 V\cap g(\Sigma_i\times\Sigma_{r-i})g^{-1},\qquad 0<i<r.
\]
Each is proper, because $V$ acts transitively on the particles whereas
a two-block subgroup preserves a nonempty proper subset. If $W<V$,
restriction $H^*BV\to H^*BW$ is surjective. The projection formula
then expresses transfer of any class as a class from $H^*BV$ times
the transfer of $1$, which is the even index $[V:W]$, hence zero.
Naturality of the mod-$2$ Thom class gives the same assertion with
the representation twist.

The normalizer of $V$ acts through $\GL_k(\F_2)$, so the restriction
image is contained in $D_k$ times the Thom class. It contains all
of that module: the Stiefel--Whitney classes of $\rho_r$ restrict to
the Dickson generators in \eqref{eq:dickson-polynomial}. More
explicitly, $\rho_r|_V$ is the sum of all the real one-dimensional
characters, and its total Stiefel--Whitney class is the product of
$1+a$ over all $a\in V^*$. Thus
$w_{2^k-2^j}(\rho_r)|_{BV}=d_j$, and polynomials in these classes,
multiplied by the Thom class, give the claimed image. Restriction
therefore gives a surjection from the transfer quotient onto $D_k$
times the Thom class. The dimension computation
\eqref{eq:primitive-degree-count} makes this surjection an
isomorphism in every degree.

Finally, split the regular representation over $BV$ as
$\mathbb R\oplus\bar\rho_r$. Pulling the Thom class back along
the trivial $n$-plane inclusion
\[
 \mathbb R^n\longrightarrow n\rho_r
\]
multiplies it by the Euler class of $n\bar\rho_r$, namely $d_0^n$,
and leaves the trivial Thom shift $\Sigma^n$. Multiplication by
$d_0^n$ is injective in the polynomial algebra. All maps used here
are stable cohomology maps; hence the resulting identification is
the Steenrod module $\Sigma^nd_0^nD_k$ with its inherited action.
In particular this also proves stability of that ideal under $A$.
Weight one is immediate from $D_1(\Sigma^nS)=\Sigma^nS$.

For compatibility with suspension, the Thom map in
Lemma~\ref{lem:weight-suspension} has complement $\bar\rho_r$.
Its pullback sends a Thom coefficient $f$ to $d_0f$. The level-$n$
identification then sends this coefficient to
$d_0^n(d_0f)=d_0^{n+1}f$, which is its image under the level-$(n+1)$
identification. Thus, after the common suspension shift, the map is
\[
 M_k(n+1)=\Sigma^{n+1}d_0^{n+1}D_k
 \longrightarrow
 \Sigma M_k(n)=\Sigma^{n+1}d_0^nD_k,
\]
given by inclusion. This proves \eqref{eq:module-suspension}, with
all degree shifts as asserted.
\end{proof}

\subsection{The weight filtration at level zero}

At positive levels the structured splitting gives a direct sum of
weights. At level zero we instead construct compatible finite weight
projections and show that they determine the first nonzero weight.
This requires a truncation of the free algebra.

For $N\geq0$, write $\mathbb P_{\leq N}(T)$ for the algebra obtained by
discarding weights greater than $N$. Its underlying spectrum is
$\bigvee_{0\leq r\leq N}D_rT$, and an operation whose output weight exceeds
$N$ is zero. More formally, one first truncates in nonnegatively graded
spectra, with the truncated tensor product. Extend the result by zero
to the full category of nonnegatively graded spectra, using the
induced lax monoidal structure, and then take its direct sum.
Nonnegative grading ensures that an operation killed
by the truncation cannot subsequently return to a retained weight. This
gives a commutative ring spectrum and a natural algebra projection
$\mathbb P(T)\to\mathbb P_{\leq N}(T)$.

