np+1}\{p\}$, 
does commute because it is known that $S$ is an $H$-map. The $2$-primary 
version in~(\ref{majorobstr}) does not commute because it does not even 
commute in homology. To see this, observe that if $x\in H_{2n-1}(S^{2n-1})$ 
and $a\in H_{2n-1}(\Omega^{2} S^{2n+1})$ are generators, then 
$(E^{2}\cdot 1)_{\ast}(x,a)=a^{2}$. Note that $a^{2}$ is the Bockstein 
of the generator $b\in H_{4n-1}(\Omega^{2} S^{2n+1})$. Under $\Omega H$ 
we have $(\Omega H)_{\ast}(b)=v$, where $v\in H_{4n-1}(\Omega^{2} S^{4n+1})$ 
is a generator. Since $S$ is a lift of $\Omega H$ through 
\(\namedright{\Omega^{2} S^{4n+1}\{2\}}{\Omega\delta}{\Omega^{2} S^{2n+1}}\) 
and $(\Omega\delta)_{\ast}$ is onto, we have $S_{\ast}(b)=\overline{v}$, 
where $(\Omega\delta)_{\ast}(\overline{v})=v$. Now $\beta\overline{v}$ 
is nonzero and generates $H_{4n-2}(\Omega^{2} S^{4n+1}\{2\})$. Thus 
$S_{\ast}(a^{2})=\beta\overline{v}$ and so 
$(S\circ(E^{2}\cdot 1))_{\ast}(x,y)\neq 0$. On the other hand, if the 
diagram above were to commute then we would have 
$(S\circ(E^{2}\cdot 1))_{\ast}(x,y)=0$. 

To get around this obstruction, we introduce a degree $2$ map on $S^{2n-1}$. 
In Lemma~\ref{k=1case} we show that 
$S\circ((E^{2}\circ\underline{2})\cdot 1)$ does factor as $S\circ\pi_{1}$, 
and then we use this as the base case in a modified induction to 
prove Proposition~\ref{Ebarextension}. 

To prepare for Lemma~\ref{k=1case}, we make three preliminary 
observations. First, in general, if $A$ 
and~$B$ are spaces, then there is a homotopy equivalence 
\(e\colon\namedright{\Sigma A\vee\Sigma B\vee(\Sigma A\wedge B)}{\simeq} 
    {\Sigma(A\times B)}\) 
given by $\Sigma i_{1}+\Sigma i_{2}+j$, where $i_{1}$ and $i_{2}$ are 
the inclusions of the left and right factors respectively, and~$j$ is the 
canonical map of a join into the suspension of a product. If $A=B$ 
and $B=\Omega X$ with loop multiplication $\mu$ then the composite 
\(\mu^{\ast}\colon\nameddright{\Sigma\Omega X\wedge\Omega X}{j} 
    {\Sigma(\Omega X\times\Omega X)}{\Sigma\mu}{\Sigma\Omega X}\) 
is the Hopf construction on $\Omega X$ and Ganea~\cite{Ga} showed that 
there is a homotopy fibration 
\(\nameddright{\Sigma\Omega X\wedge\Sigma\Omega X}{\mu^{\ast}} 
     {\Sigma\Omega X}{ev}{X}\). 

