$m_{M}(\widetilde{\psi}\wedge 1_{M})\alpha_{X\wedge
M}m_{M}(\widetilde{\psi}\wedge 1_{M})=\alpha m_{M}(\widetilde{\psi}\wedge
1_{M})=m_{M}(\widetilde{\psi}\wedge 1_{M})(1_{X}\wedge\alpha)$ so that by
(9.4.2) we have $\alpha_{X\wedge M}m_{M}(\widetilde{\psi}\wedge
1_{M})=1_{X}\wedge\alpha$ modulo $(u^{\prime})_{*}[\Sigma^{q}X\wedge M,L\wedge
K]$ = 0 , this is because $[\Sigma^{q}L\wedge K,L\wedge K]$ = 0 and
$[\Sigma^{3q}M,L\wedge K]$ = 0. Concludingly we have
(9.4.3) $(\phi\wedge 1_{M})m_{M}(\widetilde{\psi}\wedge
1_{M})=(1_{L}\wedge\alpha)(u^{\prime\prime}\wedge 1_{M}),$

$\qquad\quad\alpha_{X\wedge M}m_{M}(\widetilde{\psi}\wedge
1_{M})=1_{X}\wedge\alpha$

The cofibre of the map $\alpha_{X\wedge M}:\Sigma^{3q}M\rightarrow X\wedge M$
is $W\wedge K$ given by the cofibration
(9.4.4) $\Sigma^{3q}M\stackrel{{\scriptstyle\alpha_{X\wedge
M}}}{{\longrightarrow}}X\wedge M\stackrel{{\scriptstyle\mu_{X\wedge
M}}}{{\longrightarrow}}W\wedge K\stackrel{{\scriptstyle
j^{\prime}(j^{\prime\prime}u\wedge 1_{K})}}{{\longrightarrow}}\Sigma^{3q+1}M$
This can be seen by the following homotopy commutative diagram of $3\times
3$-Lemma

$\qquad\qquad\Sigma^{3q}M\quad\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}\quad\Sigma^{2q}M\quad\stackrel{{\scriptstyle(\phi\wedge
1_{K})i^{\prime}}}{{\longrightarrow}}\quad\Sigma L\wedge K$

$\qquad\qquad\quad\searrow\alpha_{X\wedge M}\nearrow_{m_{M}(\tilde{\psi}\wedge
1_{M})}\searrow i^{\prime}\quad\nearrow(\phi\wedge 1_{K})$
(9.4.5)$\qquad\qquad\quad X\wedge M\qquad\qquad\quad\Sigma^{2q}K$

$\qquad\qquad\quad\nearrow u^{\prime}\quad\searrow\mu_{X\wedge M}\quad\nearrow
j^{\prime\prime}u\wedge 1_{K}\searrow j^{\prime}$

$\qquad\qquad L\wedge K\quad\stackrel{{\scriptstyle w\wedge
1_{K}}}{{\longrightarrow}}\quad W\wedge K\stackrel{{\scriptstyle
j^{\prime}(j^{\prime\prime}u\wedge
1_{K})}}{{\longrightarrow}}\quad\Sigma^{3q+1}M$

By (9.2.13), $ijm_{M}(\widetilde{\psi}\cdot\widetilde{u}\wedge
1_{M})=ij(u_{2}\wedge 1_{M})=(u_{2}\wedge 1_{M})(1_{U}\wedge
ij)=m_{M}(\widetilde{\psi}\cdot\widetilde{u}\wedge 1_{M})(1_{U}\wedge
ij)=m_{M}(\widetilde{\psi}\wedge 1_{M})(1_{X}\wedge ij)(\widetilde{u}\wedge
1_{M})$, then we have $ijm_{M}(\widetilde{\psi}\wedge
1_{M})=m_{M}(\widetilde{\psi}\wedge 1_{M})(1_{X}\wedge
ij)+\lambda(j\widetilde{\psi}\wedge 1_{M})$ for some $\lambda\in Z_{p}$. It
follows that $\lambda j(j\widetilde{\psi}\wedge
1_{M})=-jm_{M}(\widetilde{\psi}\wedge 1_{M})(1_{X}\wedge
ij)=-j\widetilde{\psi}(1_{X}\wedge j)=j(j\widetilde{\psi}\wedge 1_{M})$ and so
$\lambda=1$. In addition, $i^{\prime}(\alpha_{1}\wedge
1_{M})m_{M}(\widetilde{\psi}\wedge 1_{M})=(j^{\prime\prime}\wedge
1_{K})(1_{L}\wedge i^{\prime})(\phi\wedge 1_{M})m_{M}(\widetilde{\psi}\wedge
1_{M})$ = 0, then by (9.1.23) we have $m_{M}(\widetilde{\psi}\wedge
1_{M})=m_{M}(\overline{u}\wedge 1_{M})\psi_{X\wedge M}$ , where $\psi_{X\wedge
M}\in[\Sigma^{-q+1}X\wedge M,Y\wedge M]$. In addition, $[\Sigma^{-q+1}X\wedge
M,Y\wedge M]\cong Z_{p}\\{\psi_{X\wedge M}\\}$ , this can be obtained from
$[\Sigma^{-2q}X\wedge M,M]\cong Z_{p}\\{m_{M}(\widetilde{\psi}\wedge
1_{M})\\}$ , (9.1.23) and $[\Sigma^{-q}X\wedge M,K]$ = 0. Then, by
$j^{\prime}(j^{\prime\prime}u\wedge 1_{K})\cdot\mu_{X\wedge M}$ = 0 and
(9.1.27) we have $(u\wedge 1_{K})\mu_{X\wedge
M}=\overline{\mu}_{2}(1_{Y}\wedge i^{\prime})\psi_{X\wedge M}$ (up to nonzero
scalar). Concludingly we have
(9.4.6) $j\widetilde{\psi}\wedge 1_{M}=ijm_{M}(\widetilde{\psi}\wedge
1_{M})-m_{M}(\widetilde{\psi}\wedge 1_{M})(1_{X}\wedge ij)$,

$\qquad(u\wedge 1_{K})\mu_{X\wedge M}=\overline{\mu}_{2}(1_{Y}\wedge
i^{\prime})\psi_{X\wedge M}$ (up to nonzero scalar)

$\qquad[\Sigma^{-q+1}X\wedge M,Y\wedge M]\cong Z_{p}\\{\psi_{X\wedge M}\\},$

$\qquad\quad m_{M}(\overline{u}\wedge 1_{M})\psi_{X\wedge
M}=m_{M}(\widetilde{\psi}\wedge 1_{M})$,

By the following homotopy commutative diagram of $3\times 3$-Lemma

$\qquad\qquad L\wedge K\quad\stackrel{{\scriptstyle(1_{X}\wedge
j)u^{\prime}}}{{\longrightarrow}}\quad\Sigma X\qquad\stackrel{{\scriptstyle
1_{X}\wedge p}}{{\longrightarrow}}\qquad\Sigma X$

$\qquad\qquad\quad\searrow u^{\prime}\qquad\nearrow 1_{X}\wedge
j\qquad\searrow\omega\qquad\nearrow\widetilde{u}w_{2}$

$X\wedge M\qquad\qquad\qquad\qquad Y$

$\qquad\qquad\quad\nearrow 1_{X}\wedge
i\quad\searrow^{m_{M}(\tilde{\psi}\wedge 1_{M})}\nearrow_{(1_{Y}\wedge
j)\alpha_{Y\wedge M}}\searrow^{\overline{\mu}_{2}(1_{Y}\wedge i^{\prime}i)}$

$\qquad\qquad
X\qquad\stackrel{{\scriptstyle\tilde{\psi}}}{{\longrightarrow}}\qquad\Sigma^{2q}M\qquad\stackrel{{\scriptstyle(\phi\wedge
1_{K})i^{\prime}}}{{\longrightarrow}}\qquad\Sigma L\wedge K$
we know that the cofibre of $(1_{X}\wedge j)u^{\prime}:L\wedge
K\rightarrow\Sigma X$ is $Y$ given by the cofibration
(9.4.7) $L\wedge K\stackrel{{\scriptstyle(1_{X}\wedge
j)u^{\prime}}}{{\longrightarrow}}\Sigma
X\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}Y\stackrel{{\scriptstyle\overline{\mu}_{2}(1_{Y}\wedge
i^{\prime}i)}}{{\longrightarrow}}\Sigma L\wedge K$
In addition, by the commutativity of the above rectangle we have
(9.4.8) $\omega\wedge 1_{M}=\alpha_{Y\wedge M}m_{M}(\widetilde{\psi}\wedge
1_{M})$.

Proposition 9.4.9 Under the supposition (I) of the main Thoerem C we have

(1) $Ext_{A}^{s+1,tq+r}(H^{*}K,H^{*}M)$ = 0 for $r=1,2$,

(2) $Ext_{A}^{s+1,tq+rq+1}(H^{*}K,H^{*}K)$ = 0 for $r=-1,0,1,2$.

Proof: (1) By the supposition, $Ext_{A}^{s+1,tq-q+r}(Z_{p},Z_{p})$ = 0 for
$r=-1,0,1,\\\ 2,3$, then
$(j^{\prime})_{*}Ext_{A}^{s+1,tq+r}(H^{*}K,Z_{p})\subset Ext_{A}^{s+1,tq-
q-r-1}(H^{*}M,Z_{p})$ = 0 for $r=1,2,3$ and so
$Ext_{A}^{s+1,tq+r}(H^{*}K,Z_{p})=(i^{\prime})_{*}Ext_{A}^{s+1,tq+r}(H^{*}M,Z_{p})$
= 0 for $r=1,2,3$ (cf. Prop. 9.3.0(1)) and the result follows.

(2) Consider the following exact sequence ($r=-1,0,1,2$)

$0=Ext_{A}^{s+1,tq+(r+1)q+2}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq+rq+1}(H^{*}K,H^{*}K)$

$\qquad\quad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq+rq+1}(H^{*}K,H^{*}M)$
induced by (9.1.2). The right group is zero for $r=0,1,2$( cf. (1) and Prop.
9.3.2(1)(2)) and also is zero for $r=-1$ which is obtained by the supposition
on $Ext_{A}^{s+1,tq-q+r}(Z_{p},Z_{p})$ = 0 for $r=-1,0,1,2$. The left group is
zero for $r=-1,0,1$ (cf. (1) and Prop. 9.3.2). The left group also is zero for
$r=2$, this is because $Ext_{A}^{s+1,tq+rq+u}(Z_{p},Z_{p})$ = 0 for
$r=2,3,u=0,1,2,3$ by the supposition. Then the middle group is zero as
desired. Q.E.D.

Proposition 9.4.10 Under the supposition (I) of the main Theorem C we have

(1) $Ext_{A}^{s+1,tq+q+1}(H^{*}W\wedge K,H^{*}X\wedge M)$ = 0.

(2) $Ext_{A}^{s+1,tq+2q+1}(H^{*}Y,H^{*}M)\cong Z_{p}\\{((1_{Y}\wedge
j)\alpha_{Y\wedge M})_{*}(\tilde{\sigma})\\}$,

$Ext_{A}^{s+1,tq+q}(H^{*}Y,H^{*}Y)\cong
Z_{p}\\{(\overline{u})^{*}((1_{Y}\wedge j)\alpha_{Y\wedge
M})_{*}(\tilde{\sigma})\\}$,

(3) $Ext_{A}^{s+1,tq+3q}(H^{*}X,H^{*}M)\cong Z_{p}\\{((1_{X}\wedge
j)\alpha_{X\wedge M})_{*}(\tilde{\sigma})\\}$

Proof: (1) Consider the following exact sequence

$0=Ext_{A}^{s+1,tq+3q+1}(H^{*}W\wedge K,H^{*}M)\stackrel{{\scriptstyle
m_{M}(\tilde{\psi}\wedge
1_{M})^{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq+q+1}(H^{*}W\wedge
K,H^{*}X\wedge
M)\stackrel{{\scriptstyle(u^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq+q+1}(H^{*}W\wedge
K,H^{*}L\wedge K)$
induced by (9.4.2). The right group is zero by Prop. 9.4.9(2) and
(9.1.12)(9.1.3). The left group also is zero by
$Ext_{A}^{s+1,tq+rq+1}(H^{*}K,H^{*}M)$ = 0(for $r=1,2,3$)(cf. the proof of
Prop. 9.4.9(2)) . Then the middle group is zero as desired.

(2) Since $\overline{u}(1_{Y}\wedge j)\alpha_{Y\wedge
M}\in[\Sigma^{q-1}M,M]\cong Z_{p}\\{ij\alpha,\alpha ij\\}$, then
$\overline{u}(1_{Y}\wedge j)\alpha_{Y\wedge
M}=\lambda_{1}ij\alpha+\lambda_{1}\alpha ij$ where the scalar
$\lambda_{1},\lambda_{2}\in Z_{p}$ satisfy $\lambda_{1}j\alpha
ij\alpha+\lambda_{2}j\alpha^{2}ij$ = 0. Consider the following exact sequence

$Ext_{A}^{s+1,tq+2q}(Z_{p},H^{*}M)\stackrel{{\scriptstyle(\overline{w})_{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq+2q+1}(H^{*}Y,H^{*}M)$

$\qquad\quad\stackrel{{\scriptstyle(\overline{u})_{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq+q}(H^{*}M,H^{*}M)\stackrel{{\scriptstyle(j\alpha)_{*}}}{{\longrightarrow}}$
induced by (9.1.5). The left group is zero which can be obtained by the
supposition on $Ext_{A}^{s+1,tq+2q+k}(Z_{p},Z_{p})$ = 0 (for $k=0,1$). By
Prop. 9.3.1(2), the right group has two generators
$(ij)_{*}\alpha_{*}(\tilde{\sigma})$ and $\alpha_{*}(ij)_{*}(\tilde{\sigma})$
. Then $(\overline{u})_{*}\\\ Ext_{A}^{s+1,tq+2q+1}(H^{*}Y,H^{*}M)$ has unique
generator $(\overline{u})_{*}((1_{Y}\wedge j)\alpha_{Y\wedge
M})_{*}(\tilde{\sigma})$ so that the first result follows. For the second
result, consider the following exact sequence

$Ext_{A}^{s+1,tq+q}(Z_{p},Z_{p})\stackrel{{\scriptstyle(\overline{w})_{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq+q+1}(H^{*}Y,Z_{p})$

$\qquad\quad\stackrel{{\scriptstyle(\overline{u})_{*}}}{{\longrightarrow}}Ext_{A}^{s+1,tq}(H^{*}M,Z_{p})\stackrel{{\scriptstyle(j\alpha)_{*}}}{{\longrightarrow}}$
induced by (9.1.5). By the supposition, the left group has unique generator
$h_{0}\sigma=(j\alpha i)_{*}(\sigma)$ so that im $(\overline{w})_{*}$ = 0. The
right group is zero or has (one or two) generator $i_{*}(\sigma^{\prime})$
such that $(j\alpha)_{*}i_{*}(\sigma^{\prime})=h_{0}\sigma^{\prime}\neq 0$.
Then the middle group is zero and so the second result follows.

(3) Since $\widetilde{\psi}(1_{X}\wedge j)\alpha_{X\wedge
M}\in[\Sigma^{q-1}M,M]\cong Z_{p}\\{ij\alpha,\alpha ij\\}$, then
$\widetilde{\psi}(1_{X}\wedge j)\alpha_{X\wedge
M}=\lambda_{3}ij\alpha+\lambda_{4}\alpha ij$, where the scalar
$\lambda_{3},\lambda_{4}\in Z_{p}$ satisfy $\lambda_{3}(1_{Y}\wedge
j)\alpha_{Y\wedge M}ij\alpha+\lambda_{4}(1_{Y}\wedge j)\alpha_{Y\wedge
M}\alpha ij$ = 0. Then, similar to that in (2), $(\widetilde{\psi})_{*}\\\
Ext_{A}^{s+1,tq+3q}(H^{*}X,H^{*}M)$ has unique generator
$(\widetilde{\psi})_{*}((1_{X}\wedge j)\alpha_{X\wedge
M})_{*}(\tilde{\sigma})$ so that $Ext_{A}^{s+1,tq+3q}(H^{*}X,H^{*}M)$ has
unique generator $((1_{X}\wedge j)\alpha_{X\wedge M})_{*}(\tilde{\sigma})$ ,
this is because $Ext_{A}^{s+1,tq+3q+1}(H^{*}Y,H^{*}M)$ = 0 which can be
obtained by the supposition (I)(b) on $Ext_{A}^{s+1,tq+rq+k}(Z_{p},Z_{p})$ = 0
for $r=1,2,k=-1,0,1,2$). Q.E.D.

Proposition 9.4.11 Under the supposition (I) of the main Theorem C we have

(1) $Ext_{A}^{s,tq-2q}(H^{*}M,H^{*}X\wedge M)\cong
Z_{p}\\{m_{M}(\widetilde{\psi}\wedge 1_{M})^{*}(\tilde{\sigma})\\}$.

(2) $Ext_{A}^{s,tq+rq+u}(H^{*}K,H^{*}M)$ = 0 for $r=-1,1,2,3,u=0,1,2$

$Ext_{A}^{s,tq}(H^{*}K,H^{*}K)\cong Z_{p}\\{\sigma_{K}\\}$ satisfying
$(i^{\prime})^{*}(\sigma_{K})=(i^{\prime})_{*}(\tilde{\sigma})$,

(3) $Ext_{A}^{s,tq}(H^{*}L\wedge K,H^{*}L\wedge K)\cong
Z_{p}\\{\sigma_{L\wedge K}\\}$

satisfying $(j^{\prime\prime}\wedge 1_{K})_{*}(\sigma_{L\wedge
K})=(j^{\prime\prime}\wedge 1_{K})^{*}(\sigma_{K})$

$Ext_{A}^{s,tq+rq+u}(H^{*}L\wedge K,H^{*}M)$ = 0 for $r=1,2,3,u=0,1,2$,

(4) $Ext_{A}^{s,tq+rq+u}(H^{*}W\wedge K,H^{*}M)$ = 0 for $r=1,2,3,u=0,1,2$,

$Ext_{A}^{s,tq+q}(H^{*}W\wedge K,H^{*}X\wedge M)$ = 0

Proof: (1) Consider the following exact sequence

$Ext_{A}^{s,tq}(H^{*}M,H^{*}M\wedge
M)\stackrel{{\scriptstyle(\tilde{\psi}\wedge
1_{M})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq-2q}(H^{*}M,H^{*}X\wedge M)$

$\qquad\quad\stackrel{{\scriptstyle(\tilde{u}w_{2}\wedge
1_{M})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq-2q}(H^{*}M,H^{*}Y\wedge M)$
induced by (9.2.12). By the suppopsition on $Ext_{A}^{s,tq-rq+u}(Z_{p},Z_{p})$
= 0 with ($r=1,2$, $u=0,1,2$) and the degree of the top cell of $Y\wedge M$ is
$q+3$ we know that the right group is zero. Since
$(\overline{m}_{M})^{*}Ext_{A}^{s,tq}(H^{*}M,H^{*}M\wedge M)\subset
Ext_{A}^{s,tq+1}(H^{*}M,H^{*}M)$ = 0 ( cf. Prop. 9.3.0(2)), then the left
group has unique generator $(m_{M})^{*}(\tilde{\sigma})$ and so the result
follows.

(2) Consider the following exact sequence ($r=-1,1,2,3,u=0,1,2$)

$Ext_{A}^{s,tq+rq+u}(H^{*}M,H^{*}M)\stackrel{{\scriptstyle(i^{\prime})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+rq+u}(H^{*}K,H^{*}M)$

$\qquad\quad\stackrel{{\scriptstyle(j^{\prime})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+(r-1)q+u-1}(H^{*}M,H^{*}M)\stackrel{{\scriptstyle\alpha_{*}}}{{\longrightarrow}}$
induced by (9.1.2). The left group is zero for $r=-1,1,2,3,u=0,1,2$, this is
obtained from the supposition I(c) on $Ext_{A}^{s,tq+rq+k}(Z_{p},Z_{p})$ = 0
(for $r=-1,1,2,3,k=-1,0,1,2,3$). By the supposition and Prop. 9.3.0(2),the
right group is zero except for $r=1,u=0,1$ it has unique generator
$(ij)_{*}(\tilde{\sigma})$ or $\tilde{\sigma}$ respectively. However, it
satisfies $\alpha_{*}(ij)_{*}(\tilde{\sigma})\neq 0$,
$\alpha_{*}(\tilde{\sigma})\neq 0$ then, the middle group is zero as desired.
Consider the following exact sequence

$0=Ext_{A}^{s,tq+q+1}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq}(H^{*}K,H^{*}K)$

$\qquad\quad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}$
induced by (9.1.2). The left group is zero as shown above. The right group has
unique generator $(i^{\prime})_{*}(\tilde{\sigma})$ , this is because
$(j^{\prime})_{*}Ext_{A}^{s,tq}\\\ (H^{*}K,H^{*}M)\subset
Ext_{A}^{s,tq-q-1}(H^{*}M,H^{*}M)$ = 0 and $Ext_{A}^{s,tq}(H^{*}M,H^{*}M)\cong
Z_{p}\\{\tilde{\sigma}\\}$. Then the middle has unique generatot $\sigma_{K}$
as desired.

(3) Consider the following exact sequence ($r=-1,0$)

$Ext_{A}^{s,tq+(r+1)q}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle(j^{\prime\prime}\wedge
1_{K})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+rq}(H^{*}K,H^{*}L\wedge K)$

$\qquad\quad\stackrel{{\scriptstyle(i^{\prime\prime}\wedge
1_{K})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+rq}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle(\alpha_{1}\wedge
1_{K})^{*}}}{{\longrightarrow}}$
induced by (9.1.3). The left group is zero for $r=0$ , this is because by (2)
$(i^{\prime})^{*}Ext_{A}^{s,tq+q}(H^{*}K,H^{*}K)\subset
Ext_{A}^{s,tq+q}(H^{*}K,H^{*}M)$ = 0 and $Ext_{A}^{s,tq+2q+1}\\\
(H^{*}K,H^{*}M)$ = 0. Moreover, by (2), the left group has unique generator
$\sigma_{K}$ for $r=-1$. The right group is zero $r=-1$, this is because by
(2) $(i^{\prime})^{*}Ext_{A}^{s,tq-q}(H^{*}K,H^{*}K)\subset
Ext_{A}^{s,tq-q}(H^{*}K,H^{*}M)$ = 0 and $Ext_{A}^{s,tq+1}(H^{*}K,\\\ H^{*}M)$
= 0. The right group has unique generator $\sigma_{K}$ for $r=0$ which
satisfies $(\alpha_{1}\wedge 1_{K})^{*}(\sigma_{K})\neq 0\in
Ext_{A}^{s+1,tq+q}(H^{*}K,H^{*}K)$ , this is because
$(i^{\prime})^{*}(\alpha_{1}\wedge 1_{K})^{*}(\sigma_{K})=(\alpha_{1}\wedge
1_{M})^{*}(i^{\prime})^{*}(\sigma_{K})=(\alpha_{1}\wedge
1_{M})^{*}(i^{\prime})_{*}(\tilde{\sigma})=(i^{\prime})_{*}(\alpha_{1}\wedge
1_{M})_{*}(\tilde{\sigma})\neq 0\in Ext_{A}^{s+1,tq+q}(H^{*}K,H^{*}M)$. Then
the middle group is zero for $r=0$ and has unique generator
$(j^{\prime\prime}\wedge 1_{K})^{*}(\sigma_{K})$ for $r=-1$ so that the first
result can be obtained by the following exact sequence

$0=Ext_{A}^{s,tq}(H^{*}K,H^{*}L\wedge
K)\stackrel{{\scriptstyle(i^{\prime\prime}\wedge
1_{K})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq}(H^{*}L\wedge K,H^{*}L\wedge K)$

$\qquad\quad\stackrel{{\scriptstyle(j^{\prime\prime}\wedge
1_{K})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq-q}(H^{*}K,H^{*}L\wedge
K)\stackrel{{\scriptstyle(\alpha_{1}\wedge 1_{K})_{*}}}{{\longrightarrow}}$
induced by (9.1.3). For the second result, look at the following exact
sequence ($r=1,2,3,u=0,1,2$)

$0=Ext_{A}^{s,tq+rq+u}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(i^{\prime\prime}\wedge
1_{K})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+rq+u}(H^{*}L\wedge K,H^{*}M)$

$\qquad\quad\stackrel{{\scriptstyle(j^{\prime\prime}\wedge
1_{K})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+(r-1)q+u}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(\alpha_{1}\wedge
1_{K})_{*}}}{{\longrightarrow}}$
induced by (9.1.3). By (2), the left group is zero for $r=1,2,3,u=0,1,2$ and
the right group also is zero for $r=2,3,u=0,1,2$. By Prop. 9.3.0 and the
supposition , the right group also is zero for $r=1,u=1,2$. For $r=1,u=0$, The
right group has unique generator $(i^{\prime})_{*}(\tilde{\sigma})$ which
satisfies $(\alpha_{1}\wedge 1_{K})_{*}(i^{\prime})_{*}(\tilde{\sigma})\neq
0$. Then the middle group is zero for $r=1,2,3,u=0,1,2$.