The particle monoid at level zero is
$C(S^0)=\coprod_{r\geq0}B\Sigma_r$. It is essential to distinguish
its particle generator from the reduced generator of
Proposition~\ref{prop:free-ring}. Denote the former by $t$ and the
latter by $u$, so that $t=1+u$. The symbol $1$ here is the additive
monoid unit, represented in homology by the empty-particle component
point; it is not the point in component $-1$ of $QS^0$.

\begin{lemma}\label{lem:zero-weights}
There is an equivalence of augmented commutative ring spectra
\begin{equation}\label{eq:zero-localization}
 R:=\Sigma^\infty_+QS^0
   \simeq\mathbb P(S\{u\})[(1+u)^{-1}],\qquad t=1+u,
\end{equation}
and compatible augmented commutative ring maps
\begin{equation}\label{eq:weight-projections}
 R\longrightarrow\mathbb P_{\leq N}(S\{u\}),\qquad N\geq0.
\end{equation}
The induced homology maps jointly detect all classes. Consequently a
nonzero positive-dimensional class has a well-defined smallest
nonzero weight, called its \emph{leading weight}. Multiplication by
a power of $t$ does not change its leading component.

Let $z\in H_dQ_0S^0$, $d>0$, be primitive and annihilated by positive
Steenrod operations. Its leading component $z_r$ is primitive and
$A$-annihilated for the linear coproduct calculation of
Lemma~\ref{lem:weight-modules} at $n=0$. In particular
$r=2^k$ for some $k\geq1$. The weight-$r$ component of
$\sigma_*z$ is the image of $z_r$ under
\eqref{eq:weight-suspension}; its functional on $M_k(1)$ is the
restriction, shifted by one, of the functional on $D_k$ defined by
$z_r$. If $d$ is odd, this component is nonzero.
\end{lemma}

\begin{proof}
Under Proposition~\ref{prop:free-ring}, the particle is the ring
element $t=1+u$. Group completion of $C(S^0)$ inverts this particle
component. For a commutative ring spectrum $A'$, a map from its
suspension ring is a map from the additive $E_\infty$ particle monoid
to the multiplicative $E_\infty$ space of $A'$. Such a map extends
over group completion precisely when the particle maps to a unit.
Equivalently, the resulting ring map inverts $1+u$. This is the
universal property of the localization in
\eqref{eq:zero-localization}, including its mapping spaces. The
augmentation sends $u$ to zero and $t$ to one, so the equivalence is
augmented. The underlying module of localization at a degree-zero
element is its multiplication telescope. Therefore ordinary
homology satisfies
\[
 H_*R=H_*\mathbb P(S\{u\})[(1+u)^{-1}].
\]

In $\mathbb P_{\leq N}(S\{u\})$, the element $u$ is nilpotent in
$\pi_0$. Thus $1+u$ is a unit, with inverse
$\sum_{j=0}^N(-u)^j$ in $\pi_0$, and the truncation extends across
localization to give \eqref{eq:weight-projections}. Their
compatibility follows from the localization universal property.

Every class of $H_*R$ can be multiplied by a sufficiently large
power of $t$ to become a class in the free homology algebra
$H_*\mathbb P(S\{u\})$. This algebra is polynomial, as used in
Lemma~\ref{lem:weight-modules}, and is a direct sum of weights.
Hence such a numerator has only finitely many nonzero weights.
If its smallest nonzero component has weight $r$, multiplying by
$(1+u)^a$ leaves that component unchanged: the weight-zero term is
one and every other term has positive weight. Conversely, in a
finite truncation the inverse of $1+u$ has weight-zero term one
and only positive weights otherwise. Thus a nonzero numerator
remains nonzero in every truncation retaining its leading weight.
This proves joint detection, independence of the leading component
from the chosen numerator, and invariance under powers of $t$.
The inclusion $H_*Q_0S^0\to H_*QS^0$ is injective, since it is the
inclusion of one component, so these conclusions apply to the base
component as well.