The second observation is a well-known factorization of the
$2^{nd}$-power map on $\Omega^{2} S^{2n+1}\{2\}$. Consider the homotopy
cofibration sequence
\(\namedddright{S^{2}}{\underline{2}}{S^{2}}{i}{P^{3}(2)}
   {q}{S^{3}}\),
where $P^{3}(2)$ is the mod-$2$ Moore space, $i$ is the inclusion of
the bottom cell, and $q$ is the pinch onto the top cell. This cofibration
sequence induces a homotopy fibration sequence
\[\namedddright{\mbox{Map}_{\ast}(S^{3},S^{2n+1})}{q^{\ast}}
    {\mbox{Map}_{\ast}(P^{3}(2),S^{2n+1})}{i^{\ast}}
    {\mbox{Map}_{\ast}(S^{2},S^{2n+1})}{\underline{2}^{\ast}}
    {\mbox{Map}_{\ast}(S^{2},S^{2n+1})}.\]
Since $\mbox{Map}_{\ast}(S^{2},S^{2n+1})=\Omega^{2} S^{2n+1}$ and
$\underline{2}^{\ast}$ is the $2^{nd}$-power map on $\Omega^{2} S^{2n+1}$,
this homotopy fibration sequence can be identified with the homotopy
fibration sequence
\[\namedddright{\Omega^{3} S^{2n+1}}{\Omega\partial}{\Omega^{2} S^{2n+1}\{2\}}
    {\Omega\delta}{\Omega^{2} S^{2n+1}}{2}{\Omega^{2} S^{2n+1}}.\]
In particular, $\mbox{Map}_{\ast}(P^{3}(2),S^{2n+1})=\Omega^{2} S^{2n+1}\{2\}$,
and so the $2^{nd}$-power map on $\Omega^{2} S^{2n+1}\{2\}$ is determined
by the degree~$2$ map on $P^{3}(2)$. This factors as the composite
\(\namedddright{P^{3}(2)}{q}{S^{3}}{\eta}{S^{2}}{i}{P^{3}(2)}\),
where $\eta$ is an element of Hopf invariant one. Thus we have the 
following.

\begin{lemma}
   \label{curly2power}
   There is a homotopy commutative diagram
   \[\diagram
        \Omega^{2} S^{2n+1}\{2\}\rrto^-{2}\dto^{\Omega\delta}
            & & \Omega^{2} S^{2n+1}\{2\}\ddouble \\
        \Omega^{2} S^{2n+1}\rto^-{\eta^{\ast}}
            & \Omega^{3} S^{2n+1}\rto^-{\Omega\partial}
            & \Omega^{2} S^{2n+1}\{2\}.
     \enddiagram\]
   $\qqed$
\end{lemma}

The third observation is a consequence of the James construction. 
Given a map 
\(f\colon:\namedright{A}{}{\Omega X}\), 
the James construction implies that there is a unique $H$-map 
\(\overline{f}\colon\namedright{\Omega\Sigma A}{}{\Omega X}\) 
such that $\overline{f}\circ E\simeq f$. The uniqueness assertion 
immediately implies the following lemma. 

\begin{lemma} 
   \label{Jamescase} 
   Suppose $A=\Omega A^{\prime}$ and $f$ is an $H$-map. Then there 
   is a homotopy commutative diagram 
   \[\diagram 
        \Omega\Sigma\Omega A^{\prime}\rto^-{\overline{f}}\dto^{\Omega ev} 
             & \Omega X\ddouble \\ 
        \Omega A^{\prime}\rto^-{f} & \Omega X. 
     \enddiagram\] 
   $\qqed$ 
\end{lemma} 

Now we are ready to prove the modified square. 

\begin{lemma} 
   \label{k=1case} 
   There is a homotopy commutative diagram 
   \[\diagram 
        S^{2n-1}\times\Omega^{2} S^{2n+1}
             \rto^-{(E^{2}\circ\underline{2})\cdot 1}\dto^{\pi_{2}} 
          & \Omega^{2} S^{2n+1}\dto^{S} \\ 
        \Omega^{2} S^{2n+1}\rto^-{S} & \Omega^{2} S^{4n+1}\{2\}.   
     \enddiagram\] 
\end{lemma} 

\begin{proof} 
The adjoint of the upper direction around the diagram is homotopic 
to the composite 
\(\psi\colon\lllnameddright{\Sigma(S^{2n-1}\times\Omega^{2} S^{2n+1})} 
    {\Sigma(E^{2}\circ\underline{2})\cdot 1} 
    {\Sigma\Omega^{2} S^{2n+1}}{\widetilde{S}}{\Omega S^{4n+1}\{2\}}\) 
where $\widetilde{S}$ is the adjoint of $S$. Precomposing with the 
homotopy equivalence  
\(e\colon\namedright{\Sigma