(4) Consider the following exact sequence ($r=1,2,3,u=0,1,2$)

$0=Ext_{A}^{s,tq+rq+u}(H^{*}L\wedge K,H^{*}M)\stackrel{{\scriptstyle(w\wedge
1_{K})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+rq+u}(H^{*}W\wedge K,H^{*}M)$

$\qquad\quad\stackrel{{\scriptstyle(j^{\prime\prime}u\wedge
1_{K})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+(r-2)q+u}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(\phi\wedge
1_{K})_{*}}}{{\longrightarrow}}$
induced by (9.1.12). By (3), the left group is zero for $r=1,2,3,u=0,1,2$. By
(2), the right group is zero for $r=1,3,u=0,1,2$ and by Prop. 9.3.0 and the
supposition, it also is zero for $r=2,u=1$. For $r=2,u=0$, the right group has
unique generator $(i^{\prime})_{*}(\tilde{\sigma})$ which satisfies
$(\phi\wedge 1_{K})_{*}(i^{\prime})_{*}(\tilde{\sigma})\neq 0\in
Ext_{A}^{s+1,tq+2q}(H^{*}L\wedge K,H^{*}M)$. Then the middle group is zero for
$r=1,2,3,u=0,1,2$ as desired.

Since $(\widetilde{u}w_{2}\overline{w}\wedge
1_{M})^{*}Ext_{A}^{s,tq+q}(H^{*}W\wedge K,H^{*}X\wedge M)\subset
Ext_{A}^{s,tq+q}(H^{*}W\wedge K,H^{*}M)$ = 0, then, by (9.1.5),
$(\widetilde{u}w_{2}\wedge 1_{M})^{*}Ext_{A}^{s,tq+q}(H^{*}W\wedge
K,H^{*}X\wedge M)=(\overline{u}\wedge
1_{M})^{*}Ext_{A}^{s,tq+2q+1}(H^{*}W\wedge K,H^{*}M\wedge M)$ = 0. and by
using (9.2.12) we know that $Ext_{A}^{s,tq+q}(H^{*}W\wedge K,H^{*}X\wedge
M)=(\widetilde{\psi}\wedge 1_{M})^{*}Ext_{A}^{tq+3q}(H^{*}W\wedge
K,H^{*}M\wedge M)$ = 0. Q.E.D.

The proof of the main Theorem C will be done by some argument processing in
the Adams resolution (cf. 9.2.9) of some spectra related to the sphere
spectrum $S$. Before proving the main Theorem C , we first prove the following
Lemmas.

Lemma 9.4.12 Under the supposition (I)(II) of the main Theorem C we have

(1) Let $\widetilde{h_{0}\sigma}\in[\Sigma^{tq+q+1}M,KG_{s+1}\wedge M]$ be a
$d_{1}$-cycle which represents $\alpha_{*}(\tilde{\sigma})\in
Ext_{A}^{s+1,tq+q+1}(H^{*}M,H^{*}M)$, then $(\bar{c}_{s+1}\wedge
1_{M})\widetilde{h_{0}\sigma}=(1_{E_{s+2}}\wedge\alpha)(\kappa\wedge 1_{M})$
(up to scalar), where $\kappa\in\pi_{tq+1}E_{s+2}$ such that
$\bar{a}_{s+1}\cdot\kappa=\bar{c}_{s}\cdot\sigma$ and
$\sigma\in\pi_{tq}KG_{s}\cong Ext_{A}^{s,tq}(Z_{p},Z_{p})$.

(2) $(1_{E_{s+2}}\wedge\phi\wedge 1_{M})(\kappa\wedge 1_{M})$ = 0,
$(1_{E_{s+2}}\wedge\alpha_{1}\wedge 1_{M})(\kappa\wedge 1_{M})$ = 0.

Proof: (1) Since $(1_{KG_{s+1}}\wedge i^{\prime})\widetilde{h_{0}\sigma}$ is a
$d_{1}$-boundary, then $(\bar{c}_{s+1}\wedge 1_{K})\\\ (1_{KG_{s+1}}\wedge
i^{\prime})(\widetilde{h_{0}\sigma})$ = 0 so that $(\bar{c}_{s+1}\wedge
1_{M})\widetilde{h_{0}\sigma}=(1_{E_{s+2}}\wedge\alpha)f^{\prime}$ for some
$f^{\prime}\in[\Sigma^{tq+1}M,E_{s+2}\wedge M]$. It follows that
$(\bar{a}_{s+1}\wedge 1_{M})(1_{E_{s+2}}\wedge\alpha)f^{\prime}$ = 0 and so
$(\bar{a}_{s+1}\wedge 1_{M})f^{\prime}=(1_{E_{s+1}}\wedge
j^{\prime})f^{\prime}_{2}$ with
$f^{\prime}_{2}\in[\Sigma^{tq+q+1}M,E_{s+1}\wedge K]$. The $d_{1}$-cycle
$(\bar{b}_{s+1}\wedge 1_{K})f^{\prime}_{2}$ represents an element in
$Ext_{A}^{s+1,tq+q+1}(H^{*}K,H^{*}M)$ and this group is zero by Prop.
9.3.2(2), then $(\bar{b}_{s+1}\wedge
1_{K})f^{\prime}_{2}=(\bar{b}_{s+1}\bar{c}_{s}\wedge 1_{K})g^{\prime}_{0}$ for
some $g^{\prime}_{0}\in[\Sigma^{tq+q+1}M,KG_{s}\wedge K]$. Consequently we
have, $f^{\prime}_{2}=(\bar{c}_{s}\wedge
1_{K})g^{\prime}_{0}+(\bar{a}_{s+1}\wedge 1_{K})f^{\prime}_{3}$ for some
$f^{\prime}_{3}\in[\Sigma^{tq+q+2}M,E_{s+2}\wedge K]$ and so
$(\bar{a}_{s+1}\wedge 1_{M})f^{\prime}=(\bar{a}_{s+1}\wedge
1_{M})(1_{E_{s+2}}\wedge j^{\prime})f^{\prime}_{3}+(\bar{c}_{s}\wedge
1_{M})(1_{KG_{s}}\wedge j^{\prime})g^{\prime}_{0}=(\bar{a}_{2}\wedge
1_{M})(1_{E_{s+2}}\wedge j^{\prime})f^{\prime}_{3}+(\bar{c}_{s}\wedge
1_{M})(\sigma\wedge 1_{M})=(\bar{a}_{s+1}\wedge 1_{M})(1_{E_{s+2}}\wedge
j^{\prime})f^{\prime}_{3}+(\bar{a}_{s+1}\wedge 1_{M})(\kappa\wedge 1_{M})$,
where the $d_{1}$-cycle $(1_{KG_{s}}\wedge
j^{\prime})g^{\prime}_{0}\in[\Sigma^{tq}M,KG_{s}\wedge M]$ represents an
element in $Ext_{A}^{s,tq}(H^{*}M,H^{*}M)$ and this group has unique generator
$\tilde{\sigma}$ so that it equals to $\sigma\wedge 1_{M}$ (mod
$d_{1}$-boundary). Hence we have $f^{\prime}=(1_{E_{s+1}}\wedge
j^{\prime})f^{\prime}_{3}+(\kappa\wedge 1_{M})+(\bar{c}_{s+1}\wedge
1_{M})\tilde{g}_{1}$ for some $\tilde{g}_{1}\in[\Sigma^{tq+1}M,KG_{s+1}\wedge
M]$ and so $(\bar{c}_{s+1}\wedge
1_{M})\widetilde{h_{0}\sigma}=(1_{E_{s+2}}\wedge\alpha)f^{\prime}=(1_{E_{s+1}}\wedge\alpha)(\kappa\wedge
1_{M})$ which shows the result.

(2) Since $Ext_{A}^{s+1,tq+rq}(Z_{p},Z_{p})$ is zero for $r=2$ and has unique
generator $h_{0}\sigma=(j^{\prime\prime})_{*}(\phi)_{*}(\sigma)$ for$r=1$,
then $Ext_{A}^{s+1,tq+2q}(H^{*}L,Z_{p})\cong Z_{p}\\{(\phi)_{*}(\sigma)\\}$
and $Ext_{A}^{s+1,tq+2q}(H^{*}W,Z_{p})$ = 0. By this and a similar proof as
given in (1) we know that $(1_{E_{s+2}}\wedge\phi)\kappa=(\bar{c}_{s+1}\wedge
1_{L})\sigma\phi$ (up to scalar), where
$\sigma\phi\in\pi_{tq+2q}(KG_{s+1}\wedge L)$ is a $d_{1}$-cycle which
represents $(\phi)_{*}(\sigma)\in Ext_{A}^{s+1,tq+2q}(H^{*}L,\\\ Z_{p})$.
Then, by the supposition (II) of the main Theorem C we have
$(1_{E_{s+2}}\wedge\phi\wedge 1_{M})(\kappa\wedge 1_{M})=(\bar{c}_{s+1}\wedge
1_{L\wedge M})(\sigma\phi\wedge 1_{M})$ = 0 so that the result follows. Q.E.D.

Lemma 9.4.13 Under the supposition (I) of the main Theorem C we have

(1) $Ext_{A}^{s,tq}(H^{*}X\wedge M,H^{*}X\wedge M)\cong Z_{p}\\{[\sigma\wedge
1_{X\wedge M}]\\}$.

(2) For any $d_{1}$-cycle $g_{0}\in[\Sigma^{tq+q}X,KG_{s+1}\wedge X]$,
$g_{0}=\lambda^{\prime}(h_{0}\sigma\wedge 1_{X})$ (mod $d_{1}$-boundary) with
$\lambda^{\prime}\in Z_{p}$ and $(\psi_{X\wedge M})_{*}[h_{0}\sigma\wedge
1_{X\wedge M}]\neq 0\in Ext_{A}^{s+1,tq+1}(H^{*}Y\wedge M,H^{*}X\wedge M)$.

Proof (1) Consider the following exact sequence

$Ext_{A}^{s,tq+2q}(H^{*}L\wedge K,H^{*}M)\stackrel{{\scriptstyle
m_{M}(\tilde{\psi}\wedge 1_{M})^{*}}}{{\to}}\longrightarrow
Ext_{A}^{s,tq}(H^{*}L\wedge K,H^{*}X\wedge M)$

$\qquad\quad\stackrel{{\scriptstyle(u^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq}(H^{*}L\wedge
K,H^{*}L\wedge K)\stackrel{{\scriptstyle((1_{L}\wedge i^{\prime})(\phi\wedge
1_{M}))^{*}}}{{\longrightarrow}}$
induced by (9.4.2). By Prop. 9.4.11(3), the left group is zero and the right
group has unique generator $\sigma_{L\wedge K}$ which satisfies $((1_{L}\wedge
i^{\prime})(\phi\wedge 1_{M}))^{*}(\sigma_{L\wedge K})\neq 0\in
Ext_{A}^{s+1,tq+2q}(H^{*}L\wedge K,H^{*}M)$, this is because
$(j^{\prime\prime}\wedge 1_{K})_{*}((1_{L}\wedge i^{\prime})(\phi\wedge
1_{M}))^{*}(\sigma_{L\wedge K})=((1_{L}\wedge i^{\prime})(\phi\wedge
1_{M}))^{*}(j^{\prime\prime}\wedge 1_{K})_{*}(\sigma_{L\wedge
K})=((1_{L}\wedge i^{\prime})(\phi\wedge 1_{M})^{*}(j^{\prime\prime}\wedge
1_{K})^{*}(\sigma_{K})=((\alpha_{1}\wedge
1_{K})i^{\prime})^{*}(\sigma_{K})=(\alpha_{1}\wedge
1_{M})^{*}(i^{\prime})_{*}(\tilde{\sigma})=(i^{\prime}(\alpha_{1}\wedge
1_{M}))_{*}(\tilde{\sigma})\neq 0\in Ext_{A}^{s+1,tq+q}(H^{*}K,H^{*}M)$. Then
the middle group is zero. Look at the following exact sequence

$Ext_{A}^{s,tq}(H^{*}L\wedge K,H^{*}X\wedge
M)\stackrel{{\scriptstyle(u^{\prime})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq}(H^{*}X\wedge
M,H^{*}X\wedge M)$

$\qquad\quad\stackrel{{\scriptstyle m_{M}(\tilde{\psi}\wedge
1_{M})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq-2q}(H^{*}M,H^{*}X\wedge
M)\stackrel{{\scriptstyle((1_{L}\wedge i^{\prime})(\phi\wedge
1_{M}))_{*}}}{{\longrightarrow}}$
induced by (9.4.2). As shown above, the left group is zero. By Prop.
9.4.11(1), the right group has unique generator $m_{M}(\widetilde{\psi}\wedge
1_{M})^{*}(\tilde{\sigma})\\\ =m_{M}(\widetilde{\psi}\wedge
1_{M})^{*}[\sigma\wedge 1_{M}]=[(\sigma\wedge
1_{M})m_{M}(\widetilde{\psi}\wedge 1_{M})]=[(1_{KG_{s}}\wedge
m_{M}(\widetilde{\psi}\wedge 1_{M}))(\sigma\wedge 1_{X\wedge
M})]=m_{M}(\widetilde{\psi}\wedge 1_{M})_{*}[\sigma\wedge 1_{X\wedge M}]$ and
it satisfies $((1_{L}\wedge i^{\prime})(\phi\wedge
1_{M}))_{*}m_{M}(\widetilde{\psi}\wedge
1_{M})^{*}(\tilde{\sigma})=((1_{L}\wedge i^{\prime})(\phi\wedge
1_{M}))_{*}m_{M}(\widetilde{\psi}\wedge 1_{M})_{*}[\sigma\wedge 1_{X\wedge
M}]$ = 0. Then the middle group has unique generator $[\sigma\wedge 1_{X\wedge
M}]$ as desired.

(2) Note that
$(\widetilde{\psi})_{*}(\widetilde{u}w_{2})^{*}Ext_{A}^{s+1,tq+q}(H^{*}X,H^{*}X)\subset
Ext_{A}^{s+1,tq-q-1}(H^{*}M,\\\ H^{*}Y)$. Similar to that in Prop. 9.3.0(1),
by the supposition we know that $Ext_{A}^{s+1,tq}(H^{*}M,H^{*}M)$ is zero or
has (one or two ) generator $\tilde{\sigma}^{\prime}$, then
$Ext_{A}^{s+1,tq-q-1}(H^{*}M,H^{*}Y)$ is zero or has (one or two) generator
$(\overline{u})^{*}(\tilde{\sigma}^{\prime})$ and it satisfies $((1_{Y}\wedge
j)\alpha_{Y\wedge
M})_{*}(\overline{u})^{*}(\tilde{\sigma}^{\prime})=((1_{Y}\wedge
j)\alpha_{Y\wedge M})_{*}(\overline{u})_{*}[\sigma^{\prime}\wedge
1_{Y}]=(1_{Y}\wedge\alpha_{1})_{*}[\sigma^{\prime}\wedge
1_{Y}]=[h_{0}\sigma^{\prime}\wedge 1_{Y}]\neq 0$, then
$(\widetilde{\psi})_{*}(\widetilde{u}w_{2})^{*}Ext_{A}^{s+1,tq+q}(H^{*}X,H^{*}X)$
= 0 and so we have
$(\widetilde{u}w_{2})^{*}Ext_{A}^{s+1,tq+q}(H^{*}X,H^{*}X)=(\widetilde{u}w_{2})_{*}Ext_{A}^{s+1,tq+q}(H^{*}Y,\\\
H^{*}Y)$ = 0 , this is because $Ext_{A}^{s+1,tq+q}(H^{*}Y,H^{*}Y)\cong
Z_{p}\\{((1_{Y}\wedge j)\alpha_{Y\wedge M})_{*}(\overline{u})^{*}\\\
(\tilde{\sigma})\\}$ ( cf. Prop. 9.4.10(2)). Then
$Ext_{A}^{s+1,tq+q}(H^{*}X,H^{*}X)=(\widetilde{\psi})^{*}Ext_{A}^{s+1,tq+3q}\\\
(H^{*}X,H^{*}M)$ and it has unique generator
$(\widetilde{\psi})^{*}((1_{X}\wedge j)\alpha_{X\wedge
M})_{*}(\tilde{\sigma})=((1_{X}\wedge j)\alpha_{X\wedge M})_{*}[(\sigma\wedge
1_{M})\widetilde{\psi}]=((1_{X}\wedge j)\alpha_{X\wedge
M})_{*}[(1_{KG_{s+1}}\wedge\widetilde{\psi})(\sigma\wedge
1_{X})]=((1_{X}\wedge j)\alpha_{X\wedge M})_{*}m_{M}(\widetilde{\psi}\wedge
1_{M})_{*}(1_{X}\wedge i)_{*}[\sigma\wedge 1_{X}]=(1_{X}\wedge j\alpha
i)_{*}[\sigma\wedge 1_{X}]=[h_{0}\sigma\wedge 1_{X}]$ (cf. Prop. 9.4.10(3))
Then the first result follows. For the second result , by (9.4.6), the
$d_{1}$-cycle $(1_{KG_{s+1}}\wedge m_{M}(\overline{u}\wedge
1_{M})\psi_{X\wedge M})(h_{0}\sigma\wedge 1_{X\wedge M})=(1_{KG_{s+1}}\wedge
m_{M}(\widetilde{\psi}\wedge 1_{M}))(h_{0}\sigma\wedge 1_{X\wedge
M})=(h_{0}\sigma\wedge 1_{M})m_{M}(\widetilde{\psi}\wedge 1_{M})$ and it
represents an element $m_{M}(\widetilde{\psi}\wedge
1_{M})^{*}[h_{0}\sigma\wedge 1_{M}]=m_{M}(\widetilde{\psi}\wedge
1_{M})^{*}(\alpha_{1}\wedge 1_{M})_{*}(\tilde{\sigma})\neq 0$ so that the
second result follows. Q.E.D.

Proof the main Theorem C By Lemma 9.4.12(1), it suffices to prove
$(\bar{c}_{s+1}\wedge
1_{M})\widetilde{h_{0}\sigma}=(1_{E_{s+1}}\wedge\alpha)(\kappa\wedge
1_{M})=0$. The proof is divided into the following two steps.

Step 1 To prove $(\kappa\wedge 1_{X\wedge M})(1_{X}\wedge\alpha)$ = 0.

By (9.4.3), $(\phi\wedge 1_{M})m_{M}(\widetilde{\psi}\wedge
1_{M})=(u^{\prime\prime}\wedge 1_{M})(1_{X}\wedge\alpha)$, then by Lemma
9.4.12(2) we have $(1_{E_{s+2}}\wedge u^{\prime\prime}\wedge
1_{M})(1_{E_{s+2}}\wedge 1_{X}\wedge\alpha)(\kappa\wedge 1_{X\wedge
M})=(1_{E_{s+2}}\wedge\phi\wedge 1_{M})(\kappa\wedge
1_{M})m_{M}(\widetilde{\psi}\wedge 1_{M})$ = 0. Moreover, by (9.2.16) we have
$(1_{E_{s+2}}\wedge 1_{X}\wedge\alpha)(\kappa\wedge 1_{X\wedge
M})=(1_{E_{s+2}}\wedge\widetilde{u}w_{3}\wedge 1_{M})f$ for some
$f\in[\Sigma^{tq+q+1}X\wedge M,E_{s+2}\wedge W\wedge M]\cap(kerd)$ (cf. Cor.
6.4.15). By composing $(1_{E_{s+2}}\wedge 1_{X}\wedge i^{\prime}i\wedge
1_{M})$ on the above equation we have
$(1_{E_{s+2}}\wedge\widetilde{u}w_{3}\wedge 1_{K\wedge M})(1_{E_{s+2}}\wedge
1_{W}\wedge i^{\prime}i\wedge
1_{M})f=(1_{E_{s+2}}\wedge(1_{X}\wedge(i^{\prime}i\wedge
1_{M})\alpha)(\kappa\wedge 1_{X\wedge M})=(1_{E_{s+2}}\wedge
1_{X}\wedge\overline{m}_{K}i^{\prime}(\alpha_{1}\wedge 1_{M})))(\kappa\wedge
1_{X\wedge M})$ = 0 ,where we use the result on
$(1_{E_{s+2}}\wedge\alpha_{1}\wedge 1_{M})(\kappa\wedge 1_{M})$ = 0 in Lemma
9.4.12(2). Consequently, by (9.2.16), $(1_{E_{s+2}}\wedge 1_{W}\wedge
i^{\prime}i\wedge 1_{M})f=(1_{E_{s+2}}\wedge w^{\prime}(\pi\wedge 1_{L})\wedge
1_{K\wedge M})f_{2}$ = 0 (for some $f_{2}\in[\Sigma^{tq+1}X\wedge
M,E_{s+2}\wedge L\wedge K\wedge M]$), this is because $\pi\wedge 1_{K}$ = 0.
Then by (9.1.4) we have $f=(1_{E_{s+2}}\wedge 1_{W}\wedge\epsilon\wedge
1_{M})f_{3}=(1_{E_{s+2}}\wedge 1_{W}\wedge\alpha m_{M}(\overline{u}\wedge
1_{M})f_{3}$ for some $f_{3}\in[\Sigma^{tq+q+2}X\wedge M,E_{s+2}\wedge W\wedge
Y\wedge M]\cap(kerd)$ ( cf. Cor. 6.4.15) and so
(9.4.14) $(1_{E_{s+2}}\wedge 1_{X}\wedge\alpha)(\kappa\wedge 1_{X\wedge M})$

$\quad=(1_{E_{s+2}}\wedge\widetilde{u}w_{3}\wedge 1_{M})(1_{E_{s+2}}\wedge
1_{W}\wedge\alpha m_{M}(1_{M}\wedge\overline{u}))f_{3}$

$\quad=(1_{E_{s+2}}\wedge\alpha_{X\wedge M}(j^{\prime\prime}u\wedge
1_{M}))(1_{E_{s+2}}\wedge 1_{W}\wedge m_{M}(1_{M}\wedge\overline{u}))f_{3}$ (
cf. (9.4.3))

By (9.4.14), $(\bar{a}_{s+1}\wedge 1_{X\wedge
M})(1_{E_{s+2}}\wedge\widetilde{u}w_{3}\wedge
1_{M})(1_{E_{s+2}}\wedge(1_{W}\wedge\alpha m_{M}(\overline{u}\wedge
1_{M}))f_{3}$ = $(\bar{a}_{s+1}\wedge 1_{X\wedge M})(1_{E_{s+2}}\wedge
1_{X}\wedge\alpha)(\kappa\wedge 1_{X\wedge M})=(\bar{c}_{s}\wedge 1_{X\wedge
M})(1_{KG_{s}}\wedge 1_{X}\wedge\alpha)(\sigma\wedge 1_{X\wedge M})$ = 0 ,this
is because $\alpha$ induces zero homomorphism in $Z_{p}$-cohomology. Then, by
(9.2.16) and $w^{\prime}(\pi\wedge 1_{L})\wedge 1_{M}=(w\wedge
1_{M})(1_{L}\wedge\alpha)$ we have
(9.4.15) $(\bar{a}_{s+1}\wedge 1_{W\wedge M})(1_{E_{s+2}}\wedge
1_{W}\wedge\alpha m_{M}(\overline{u}\wedge 1_{M}))f_{3}$

$=(1_{E_{s+1}}\wedge(1_{W}\wedge\alpha)(w\wedge 1_{M}))f_{5}$
with $f_{5}\in[\Sigma^{tq}X\wedge M,E_{s+1}\wedge L\wedge M]\cap(kerd)$ ( cf.
Cor. 6.4.15 ).