Before localization, the actual space diagonal on the reduced
generator is
\begin{equation}\label{eq:zero-diagonal}
 \Delta u=u\otimes1+1\otimes u+u\otimes u.
\end{equation}
Consequently, a weight-$r$ class maps only to terms of total weight
at least $r$, and its terms of total weight $r$ are given by the
linear coproduct. This follows by applying extended powers to
\eqref{eq:zero-diagonal} and then multiplying. It remains true
after localization: the diagonal of $t$ is $t\otimes t$, whose
inverse has constant term one and only positive total weights
otherwise. These statements may all be made in finite truncations
in the two target factors, so no convergence of an infinite sum
is required.

The kernels of the projections through weight $r-1$ define a decreasing
weight filtration. The preceding calculation identifies its associated
graded coproduct with the linear coproduct.
Since $z$ is primitive, comparison in its leading total weight gives
\[
 \Delta_{\mathrm{lin}}z_r=z_r\otimes1+1\otimes z_r.
\]
The weight projections are maps of spectra, so they commute with
the stable Steenrod action. Thus $z_r$ is also $A$-annihilated.
Lemma~\ref{lem:weight-modules} makes $r$ a power of two; weight
one has no positive-dimensional homology, so $r=2^k$ with $k\geq1$.

For the suspension assertion, choose $a\geq0$ such that $t^az$
has a finite numerator in the particle monoid homology. The
translation rule in Lemma~\ref{lem:weight-suspension} gives
\[
 \sigma_*(t^az)=\sigma_*z.
\]
The numerator has the same leading component $z_r$ and no smaller
weights. Apply the weight-preserving family of
Lemma~\ref{lem:weight-suspension} before group completion.
It follows that the weight-$r$ component of the suspension is
exactly the image of $z_r$ under
\eqref{eq:weight-suspension}. By
\eqref{eq:module-suspension}, the corresponding functional is the
restriction from $D_k$ to $d_0D_k$, with one suspension shift.

Finally, all the generators of $D_k/(d_0)$ have even degrees:
$|d_j|=2^k-2^j$ is even for $j\geq1$ (and for $k=1$ the quotient
is just $\F_2$). A nonzero odd-degree functional on $D_k$ therefore
cannot vanish on $d_0D_k$. This proves that the leading component
survives suspension when $d$ is odd.
\end{proof}

\subsection{The suspension kernel and square roots of primitives}

Let $H^+$ denote the positive-degree part of a connected homology
algebra, and call $(H^+)^2$ its subspace of decomposables.

\begin{lemma}\label{lem:suspension-kernel}
For $n\geq1$, a positive-dimensional class in the kernel of
\[
 \sigma_*:H_*QS^n\longrightarrow H_{*+1}QS^{n+1}
\]
is decomposable. The same assertion holds for
$\sigma_*:H_*Q_0S^0\to H_{*+1}QS^1$. In each case the kernel in
positive dimensions is precisely the subspace of decomposables.
\end{lemma}

\begin{proof}
The product rule of Lemma~\ref{lem:weight-suspension} kills the
decomposables. For the reverse inclusion, use the polynomial
generators in \eqref{eq:free-homology-basis}. Suspension commutes
with upper-index operations. When the bottom class changes from
$g_n$ to $g_{n+1}$, every intermediate word with those same upper
indices increases in degree by one. Thus each lower index of a
nonempty polynomial generator decreases by one. The original
indices satisfy $0<j_1\leq\cdots\leq j_k$, so the resulting
indices satisfy
\[
 0\leq j_1-1\leq\cdots\leq j_k-1.
\]
These are distinct nonzero primitive words in the target. The
empty-word generator $g_n$ suspends to $g_{n+1}$, also nonzero.
Hence suspension is injective on the span of the polynomial
generators modulo decomposables.