By (9.4.15)(9.1.2) , $(\bar{a}_{s+1}\wedge 1_{W\wedge M})(1_{E_{s+2}}\wedge
1_{W}\wedge m_{M}(\overline{u}\wedge 1_{M}))f_{3}=(1_{E_{s+1}}\wedge w\wedge
1_{M})f_{5}+(1_{E_{s+1}}\wedge 1_{W}\wedge j^{\prime})f_{6}$ for some
$f_{6}\in[\Sigma^{tq+q+1}X\wedge M,E_{s+1}\wedge W\wedge K]\cap(kerd)$ ( cf.
Prop. 6.5.26). Since
$(1_{W}\wedge\alpha_{1})w=w(1_{L}\wedge\alpha_{1})=w\cdot\phi
j^{\prime\prime}$ = 0, then $w=(1_{W}\wedge j^{\prime\prime})\psi_{W}$ , where
$\psi_{W}\in[\Sigma^{q}L,W\wedge L]$. So we have $w\wedge 1_{M}=(1_{W}\wedge
j^{\prime\prime})\psi_{W}\wedge 1_{M}=(1_{W}\wedge m_{M}(\overline{u}\wedge
1_{M}))((1_{W}\wedge\bar{h})\psi_{W}\wedge 1_{M})$. Hence,
$-(\bar{a}_{s+1}\wedge 1_{W\wedge Y\wedge
M})f_{3}=(1_{E_{s+2}}\wedge(1_{W}\wedge h)\psi_{W}\wedge
1_{M})f_{5}+(1_{E_{s+1}}\wedge 1_{W}\wedge(1_{Y}\wedge
i)r)f_{6}+(1_{E_{s+1}}\wedge 1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})f_{7}$
and by Prop. 6.5.26, $f_{7}=f_{8}(1_{X}\wedge i^{\prime})+f_{9}(1_{X}\wedge
i^{\prime}ij)$, where $f_{8}\in[\Sigma^{tq+q}X\wedge K,E_{s+1}\wedge W\wedge
K]\cap(kerd)$ and $f_{9}\in[\Sigma^{tq+q+1}X\wedge K,E_{s+1}\wedge W\wedge
K]\cap(kerd)$. Since $d((1_{Y}\wedge i)r)=((r\wedge 1_{M})d(1_{K}\wedge
i)=(r\wedge 1_{M})(1_{K}\wedge m_{M})(T_{K,M}\wedge 1_{M})(1_{M}\wedge
1_{K}\wedge i)\overline{m}_{K}=(r\wedge 1_{M})(1_{K}\wedge m_{M}(1_{M}\wedge
i)\overline{m}_{K}=(r\wedge 1_{M})\overline{m}_{K}$, by applying the
derivation $d$ using Theorem 6.4.8(1) we have $-(1_{E_{s+1}}\wedge
1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})f_{6}-(1_{E_{s+1}}\wedge
1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})f_{9}(1_{X}\wedge i^{\prime})$ = 0
(Note : $f_{6}$ has odd degree) and so
(9.4.16) $-(\bar{a}_{s+1}\wedge 1_{W\wedge Y\wedge
M})f_{3}=(1_{E_{s+1}}\wedge(1_{W}\wedge\bar{h})\psi_{W}\wedge 1_{M})f_{5}$

$\quad+(1_{E_{s+1}}\wedge 1_{W}\wedge(1_{Y}\wedge i)r)f_{6}+(1_{E_{s+1}}\wedge
1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})f_{8}(1_{X}\wedge i^{\prime})$

$\quad-(1_{E_{s+1}}\wedge 1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})f_{6}(1_{X}\wedge ij)$

Note that the $d_{1}$-cycle $(\bar{b}_{s+1}\wedge 1_{W\wedge
K})f_{6}\in[\Sigma^{tq+q+1}X\wedge M,KG_{s+1}\wedge W\wedge K]\cap(kerd)$
represents an element in $Ext_{A}^{s+1,tq+q+1}(H^{*}W\wedge K,H^{*}X\wedge M)$
and by Prop. 9.4.10(1) this group is zero, then $(\bar{b}_{s+1}\wedge
1_{W\wedge K})f_{6}=(\bar{b}_{s+1}\bar{c}_{s}\wedge 1_{W\wedge K})g$ for some
$g\in[\Sigma^{tq+q+1}X\wedge M,KG_{s}\wedge W\wedge K]\cap(kerd)$ (cf. Prop.
6.5.26) and so $f_{6}=(\bar{c}_{s}\wedge 1_{W\wedge K})g+(\bar{a}_{s+1}\wedge
1_{W\wedge K})f^{\prime}$ with $f^{\prime}\in[\Sigma^{tq+q+2}X\wedge
M,E_{s+2}\wedge W\wedge K]\cap(kerd)$ ( cf. Prop. 6.5.26). Then we have
(9.4.17) $-(\bar{a}_{s+1}\wedge 1_{W\wedge Y\wedge
M})f_{3}=(1_{E_{s+1}}\wedge(1_{W}\wedge\bar{h})\psi_{W}\wedge 1_{M})f_{5}$

$\qquad+(\bar{a}_{s+1}\wedge 1_{W\wedge Y\wedge M})(1_{E_{s+2}}\wedge
1_{W}\wedge(1_{Y}\wedge i)r)f^{\prime}$

$\qquad+(\bar{c}_{s}\wedge 1_{W\wedge Y\wedge M})(1_{KG_{s}}\wedge
1_{W}\wedge(1_{Y}\wedge i)r)g$

$\qquad+(\bar{a}_{s+1}\wedge 1_{W\wedge Y\wedge M})(1_{E_{s+2}}\wedge
1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})f^{\prime}(1_{X}\wedge ij)$

$\qquad-(\bar{c}_{s}\wedge 1_{W\wedge Y\wedge M})(1_{KG_{s}}\wedge
1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})g(1_{X}\wedge ij)$

$\qquad+(1_{E_{s+1}}\wedge 1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})f_{8}(1_{X}\wedge i^{\prime})$

Let $P$ be the cofibre of
$(1_{W}\wedge\bar{h})\psi_{W}:\Sigma^{q+1}L\rightarrow W\wedge Y$ given by the
cofibration
(9.4.18)
$\Sigma^{q+1}L\stackrel{{\scriptstyle(1_{W}\wedge\bar{h})\psi_{W}}}{{\longrightarrow}}W\wedge
Y\stackrel{{\scriptstyle w_{5}}}{{\longrightarrow}}P\stackrel{{\scriptstyle
u_{5}}}{{\longrightarrow}}\Sigma^{q+2}L$
Then the cofibre of $w_{5}(1_{W}\wedge r):W\wedge K\rightarrow P$ is $\Sigma
X$ given by the cofibration
(9.4.19) $W\wedge K\stackrel{{\scriptstyle w_{5}(1_{W}\wedge
r)}}{{\longrightarrow}}V\stackrel{{\scriptstyle
w_{6}}}{{\longrightarrow}}\Sigma X\stackrel{{\scriptstyle
u_{6}}}{{\longrightarrow}}\Sigma W\wedge K$
This can be seen by the following homotopy commutative diagram of $3\times
3$-Lemma

$\qquad\qquad W\wedge K\quad\stackrel{{\scriptstyle w_{5}(1_{W}\wedge
r)}}{{\longrightarrow}}\quad P\quad\stackrel{{\scriptstyle
u_{5}}}{{\longrightarrow}}\quad\Sigma^{q+2}L$

$\qquad\qquad\quad\searrow 1_{W}\wedge r\quad\nearrow w_{5}\quad\searrow
w_{6}\quad\nearrow u^{\prime\prime}$

$\qquad\qquad\qquad\quad W\wedge Y\qquad\qquad\qquad\Sigma X$

$\qquad\qquad\quad\nearrow^{(1_{W}\wedge\bar{h})\psi_{W}}\searrow
1_{W}\wedge\epsilon\nearrow\widetilde{u}w_{3}\searrow u_{6}$

$\qquad\qquad\Sigma^{q+1}L\stackrel{{\scriptstyle w^{\prime}(\pi\wedge
1_{L})}}{{\longrightarrow}}\quad\Sigma W\quad\stackrel{{\scriptstyle
1_{W}\wedge i^{\prime}i}}{{\longrightarrow}}\quad\Sigma W\wedge K$

Note that $u_{6}=\mu_{X\wedge M}(1_{X}\wedge i)$, then by composing
$(\bar{b}_{s+1}\wedge 1_{P})(1_{E_{s+1}}\wedge w_{5}\wedge j)$ on the left
hand side of (9.4.17) and composing $(1_{X}\wedge i)$ on the right hand side
we have $(\bar{b}_{s+1}\wedge 1_{P})(1_{E_{s+1}}\wedge w_{5}(1_{W}\wedge
r))f_{8}(1_{X}\wedge i^{\prime}i)$ = 0 and so $(\bar{b}_{s+1}\wedge 1_{W\wedge
K})f_{8}(1_{X}\wedge i^{\prime}i)=(1_{KG_{s+1}}\wedge
u_{6})g_{0}=(1_{KG_{s+1}}\wedge\mu_{X\wedge M}(1_{X}\wedge
i))g_{0}=(1_{KG_{s+1}}\wedge\mu_{X\wedge M})(g_{0}\wedge 1_{M})(1_{X}\wedge
i)$ with $d_{1}$-cycle $g_{0}\in[\Sigma^{tq+q}X,KG_{s+1}\wedge X]$. Moreover ,
by Lemma 9.4.13(2), $g_{0}=\lambda_{1}(h_{0}\sigma\wedge 1_{X})$ (mod
$d_{1}$-boundary), where $\lambda_{1}\in Z_{p}$. On the other hand, by
applying the derivation $d$ to $(\bar{b}_{s+1}\wedge 1_{W\wedge
K})f_{8}(1_{X}\wedge i^{\prime}ij)=(1_{KG_{s+1}}\wedge\mu_{X\wedge
M})(g_{0}\wedge 1_{M})(1_{X}\wedge ij)$ we have

(9.4.20) $(\bar{b}_{s+1}\wedge 1_{W\wedge K})f_{8}(1_{X}\wedge
i^{\prime})==(1_{KG_{s+1}}\wedge\mu_{X\wedge M})(g_{0}\wedge 1_{M})$ ,

$\qquad\qquad g_{0}=\lambda_{1}(h_{0}\sigma\wedge
1_{X})\in[\Sigma^{tq+q}X,KG_{s+1}\wedge X]$ (mod $d_{1}$-boundary)

Consider the following commutative diagram of exact sequences

$\Sigma^{q+1}L\wedge M\quad\stackrel{{\scriptstyle w\wedge
1_{M}}}{{\longrightarrow}}\quad\Sigma^{q+1}W\wedge
M\quad\stackrel{{\scriptstyle j^{\prime\prime}u\wedge
1_{M}}}{{\longrightarrow}}\quad\Sigma^{3q+1}M\stackrel{{\scriptstyle\phi\wedge
1_{M}}}{{\longrightarrow}}\Sigma^{q+2}L\wedge M$

$\qquad\big{\uparrow}1_{L\wedge
M}\qquad\qquad\quad\qquad\big{\uparrow}1_{W}\wedge m_{M}(\overline{u}\wedge
1_{M})\qquad\big{\uparrow}u_{7}\qquad\qquad\big{\uparrow}1_{L\wedge M}$

$\Sigma^{q+1}L\wedge
M\stackrel{{\scriptstyle(1_{W}\wedge\bar{h})\psi_{W}\wedge
1_{M}}}{{\longrightarrow}}W\wedge Y\wedge M\quad\stackrel{{\scriptstyle
w_{5}\wedge 1_{M}}}{{\longrightarrow}}\quad P\wedge M\stackrel{{\scriptstyle
u_{5}\wedge 1_{M}}}{{\longrightarrow}}\Sigma^{q+2}L\wedge M$
of the cofibrations (9.1.12)(9.4.18). Since the left rectangle homotopy
commutes then there exists $u_{7}\in[\Sigma^{-3q-1}P\wedge M,M]$ such that all
the above rectangle homotopy commute. That is we have
(9.4.21) $u_{7}(w_{5}\wedge 1_{M})=(j^{\prime\prime}u\wedge 1_{M})(1_{W}\wedge
m_{M}(\overline{u}\wedge 1_{M})),\quad$

$\quad\qquad(\phi\wedge 1_{M})u_{7}=\pm\quad u_{5}\wedge 1_{M}$
where $u_{7}\in[\Sigma^{-3q-1}P\wedge M,M]$. By the above two equations , we
have the following homotopy commutative diagram of $3\times 3$-Lemma in which
we use the cofibrations (9.2.12)(9.4.18)(9.1.23)

$\qquad P\wedge M\quad\stackrel{{\scriptstyle u_{5}\wedge
1_{M}}}{{\longrightarrow}}\quad\Sigma^{q+2}L\wedge
M\quad\stackrel{{\scriptstyle w\wedge
1_{M}}}{{\longrightarrow}}\quad\Sigma^{q+2}W\wedge M$

$\qquad\qquad\searrow u_{7}\quad\nearrow\phi\wedge
1_{M}\quad\searrow^{((1_{W}\wedge\bar{h})\psi_{W}\wedge
1_{M}}\nearrow_{1_{W}\wedge m_{M}(\overline{u}\wedge
1_{M})}\searrow^{j^{\prime\prime}u\wedge 1_{M}}$
(9.4.22)$\qquad\qquad\Sigma^{3q+1}M\quad\qquad\qquad\qquad\Sigma W\wedge
Y\wedge M\qquad\qquad\quad\Sigma^{3q+2}M$

$\qquad\qquad\nearrow j^{\prime\prime}u\wedge 1_{M}\searrow^{(\phi_{W}\wedge
1_{K})i^{\prime}}\nearrow_{1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K}}\searrow
w_{5}\wedge 1_{M}\quad\nearrow u_{7}$

$\quad\Sigma^{q+1}W\wedge
M\stackrel{{\scriptstyle\tilde{\lambda}(1_{W}\wedge\alpha^{\prime}i^{\prime})}}{{\longrightarrow}}\Sigma^{2}W\wedge
K\qquad\quad\longrightarrow\qquad\quad\Sigma P\wedge M$
Then there is a cofibration
(9.4.23) $\Sigma^{3q-1}M\stackrel{{\scriptstyle(\phi_{W}\wedge
1_{K})i^{\prime}}}{{\longrightarrow}}W\wedge
K\stackrel{{\scriptstyle(w_{5}\wedge 1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})}}{{\longrightarrow}}$

$\qquad\qquad\Sigma^{-1}P\wedge M\stackrel{{\scriptstyle
u_{7}}}{{\longrightarrow}}\Sigma^{3q}M$
in which $\phi_{W}\in[\Sigma^{3q-1}S,W]$ such that
$u\cdot\phi_{W}=\phi\in[\Sigma^{2q-1}S,L]$. Since $(\phi\wedge
1_{K})i^{\prime}\cdot u_{7}=(u\cdot\phi_{W}\wedge 1_{K})i^{\prime}\cdot u_{7}$
= 0, then by (9.4.2) we have
(9.4.24) $u_{7}=m_{M}(\widetilde{\psi}\wedge 1_{M})u_{8}$
where $u_{8}\in[\Sigma^{-q-1}P\wedge M,X\wedge M]$. On the other hand, by
(9.4.8), $(\omega\wedge 1_{M})u_{8}(w_{5}(1_{W}\wedge r)\wedge
1_{M})=\alpha_{Y\wedge M}m_{M}(\widetilde{\psi}\wedge
1_{M})u_{8}(w_{5}(1_{W}\wedge r)\wedge 1_{M})=\alpha_{Y\wedge M}u_{7}\\\
(w_{5}(1_{W}\wedge r)\wedge 1_{M})=\alpha_{Y\wedge
M}j^{\prime}(j^{\prime\prime}u\wedge 1_{K})(1_{W}\wedge m_{K})$ = 0 (cf.
(9.1.27)). Then, by (9.4.7), $u_{8}(w_{5}(1_{W}\wedge r)\wedge
1_{M})=((1_{X}\wedge j)u^{\prime}\wedge 1_{M})\Delta_{1}$ with
$\Delta_{1}\in[\Sigma^{-q}W\wedge K\wedge M,L\wedge K\wedge M]\cap(kerd)$. By
composing $\mu_{X\wedge M}(1_{X}\wedge i)\wedge 1_{M}$ on the above equation
and using (9.4.19) we have $((1_{X}\wedge j)u^{\prime}\wedge
1_{M})\Delta_{1}(\mu_{X\wedge M}(1_{X}\wedge i)\wedge 1_{M})$ = 0 and so by
(9.4.7)(9.4.6), $\Delta_{1}(\mu_{X\wedge M}(1_{X}\wedge i)\wedge
1_{M})=(\overline{\mu}_{2}(1_{Y}\wedge i^{\prime}i)\wedge 1_{M})\psi_{X\wedge
M}$. Then $(j^{\prime\prime}\wedge 1_{K\wedge M})\Delta_{1}(\mu_{X\wedge
M}(1_{X}\wedge i)\wedge 1_{M})=((j^{\prime\prime}\wedge
1_{K})\overline{\mu}_{2}(1_{Y}\wedge i^{\prime}i)\wedge 1_{M})\psi_{X\wedge
M}=(i^{\prime}\overline{u}\wedge 1_{M})\psi_{X\wedge M}=(i^{\prime}i\wedge
1_{M})m_{M}(\overline{u}\wedge 1_{M})\psi_{X\wedge M}+(i^{\prime}\wedge
1_{M})\overline{m}_{M}(j\overline{u}\wedge 1_{M})\psi_{X\wedge M}$ and so
$(j^{\prime\prime}\wedge 1_{K\wedge M})\Delta_{1}(\mu_{X\wedge M}(1_{X}\wedge
i)\cdot\widetilde{u}w_{2}\wedge 1_{M})$ = 0. Consequently we have
$(j^{\prime\prime}\wedge 1_{K\wedge M})\Delta_{1}(\mu_{X\wedge
M}(\widetilde{u}w_{2}\wedge 1_{M})\wedge 1_{M})\in(1_{Y}\wedge j\wedge
1_{M})^{*}[\Sigma^{-2q}Y\wedge M,K\wedge M]$ = 0, this is because the degree
of the top cell of $Y\wedge M$ is $q+3$. Then $(j^{\prime\prime}\wedge
1_{K\wedge M})\Delta_{1}(\mu_{X\wedge M}\wedge
1_{M})\in(\widetilde{\psi}\wedge 1_{M\wedge M})^{*}[M\wedge M\wedge M,K\wedge
M]$ and so $(j^{\prime}(j^{\prime\prime}\wedge 1_{K})\wedge
1_{M})\Delta_{1}(\mu_{X\wedge M}\wedge 1_{M})$ = 0 and by (9.4.4) we have
$(j^{\prime}(j^{\prime\prime}\wedge 1_{K})\wedge
1_{M})\Delta_{1}=\Delta_{2}(j^{\prime}(j^{\prime\prime}u\wedge 1_{K})\wedge
1_{M})=\lambda(j^{\prime}(j^{\prime\prime}u\wedge 1_{K})\wedge 1_{M})$ with
$\lambda\in Z_{p}$, this is because $\Delta_{2}\in[M\wedge M,M\wedge
M]\cap(kerd)\cong Z_{p}\\{1_{M\wedge M}\\}$. Hence, $(\widetilde{\psi}\wedge
1_{M})u_{8}(w_{5}(1_{W}\wedge r)\wedge 1_{M})=(\widetilde{\psi}(1_{X}\wedge
j)u^{\prime}\wedge 1_{M})\Delta_{1}=(j^{\prime}(j^{\prime\prime}\wedge
1_{K})\wedge 1_{M})\Delta_{1}=\lambda(j^{\prime}(j^{\prime\prime}u\wedge
1_{K})\wedge 1_{M})$ and by (9.4.21)(9.4.24) we know that $\lambda=1$ so that
(9.4.25) $m_{M}(\widetilde{\psi}\wedge 1_{M})u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)=j^{\prime}(j^{\prime\prime}u\wedge 1_{K})$

$\qquad\quad=(j\widetilde{\psi}\wedge 1_{M})u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})$ ,

$\qquad\quad(j\widetilde{\psi}\wedge 1_{M})u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)=ijj^{\prime}(j^{\prime\prime}u\wedge
1_{K})$
where we use $(jj^{\prime}\wedge 1_{M})\overline{m}_{K}=j^{\prime}$ in the
above equation. By composing $(1_{E_{s+1}}\wedge u_{8}(w_{5}\wedge 1_{M}))$
(it has odd degree) on (9.4.17) we have
(9.4.26) $(\bar{a}_{s+1}\wedge 1_{X\wedge M})(1_{E_{s+2}}\wedge
u_{8}(w_{5}\wedge 1_{M}))f_{3}$

$=-(\bar{a}_{s+1}\wedge 1_{X\wedge M})(1_{E_{s+2}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)f^{\prime}$

$\qquad-\overline{\lambda}(\bar{a}_{s+1}\wedge 1_{X\wedge
M})(1_{E_{s+2}}\wedge u^{\prime}(u\wedge 1_{K}))f^{\prime}(1_{X}\wedge ij)$

$\qquad+(\bar{c}_{s}\wedge 1_{X\wedge M})(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)g$

$\qquad-(\bar{c}_{s}\wedge 1_{X\wedge M})(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K}))g(1_{X}\wedge ij)$

$\qquad+\overline{\lambda}(1_{E_{s+1}}\wedge u^{\prime}(u\wedge
1_{K}))f_{8}(1_{X}\wedge i^{\prime})$
where we use $u_{8}(w_{5}\wedge 1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})=\overline{\lambda}u^{\prime}(u\wedge 1_{K})$, for some
nonzero $\overline{\lambda}\in Z_{p}$. Moreover, by (9.4.20)(9.4.6),
$(\bar{b}_{s+1}\wedge 1_{L\wedge K})(1_{E_{s+1}}\wedge u\wedge
1_{K})f_{8}(1_{X}\wedge i^{\prime})=(1_{KG_{s+1}}\wedge(u\wedge
1_{K})\mu_{X\wedge M})(g_{0}\wedge
1_{M})=(1_{KG_{s+1}}\wedge\overline{\mu}_{2}(1_{Y}\wedge
i^{\prime})\psi_{X\wedge M})(g_{0}\wedge
1_{M})=\lambda_{1}(1_{KG_{s+1}}\wedge\overline{\mu}_{2}(1_{Y}\wedge
i^{\prime})\psi_{X\wedge M})(h_{0}\sigma\wedge 1_{X\wedge
M})=\lambda_{1}(h_{0}\sigma\wedge 1_{L\wedge K})\overline{\mu}_{2}(1_{Y}\wedge
i^{\prime})\psi_{X\wedge M}$ (mod $d_{1}$-boundary). Then
$[(\bar{b}_{s+1}\wedge 1_{L\wedge K})(1_{E_{s+1}}\wedge u\wedge
1_{K})f_{8}(1_{X}\wedge i^{\prime})]=\lambda_{1}(\phi\wedge
1_{K})_{*}(j^{\prime\prime}\wedge 1_{K})_{*}[(\sigma\wedge 1_{L\wedge
K})\overline{\mu}_{2}(1_{Y}\wedge i^{\prime})\psi_{X\wedge
M}]=\lambda_{1}(\phi\wedge 1_{K})_{*}(j^{\prime\prime}\wedge
1_{K})_{*}(\overline{\mu}_{2}(1_{Y}\wedge i^{\prime}))_{*}(\psi_{X\wedge
M})_{*}[\sigma\wedge 1_{X\wedge M}]=\lambda_{1}(\phi\wedge
1_{K})_{*}(i^{\prime})_{*}(m_{M}(\overline{u}\wedge 1_{M}))_{*}(\psi_{X\wedge
M})_{*}[\sigma\wedge 1_{X\wedge M}]=\lambda_{1}((1_{L}\wedge
i^{\prime})(\phi\wedge 1_{M}))_{*}(m_{M}(\widetilde{\psi}\wedge
1_{M})_{*}[\sigma\wedge 1_{X\wedge M}]$ = 0 $\in Ext_{A}^{s+1,tq}(H^{*}L\wedge
K,H^{*}X\wedge M)$. That is we have $(\bar{b}_{s+1}\wedge 1_{L\wedge
K})(1_{E_{s+1}}\wedge u\wedge 1_{K})f_{8}(1_{X}\wedge
i^{\prime})=(\bar{b}_{s+1}\bar{c}_{s}\wedge 1_{L\wedge K})g_{3}$ with
$g_{3}\in[\Sigma^{tq}X\wedge M,KG_{s}\wedge L\wedge K]\cap(kerd)$ ( cf. Prop.
9.5.26) and so $(1_{E_{s+1}}\wedge u\wedge 1_{K})f_{8}(1_{X}\wedge
i^{\prime})=(\bar{c}_{s}\wedge 1_{L\wedge K})g_{3}+(\bar{a}_{s+1}\wedge
1_{L\wedge K})f^{\prime}_{2}$ with $f^{\prime}_{2}\in[\Sigma^{tq+1}X\wedge
M,E_{s+2}\wedge L\wedge K]\cap(kerd)$ (cf. Prop. 6.5.26). Hence, (9.4.26)
becomes
(9.4.27) $(\bar{a}_{s+1}\wedge 1_{X\wedge M})(1_{E_{s+2}}\wedge
u_{8}(w_{5}\wedge 1_{M}))f_{3}$

$=-(\bar{a}_{s+1}\wedge 1_{X\wedge M})(1_{E_{s+2}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r))f^{\prime}$

$\quad-\overline{\lambda}(\bar{a}_{s+1}\wedge 1_{X\wedge M})(1_{E_{s+2}}\wedge
u^{\prime}(u\wedge 1_{K}))f^{\prime}(1_{X}\wedge ij)$

$\quad+(\bar{c}_{s}\wedge 1_{X\wedge M})(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r))g$

$\quad-(\bar{c}_{s}\wedge 1_{X\wedge M})(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K}))g(1_{X}\wedge ij)$

$\quad+\overline{\lambda}(\bar{c}_{s}\wedge 1_{X\wedge M})(1_{KG_{s}}\wedge
u^{\prime})g_{3}+\overline{\lambda}(\bar{a}_{s+1}\wedge 1_{X\wedge
M})(1_{E_{s+2}}\wedge u^{\prime})f^{\prime}_{2}$

By (9.4.27), $(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)g-(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})g(1_{X}\wedge
ij)+\overline{\lambda}(1_{KG_{s}}\wedge u^{\prime})g_{3}\in[\Sigma^{tq}X\wedge
M,KG_{s}\wedge X\wedge M]$ is a $d_{1}$-cycle which represents an element in
$Ext_{A}^{s,tq}(H^{*}X\wedge M,H^{*}X\wedge M)\cong Z_{p}\\{[\sigma\wedge
1_{X\wedge M}]\\}$ ( cf. Lemma 9.4.13). Then we have
(9.4.28) $(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge 1_{M})(1_{W}\wedge(1_{Y}\wedge
i)r)g+\overline{\lambda}(1_{KG_{s}}\wedge u^{\prime})g_{3}$

$\quad-(1_{KG_{s}}\wedge u_{8}(w_{5}\wedge 1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K}))g(1_{X}\wedge ij)$

$\qquad=\bar{\lambda}_{0}(\sigma\wedge 1_{X\wedge M})$ (mod $d_{1}$-boundary).
Now we consider the cases of $\bar{\lambda}_{0}\neq 1$ or
$\bar{\lambda}_{0}=1$ separately .

If $\bar{\lambda}_{0}\neq 1$, then by (9.4.27) and
$\bar{c}_{s}\cdot\sigma=\bar{a}_{s+1}\cdot\kappa$ we have

$\qquad(1_{E_{s+2}}\wedge u_{8}(w_{5}\wedge 1_{M}))f_{3}=-(1_{E_{s+2}}\wedge
u_{8}(w_{5}\wedge 1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)f^{\prime}$

$\qquad\quad-\bar{\lambda}(1_{E_{s+2}}\wedge u^{\prime}(u\wedge
1_{K}))f^{\prime}(1_{X}\wedge ij)+\bar{\lambda}(1_{E_{s+2}}\wedge
u^{\prime})f^{\prime}_{2}$

$\qquad\quad+\bar{\lambda}_{0}(\kappa\wedge 1_{X\wedge
M})+(\bar{c}_{s+1}\wedge 1_{X\wedge M})g_{4}$
with $g_{4}\in[\Sigma^{tq+1}X\wedge M,KG_{s+1}\wedge X\wedge M]$ and by
composing $(1_{E_{s+2}}\wedge
1_{X}\wedge\alpha)=(1_{E_{s+2}}\wedge\alpha_{X\wedge
M}m_{M}(\widetilde{\psi}\wedge 1_{M}))$ we obtain that $(1_{E_{s+2}}\wedge
1_{X}\wedge\alpha)(\kappa\wedge 1_{X\wedge
M})=(1_{E_{s+2}}\wedge\alpha_{X\wedge M}(j^{\prime\prime}u\wedge
1_{M})(1_{W}\wedge m_{M}(\overline{u}\wedge
1_{M}))f_{3}=(1_{E_{s+2}}\wedge\alpha_{X\wedge M}\cdot
m_{M}(\widetilde{\psi}\wedge 1_{M})u_{8}(w_{5}\wedge
1_{M}))f_{3}=\bar{\lambda}_{0}(1_{E_{s+2}}\wedge
1_{X}\wedge\alpha)(\kappa\wedge 1_{X\wedge M})$ so that the result of the step
1 follows.

If $\bar{\lambda}_{0}=1$, then by composing $(1_{KG_{s}}\wedge
m_{M}(\widetilde{\psi}\wedge 1_{M}))$ on (9.4.28) and using (9.4.25) we have
$(1_{KG_{s}}\wedge j^{\prime}(j^{\prime\prime}u\wedge
1_{K}))g=(1_{KG_{s}}\wedge m_{M}(\widetilde{\psi}\wedge
1_{M})u_{8}(w_{5}\wedge 1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)g=(\sigma\wedge
1_{M})m_{M}(\widetilde{\psi}\wedge 1_{M})$ (mod $d_{1}$-boundary). Moreover,
by composing $(1_{KG_{s}}\wedge j\widetilde{\psi}\wedge 1_{M})$ on (9.4.28)
and using (9.4.25) we have

$(1_{KG_{s}}\wedge j\widetilde{\psi}\wedge 1_{M})(\sigma\wedge 1_{X\wedge M})$

$=(1_{KG_{s}}\wedge(j\widetilde{\psi}\wedge 1_{M})u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)g$

$\quad-(1_{KG_{s}}\wedge(j\widetilde{\psi}\wedge 1_{M})u_{8}(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K}))g(1_{X}\wedge ij)$

$\quad+\overline{\lambda}(1_{KG_{s}}\wedge(j\widetilde{\psi}\wedge
1_{M})u^{\prime})g_{3}$

$=(1_{KG_{s}}\wedge ij(j^{\prime}(j^{\prime\prime}\wedge 1_{K})(u\wedge
1_{K}))g$

$\quad-(1_{KG_{s}}\wedge j^{\prime}(j^{\prime\prime}u\wedge
1_{K}))g(1_{X}\wedge ij)+\bar{\lambda}(1_{KG_{s}}\wedge
j^{\prime}(j^{\prime\prime}\wedge 1_{K}))g_{3}$ by (9.4.25)

$=(1_{KG_{s}}\wedge ij)(\sigma\wedge 1_{M})m_{M}(\widetilde{\psi}\wedge
1_{M})-(\sigma\wedge 1_{M})m_{M}(\widetilde{\psi}\wedge 1_{M})(1_{X}\wedge
ij)$

$\quad+\overline{\lambda}(1_{KG_{s}}\wedge j^{\prime}(j^{\prime\prime}\wedge
1_{K}))g_{3}$

$=(1_{KG_{s}}\wedge j\widetilde{\psi}\wedge 1_{M})(\sigma\wedge 1_{X\wedge
M})+\overline{\lambda}(1_{KG_{s}}\wedge j^{\prime}(j^{\prime\prime}\wedge
1_{K}))g_{3}$ by (9.4.6)
(mod $d_{1}$-boundary), then $(1_{KG_{s}}\wedge
j^{\prime}(j^{\prime\prime}\wedge 1_{K}))g_{3}$ = 0 and so
$g_{3}=(1_{KG_{s}}\wedge\overline{\mu}_{2}(1_{Y}\wedge i^{\prime}))g_{5}$ (mod
$d_{1}$-boundary) for some $g_{5}\in[\Sigma^{tq+q+1}X\wedge M,KG_{s}\wedge
Y\wedge M]$. So, by (9.4.6)(9.4.20)
$(1_{KG_{s+1}}\wedge\overline{\mu}_{2}(1_{Y}\wedge i^{\prime})\psi_{X\wedge
M})(g_{0}\wedge 1_{M})=(1_{KG_{s+1}}\wedge(u\wedge 1_{K})\mu_{X\wedge
M})(g_{0}\wedge 1_{M})=(\bar{b}_{s+1}\wedge 1_{L\wedge K})(1_{E_{s+1}}\wedge
u\wedge 1_{K})f_{8}(1_{X}\wedge i^{\prime})=(\bar{b}_{s+1}\bar{c}_{s}\wedge
1_{L\wedge K})g_{3}=(\bar{b}_{s+1}\bar{c}_{s}\wedge 1_{L\wedge
K})(1_{KG_{s}}\wedge\overline{\mu}_{2}(1_{Y}\wedge i^{\prime}))g_{5}$ so that
$(1_{KG_{s+1}}\wedge\psi_{X\wedge M})(g_{0}\wedge
1_{M})=(\bar{b}_{s+1}\bar{c}_{s}\wedge 1_{Y\wedge M})g_{5}$ , this shows
$\lambda_{1}(\psi_{X\wedge M})_{*}[h_{0}\sigma\wedge 1_{X\wedge
M}]=(\psi_{X\wedge M})_{*}[g_{0}\wedge 1_{M}]=0\in
Ext_{A}^{s+1,tq+1}(H^{*}Y\wedge M,.H^{*}X\wedge M)$ and by Lemma 9.4.13(2) we
have $\lambda_{1}$ = 0 . Then $[g_{0}\wedge 1_{M}]$ = 0 and so
$(\bar{b}_{s+1}\wedge 1_{W\wedge K})f_{8}(1_{X}\wedge
i^{\prime})=(\bar{b}_{s+1}\bar{c}_{s}\wedge 1_{W\wedge K})g_{6}$ for some
$g_{6}\in[\Sigma^{tq+q}X\wedge M,KG_{s}\wedge W\wedge K]$ and
$f_{8}(1_{X}\wedge i^{\prime})=(\bar{c}_{s}\wedge 1_{W\wedge
K})g_{6}+(\bar{a}_{s+1}\wedge 1_{W\wedge K})f^{\prime}_{3}$ with
$f^{\prime}_{3}\in[\Sigma^{tq+q+1}X\wedge M,E_{s+2}\wedge W\wedge K]$. Then,
by composing $(1_{E_{s+1}}\wedge w_{5}\wedge 1_{M})$ on (9.4.17) we have

$-(\bar{a}_{s+1}\wedge 1_{P\wedge M})(1_{E_{s+2}}\wedge w_{5}\wedge
1_{M})f_{3}$

$=(\bar{a}_{s+1}\wedge 1_{P\wedge M})(1_{E_{s+2}}\wedge(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)f^{\prime}$

$\quad+(\bar{a}_{s+1}\wedge 1_{P\wedge M})(1_{E_{s+2}}\wedge(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})f^{\prime}(1_{X}\wedge ij)$

$\quad+(\bar{a}_{s+1}\wedge 1_{P\wedge M})(1_{E_{s+2}}\wedge(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})f^{\prime}_{3}+(\bar{c}_{s}\wedge 1_{P\wedge M})g_{7}$
where the $d_{1}$-cycle $g_{7}=(1_{KG_{s}}\wedge(w_{5}\wedge
1_{M})(1_{W}\wedge(1_{Y}\wedge i)r)g-(1_{KG_{s}}\wedge(w_{5}\wedge
1_{M})(1_{W}\wedge(r\wedge 1_{M})\overline{m}_{K})g(1_{X}\wedge
ij)+(1_{KG_{s}}\wedge(w_{5}\wedge 1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})g_{6}\in[\Sigma^{tq+q+1}X\wedge M,KG_{s}\wedge P\wedge
M]$ which represents an element in $Ext_{A}^{s,tq+q+1}(H^{*}P\wedge
M,H^{*}X\wedge M)$. However, this group is zero , this can be obtained by the
following exact sequence

$0=Ext_{A}^{s,tq+q}(H^{*}W\wedge K,H^{*}X\wedge
M)\stackrel{{\scriptstyle((w_{5}\wedge 1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})_{*}}}{{\longrightarrow}}$

$\qquad Ext_{A}^{s,tq+q+1}(H^{*}P\wedge M,H^{*}X\wedge
M)\stackrel{{\scriptstyle(u_{7})_{*}}}{{\longrightarrow}}$

$\qquad Ext_{A}^{s,tq-2q}(H^{*}M,H^{*}X\wedge
M)\stackrel{{\scriptstyle((1_{W}\wedge i^{\prime})(\phi_{W}\wedge
M))_{*}}}{{\longrightarrow}}$
induced by (9.4.23), where the left group is zero by Prop. 9.4.11(4) and by
Prop. 9.4.11(1) the right group has unique generator
$m_{M}(\widetilde{\psi}\wedge 1_{M})^{*}(\tilde{\sigma})$, which satisfies
$((1_{W}\wedge i^{\prime})(\phi_{W}\wedge
1_{M}))_{*}m_{M}(\widetilde{\psi}\wedge 1_{M})^{*}(\tilde{\sigma})\neq 0\in
Ext_{A}^{s+1,tq+q}(H^{*}W\wedge K,H^{*}X\wedge M)$.

Then, $(\bar{c}_{s}\wedge 1_{P\wedge M})g_{7}$ = 0 and so $-(1_{E_{s+2}}\wedge
w_{5}\wedge 1_{M})f_{3}=(1_{E_{s+2}}\wedge(w_{5}\wedge
1_{M})u_{8}(1_{W}\wedge(1_{Y}\wedge
i)r)f^{\prime}-(1_{E_{s+2}}\wedge(w_{5}\wedge 1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K})f^{\prime}(1_{X}\wedge
ij)+(1_{E_{s+2}}\wedge(w_{5}\wedge 1_{M})(1_{W}\wedge(r\wedge
1_{M})\overline{m}_{K}))f^{\prime}_{3}+(\bar{c}_{s+1}\wedge 1_{P\wedge
M})g_{8}$ for some $g_{8}\in[\Sigma^{tq+q+2}X\wedge M,KG_{s+1}\wedge P\wedge
M]$. By composing $(1_{E_{s+2}}\wedge\alpha_{X\wedge M}\cdot u_{7})$ we have
$(1_{E_{s+2}}\wedge 1_{X}\wedge\alpha)(\kappa\wedge 1_{X\wedge
M})=(1_{E_{s+2}}\wedge\alpha_{X\wedge M}(j^{\prime\prime}u\wedge
1_{M})(1_{W}\wedge m_{M}(\overline{u}\wedge
1_{M}))f_{3}=(1_{E_{s+2}}\wedge\alpha_{X\wedge M}\cdot u_{7}(w_{5}\wedge
1_{M}))f_{3}$ = 0 . This shows the result of step 1.

Step 2 To prove $(\bar{c}_{s+1}\wedge
1_{M})\widetilde{h_{0}\sigma}=(\kappa\wedge 1_{M})\alpha$ = 0.

By (9.4.3)(9.4.4), $\mu_{X\wedge M}(1_{X}\wedge\alpha i)=\mu_{X\wedge
M}\alpha_{X\wedge M}\widetilde{\psi}$ = 0 and so by (9.1.15) $\mu_{X\wedge
M}=\mu_{X\wedge K^{\prime}}(1_{X}\wedge v)$ , where $\mu_{X\wedge
K^{\prime}}\in[X\wedge K^{\prime},W\wedge K]$. We claim that $X\wedge
K^{\prime}$ splits into $W\wedge K\vee\Sigma^{q}Y$, that is, there is a split
cofibration $\Sigma^{q}Y\rightarrow X\wedge K^{\prime}\rightarrow W\wedge K$,
this can be seen by the following homotopy commutative diagram of $3\times
3$-Lemma and using $(1_{Y}\wedge j)\alpha_{Y\wedge
M}j^{\prime}=r(1_{K}\wedge\alpha_{1})$

$\qquad\qquad X\wedge M\quad\stackrel{{\scriptstyle\mu_{X\wedge
M}}}{{\longrightarrow}}\quad W\wedge K\quad\stackrel{{\scriptstyle
0}}{{\longrightarrow}}\quad\Sigma^{q+1}Y$

$\qquad\qquad\quad\searrow 1_{X}\wedge v\nearrow\mu_{X\wedge
K^{\prime}}\searrow^{j^{\prime}(j^{\prime\prime}u\wedge
1_{K})}\nearrow_{(1_{Y}\wedge j)\alpha_{Y\wedge M}}$

$\qquad\qquad\qquad X\wedge K^{\prime}\qquad\qquad\qquad\Sigma^{3q+1}M$

$\qquad\qquad\quad\nearrow\widetilde{\tau}_{X\wedge K^{\prime}}\searrow
1_{X}\wedge\quad\nearrow\widetilde{\psi}\quad\searrow\alpha_{X\wedge M}$

$\qquad\qquad\Sigma^{q}Y\quad\stackrel{{\scriptstyle\tilde{u}w_{2}}}{{\longrightarrow}}\qquad\Sigma^{q+1}X\quad\stackrel{{\scriptstyle
1_{X}\wedge\alpha i}}{{\longrightarrow}}\quad\Sigma X\wedge M$
Hence, there is a split cofibration
$\Sigma^{q}Y\stackrel{{\scriptstyle\tau_{X\wedge
K^{\prime}}}}{{\longrightarrow}}X\wedge
K^{\prime}\stackrel{{\scriptstyle\mu_{X\wedge
K^{\prime}}}}{{\longrightarrow}}W\wedge K$ and so there are $\nu_{X\wedge
K^{\prime}}:X\wedge K^{\prime}\rightarrow\Sigma^{q}Y$ and
$\widetilde{\nu}_{X\wedge K^{\prime}}:W\wedge K\rightarrow X\wedge K^{\prime}$
such that $\nu_{X\wedge K^{\prime}}\cdot\tau_{X\wedge
K^{\prime}}=1_{Y},\quad\mu_{X\wedge K^{\prime}}\cdot\widetilde{\nu}_{X\wedge
K^{\prime}}=1_{W\wedge K},$ $\widetilde{\tau}_{X\wedge
K^{\prime}}\cdot\nu_{X\wedge K^{\prime}}+\widetilde{\nu}_{X\wedge
K^{\prime}}\cdot\mu_{X\wedge K^{\prime}}=1_{X\wedge K^{\prime}}$.

By the result of step 1 we have $(\kappa\wedge 1_{M\wedge X\wedge
K^{\prime}})(\alpha\wedge 1_{X\wedge K^{\prime}})$ = 0, then $(\kappa\wedge
1_{M\wedge Y})(\alpha\wedge 1_{Y})=(1_{E_{s+2}}\wedge 1_{M}\wedge\nu_{X\wedge
K^{\prime}})(\kappa\wedge 1_{M\wedge X\wedge K^{\prime}})(\alpha\wedge
1_{X\wedge K^{\prime}})(1_{M}\wedge\tau_{X\wedge K^{\prime}})$ = 0. Moreover,
by using the splitness in (9.1.32) we have $(\bar{c}_{s+1}\wedge
1_{M})\widetilde{h_{0}\sigma}=(\kappa\wedge 1_{M})\alpha=(1_{E_{s+2}}\wedge
1_{M}\wedge\widetilde{\nu})(\kappa\wedge 1_{M\wedge Y\wedge
K^{\prime}})(\alpha\wedge 1_{Y\wedge
K^{\prime}})(1_{M}\wedge\widetilde{\tau})$ = 0 which shows the main Theorem C.
Q.E.D.

Remark. In the proof of the main Theorem C, We only use the supposition (II)
for our geometric input to obtain that $(1_{E_{s+2}}\wedge\phi\wedge
1_{M})(\kappa\wedge 1_{M})m_{M}(\widetilde{\psi}\wedge 1_{M})$ = 0. Then , the
geometric supposition (II) of the main Theorem C can be weakened to be the
supposition on $m_{M}(\widetilde{\psi}\wedge 1_{M})^{*}(\phi\wedge
1_{M})_{*}(\tilde{\sigma})\in Ext_{A}^{s+1,tq}(H^{*}L\wedge M,H^{*}X\wedge M)$
is a permanent cycle in the ASS.

Using some new cofibrations in this section , we also can give an alternative
proof of Theorem 9.3.9( and so the main Theorem B). We first do some
preminalaries.