At level zero, localization at the particle gives
\[
 H_*QS^0=\F_2\bigl[t,t^{-1},
 Q_{j_1}\cdots Q_{j_k}t:
 k\geq1,\ 0<j_1\leq\cdots\leq j_k\bigr].
\]
A word of length $k$ has component degree $2^k$. Replacing each
such generator by its product with $t^{-2^k}$ identifies this algebra
with the Laurent polynomial algebra on $t$ over a polynomial algebra
concentrated in component zero. Thus $H_*Q_0S^0$ is polynomial,
with generators the translates to component zero
of the nonempty positive-excess particle generators
$Q_{j_1}\cdots Q_{j_k}t$ with
$0<j_1\leq\cdots\leq j_k$. A word of length $k$ is translated
by $t^{-2^k}$. Suspension ignores that translation and commutes
with its upper-index operations, giving the same distinct
nonzero words on $g_1$ with all lower indices decreased by one.
One may equivalently start with the reduced generator $u=t-1$:
positive operations on the additive monoid unit vanish, so the
positive operations on $u$ and $t$ agree. This proves injectivity
on indecomposables at level zero as well.

The injectivity here is into the full homology of the target.
Some of the resulting words have initial lower index zero and
are themselves squares; they must not be discarded by projecting
the target to indecomposables. Their independence as primitive
words is exactly what gives the asserted kernel description.
\end{proof}

\begin{lemma}\label{lem:primitive-roots}
Let $H$ be a connected graded commutative Hopf algebra over $\F_2$
which is polynomial as an algebra on positive-degree generators.
Every decomposable primitive of $H$ is a square, and its unique
square root is primitive. If $H$ also has a Steenrod action
satisfying the homology Cartan formula, an $A$-annihilated square
has an $A$-annihilated root.

These statements apply to $H_*Q_0S^0$ and to $H_*QS^n$ for
$n\geq1$.
\end{lemma}

\begin{proof}
Put $QH=H^+/(H^+)^2$, let $\pi:H\to QH$ send $1$ to zero,
and define
\[
 \partial=(1\otimes\pi)\Delta:H\longrightarrow H\otimes QH.
\]
The identity
$\pi(ab)=\epsilon(a)\pi(b)+\epsilon(b)\pi(a)$ shows that
$\partial$ is a derivation. If $y_i$ are homogeneous polynomial
generators, connectedness and grading give
\[
 \partial y_i=1\otimes[y_i]
  +\text{terms with smaller positive degree in the second factor}.
\]
For a polynomial $f$, the derivation rule yields
\[
 \partial f=\sum_i\frac{\partial f}{\partial y_i}\,\partial y_i.
\]
Only finitely many variables occur in $f$. If $\partial f=0$,
inspect the coefficient of $[y_i]$ for variables of largest
degree occurring in $f$. There is no contribution from a
higher-degree generator, so their ordinary polynomial partial
derivatives vanish. Descending through generator degrees shows
that every $\partial f/\partial y_i$ vanishes. Over $\F_2$, this
means that every exponent in every monomial of $f$ is even;
equivalently, $f=z^2$ for a unique polynomial $z$.

If $f$ is a decomposable primitive, then
$\partial f=1\otimes\pi(f)=0$, so this argument applies.
Moreover,
\[
 \bigl(\Delta z-z\otimes1-1\otimes z\bigr)^2=0.
\]
The tensor product $H\otimes H$ is again a polynomial algebra,
so its Frobenius map is injective. Hence the displayed reduced
coproduct is zero and $z$ is primitive.

Finally, the homology Cartan formula in characteristic two gives
\[
 \Sq_*^{2a}(z^2)=(\Sq_*^a z)^2,
 \qquad \Sq_*^{2a+1}(z^2)=0.
\]
If $z^2$ is $A$-annihilated, injectivity of Frobenius forces
$\Sq_*^a z=0$ for every $a>0$. Since these squares generate the
positive Steenrod action, the root is $A$-annihilated. The
polynomial homology descriptions above verify the algebra
hypotheses for the infinite loop spaces under consideration.
\end{proof}
==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//weights.tex ====

==== BEGIN The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//references.bib ====

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==== END The-Stable-Hurewicz-Image-of-the-Sphere-at-Two//references.bib ====