Since $\alpha^{\prime}\alpha^{\prime}i^{\prime}=0$, then by (9.1.23), there
exists $\alpha^{\prime\prime}_{Y\wedge M}\in[\Sigma^{q-2}Y\wedge M,K]$ such
that $\alpha^{\prime\prime}_{Y\wedge M}(r\wedge
1_{M})\overline{m}_{K}=\alpha^{\prime}$. By applying the derivation $d$,
$d(\alpha^{\prime\prime}_{Y\wedge M})(r\wedge
1_{M})\overline{m}_{K}=-d(\alpha^{\prime})=0$ and so
$d(\alpha^{\prime\prime}_{Y\wedge M})\in(m_{M}(\overline{u}\wedge
1_{M}))^{*}[\Sigma^{2q}M,K]$ = 0. $\alpha_{Y\wedge M}(1_{Y}\wedge
i)r\in[\Sigma^{q-2}K,K]\cong Z_{p}\\{\alpha^{\prime\prime}\\}$ and so
$\alpha^{\prime\prime}_{Y\wedge M}(1_{Y}\wedge
i)r=\lambda\alpha^{\prime\prime}$ for some $\lambda\in Z_{p}$. Note that
$d((1_{Y}\wedge i)r)=(r\wedge 1_{M})d(1_{K}\wedge i)=(r\wedge
1_{M})\overline{m}_{K}$, then by applying the derivation $d$, we have
$\alpha^{\prime}=\alpha^{\prime\prime}_{Y\wedge M}(r\wedge
M)\overline{m}_{K}=\lambda d(\alpha^{\prime\prime})=-\lambda\alpha^{\prime}$
and so $\lambda=-1$. By (9.1.8), $\bar{h}i^{\prime\prime}=\overline{w}$,
$ri^{\prime}=\overline{w}\cdot j$( up to sign), then $(r\wedge
1_{M})\overline{m}_{K}i^{\prime}=-(ri^{\prime}\wedge
1_{M})\overline{m}_{M}=\pm(\overline{w}\wedge
1_{M})=\pm(\bar{h}i^{\prime\prime}\wedge 1_{M})$ and so
$\alpha^{\prime\prime}_{Y\wedge M}(\bar{h}i^{\prime\prime}\wedge
1_{M})=\lambda_{0}\alpha^{\prime\prime}_{Y\wedge M}(r\wedge
1_{M})\overline{m}_{K}i^{\prime}=\lambda_{0}\alpha^{\prime}i^{\prime}=\lambda_{0}i^{\prime}((\alpha_{1})_{L}i^{\prime\prime}\wedge
1_{M})$ and we have $\alpha^{\prime\prime}_{Y\wedge M}(\bar{h}\wedge
1_{M})=\lambda_{0}i^{\prime}((\alpha_{1})_{L}\wedge 1_{M})$, where
$\lambda_{0}=\pm 1$. On the other hand, $i^{\prime}((\alpha_{1})_{L}\wedge
1_{M})(1_{L}\wedge j^{\prime})(i^{\prime\prime}\wedge
1_{K})=i^{\prime}(\alpha_{1}\wedge 1_{M})j^{\prime}=i^{\prime}(ij\alpha-\alpha
ij)j^{\prime}$ = 0, then $i^{\prime}((\alpha_{1})_{L}\wedge 1_{M})(1_{L}\wedge
j^{\prime})=\lambda^{\prime}\alpha^{\prime\prime}(j^{\prime\prime}\wedge
1_{K})$ with $\lambda^{\prime}\in Z_{p}$. By composing the map
$\widetilde{\Delta}$ in Theorem 6.5.18 we have
$\lambda^{\prime}\alpha^{\prime}i^{\prime}ijj^{\prime}=\lambda^{\prime}\alpha^{\prime\prime}i^{\prime}j^{\prime}=\lambda^{\prime}\alpha^{\prime\prime}(j^{\prime\prime}\wedge
1_{K})\widetilde{\Delta}=i^{\prime}((\alpha_{1})_{L}\wedge 1_{M})(1_{L}\wedge
j^{\prime})\widetilde{\Delta}=-i^{\prime}((\alpha_{1})_{L}i^{\prime\prime}\wedge
1_{M})ijj^{\prime}=-\alpha^{\prime}i^{\prime}ijj^{\prime}$ so that
$\lambda^{\prime}=-1$. Concludingly , there is $\alpha^{\prime\prime}_{Y\wedge
M}\in[\Sigma^{q-2}Y\wedge M,K]$ such that
(9.4.29) $\alpha^{\prime\prime}_{Y\wedge M}(r\wedge
1_{M})\overline{m}_{K}=\alpha^{\prime},\quad\qquad\alpha^{\prime\prime}_{Y\wedge
M}(1_{Y}\wedge i)r=-\alpha^{\prime\prime},$

$\qquad\qquad d(\alpha^{\prime\prime}_{Y\wedge
M})=0,\quad\qquad\alpha^{\prime\prime}_{Y\wedge M}(\bar{h}\wedge
1_{M})=\lambda_{0}i^{\prime}((\alpha_{1})_{L}\wedge 1_{M}),$

$\qquad\qquad i^{\prime}((\alpha_{1})_{L}\wedge 1_{M})(1_{L}\wedge
j^{\prime})=-\alpha^{\prime\prime}(j^{\prime\prime}\wedge 1_{K})$
where $\lambda_{0}=\pm 1$.

Note that the cofibre of $\alpha^{\prime\prime}_{Y\wedge
M}:\Sigma^{q-2}Y\wedge M\to K$ is $X\wedge M$ given by the cofibration
(9.4.30) $\Sigma^{q-2}Y\wedge
M\stackrel{{\scriptstyle\alpha^{\prime\prime}_{Y\wedge
M}}}{{\longrightarrow}}K\stackrel{{\scriptstyle
u^{\prime}(i^{\prime\prime}\wedge 1_{K})}}{{\longrightarrow}}X\wedge
M\stackrel{{\scriptstyle\psi_{X\wedge
M}}}{{\longrightarrow}}\Sigma^{q-1}Y\wedge M$
and the above map $\psi_{X\wedge M}\in[X\wedge M,\Sigma^{q-1}Y\wedge M]$ and
$u^{\prime}\in[L\wedge K,X\wedge M]$ is just the map in (9.4.2) and (9.4.6).
This can be seen by the equation $m_{M}(\overline{u}\wedge 1_{M})\psi_{X\wedge
M}=m_{M}(\widetilde{\psi}\wedge 1_{M})$ in (9.4.6),(9.4.2) and the following
homotopy commutative diagram of $3\times 3$-Lemma

$\qquad\quad X\wedge M\quad\stackrel{{\scriptstyle m_{M}(\tilde{\psi}\wedge
1_{M})}}{{\longrightarrow}}\quad\Sigma^{2q}M\qquad\stackrel{{\scriptstyle\alpha^{\prime}i^{\prime}}}{{\longrightarrow}}\qquad\Sigma^{q+1}K$

$\qquad\qquad\quad\searrow\psi_{X\wedge
M}\quad\nearrow^{m_{M}(\overline{u}\wedge 1_{M})}\quad\searrow^{(\phi\wedge
1_{K})i^{\prime}}\quad\nearrow j^{\prime\prime}\wedge 1_{K}\searrow^{(r\wedge
1_{M})\overline{m}_{K}}$
(9.4.31) $\Sigma^{q-1}Y\wedge M\qquad\qquad\qquad\quad\Sigma L\wedge
K\qquad\qquad\qquad\Sigma^{q}Y\wedge M$

$\qquad\qquad\nearrow^{(r\wedge
1_{M})\overline{m}_{K}}\quad\searrow^{\alpha^{\prime\prime}_{Y\wedge
M}}\quad\nearrow i^{\prime\prime}\wedge 1_{K}\quad\searrow
u^{\prime}\qquad\quad\nearrow\psi_{X\wedge M}$

$\qquad\quad\Sigma^{q}K\qquad\stackrel{{\scriptstyle\alpha^{\prime}}}{{\longrightarrow}}\qquad\quad\Sigma
K\qquad\stackrel{{\scriptstyle u^{\prime}(i^{\prime\prime}\wedge
1_{K})}}{{\longrightarrow}}\qquad\Sigma X\wedge M$
and by this we have the following relation
(9.4.32) $\psi_{X\wedge M}u^{\prime}=(r\wedge
1_{M})\overline{m}_{K}(j^{\prime\prime}\wedge 1_{K})$.

Proposition 9.4.33 Let $p\geq 5$ and $V$ be any spectrum, then for any map
$f\in[\Sigma^{*}K,V\wedge K]$ we have
$(1_{V}\wedge\alpha^{\prime})d(f)=d(f)\alpha^{\prime}=0$.

Proof: By (6.5.12), $\alpha\wedge
1_{K}=\overline{m}_{K}^{\prime}\alpha^{\prime}m_{K}^{\prime}$, where
$m_{K}^{\prime}=m_{K}T:M\wedge K\to K$,
$\overline{m}_{K}^{\prime}=T\overline{m}_{K}:\Sigma K\to M\wedge K$.
$d(f)\alpha^{\prime}m_{K}^{\prime}=(1_{V}\wedge
m_{K}^{\prime})(T^{\prime}\wedge 1_{K})(1_{M}\wedge
f)\overline{m}_{K}^{\prime}\alpha^{\prime}m_{K}^{\prime}=(1_{V}\wedge
m_{K}^{\prime})(T^{\prime}\wedge 1_{K})(1_{M}\wedge f)(\alpha\wedge
1_{K})=(1_{V}\wedge m_{K}^{\prime})(T^{\prime}\wedge 1_{K})(\alpha\wedge
1_{V}\wedge 1_{K})(1_{M}\wedge f)=(1_{V}\wedge m_{K}^{\prime}(\alpha\wedge
1_{K}))(T^{\prime}\wedge 1_{K})(1_{M}\wedge f)$ = 0, where $T^{\prime}:M\wedge
V\to V\wedge M$is the switching map. Q.E.D.

Proposition 9.4.34 Under the supposition (I) of the main Theorem B we have

(1) $Ext_{A}^{s,tq-1}(H^{*}K,H^{*}K)$ = 0.

(2) $Ext_{A}^{s,tq}(H^{*}Y\wedge M,H^{*}K)$ has unique generator $(1_{Y}\wedge
i)_{*}r_{*}[\sigma\wedge 1_{K}]$.

Proof: (1) Consider the following exact sequence

$\qquad
Ext_{A}^{s,tq+q}(H^{*}M,H^{*}M)\stackrel{{\scriptstyle(i^{\prime})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq+q}(H^{*}K,H^{*}M)$

$\qquad\qquad\stackrel{{\scriptstyle(j^{\prime})_{*}}}{{\longrightarrow}}Ext_{A}^{s,tq-1}(H^{*}M,H^{*}M)\stackrel{{\scriptstyle\alpha_{*}}}{{\longrightarrow}}$
induced by (9.1.2). By the supposition (I), the right group has unique
generator $j^{*}i_{*}(\sigma)$ which satisfies
$\alpha_{*}j^{*}i_{*}(\sigma)=(ij)_{*}\alpha_{*}(\tilde{\sigma})\neq 0$. Then
im $(j^{\prime})_{*}$ = 0. By the supposition (I), the left group is zero or
has two generators
$(ij)_{*}\alpha_{*}(\widetilde{\tau}^{\prime}),(ij)^{*}\alpha_{*}(\widetilde{\tau}^{\prime})$
(this can be obtained by a similar proof as given in Prop. 9.3.1(2)), then
$Ext_{A}^{s,tq+q}(H^{*}K,H^{*}M)=(i^{\prime})_{*}Ext_{A}^{s,tq+q}(H^{*}M,H^{*}M)$
is zero or has unique generator $(i^{\prime})_{*}(\alpha_{1}\wedge
1_{M})_{*}(\widetilde{\tau}^{\prime})$. Look at the following exact sequence

$\qquad
Ext_{A}^{s,tq+q}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq-1}(H^{*}K,H^{*}K)$

$\qquad\qquad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s,tq-1}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}$
induced by (9.1.2). By the supposition (I), the right group has unique
generator $j^{*}(i^{\prime}i)_{*}(\sigma)$ which satisfies
$\alpha^{*}j^{*}(i^{\prime}i)_{*}(\sigma)=(i^{\prime})_{*}(ij)_{*}\alpha_{*}(\tilde{\sigma})\neq
0\in Ext_{A}^{s+1,tq+q}(H^{*}K,H^{*}M)$ so that im $(i^{\prime})^{*}$ = 0. The
left group is zero or has unique generator $(i^{\prime})_{*}(\alpha_{1}\wedge
1_{M})_{*}(\widetilde{\tau}^{\prime})$ and so im $(j^{\prime})^{*}$ = 0 and
the middle group is zero as desired.

(2) For any $g\in Ext_{A}^{s,tq}(H^{*}Y\wedge
M,H^{*}K)$,$m_{M}(\overline{u}\wedge 1_{M}))_{*}(g)\in Ext_{A}^{s,tq-q-1}\\\
(H^{*}M,H^{*}K)\cong Z_{p}\\{(j^{\prime})^{*}(\tilde{\sigma})\\}$, this can be
obtained from $Ext_{A}^{s,tq}(H^{*}M,H^{*}M)\\\ \cong
Z_{p}\\{\tilde{\sigma}\\}$ in Prop. 9.3.0(2) and
$Ext_{A}^{s,tq-q-1}(H^{*}M,H^{*}M)$ = 0, where the last is obtained by the
supposition (I) on $Ext_{A}^{s,tq-q+u}\\\ (Z_{p},Z_{p})$ = 0(for $u=0,-1,1$).
Then $(m_{M}(\overline{u}\wedge
1_{M}))_{*}(g)=\lambda^{\prime}(j^{\prime})^{*}[\sigma\wedge
1_{M}]=\lambda^{\prime}[(\sigma\wedge
1_{M})j^{\prime}]=\lambda^{\prime}[(1_{KG_{s}}\wedge j^{\prime})(\sigma\wedge
1_{K})]=\lambda^{\prime}(j^{\prime})_{*}[\sigma\wedge
1_{K}]=\lambda^{\prime}(m_{M}(\overline{u}\wedge 1_{M}))_{*}(1_{Y}\wedge
i)_{*}r_{*}[\sigma\wedge 1_{K}]$ and so $g=\lambda^{\prime}(1_{Y}\wedge
i)_{*}r_{*}[\sigma\wedge 1_{K}]$ ( with $\lambda^{\prime}\in Z_{p}$) modulo
$((r\wedge 1_{M})\overline{m}_{K})_{*}Ext_{A}^{s,tq-1}(H^{*}K,H^{*}K)$ = 0.
Q.E.D.

An alternative proof of Theorem 9.3.9: By the supposition (II) of the main
Theorem B we have $(1_{E_{s+2}}\wedge\alpha^{\prime})(\kappa\wedge
1_{K})=(\bar{c}_{s+1}\wedge 1_{K})(h_{0}\sigma\wedge 1_{K})$ = 0, then
$(\kappa\wedge 1_{K})=(1_{E_{s+2}}\wedge j^{\prime\prime}\wedge 1_{K})f$ and
we have $d((1_{E_{s+2}}\wedge j^{\prime\prime}\wedge 1_{K})f)$ = 0. That is we
have
(9.4.35) $\kappa\wedge 1_{K}=(1_{E_{s+2}}\wedge j^{\prime\prime}\wedge
1_{K})f$ $d(f)=(1_{E_{s+2}}\wedge i^{\prime\prime}\wedge 1_{K})f^{\prime}$
for some $f\in[\Sigma^{tq+q+1}K,E_{s+2}\wedge L\wedge
K],f^{\prime}\in[\Sigma^{tq+q+2}K,E_{s+2}\wedge K]$.

By (9.4.29)(9.4.35),$(1_{E_{s+2}}\wedge\alpha^{\prime\prime}_{Y\wedge
M}(\bar{h}\wedge 1_{M})(1_{L}\wedge
j^{\prime}))f=\lambda_{0}(1_{E_{s+2}}\wedge i^{\prime}((\alpha_{1})_{L}\wedge
1_{M})(1_{L}\wedge
j^{\prime}))f=-\lambda_{0}(1_{E_{s+2}}\wedge\alpha^{\prime\prime}(j^{\prime\prime}\wedge
1_{K}))f=-\lambda_{0}(1_{E_{s+2}}\wedge\alpha^{\prime\prime})(\kappa\wedge
1_{K})$, where $\lambda_{0}=\pm 1$. That is we have
(9.4.36) $(1_{E_{s+2}}\wedge\alpha^{\prime\prime}_{Y\wedge M}(\bar{h}\wedge
1_{M})(1_{L}\wedge
j^{\prime}))f=-\lambda_{0}(1_{E_{s+2}}\wedge\alpha^{\prime\prime})(\kappa\wedge
1_{K})$ where $\lambda_{0}=\pm 1$

It follows that $(\bar{a}_{s+1}\wedge
1_{K})(1_{E_{s+2}}\wedge\alpha^{\prime\prime}_{Y\wedge M}(\bar{h}\wedge
1_{M})(1_{L}\wedge j^{\prime}))f=-\lambda_{0}(\bar{a}_{s+1}\wedge
1_{K})(1_{E_{s+2}}\wedge\alpha^{\prime\prime})(\kappa\wedge
1_{K})=-\lambda_{0}(\bar{c}_{s}\wedge
1_{K})(1_{KG_{s}}\wedge\alpha^{\prime\prime})(\sigma\wedge 1_{K})$ = 0 since
$\alpha^{\prime\prime}$ induces zero homomorphism in $Z_{p}$-cohomology. Then,
by (9.4.30),$(\bar{a}_{s+1}\wedge 1_{Y\wedge
M})(1_{E_{s+2}}\wedge(\bar{h}\wedge 1_{M})(1_{L}\wedge
j^{\prime}))f=(1_{E_{s+1}}\wedge\psi_{X\wedge M})f_{2}$ with
$f_{2}\in[\Sigma^{tq+q-1}K,\\\ E_{s+1}\wedge X\wedge M]$. Consequently,
$(\bar{b}_{s+1}\wedge 1_{Y\wedge M})(1_{E_{s+1}}\wedge\psi_{X\wedge M})f_{2}$
= 0 and so by (9.4.30) we have $(\bar{b}_{s+1}\wedge 1_{X\wedge
M})f_{2}=(1_{KG_{s+1}}\wedge u^{\prime}(i^{\prime\prime}\wedge 1_{K}))g$, with
$d_{1}$-cycle $g\in[\Sigma^{tq+q-1}K,KG_{s+1}\wedge K]$ and this $d_{1}$-cycle
represents an element in $Ext_{A}^{s+1,tq+q-1}(H^{*}K,H^{*}K)\cong
Z_{p}\\{(h_{0}\sigma)^{\prime\prime}\\}$. Then
$[g]=\lambda^{\prime}(h_{0}\sigma)^{\prime\prime}=\lambda^{\prime}(\alpha^{\prime\prime})_{*}[\sigma\wedge
1_{K}]$ for some $\lambda^{\prime}\in Z_{p}$ so that

$\quad[(\bar{b}_{s+1}\wedge 1_{X\wedge
M})f_{2}]=(u^{\prime}(i^{\prime\prime}\wedge
1_{K}))_{*}[g]=\lambda^{\prime}(u^{\prime}(i^{\prime\prime}\wedge
1_{K}))_{*}(\alpha^{\prime\prime})_{*}[\sigma\wedge 1_{K}]$

$=\lambda^{\prime}(u^{\prime}(i^{\prime\prime}\wedge
1_{K}))_{*}(\alpha^{\prime\prime}_{Y\wedge M})_{*}((1_{Y}\wedge
i)r)_{*}[\sigma\wedge 1_{K}]$ = 0
Hence, $(\bar{b}_{s+1}\wedge 1_{X\wedge
M})f_{2}=(\bar{b}_{s+1}\bar{c}_{s}\wedge 1_{X\wedge M})g_{2}$ for some
$g_{2}\in[\Sigma^{tq+q-1}K,KG_{s}\wedge X\wedge M]$ and so
$f_{2}=(\bar{c}_{s}\wedge 1_{X\wedge M})g_{2}+(\bar{a}_{s+1}\wedge 1_{X\wedge
M})f_{3}$ with $f_{3}\in[\Sigma^{tq+q-1}K,\\\ E_{s+2}\wedge X\wedge M]$ and we
have

$(\bar{a}_{s+1}\wedge 1_{Y\wedge M})(1_{E_{s+2}}\wedge(\bar{h}\wedge
1_{M})(1_{L}\wedge j^{\prime}))f$

$=(\bar{a}_{s+1}\wedge 1_{Y\wedge M})(1_{E_{s+2}}\wedge\psi_{X\wedge
M})f_{3}+(\bar{c}_{s}\wedge 1_{Y\wedge M})(1_{KG_{s}}\wedge\psi_{X\wedge
M})g_{2}$

$=(\bar{a}_{s+1}\wedge 1_{Y\wedge M})(1_{E_{s+2}}\wedge\psi_{X\wedge
M})f_{3}+\lambda(\bar{c}_{s}\wedge 1_{Y\wedge M})(1_{KG_{s}}\wedge(1_{Y}\wedge
i)r)(\sigma\wedge 1_{K})$

$=(\bar{a}_{s+1}\wedge 1_{Y\wedge M})(1_{E_{s+2}}\wedge\psi_{X\wedge M})f_{3}$

$\qquad+\lambda(\bar{a}_{s+1}\wedge 1_{Y\wedge
M})(1_{E_{s+2}}\wedge(1_{Y}\wedge i)r)(\kappa\wedge 1_{K})$
with $\lambda\in Z_{p}$, where the $d_{1}$-cycle
$(1_{KG_{s}}\wedge\psi_{X\wedge M})g_{2}\in[\Sigma^{tq}K,KG_{s}\wedge Y\wedge
M]$ represents an element $\lambda((1_{Y}\wedge i)r)_{*}[\sigma\wedge
1_{K}]\in Ext_{A}^{s,tq}(H^{*}Y\wedge M,H^{*}K)$( cf. Prop. 9.4.34(2)) and so
it equals to $\lambda(1_{KG_{s}}\wedge(1_{Y}\wedge i)r)(\sigma\wedge 1_{K})$ (
mod $d_{1}$-boundary). Then we have $(1_{E_{s+2}}\wedge(\bar{h}\wedge
1_{M})(1_{L}\wedge j^{\prime}))f=(1_{E_{s+2}}\wedge\psi_{X\wedge
M})f_{3}+\lambda(1_{E_{s+2}}\wedge(1_{Y}\wedge i)r)(\kappa\wedge
1_{K})+(\bar{c}_{s+1}\wedge 1_{Y\wedge M})g_{3}$ for some
$g_{3}\in[\Sigma^{tq+1}K,KG_{s+1}\wedge Y\wedge M]$. By composing
$1_{E_{s+2}}\wedge 1_{Y}\wedge\alpha$ and using(9.4.8) we have
$(1_{E_{s+2}}\wedge\omega\wedge 1_{M})f_{3}=(1_{E_{s+2}}\wedge\alpha_{Y\wedge
M}m_{M}(\widetilde{\psi}\wedge
1_{M}))f_{3}=(1_{E_{s+2}}\wedge(1_{Y}\wedge\alpha)\psi_{X\wedge
M})f_{3}=-\lambda(1_{E_{s+2}}\wedge(1_{Y}\wedge\alpha i)r)(\kappa\wedge
1_{K})=-\lambda(1_{E_{s+2}}\wedge(r\wedge
1_{M})\overline{m}_{K}\alpha^{\prime})(\kappa\wedge 1_{K})$ = 0 and by (9.4.7)
we have $f_{3}=(1_{E_{s+2}}\wedge(1_{x}\wedge j)u^{\prime}\wedge 1_{M})f_{4}$
with $f_{4}\in[\Sigma^{tq+q+1}K,E_{s+2}\wedge L\wedge K\wedge M]$. That is we
have
(9.4.37) $(1_{E_{s+2}}\wedge(\bar{h}\wedge 1_{M})(1_{L}\wedge
j^{\prime}))f=(1_{E_{s+2}}\wedge\psi_{X\wedge M}((1_{X}\wedge
j)u^{\prime}\wedge 1_{M}))f_{4}$

$\qquad+\lambda(1_{E_{s+2}}\wedge(1_{Y}\wedge i)r)(\kappa\wedge
1_{K})+(\bar{c}_{s+1}\wedge 1_{Y\wedge M})g_{3}$

$=(1_{E_{s+2}}\wedge\psi_{X\wedge
M}u^{\prime})f_{5}+(1_{E_{s+2}}\wedge\psi_{X\wedge M}(1_{X}\wedge
ij)u^{\prime}(1_{L}\wedge m_{K}))f_{4}$

$\qquad+\lambda(1_{E_{s+2}}\wedge(1_{Y}\wedge i)r)(\kappa\wedge
1_{K})+(\bar{c}_{s+1}\wedge 1_{Y\wedge M})g_{3}$
where we use $f_{4}=(1_{E_{s+2}}\wedge(1_{L}\wedge\overline{m}_{K})(1_{L\wedge
K}\wedge j))f_{4}+(1_{E_{s+2}}\wedge(1_{L\wedge K}\wedge i)(1_{L}\wedge
m_{K}))f_{4}$ and write $(1_{E_{s+2}}\wedge 1_{L\wedge K}\wedge
j)f_{4}=f_{5}$.

By composing $1_{E_{s+2}}\wedge\alpha^{\prime\prime}_{Y\wedge M}$ on (9.4.37)
and using (9.4.36)(9.4.30) we have $-\lambda_{0}(1_{E_{s+2}}\\\
\wedge\alpha^{\prime\prime})(\kappa\wedge
1_{K})=(1_{E_{s+2}}\wedge\alpha^{\prime\prime}_{Y\wedge M}(\bar{h}\wedge
1_{M})(1_{L}\wedge
j^{\prime}))f=\lambda(1_{E_{s+2}}\wedge\alpha^{\prime\prime}_{Y\wedge
M}(1_{Y}\wedge i)r)(\kappa\wedge
1_{K})=-\lambda(1_{E_{s+2}}\wedge\alpha^{\prime\prime})(\kappa\wedge 1_{K})$.
If $\lambda\neq\lambda_{0}$, then
$(1_{E_{s+2}}\wedge\alpha^{\prime\prime})(\kappa\wedge 1_{K})$ = 0 and the
Theorem follows. So, we suppose that $\lambda=\lambda_{0}$.

By (9.1.8) we have $\overline{u}\bar{h}=i\cdot j^{\prime\prime}$ so that
$m_{M}(\overline{u}\wedge 1_{M})(\bar{h}\wedge 1_{M})=j^{\prime\prime}\wedge
1_{M}$( up to sign). Then, what happen is either $m_{M}(\overline{u}\wedge
1_{M})(\bar{h}\wedge 1_{M})=\lambda_{0}(j^{\prime\prime}\wedge 1_{M})$ or
$m_{M}(\overline{u}\wedge 1_{M})(\bar{h}\wedge
1_{M})=-\lambda_{0}(j^{\prime\prime}\wedge 1_{M})$. Now we consider this two
cases separately.

Case 1 $m_{M}(\overline{u}\wedge 1_{M})(\bar{h}\wedge
1_{M})=\lambda_{0}(j^{\prime\prime}\wedge 1_{M})$.

In this case, by composing $1_{E_{s+2}}\wedge m_{M}(\overline{u}\wedge 1_{M})$
on (9.4.37) we have
(9.4.38) $\lambda_{0}(1_{E_{s+2}}\wedge j^{\prime})(\kappa\wedge
1_{K})=\lambda_{0}(1_{E_{s+2}}\wedge j^{\prime}(j^{\prime\prime}\wedge
1_{K}))f$

$=\lambda_{0}(1_{E_{s+2}}\wedge(j^{\prime\prime}\wedge 1_{M})(1_{L}\wedge
j^{\prime}))f=(1_{E_{s+2}}\wedge m_{M}(\overline{u}\wedge 1_{M})(\bar{h}\wedge
1_{M})(1_{L}\wedge j^{\prime}))f$

$=(1_{E_{s+2}}\wedge m_{M}(\overline{u}\wedge 1_{M})\psi_{X\wedge
M}(1_{X}\wedge ij)u^{\prime}(1_{L}\wedge m_{K}))f_{4}$

$\qquad+\lambda_{0}(1_{E_{s+2}}\wedge j^{\prime})(\kappa\wedge
1_{K})+(\bar{c}_{s+1}\wedge 1_{M})(1_{KG_{s+1}}\wedge m_{M}(\overline{u}\wedge
1_{M}))g_{3}$

$=-(1_{E_{s+2}}\wedge j^{\prime}(j^{\prime\prime}\wedge 1_{K})(1_{L}\wedge
m_{K}))f_{4}$

$\qquad+\lambda_{0}(1_{E_{s+2}}\wedge j^{\prime})(\kappa\wedge
1_{K})+(\bar{c}_{s+1}\wedge 1_{M})(1_{KG_{s+1}}\wedge m_{M}(\overline{u}\wedge
1_{M}))g_{3}$
and so $(1_{E_{s+2}}\wedge j^{\prime}(j^{\prime\prime}\wedge
1_{K})(1_{L}\wedge m_{K}))f_{4}=(\bar{c}_{s+1}\wedge 1_{M})(1_{KG_{s+1}}\wedge
m_{M}(\overline{u}\wedge 1_{M}))g_{3}$, where we use $m_{M}(\overline{u}\wedge
1_{M})\psi_{X\wedge M}(1_{X}\wedge ij)u^{\prime}=m_{M}(\widetilde{\psi}\wedge
1_{M})(1_{X}\wedge ij)u^{\prime}=-(j\widetilde{\psi}\wedge
1_{M})u^{\prime}=-j^{\prime}(j^{\prime\prime}\wedge 1_{K})$ which is obtained
by (9.4.6) and the right rectangle of the diagram (9.4.1). Moreover, by
applying the derivation $d$ to the equation (9.4.37) we have
(9.4.39) $(1_{E_{s+2}}\wedge(\bar{h}\wedge 1_{M})(1_{L}\wedge
j^{\prime})(i^{\prime\prime}\wedge
1_{K}))f^{\prime}=(1_{E_{s+2}}\wedge\psi_{X\wedge M}u^{\prime})d(f_{5})$

$\quad+(1_{E_{s+2}}\wedge\psi_{X\wedge M}(1_{X}\wedge
ij)u^{\prime})d((1_{E_{s+2}}\wedge 1_{L}\wedge m_{K})f_{4})$

$\quad+(1_{E_{s+2}}\wedge\psi_{X\wedge M}u^{\prime}(1_{L}\wedge
m_{K}))f_{4}-\lambda_{0}(1_{E_{s+2}}\wedge(r\wedge
1_{M})\overline{m}_{K})(\kappa\wedge 1_{K})$

$\quad+(\bar{c}_{s+1}\wedge 1_{Y\wedge M})d(g_{3})$
By (9.4.32) we have $\psi_{X\wedge M}u^{\prime}=(r\wedge
1_{M})\overline{m}_{K}(j^{\prime\prime}\wedge 1_{K})$ so that
$(j\overline{u}\wedge 1_{M})\psi_{X\wedge
M}u^{\prime}=j^{\prime}(j^{\prime\prime}\wedge 1_{K})$. Then, by composing
$1_{E_{s+2}}\wedge\phi\cdot j\overline{u}\wedge 1_{M}$ on (9.4.39), it becomes
(9.4.40) $\lambda_{0}(1_{E_{s+2}}\wedge(\phi\wedge
1_{M})j^{\prime})(\kappa\wedge 1_{K})=(1_{E_{s+2}}\wedge(\phi\wedge
1_{M})j^{\prime}(j^{\prime\prime}\wedge 1_{K}))d(f_{5})$

$\quad+(1_{E_{s+2}}\wedge(\phi\wedge 1_{M})ijj^{\prime}(j^{\prime\prime}\wedge
1_{K}))d((1_{E_{s+2}}\wedge 1_{L}\wedge m_{K})f_{4})=0$
here we use $(1_{E_{s+2}}\wedge(\phi\cdot j\overline{u}\wedge
1_{M})\psi_{X\wedge M}u^{\prime}(1_{L}\wedge m_{K}))f_{4}=(1_{E_{s+2}}\wedge
j^{\prime}(j^{\prime\prime}\wedge 1_{K})(1_{L}\wedge
m_{K}))f_{4}=(\bar{c}_{s+1}\wedge 1_{L\wedge M})(1_{KG_{s+1}}\wedge(\phi\wedge
1_{M})m_{M}(\overline{u}\wedge 1_{M}))g_{3}$ = 0 and by
$1_{L}\wedge\alpha_{1}=\phi\cdot j^{\prime\prime}$(up to nonzero scalar) we
obtain that $(1_{E_{s+2}}\wedge(\phi\wedge
1_{M})j^{\prime}(j^{\prime\prime}\wedge
1_{K}))d(f_{5})=(1_{E_{s+2}}\wedge(1_{L}\wedge
j^{\prime}\alpha^{\prime}))d(f_{5})$ = 0 ( cf. Prop. 9.4.33) and so the first
term of the right hand side of (9.4.40) is zero. the second term of the right
hand side of (9.4.40) is zero by the same reason.

It follows from (9.4.40) that $(1_{E_{s+2}}\wedge\phi\wedge
1_{K})(\kappa\wedge 1_{K})=(1_{E_{s+2}}\wedge(1_{L}\wedge\mu(i^{\prime}i\wedge
1_{K}))(\phi\wedge 1_{K})(\kappa\wedge 1_{K})(jj^{\prime}\wedge 1_{K})\nu$ = 0
and so we have $(1_{E_{s+2}}\wedge\alpha^{\prime\prime})(\kappa\wedge
1_{K})=(1_{E_{s+2}}\wedge\overline{\Delta}(\phi\wedge 1_{K}))(\kappa\wedge
1_{K})$ = 0 and the Theorem follows.

Case 2 $m_{M}(\overline{u}\wedge 1_{M})(\bar{h}\wedge
1_{M})=-\lambda_{0}(j^{\prime\prime}\wedge 1_{M})$.

In this case, the left hand side of (9.4.38) changes sign, then
$-(1_{E_{s+2}}\wedge j^{\prime}(j^{\prime\prime}\wedge 1_{K})(1_{L}\wedge
m_{K}))f_{4}+(\bar{c}_{s+1}\wedge 1_{M})(1_{KG_{s+1}}\wedge
m_{M}(\overline{u}\wedge 1_{M}))g_{3}=-2\lambda_{0}(1_{E_{s+2}}\wedge
j^{\prime})(\kappa\wedge 1_{K})$ and by composing $1_{E_{s+2}}\wedge\phi\cdot
j\overline{u}\wedge 1_{M}$ on (9.4.39) we have

$\qquad(\lambda_{0}-2\lambda_{0})(1_{E_{s+2}}\wedge(\phi\wedge
1_{M})j^{\prime})(\kappa\wedge 1_{K})$

$=\lambda_{0}(1_{E_{s+2}}\wedge(\phi\wedge 1_{M})j^{\prime})(\kappa\wedge
1_{K})-(1_{E_{s+2}}\wedge(\phi\wedge 1_{M})j^{\prime}(j^{\prime\prime}\wedge
1_{K})(1_{L}\wedge m_{K}))f_{4}$

$=(1_{E_{s+2}}\wedge(\phi\wedge 1_{M})j^{\prime}(j^{\prime\prime}\wedge
1_{K}))d(f_{5})$

$\qquad+(1_{E_{s+2}}\wedge(\phi\wedge
1_{M})ijj^{\prime}(j^{\prime\prime}\wedge 1_{K}))d((1_{E_{s+2}}\wedge
1_{L}\wedge m_{K})f_{4})$ = 0
so that the Theorem follows by the same reason. Q.E.D.

§5. A sequence of $h_{0}\sigma$ new families in the stable homotopy groups of
spheres

In this section, the convergence of a sequence of $h_{0}\sigma$ and
$h_{0}\sigma^{\prime}$ new families will be derived by the main Theorem A in
§2 and the main Theorem C in §4, where $\sigma$ and $\sigma^{\prime}$ is a
pair of $a_{0}$-related elements.

Theorem 9.5.1 Let $p\geq 7,n\geq 2$, then

$h_{0}h_{n}\in Ext_{A}^{2,p^{n}q+q}(Z_{p},Z_{p}),\quad h_{0}b_{n-1}\in
Ext_{A}^{3,p^{n}q+q}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge in the ASS to homotopy
elements of order $p$ in $\pi_{p^{n}q+q-2}S,\pi_{p^{n}q+q-3}S$ respectively.

Proof : By [12] Theorem 1.2.14 we have $d_{2}(h_{n})=a_{0}b_{n-1}\in
Ext_{A}^{3,p^{n}q+1}\\\ (Z_{p},Z_{p}),n\geq 1$, where
$d_{2}:Ext_{A}^{1,p^{n}q}(Z_{p},Z_{p})\to Ext_{A}^{3,p^{n}q+1}(Z_{p},Z_{p})$
is a secondary differential in the ASS. That is, $h_{n}$ and $b_{n-1}$ is a
pair of $a_{0}$-related elements so that the main Theorem A can apply to
$(\sigma,\sigma^{\prime})=(h_{n},b_{n-1}),(s,tq)=(1,p^{n}q)$. We only need to
check the supposition (I)(II)(III) in the main Theorem A hold. By knowledge on
the $Z_{p}$-base of $Ext_{A}^{s,*}(Z_{p},Z_{p})$ for $(s\leq 3)$ we know that
the the supposition (I)(II) of the main Theorem A hold for
$(\sigma,\sigma^{\prime})=(h_{n},b_{n-1}),(s,tq)=(1,p^{n}q)$. On the other
hand, from some results on $Ext_{A}^{4,*}(Z_{p},Z_{p})$ in [17] we know that
the following hold.

$Ext_{A}^{4,p^{n}q+rq+1}(Z_{p},Z_{p})=0(r=1,3,4),$

$Ext_{A}^{4,p^{n}q+rq}(Z_{p},Z_{p})=0(r=2,3)$,

$Ext_{A}^{4,p^{n}q+2q+1}(Z_{p},Z_{p})\cong
Z_{p}\\{\widetilde{\alpha}_{2}b_{n-1}\\}$,

$Ext_{A}^{4,p^{n}q+2}(Z_{p},Z_{p})\cong
Z_{p}\\{a_{0}^{2}b_{n-1}\\},Ext_{A}^{4,p^{n}q+1}(Z_{p},Z_{p})=0$.
That is, the supposition )III) of the main Theorem A hold for
$(\sigma,\sigma^{\prime})=(h_{n},b_{n-1}),(s,tq)=(1,p^{n}q)$. Then, by the
main Theorem A we obtain that $h_{0}b_{n-1}\in
Ext_{A}^{3,p^{n}q+q}(Z_{p},Z_{p})$, $i_{*}(h_{0}h_{n})\in
Ext_{A}^{2,p^{n}q+q}\\\ (H^{*}M,Z_{p})$ are permanent cycles in the ASS. By
Remark 9.2.35, the main Theorem A also obtains that $(1_{L}\wedge
i)_{*}\phi_{*}(h_{n})\in Ext_{A}^{2,p^{n}q+2q}(H^{*}L,Z_{p})$ is a permanent
cycle in the ASS so that the main Theorem C can apply to obtain the result of
the Theorem, this is because by knowledge on the $A_{p}$-base of
$Ext_{A}^{s,*}(Z_{p},Z_{p})$ for $s=1,2,3$ we can easy to see that the
supposition (I) of the main Theorem C hold for
$(\sigma,\sigma^{\prime},s,tq)=(h_{n},b_{n-1},1,p^{n}q)$. Q.E.D.

Now we apply the main Theorem A and the main Theorem C to
$(\sigma,\sigma^{\prime})=(h_{n}h_{m},h_{n}b_{m-1}-h_{m}b_{n-1}),(s,tq)=(2,p^{n}q+p^{m}q)$
to obtain another sequence of $h_{0}\sigma$ families in the stable homotopy
groups of spheres. For checking the supposition (I)(II)(III) of the main
Theorem A, we first prove the following Proposition.

Proposition 9.5.2 Let $p\geq 7,n\geq m+2\geq 4,tq=p^{n}q+p^{m}q$, then
(1) $Ext_{A}^{4,tq+rq+u}(Z_{p},Z_{p})$ = 0 for $r=2,3,4,u=-1,0$ or
$r=3,4,u=1$,

$Ext_{A}^{4,tq+q}(Z_{p},Z_{p})\cong
Z_{p}\\{h_{0}h_{n}b_{m-1},h_{0}h_{m}b_{n-1}\\},$

$Ext_{A}^{4,tq}(Z_{p},Z_{p})\cong Z_{p}\\{b_{n-1}b_{m-1}\\},$

$Ext_{A}^{4,tq+1}(Z_{p},Z_{p})\cong
Z_{p}\\{a_{0}h_{n}b_{m-1},a_{0}h_{m}b_{n-1}\\}$
(2) $Ext_{A}^{5,tq+rq+1}(Z_{p},Z_{p})=0$ for $r=1,3,4$,

$Ext_{A}^{5,tq+rq}(Z_{p},Z_{p})$ = 0 for $r=2,3$,

$Ext_{A}^{5,tq+2q+1}(Z_{p},Z_{p})\cong
Z_{p}\\{\widetilde{\alpha}_{2}h_{n}b_{m-1},\widetilde{\alpha}_{2}h_{m}b_{n-1}\\}$,

$Ext_{A}^{5,tq+2}(Z_{p},Z_{p})\cong
Z_{p}\\{a_{0}^{2}h_{n}b_{m-1},a_{0}^{2}h_{m}b_{n-1}\\}$,

$Ext_{A}^{5,tq+1}(Z_{p},Z_{p})\cong Z_{p}\\{a_{0}b_{n-1}b_{m-1}\\}$,

$a_{0}^{2}b_{n-1}b_{m-1}\neq 0\in Ext_{A}^{6,tq+2}(Z_{p},Z_{p})$

Proof : By Theorem 5.5.3, there is a May spectral sequence (MSS)
$\\{E_{r}^{s,t,*},d_{r}\\}$ which converges to $Ext_{A}^{s,t}(Z_{p},Z_{p})$
and whose $E_{1}$-term is

$\quad E_{1}^{*,*,*}=E(h_{i,j}\mid i>0,j\geq 0)\otimes P(b_{i,j}\mid i>0,j\geq
0)\otimes P(a_{i}\mid i\geq 0)$,
where $E$ denotes the exterior algebra and $P$ denotes a polynomial algebra, ,
$h_{i,j}\in$ $E_{1}^{1,2(p^{i}-1)p^{j},2i-1}$, $b_{i,j}\in\\\
E_{1}^{2,2(p^{i}-1)p^{j+1},p(2i-1)},a_{i}\in E_{1}^{1,2p^{i}-1,2i+1}$.
Consider the following second degrees (mod $p^{n}q$) of the generators in the
$E_{1}^{*,*,*}$-term, where $0\leq i\leq n,n\geq m+2\geq 4$

$\qquad\mid h_{s,i}\mid=(p^{s+i-1}+\cdots+p^{i})q$ (mod $p^{n}q),\quad 0\leq
i<s+i-1<n$

$\qquad\qquad\quad=(p^{n-1}+\cdots+p^{i})q$ (mod $p^{n}q$ ), $0\leq
i<s+i-1=n$,

$\qquad\mid b_{s,i-1}\mid=(p^{s+i-1}+\cdots+p^{i})q$ (mod $p^{n}q$), $1\leq
i<s+i-1<n$,

$\qquad\qquad\quad=(p^{n-1}+\cdots+p^{i})q$ (mod $p^{n}q$), $1\leq i<s+i-1=n$.

$\qquad\mid a_{i+1}\mid=(p^{i}+\cdots+1)q+1$ (mod $p^{n}q$ ), $1\leq i<n$

$\qquad\mid a_{i+1}\mid=(p^{n-1}+\cdots+1)q+1$ (mod $p^{n}q$ ), $i=n$.
For degree $k=tq+rq+u$ such that $0\leq r\leq 4,-1\leq u\leq 2$ we have
$k\equiv p^{m}q+rq+u$ (mod $p^{n}q$ ). Then, for $3\leq w\leq 5$,
$E_{1}^{w,tq+rq+u,*}$ has no such generators which have one of the above
elements as a factor, this is because such a generator will have second degree
$(c_{n}p^{n-1}+\cdots+c_{1}p+c_{0})q+d$ (mod $p^{n}q$ ), where $c_{i}\neq
0(1\leq i\leq m-1$ or $m<i<n$), $0\leq c_{l}<p,l=0,\cdots,n,0\leq d\leq 5$. In
addition, the second degree $\mid b_{1,i-1}\mid=p^{i}q$ (mod $p^{n}q$)( $1\leq
i\leq n$),$\mid h_{1,i}\mid=p^{i}q$ (mod $p^{n}q$)( $0\leq i\leq n$). Then ,
exclude the above factor and the factor which has second degree $\geq tq+pq$,
we know that the only possibility of the factor of the generators in
$E_{1}^{w,tq+rq+u,*}$ are $a_{1},a_{0},h_{1,0}$ , $h_{1,n},h_{1.m},$
$b_{1,n-1},b_{1,m-1}$ .

Then, by degree reasons we have

$E_{1}^{4,tq+rq+1,*}$ = 0 for $r=3,4$, $E_{1}^{4,tq+rq+u,*}$ = 0 for
$r=2,3,4,u=-1,0$

$E_{1}^{4,tq,*}=Z_{p}\\{b_{1,n-1}b_{1,m-1}\\},\quad E_{1}^{4,tq+1,*}\cong
Z_{p}\\{a_{0}h_{1,n}b_{1,m-1},a_{0}h_{1,m}b_{1,n-1}\\}$,

$E_{1}^{4,tq+2,*}=Z_{p}\\{a_{0}^{2}h_{1,n}h_{1,m}\\}$,

$E_{1}^{4,tq+2q+1,*}=Z_{p}\\{h_{1,0}a_{1}h_{1,n}h_{1,m}\\}$,

$E_{1}^{4,tq+q,*}=Z_{p}\\{h_{1,0}h_{1,n}b_{1,m-1},h_{1,0}h_{1,m}b_{1,n-1}\\}$,

$E_{1}^{3,tq+1,*}=Z_{p}\\{a_{0}h_{1,n}h_{1,m}\\}$,
$E_{1}^{3,tq.*}=Z_{p}\\{h_{1,n}b_{1,m-1},h_{1,m}b_{1,n-1}\\}$,

$E_{1}^{3,tq+q,*}=Z_{p}\\{h_{1,0}h_{1,n}h_{1,m}\\}$, $E_{1}^{3,tq+2q+1,*}=0$
Note that the differentials in the MSS is derivative, that is,
$d_{r}(xy)=d_{r}(x)y+(-1)^{s}xd_{r}(y)$ for $x\in E_{1}^{s,t,*},y\in
E_{1}^{s^{\prime},t^{\prime},*}$. In addition,
,$a_{0},h_{1,n},b_{1,n-1},h_{1,0}a_{1}$ are permanent cycles in the MSS which
converge to $a_{0},h_{n},b_{n-1},\widetilde{\alpha}_{2}\in Ext_{A}^{*,*}\\\
(Z_{p},Z_{p})$ respectively.

Then, the differential $d_{r}E_{r}^{3,tq+sq+u,*}$ = 0 for all $r\geq 1$ and
$s=u=0$ or $s=1,u=0$ or $s=0,u=1$ or $s=2,u=1$ so that
$b_{1,n-1}b_{1,m-1},a_{0}h_{1,n}b_{1,m-1},$
$a_{0}h_{1,m}b_{1,n-1},h_{1,0}h_{1,n}b_{1,m-1},h_{1,0}h_{1,m}b_{1,n-1}\in
E_{r}^{4,*,*}$ are not $d_{r}$-boundary in the MSS and so
$b_{n-1}b_{m-1},a_{0}h_{n}b_{m-1},a_{0}h_{m}b_{n-1},h_{0}h_{n}b_{m-1},\\\
h_{0}h_{m}b_{n-1}$ are all nontrivial in $Ext_{A}^{4,*}(Z_{p},Z_{p})$. This
shows (1).

Similarly, by degree reasons we have

$E_{1}^{5,tq+q+1,*}\cong
Z_{p}\\{a_{0}h_{1,0}h_{1,n}b_{1,m-1},a_{0}h_{1,0}h_{1,m}b_{1,n-1},a_{1}b_{1,n-1}b_{1,m-1}\\}$

$E_{1}^{5,tq+rq+1,*}$ = 0 for $r=3,4$, $E_{1}^{5,tq+rq,*}=0$ for $r=2,3$

$E_{1}^{5,tq+2q+1,*}\cong
Z_{p}\\{h_{1,0}a_{1}h_{1,n}b_{1,m-1},h_{1,0}a_{1}h_{1,m}b_{1,n-1}\\}$

$E_{1}^{5,tq+1,*}=Z_{p}\\{a_{0}b_{1,n-1}b_{1,m-1}\\},\quad
E_{1}^{4,tq+2q+1,*}\cong Z_{p}\\{h_{1,0}a_{1}h_{1,n}h_{1,m}\\}$

$E_{1}^{5,tq+2,*}=Z_{p}\\{a_{0}^{2}h_{1,m}b_{1,n-1},a_{0}^{2}h_{1,n}b_{1,n-1}\\}$
All the generators of $E_{1}^{5,tq+q+1,*}$ dy in the MSS, this is because
$a_{0}h_{1,0}h_{1,n}b_{1,m-1}\\\ =-d_{1}(a_{1}h_{1,n}b_{1,m-1})$,
$a_{0}h_{1,0}h_{1,m}b_{1,n-1}=-d_{1}(a_{1}h_{1,m}b_{1,n-1})$ and

$d_{1}(a_{1}b_{1,n-1}b_{1,m-1})=-a_{0}h_{1,0}b_{1,n-1}b_{1,m-1}\neq 0\in
E_{1}^{5,tq+q+1,*}$

Then, $Ext_{A}^{5,tq+q+1}(Z_{p},Z_{p})$ = 0. In addition, similar to that
given in (1) we have, $d_{r}E_{r}^{4,tq+u,*}=0$, $d_{r}E_{r}^{4,tq+2q+1,*}=0$
for all $r\geq 1$, $u=1,2$ . Then, the generators in $E_{1}^{5,*,*}$ converge
in the MSS to
$\widetilde{\alpha}_{2}h_{n}b_{m-1},\widetilde{\alpha}_{2}h_{m}b_{n-1},a_{0}b_{n-1}b_{m-1},\\\
a_{0}^{2}h_{m}b_{n-1},a_{0}^{2}h_{n}b_{m-1}$ respectively. For the last
result, note that $d_{r}E_{r}^{5,tq+2,*}$ = 0 for all $r\geq 1$ , then
$a_{0}^{2}b_{n-1}b_{m-1}\neq 0\in Ext_{A}^{6,tq+2}(Z_{p},Z_{p})$. This shows
(2). Q.E.D.

Theorem 9.5.3 Let $p\geq 7,n\geq m+2\geq 4$, then

$h_{0}h_{n}h_{m}\in Ext_{A}^{3,p^{n}q+p^{m}q+q}(Z_{p},Z_{p}),$

$h_{0}(h_{n}b_{m-1}-h_{m}b_{n-1})\in Ext_{A}^{4,p^{n}q+p^{m}q+q}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge to homotopy elements of
order $p$ in $\pi_{p^{n}q+p^{m}q+q-3}S$ and $\pi_{p^{n}q+p^{m}q+q-4}S$
respectively.

Proof : By [12]p.11 Theorem 1.2.14, there is a nontrivial secondary
differential $d_{2}(h_{n})=a_{0}b_{n-1}(n\geq 1$ ) and it follows that
$d_{2}(h_{n}h_{m})=d_{1}(h_{n})h_{m}+(-1)^{1+p^{n}q}h_{n}d_{2}(h_{m})=a_{0}h_{m}b_{n-1}-a_{0}h_{n}b_{m-1}$.
That is, $(h_{n}h_{m},h_{m}b_{n-1}-h_{n}b_{m-1})$ is a pair of $a_{0}$-related
elements. By applying the main Theorem A to
$(\sigma,\sigma^{\prime})=(h_{n}h_{m},h_{m}b_{n-1}-h_{n}b_{m-1}),(s,tq)=(2,p^{n}q+p^{m}q)$
we have $h_{0}(h_{m}b_{n-1}-h_{n}b_{m-1})\\\ \in
Ext_{A}^{4,p^{n}q+p^{m}q+q}(Z_{p},Z_{p})$ and $i_{*}(h_{0}h_{n}h_{m})\in
Ext_{A}^{3,p^{n}q+p^{m}q+q}(Z_{p},Z_{p})$ are permanent cycles in the ASS,
this is because by knowledge of $Z_{p}$-base of $Ext_{A}^{s,*}(Z_{p},\\\
Z_{p})$ for $s\leq 3$ we know that the supposition (I)(II)(III) of the main
Theorem A hold. By Remark 9.2.35, the main Theorem A also obtains that
$(1_{L}\wedge i)_{*}\phi_{*}(h_{n}h_{m})\in
Ext_{A}^{3,p^{n}q+p^{m}q+2q}(H^{*}L,Z_{p})$ is a permanent cycle in the ASS so
that by the main Theorem C , the result of the Theorem follows. This is
because the supposition (I) of the main Theorem C hold by the knowledge of the
$Z_{p}$-base of $Ext_{A}^{s,*}(Z_{p},Z_{p})$ for $s=1,2,3$. Q.E.D.

From Theorem 9.5.1 and Theorem 9.5.3 , we obtain four families of
$h_{0}\sigma$ new families. In fact, there are many pairs of $a_{0}$-related
elements so that we can expect to obtain some other sequence of $h_{0}\sigma$
new families in the stable homotopy groups of speheres. We have the following
conjectures.

Conjecture 9.5.4 Let $p\geq 7,n\geq 3$, then there is a secondary differential
$d_{2}(g_{n})=a_{0}l_{n}\in Ext_{A}^{4,p^{n+1}q+2p^{n}q+1}(Z_{p},Z_{p}),n\geq
3$ (up to nonzero scalar) and

$h_{0}g_{n}\in Ext_{A}^{3,p^{n+1}q+2p^{n}q+q}(Z_{p},Z_{p})$

$h_{0}l_{n}\in Ext_{A}^{4,p^{n+1}q+2p^{n}q+q}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge to homotopy elements of
order $p$ in $\pi_{p^{n+1}q+2p^{n}q+q-3}S$ and $\pi_{p^{n+1}q+2p^{n}q+q-4}S$
respectively, where $g_{n}\in
Ext_{A}^{2,p^{n+1}q+2p^{n}q}(Z_{p},Z_{p}),l_{n}\in
Ext_{A}^{3,p^{n+1}q+2p^{n}q}(Z_{p},Z_{p})$.

Conjecture 9.5.5 Let $p\geq 7,n\geq 3$, then there is a secondary differential
$d_{2}(k_{n})=a_{0}l^{\prime}_{n}\in
Ext_{A}^{4,2p^{n+1}q+p^{n}q+1}(Z_{p},Z_{p})$ ( up to nonzero scalar), $n\geq
3$ and

$h_{0}k_{n}\in Ext_{A}^{3,2p^{n+1}q+p^{n}q+q}(Z_{p},Z_{p})$

$h_{0}l^{\prime}_{n}\in Ext_{A}^{4,2p^{n+1}q+p^{n}q+q}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge to homotopy elementws of
order $p$ in $\pi_{2p^{n+1}+p^{n}q+q-3}S$ and $\pi_{2p^{n+1}q+p^{n}q+q-4}S$ ,
where $k_{n}\in Ext_{A}^{2,2p^{n+1}q+p^{n}q}(Z_{p},Z_{p}),l^{\prime}_{n}\in
Ext_{A}^{3,2p^{n+1}q+p^{n}q}(Z_{p},Z_{p})$.

Remark 9.5.6 By [10][25], there is Thom map
$\Phi:Ext_{BP_{*}BP}^{s,*}(BP_{*},BP_{*})\\\ \rightarrow
Ext_{A}^{s,*}(Z_{p},Z_{p})$( $s=2,3$) such that
$\Phi(\beta_{p^{n-1}/p^{n-1}-1})=h_{0}h_{n}$,
$\Phi(\beta_{p^{n-1}/p^{n-1}})=b_{n-1}$,
$\Phi(\beta_{p^{m-1}/p^{m-1}-1}\beta_{p^{n-1}/p^{n-1}}-\beta_{p^{n-1}/p^{n-1}-1}\beta_{p^{m-1}/p^{m-1}})=h_{0}(h_{m}b_{n-1}\\\
-h_{n}b_{m-1})$,
$\Phi(\gamma_{p^{n-2}/p^{n-2}-p^{m-1},p^{m-1}-1})=h_{0}h_{n}h_{m}$. Then, the
$h_{0}h_{n},h_{0}b_{n-1},\\\
h_{0}(h_{m}b_{n-1}-h_{n}b_{m-1}),h_{0}h_{n}h_{m}$-map obtained by Theorem
9.5.1 and Theorem 9.5.3 are represented by $\beta_{p^{n-1}/p^{n-1}-1}$ \+
other terms $\in Ext_{BP_{*}BP}^{2,p^{n}q+q}(BP_{*},BP_{*})$,
$\alpha_{1}\beta_{p^{n-1}/p^{n-1}}$ \+ other terms $\in
Ext_{BP_{*}BP}^{3,p^{n}q+q}(BP_{*},BP_{*})$,
$\beta_{p^{m-1}/p^{m-1}-1}\beta_{p^{n-1}/p^{n-1}}-\beta_{p^{n-1}/p^{n-1}-1}\cdot\beta_{p^{m-1}/p^{m-1}}$
\+ other terms $\in Ext_{BP_{*}BP}^{4,p^{n}q+p^{m}q+q}(BP_{*},BP_{*})$,
$\gamma_{p^{n-2}/p^{n-2}-p^{m-1},p^{m-1}-1}$ \+ other terms $\in
Ext_{BP_{*}BP}^{3,p^{n}q+p^{m}q+q}(BP_{*},BP_{*})$ respectively in the Adams-
Novikov spectral sequence.

§6. A sequence of
$h_{0}\sigma\widetilde{\gamma}_{s},g_{0}\sigma\widetilde{\gamma}_{s}$ new
families in the stable homotopy groups of spheres

In this section, we use the main Theorem B to obtain
$i^{\prime}_{*}i_{*}(g_{0}h_{n}),\\\
i^{\prime}_{*}i_{*}(g_{0}b_{n-1}),i^{\prime}_{*}i_{*}(g_{0}h_{n}h_{m}),i^{\prime}_{*}i_{*}(g_{0}(h_{n}b_{m-1}-b_{m}b_{n-1}))$
et al converge to the corresponding nontrivial homotopy elements in the
homotopy groups of Smith-Toda spectrum $V(1)$. In base of these results, we
obtain a sequence of
$g_{0}\sigma\widetilde{\gamma}_{s},h_{0}\sigma\widetilde{\gamma}_{s}$ new
families in the stable homotopy groups of spheres.

Theorem 9.6.1 Let $p\geq 5,n\geq 2$, then

$i^{\prime}_{*}i_{*}(g_{0}h_{n})\in Ext_{A}^{3,p^{n}q+pq+2q}(H^{*}K,Z_{p})$

$i^{\prime}_{*}i_{*}(g_{0}b_{n-1})\in Ext_{A}^{4,p^{n}q+pq+2q}(H^{*}K,Z_{p})$

are permanent cycles in the ASS and they converge to the corresponding
homotopy element in $\pi_{p^{n}q+pq+2q-3}K,\pi_{p^{n}q+pq+2q-4}K$
respectively.

Proof : We first apply the main Theorem B to $(\sigma,s,tq)=(h_{n},1,p^{n}q)$.
By Theorem 9.5.1, the supposition (II) of the main Theorem B holds. Moreover,
by knowledge on the $Z_{p}$-base of $Ext_{A}^{s,*}(Z_{p},Z_{p})$ for $s=1,2,3$
we know that the supposition (I) of the main Theorem B holds, then the first
result of the Theorem follows by the main Theorem B.

For the second result, we apply the main Theorem B to
$(\sigma,s,tq)=(b_{n-1},2,p^{n}q)$. Similarly by Theorem 9.5.1, the
supposition (II) of the main Theorem B holds. Noreover, by knowledge on the
$Z_{p}$-base of $Ext_{A}^{s,*}(Z_{p},Z_{p})$ for $s=2,3$ and some result in
[17] on $Ext_{A}^{4,*}(Z_{p},Z_{p})$ we know that the supposition (I) of the
main Theorem B holds. Then , the second result also follows by the main
Theorem B. Q.E.D.

alternative Proof : It is known from the proof of Theorem 9.5.1 that the
supposition (I)(II)(III) if the main Theorem A hold for
$(\sigma,\sigma^{\prime})=(h_{n},b_{n-1}),(s,tq)=(1,p^{n}q)$. Then , applying
the main Theorem B’ in $\S 3$ to
$(\sigma,\sigma^{\prime})=(h_{n},b_{n-1}),(s,tq)=(1,p^{n}q)$ we obtain the two
results of the Theorem. Q.E.D.

Theorem 9.6.2 Let $p\geq 7,n\geq m+2\geq 4$, then

$i^{\prime}_{*}i_{*}(g_{0}h_{n}h_{m})\in
Ext_{A}^{4,p^{n}q+p^{m}q+pq+2q}(H^{*}K,Z_{p})$

$i^{\prime}_{*}i_{*}(g_{0}(h_{n}b_{m-1}-h_{m}b_{n-1}))\in
Ext_{A}^{5,p^{n}q+p^{m}q+pq+2q}(H^{*}K,Z_{p})$

are permanent cycles in the ASS and they converge to nontrivial homotopy
elements in $\pi_{p^{n}q+p^{m}q+pq+2q-4}K,\pi_{p^{n}q+p^{m}q+pq+2q-5}K$
respectively.

Proof : We first apply the main Theorem B to
$(\sigma,s,tq)=(h_{n}h_{m},2,p^{n}q+p^{m}q)$. By Theorem 9.5.3, the
supposition (II) of the main Theorem B holds. By knowledge on the $Z_{p}$-base
of $Ext_{A}^{s,*}(Z_{p},Z_{p})$ for $s=2,3$ and some result in [17] on
$Ext_{A}^{4,*}(Z_{p},Z_{p})$ we know that the supposition (I) of the main
Theorem B also holds. Then the first result follows by the main Theorem B.
Moreover, we apply the main Theorem B to
$(\sigma,s,tq)=(h_{n}b_{m-1}-h_{m}b_{n-1},3,p^{n}q+p^{m}q)$. Similarly by
Theorem 9.5.3, the supposition (II) of the main Theorem B holds. By knowledge
on the $Z_{p}$-base of $Ext_{A}^{3,*}(Z_{p},Z_{p})$ for $s=3,4$ and the result
on $Ext_{A}^{5,p^{n}q+p^{m}q+2q+1}(Z_{p},Z_{p})\cong
Z_{p}\\{\widetilde{\alpha}_{2}h_{n}b_{m-1},\widetilde{\alpha}_{2}h_{m}b_{n-1}\\}$
in Prop. 9.5.2 we know that the supposition (I) of the main Theorem B holds.
Then, the second result follows immediately by the main Theorem B. Q.E.D.

alternative Proof : It is known from the proof of Theorem 9.5.2 that the
supposition (I)(II)(III) of the main Theorem A hold for
$(\sigma,\sigma^{\prime})=(h_{n}h_{m},h_{n}b_{m-1}-h_{m}b_{n-1}),(s,tq)=(2,p^{n}q+p^{m}q)$.
Then by applying the main Theorem B’ in §3 to
$(\sigma,\sigma^{\prime})=(h_{n}h_{m},h_{n}b_{m-1}-h_{m}b_{n-1}),(s,tq)=(2,p^{n}q+p^{m}q)$,
we obtain the two result of the Theorem. Q.E.D.

Using the notation in the cofibration (6.2.7)–(6.2.10), we know that

$\widetilde{\gamma}_{s}=((j_{1}j_{2}j_{3})_{*}(\gamma)_{*}^{s}(i_{3}i_{2}i_{1})_{*}(1)\in
Ext_{A}^{s,sp^{2}q+(s-1)pq+(s-2)q+s-3}(Z_{p},Z_{p})$

converges to the following third periodicity element in the ASS

$\gamma_{s}=j_{1}j_{2}j_{3}\gamma^{s}i_{3}i_{2}i_{1}\in\pi_{sp^{2}q+(s-1)pq+(s-2)q-3}S$

where $3\leq s<p$ and $1\in Ext_{A}^{0,0}(Z_{p},Z_{p})$. Now we consider the
products
$g_{0}\sigma\widetilde{\gamma}_{s},h_{0}\sigma\widetilde{\gamma}_{s}$, in
$Ext_{A}^{*,*}(Z_{p},Z_{p})$ and we will prove that they converge to the
corresponding homotopy element of order $p$ in the stable homotopy groups of
spheres, where $\sigma=h_{n},b_{n-1},h_{n}h_{m},$ or
$h_{n}b_{m-1}-h_{m}b_{n-1}$.

Theorem 9.6.3 Let $p\geq 7,n\geq 3,3\leq s<p$, then the products

$g_{0}h_{n}\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+3,p^{n}q+sp^{2}q+spq+sq+s-3}(Z_{p},Z_{p})$

$g_{0}b_{n-1}\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+4,p^{n}q+sp^{2}q+spq+sq+s-3}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge to the corresponding
homotopy elements of order $p$ in the stable homotopy groups of spheres.

Proof: By Theorem 9.6.1, there is a nontrivial $f\in\pi_{p^{n}q+pq+2q-3}K$
such that it is represented by $i^{\prime}_{*}i_{*}(g_{0}h_{n})\in
Ext_{A}^{3,p^{n}q+pq+2q}(H^{*}K,Z_{p})$ in the ASS. Let
$\tilde{f}=j_{1}j_{2}j_{3}\gamma^{s}i_{3}f$ be the following composition
($tq=p^{n}q+pq+2q-3$)

$\qquad\tilde{f}:\Sigma^{tq}S\stackrel{{\scriptstyle
f}}{{\longrightarrow}}V(1)\stackrel{{\scriptstyle
i_{3}}}{{\longrightarrow}}V(2)\stackrel{{\scriptstyle\gamma^{s}}}{{\longrightarrow}}\Sigma^{-s(p^{2}q+pq+q)}V(2)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
j_{1}j_{2}j_{3}}}{{\longrightarrow}}\Sigma^{-s(p^{2}+p+1)q+(p+2)q+q+3}S$
Since $f$ is represented by $(i_{2})_{*}(i_{1})_{*}(g_{0}h_{n})\in
Ext_{A}^{3,p^{n}q+pq+2q}(H^{*}K,Z_{p})$ in the ASS, then the above $\tilde{f}$
is represented by

$c=(j_{1}j_{2}j_{3})_{*}(\gamma_{*})^{s}(i_{3}i_{2}i_{1})_{*}(g_{0}h_{n})\in
Ext_{A}^{s+3,p^{n}q+s(p^{2}+p+1)q+s-3}(Z_{p},Z_{p})$

By knowledge of Yoneda products we know that the above element $c$ is just the
products $g_{0}h_{n}\widetilde{\gamma}_{s}\in
Ext_{A}^{s+3,p^{n}q+s(p^{2}+p+1)q+s-3}(Z_{p},Z_{p})$. Then, to obtain the
first result, it suffices to prove the product
$g_{0}h_{n}\widetilde{\gamma}_{s}$ is nonzero in the Ext group and it is not a
$d_{r}$-boundary in the ASS, that is, we still need to prove
$Ext_{A}^{s+3-r,p^{n}q+s(p^{2}+p+1)q+s-2-r}(Z_{p},Z_{p})$ is zero for $r\geq
2$. We may prove this two facts by an argument in the May spectral sequence.
By degree reasons, $h_{n},g_{0},\widetilde{\gamma}_{s}$ is represented by
$h_{1,n},h_{2,0}h_{1,0},h_{2,1}h_{1,2}h_{3,0}a_{3}^{s-3}\in E_{1}^{*,*,*}$ in
the MSS respectively. Then, the products $g_{0}h_{n}\widetilde{\gamma}_{s}$ is
represented by

$h_{1,n}h_{2,0}h_{1,0}h_{2,1}h_{1,2}h_{3,0}a_{3}^{s-3}\in
E_{1}^{s+3,p^{n}q+s(p^{2}+p+1)q+s-3,*}$

in the MSS and so we can do some computation in the degree to prove
$E_{1}^{s+2,p^{n}q+s(p^{2}+p+1)q+s-3,*}=0$ and
$E_{1}^{s+3-r,p^{n}q+s(p^{2}+p+1)q+s-2-r,*}=0(r\geq 2)$ so that the first
result follows. We leave this computation to the reader. The proof and
computation for the second result is similar. Q.E.D.

By using Theorem 9.6.2, Theorem 9.5.1 and Theorem 9.5.3, similar to that given
in the proof of Theorem 9.6.3, we can obtain the following Theorem
9.6.4–9.6.6.

Theorem 9.6.4 Let $p\geq 7,n\geq m+2\geq 5,3\leq s<p$, then

$g_{0}h_{n}h_{m}\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+4,p^{n}q+p^{m}q+s(p^{2}+p+1)q+s-3}(Z_{p},Z_{p})$

$g_{0}(h_{n}b_{m-1}-h_{m}b_{n-1})\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+5,p^{n}q+p^{m}q+s(p^{2}+p+1)q+s-3}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge to the corresponding
homotopy elements of order $p$ in the stable homotopy groups of spheres.

Theorem 9.6.5 Let $p\geq 7,n\geq 3,3\leq s<p$, then the products

$h_{0}h_{n}\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+2,p^{n}q+sp^{2}q+(s-1)(p+1)q+s-3}(Z_{p},Z_{p})$

$h_{0}b_{n-1}\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+3,p^{n}q+sp^{2}q+(s-1)(p+1)q+s-3}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge to the corresponding
elements of order $p$ in the stable homotopy groups of spheres.

Theorem 9.6.6 Let $p\geq 7,n\geq m+2\geq 5,3\leq s<p$, then the products

$h_{0}h_{n}h_{m}\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+3,p^{n}q+p^{m}q+sp^{2}q+(s-1)(p+1)q+s-3}(Z_{p},Z_{p})$

$h_{0}(h_{n}b_{m-1}-h_{m}b_{n-1})\widetilde{\gamma}_{s}\neq 0\in
Ext_{A}^{s+4,p^{n}q+p^{m}q+sp^{2}q+(s-1)(p+1)q+s-3}(Z_{p},Z_{p})$

are permanent cycles in the ASS and they converge to the corresponding
homotopy elements of order $p$ in the stable homotopy groups of spheres.

Remark 9.6.7 The new families obtained in Theorem 9.6.5 and Theorem 9.6.6

are the composition products of $h_{0}h_{n}$-element , $h_{0}b_{n-1}$-element
in Theorem 9.5.1,
$h_{0}h_{n}h_{m}$-element,$h_{0}(h_{n}b_{m-1}-h_{m}b_{n-1})$\- element in
Theorem 9.5.3 and
$\gamma_{s}=j_{1}j_{2}j_{3}\gamma^{s}i_{3}i_{2}i_{1}\in\pi_{sp^{2}q+(s-1)pq+(s-2)q-3}S$.
However, the new families obtained in Theorem 9.6.3 and Theorem 9.6.4 are
indecomposable elements in the stable homotopy groups of spheres, that is,
they are not compositions of some other elements of lower degrees in the
stable homotopy groups of spheres. This is because $g_{0}\in
Ext_{A}^{2,pq+2q}(Z_{p},Z_{p})$ dies in the ASS, that is, it support a
nontrivial differential in the Adams spectral sequence :
$d_{2}(g_{0})=b_{0}\widetilde{\alpha}_{2}$ (up to nonzero scalar) $\in
Ext_{A}^{4,pq+2q+1}(Z_{p},Z_{p})$ which can be easily proved as follows. Since
$\widetilde{\alpha}_{2},b_{0}$ converge in the ASS to
$\alpha_{2}=j\alpha^{2}i,\beta_{1}=jj^{\prime}\beta
i^{\prime}i\in\pi_{*}S$,then the composition products of
$\beta_{1}\alpha_{2}\in\pi_{pq+2q-3}S$ must be represented by
$b_{0}\widetilde{\alpha}_{2}\in Ext_{A}^{4,pq+2q+1}(Z_{p},Z_{p})$ in the ASS.
However, it is easily seen that
$\beta_{1}\alpha_{2}=jj^{\prime}\beta\i^{\prime}ij\alpha^{2}i=0$ and
$b_{0}\widetilde{\alpha}_{2}\neq 0\in Ext_{A}^{4,pq+2q+1}(Z_{p},Z_{p})$, then
$b_{0}\widetilde{\alpha}_{2}$ must be a $d_{r}$-boundary. By degree reason,
the only possibility is $b_{0}\widetilde{\alpha}_{2}=d_{2}(g_{0})$(up to
nonzero scalar).

Conjecture 9.6.8 By the conjecture 9.5.4–9.5.5, we can conjecture that, for
$p\geq 7,n\geq 3,3\leq s<p$, the products
$h_{0}g_{n}\widetilde{\gamma}_{s},h_{0}l_{n}\widetilde{\gamma}_{s},g_{0}g_{n}\widetilde{\gamma}_{s},g_{0}l_{n}\widetilde{\gamma}_{s},\\\
h_{0}k_{n}\widetilde{\gamma}_{s},h_{0}l^{\prime}_{n}\widetilde{\gamma}_{s},g_{0}k_{n}\widetilde{\gamma}_{s},g_{0}l^{\prime}_{n}\widetilde{\gamma}_{s}$
are permanent cycles in the ASS and they converge to the corresponding
homotopy elements of order $p$ in the stable homotopy groups of spheres. In
addition, all results or conjectures in this section also hold when we replace
the products with $\widetilde{\gamma}_{s}$ to be the products with
$\widetilde{\beta}_{s}(2\leq s<p)$. That is, we can obtain a sequence of
$h_{0}\sigma\widetilde{\beta}_{s},g_{0}\sigma\widetilde{\beta}_{s}$-elements,
where
$\widetilde{\beta}_{s}=(j_{1}j_{2})_{*}\beta_{*}^{s}(i_{1}i_{1})_{*}(1)\in
Ext_{A}^{s,spq+(s-1)q+s-2}(Z_{p},Z_{p}),2\leq s<p$.

§7. Third periodicity families in the stable homotopy groups of spheres

In this section, we will first prove the convergence of $h_{n}$-elements in
the homotopy groups of Smith-Toda spectrum $V(1)$ and in base of this we
obtain the convergence of third periodicity $\gamma_{p^{n}/s}$ families
$(1\leq s\leq p^{n}-1)$ in the Adams-Novikov spectral sequence.

Theorem 9.7.1 ([9] Theorem II) Let $p\geq 5,n\geq 0$,

$h_{n}\in Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},BP_{*}K)$

be the element represented by $[t_{1}^{p^{n}}]$ in the cobar complex. Then
this $h_{n}$ is a permanent cycle in the Adams-Novikov spectral sequence and
it converges to a nontrivial homotopy element in $\pi_{p^{n}q-1}K$.

The proof of the above $h_{n}$-Theorem will be the main content of this
section. By Theorem 8.1.6(b)(ii), there is a relation
(9.7.2) $h_{n}=c_{2}(p^{n-2})+v_{2}^{p^{n-2}}h_{n-2}\in
Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*}BP_{*}K)$
By [10].p.502 Cor. 7.8, the image of $v_{2}^{s}c_{2}(p^{n-2})(p^{n-2}>s\geq
1)$ under the boundary homomorphism (or connecting homomorphism)

$j^{\prime}_{*}:Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*}K)\to
Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*}M)$

and

$j_{*}:Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*}M)\to
Ext_{BP_{*}BP}^{3,*}(BP_{*},BP_{*})$

is just the third periodicity family $\gamma_{p^{n-2}/p^{n-2}-s}\neq 0\in
Ext_{BP_{*}BP}^{3,*}(BP_{*},\\\ BP_{*})$. Then, by Theorem 9.7.1, the relation
(9.7.2) and Theorem 7.3.2, we can obtain the following convergence Theorem of
third periodicity families in the stable homotopy groups of spheres
immediately.

Theorem 9.7.3 ([9] Theorem I) Let $p\geq 5,n\geq 1$ and $1\leq s\leq p^{n}-1$,
then the following third periodicity family

$\gamma_{p^{n}/s}\in Ext_{BP_{*}BP}^{3,*}(BP_{*},BP_{*})$

is a permanent cycle in tha ASS and it converge to an element of order $p$ in
$\pi_{*}S$ which has degree $p^{n+2}q+(p^{n}-s)(p+1)q-q-3$.

To prove Theorem 9.7.1, we first prove the following weaker Theorem.

Theorem 9.7.4 ([9] Theorem 4.1) Let $p\geq 5,n\geq 0$, $h_{n}\in
Ext_{A}^{1,p^{n}q}(Z_{p},\\\ Z_{p})$ be the element represented by
$\xi^{p^{n}}$ in the cobar complex, then $i^{\prime}_{*}i_{*}(h_{n})\in
Ext_{A}^{1,p^{n}q}(H^{*}K,Z_{p})$ is a permanent cycle in the ASS and it
converge to a nontrivial homotopy element in $\pi_{p^{n}q-1}K$.

The proof of Theorem 9.7.4 will be the main content of the rest of this
section. The proof need some preminilaries on low dimensional Ext groups and
an argument processing in the Adams resolution of some spectra related to $S$.
We first prove some results on Ext groups.

Theorem 9.7.5 Let $p\geq 3,n\geq 2,$ , then

(1) $Ext_{A}^{s,p^{n}q+r}(H^{*}K,Z_{p})=0$ for $s=2,3,r=1,2$,

$Ext_{A}^{s,p^{n}q+1}(H^{*}K,H^{*}M)=0$,

$Ext_{A}^{3,p^{n}q+q}(H^{*}K,H^{*}K)\cong Z_{p}\\{(h_{0}b_{n-1})^{\prime}\\}$.

(2) $Ext_{A}^{s-1,p^{n}q+q+s-3}(H^{*}Y,Z_{p})$ = 0 for s = 1,2,3.

(3) $Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}Y)$ = 0,
where $Y$ is the spectrum in the cofibration (9.1.4).

Proof : (1) Consider the following exact sequence ($s=1,2,3,r=1,2)$

$Ext_{A}^{s,p^{n}q+r}(Z_{p},Z_{p})\stackrel{{\scriptstyle
i_{*}}}{{\rightarrow}}Ext_{A}^{s,p^{n}q+r}(H^{*}M,Z_{p})\stackrel{{\scriptstyle
j_{*}}}{{\rightarrow}}Ext_{A}^{s,p^{n}q+r-1}(Z_{p},Z_{p})\stackrel{{\scriptstyle
p_{*}}}{{\rightarrow}}$
induced by (9.1.1). By knowledge of $Z_{p}$-base of
$Ext_{A}^{s,*}(Z_{p},Z_{p})$ for $s=1,2,3$ we know that the right group is
zero except for $(s,r)=(1,1),(2,1),(2,2),(3,2)$ it has unique generator
$h_{n},b_{n-1},a_{0}h_{n},a_{0}b_{n-1}$. However,$p_{*}(h_{n})=a_{0}h_{n}\neq
0,p_{*}(b_{n-1})=a_{0}b_{n-1}\neq 0,p_{*}(a_{0}h_{n})=a_{0}^{2}h_{n}\neq
0,p_{*}(a_{0}b_{n-1})=a_{0}^{2}b_{n-1}\neq 0$, then the above $p_{*}$ is monic
and so im $j_{*}$ = 0. In addition, the left group is zero except for
$(s,r)=(2,1),(3,1),(3,2)$ it has unique generator
$a_{0}h_{n}=p_{*}(h_{n}),a_{0}b_{n-1}=p_{*}(b_{n-1}),a_{0}^{2}h_{n}=p_{*}(a_{0}h_{n})$
respectively. Then we have im $i_{*}$ = 0 and obtain that
$Ext_{A}^{s,p^{n}q+r}(H^{*}M,Z_{p})=0$ for $s=1,2,3,r=1,2$.

Look at the following exact sequence ($s=2,3,r=1,2)$

$0=Ext_{A}^{s,p^{n}q+r}(H^{*}M,Z_{p})\stackrel{{\scriptstyle
i^{\prime}_{*}}}{{\longrightarrow}}Ext_{A}^{s,p^{n}q+r}(H^{*}K,Z_{p})$

$\qquad\qquad\quad\stackrel{{\scriptstyle
j^{\prime}_{*}}}{{\longrightarrow}}Ext_{A}^{s,p^{n}q-q+r-1}(H^{*}M,Z_{p})$
induced by (9.1.2). The left group is zero as shown above. The right group
also is zero, this is because $Ext_{A}^{s,p^{n}q-q+r-1}(Z_{p},Z_{p})$ = 0 for
$s=2,3,r=1,2,3$(cf. Chap. 5). Then, the middle group is zero for $s=2,3,r=1,2$
and so $Ext_{A}^{s,p^{n}q+1}(H^{*}K,H^{*}M)$ = 0 $(s=2,3$).

For the last result, consider the following exact sequence

$Ext_{A}^{3,p^{n}q+2q+1}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+q}(H^{*}K,H^{*}K)$

$\qquad\qquad\quad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+q}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}$
induced by (9.1.2). The left group is zero by Prop. 9.3.2(1) and the right
group has unique generator $i^{\prime}_{*}(\alpha_{1}\wedge
1_{M})_{*}(\tilde{b}_{n-1}))$ (cf. Prop. 9.3.1) such that
$\alpha^{*}(i^{\prime})_{*}(\alpha_{1}\wedge 1_{M})_{*}(\tilde{b}_{n-1})$ = 0.
Then, the middle group has unique generator $(h_{0}b_{n-1})^{\prime}$ such
that $(i^{\prime})^{*}(h_{0}b_{n-1})^{\prime}=i^{\prime}_{*}(\alpha_{1}\wedge
1_{M})(\tilde{b}_{n-1})$. Q.E.D.

(2) The result is obvious for $s=1$. For $s=2,3$, consider the following exact
sequence

$\stackrel{{\scriptstyle(i^{\prime}i)_{*}}}{{\longrightarrow}}Ext_{A}^{s-1,p^{n}q+q+s-3}(H^{*}K,Z_{p})\stackrel{{\scriptstyle
r_{*}}}{{\longrightarrow}}Ext_{A}^{s-1,p^{n}q+q+s-3}(H^{*}Y,Z_{p})$

$\qquad\qquad\stackrel{{\scriptstyle\epsilon_{*}}}{{\longrightarrow}}Ext_{A}^{s,p^{n}q+q+s-3}(Z_{p},Z_{p})\stackrel{{\scriptstyle(i^{\prime}i)_{*}}}{{\longrightarrow}}$
induced by (9.1.4). The left group is zero for $s=2$, this is because
$Ext_{A}^{1,t}(Z_{p},Z_{p})$ = 0 for $t=-1,-2$ (mod $q$). The left group has
unique genertor $(i^{\prime}i)_{*}(h_{0}h_{n})$ for $s=3$ so that im $r_{*}$ =
0. The right group is zero for $s=2$ and has unique generator $h_{0}b_{n-1}$
for $s=3$ which satisfies $(i^{\prime}i)_{*}(h_{0}b_{n-1})\neq 0$, then im
$\epsilon_{*}$ = 0 and so the middle group is zero for $s=1,2,3$.

(3) Observe the following exact sequence

$0=Ext_{A}^{0,p^{n}q}(H^{*}K,Z_{p})\stackrel{{\scriptstyle\epsilon^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}Y)$

$\qquad\qquad\stackrel{{\scriptstyle
r^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle(i^{\prime}i)^{*}}}{{\longrightarrow}}$
induced by (9.1.4). The left group clearly is zero and the right group has
unique generator $(h_{n})^{\prime}$ (cf. Prop. 9.3.6) which satisfies
$(i^{\prime}i)^{*}(h_{n})^{\prime}=(i^{\prime}i)_{*}(h_{n})\neq 0\in
Ext_{A}^{1,p^{n}q}(H^{*}K,Z_{p})$ , then the middle group is zero as desired.
Q.E.D.

Prop. 9.7.6 Let $p\geq 3,n\geq 2,$ then

(1) $\quad Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}M)$ = 0,
$Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}M)$ = 0.

(2) $\quad Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}K)$ = 0 ,
$Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}K)$ = 0.

(3) $\quad Ext_{A}^{2,p^{n}q+q-u}(Z_{p},H^{*}K)=0$ for $u=0,1$ ,

$Ext_{A}^{3,p^{n}q+q}(Z_{p},H^{*}K)$ = 0.

Proof: (1) Consider the following exact sequences

$Ext_{A}^{2,p^{n}q+1}(Z_{p},Z_{p})\stackrel{{\scriptstyle
j^{*}}}{{\rightarrow}}Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}M)\stackrel{{\scriptstyle
i^{*}}}{{\rightarrow}}Ext_{A}^{2,p^{n}q}(Z_{p},Z_{p})\stackrel{{\scriptstyle
p^{*}}}{{\rightarrow}}$
$Ext_{A}^{3,p^{n}q+2}(Z_{p},Z_{p})\stackrel{{\scriptstyle
j^{*}}}{{\rightarrow}}Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}M)\stackrel{{\scriptstyle
i^{*}}}{{\rightarrow}}Ext_{A}^{3,p^{n}q+1}(Z_{p},Z_{p})\stackrel{{\scriptstyle
p^{*}}}{{\rightarrow}}$

induced by (9.1.1). The upper left group has unique generator $a_{0}h_{n}$
which satisfies $j^{*}(a_{0}h_{n})=j^{*}p^{*}(h_{n})$ = 0 and the upper right
group has unique generator $b_{n-1}$ satisfying
$p^{*}(b_{n-1})=a_{0}b_{n-1}\neq 0\in Ext_{A}^{3,p^{n}q+1}(Z_{p},Z_{p})$ (cf.
Theorem 5.4.1), then we have $Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}M)$ = 0. The lower
left group has unique generator $a_{0}^{2}h_{n}$ satisfying
$j^{*}(a_{0}^{2}h_{n})=j^{*}p^{*}$ $(a_{0}h_{n})$ = 0 and the lower right
group has unique generator $a_{0}b_{n-1}$ such that
$p^{*}(a_{0}b_{n-1})=a_{0}^{2}b_{n-1}\neq 0\in
Ext_{A}^{4,p^{n}q+2}(Z_{p},Z_{p})$ (cf. Prop. 9.5.2(2)) , then
$Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}M)$ = 0.

(2) Consider the following exact sequences

$0=Ext_{A}^{2,p^{n}q+q+1}(Z_{p},H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}K)$

$\qquad\qquad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}M)=0$

$0=Ext_{A}^{3,p^{n}q+q+2}(Z_{p},H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}K)$

$\qquad\qquad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}M)=0$
induced by (9.1.2). Both two right groups are zero by (1) and both two left
groups are also zero , this is because $Ext_{A}^{2,p^{n}q+q+r}(Z_{p},Z_{p})$ =
0 for $r=1,2$ (cf. Chapter 5) and $Ext_{A}^{3,p^{n}q+q+r}(Z_{p},Z_{p})$ = 0
for $r=2,3$ (cf. Theorem 5.4.1), then the result follows.

(3) Consider the following exact sequence

$\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+2q}(Z_{p},H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+q-1}(Z_{p},H^{*}K)$

$\qquad\qquad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+q-1}(Z_{p},H^{*}M)\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}$
induced by (9.1.2). The left group is zero, this is because
$Ext_{A}^{2,p^{n}q+2q+r}(Z_{p},Z_{p})$ = 0 for $r=0,1$ (cf. Chapter 5). The
right group has unique generator $j^{*}(h_{0}h_{n})$ since
$Ext_{A}^{2,p^{n}q+q-1}(Z_{p},Z_{p})$ = 0 and
$Ext_{A}^{2,p^{n}q+q}(Z_{p},Z_{p})\cong Z_{p}\\{h_{0}h_{n}\\}$. In addition,
we claim that $\alpha^{*}j^{*}(h_{0}h_{n})$ = $\frac{1}{2}\cdot
j^{*}(\widetilde{\alpha}_{2}h_{n})\neq 0\in Ext_{A}^{3,p^{n}q+2q}(Z_{p},\\\
H^{*}M)$ . To prove this, it suffices to prove
$\alpha^{*}j^{*}(h_{0})=\frac{1}{2}j^{*}(\widetilde{\alpha}_{2})\in
Ext_{A}^{2,2q}(Z_{p},\\\ H^{*}M)$. Since
$i^{*}\alpha^{*}j^{*}(h_{0})=\alpha_{1}^{*}(h_{0})=h_{0}^{2}=0$,
then$\alpha^{*}j^{*}(h_{0})=\lambda j^{*}(\widetilde{\alpha}_{2})$ for some
scalar $\lambda\in Z_{p}$. Since both sides of the equation detect the
corresponding homotopy elements, then the relation
$\alpha_{1}j\alpha=\frac{1}{2}\alpha_{2}j$ implies $\lambda=\frac{1}{2}$. This
shows the above claim and so the above $\alpha^{*}$ is monic, im
$(i^{\prime})^{*}$ = 0 and we have $Ext_{A}^{2,p^{n}q+q-1}(Z_{p},H^{*}K)$ = 0.
The proof of the case for $u=0$ is similar.

For the second result, consider the following exact sequence

$Ext_{A}^{3,p^{n}q+2q+1}(Z_{p},H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+q}(Z_{p},H^{*}K)$

$\qquad\qquad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+q}(Z_{p},H^{*}M)\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}$
induced by (9.1.2). The left group has unique generator
$j_{*}\alpha_{*}\alpha_{*}(\tilde{h}_{n})=j_{*}\alpha_{*}\alpha^{*}(\tilde{h}_{n})$,
this is because $Ext_{A}^{3,p^{n}q+2q+1}(Z_{p},Z_{p})$ has unique generator
$\widetilde{\alpha}_{2}h_{n}\\\
=j_{*}\alpha_{*}\alpha_{*}i_{*}(h_{n})=i^{*}j_{*}\alpha_{*}\alpha_{*}(\tilde{h}_{n})$
and $Ext_{A}^{3,p^{n}q+2q+2}(Z_{p},Z_{p})$ = 0 (cf. Theorem 5.4.1), then im
$(j^{\prime})^{*}$ = 0. The right group has unique generator
$j_{*}\alpha_{*}(\tilde{b}_{n-1})$ since $Ext_{A}^{3,p^{n}q+q}(Z_{p},Z_{p})$
has unique generator
$h_{0}b_{n-1}=j_{*}\alpha_{*}i_{*}(b_{n-1})=j_{*}\alpha_{*}i^{*}(\tilde{b}_{n-1})$
and $Ext_{A}^{3,p^{n}q+q+1}(Z_{p},Z_{p})$ = 0 (cf. Theorem 5.4.1). In
addition,
$\alpha^{*}j_{*}\alpha_{*}\cdot(\tilde{b}_{n-1})=j_{*}\alpha_{*}\alpha_{*}(\tilde{b}_{n-1})\neq
0\in Ext_{A}^{4,p^{n}q+2q+1}(Z_{p},H^{*}M)$, this is because
$i^{*}j_{*}\alpha_{*}\alpha_{*}\cdot(\tilde{b}_{n-1})=\widetilde{\alpha}_{2}b_{n-1}\neq
0$ (cf. Prop. 9.5.2(2)). Then the above $\alpha^{*}$ is monic, im
$(i^{\prime})^{*}$ = 0 and so the middle group is zero as desired. Q.E.D.

Proposition 9.7.7 Let $p\geq 3,n\geq 2$, then

(1) $Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}X)$ = 0,
$Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}X)$ = 0.

(2) $Ext_{A}^{3,p^{n}q+q}(H^{*}X,H^{*}K)\cong
Z_{p}\\{w_{*}(h_{0}b_{n-1})^{\prime}\\}.$

(3) $Ext_{A}^{1,p^{n}q+q-1}(H^{*}X,Z_{p})\cong Z_{p}\\{\tau_{*}(h_{n})\\}$,
where $X$ is the spectrum in the cofibration (9.3.7) ,
$\tau:\Sigma^{q-1}S\rightarrow X$ is a map satisfying
$u\tau=i^{\prime}i:S\rightarrow K$ which is obtained by
$\alpha^{\prime\prime}i^{\prime}i=0$ and (9.3.7).

Proof : (1) Consider the following exact sequences

$0=Ext_{A}^{2,p^{n}q+q-1}(Z_{p},H^{*}K)\stackrel{{\scriptstyle
u^{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}X)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
w^{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}K)=0$

$0=Ext_{A}^{3,p^{n}q+q}(Z_{p},H^{*}K)\stackrel{{\scriptstyle
u^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}X)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
w^{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}K)=0$
induced by (9.3.7). By Prop. 9.7.6(2)(3), Both sides four groups are zero so
that the result follows.

(2) We first claim that $Ext_{A}^{s,p^{n}q+1}(H^{*}K,H^{*}K)$ = 0( $s=2,3$),
then the result follows by the following exact sequence

$\stackrel{{\scriptstyle(\alpha^{\prime\prime})_{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+q}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle
w_{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+q}(H^{*}X,H^{*}K)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
u_{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+1}(H^{*}K,H^{*}K)=0$
induced by (9.3.7), where the left group has unique generator
$(h_{0}b_{n-1})^{\prime}$(cf. Prop. 9.7.5(1)). To prove the above claim,
consider the following exact sequence

$\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}Ext_{A}^{s,p^{n}q+q+2}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s,p^{n}q+1}(H^{*}K,H^{*}K)$

$\qquad\qquad\quad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{s,p^{n}q+1}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}$
induced by (9.1.2)¿ The right group is zero for $s=2,3$ (cf. Prop. 9.7.5(1))
and the left group is zero by Prop. 9.3.2(2). This shows the above claim.

(3) Since $\alpha^{\prime\prime}i^{\prime}i=0$, then, by (9.3.7), there is
$\tau\in\pi_{q-1}X$ such that $u\tau=i^{\prime}i:S\rightarrow K$. Consider the
following exact sequence

$0=Ext_{A}^{1,p^{n}q+q-1}(H^{*}K,Z_{p})\stackrel{{\scriptstyle
w_{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q+q-1}(H^{*}X,Z_{p})$

$\qquad\qquad\quad\stackrel{{\scriptstyle
u_{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q}(H^{*}K,Z_{p})\stackrel{{\scriptstyle\alpha^{\prime\prime}_{*}}}{{\longrightarrow}}$
induced by (9.3.7). The left group is zero since $Ext_{A}^{1,t}(Z_{p},Z_{p})$
= 0 for $t\equiv-1,-2$ (mod $q$). The right group has unique generator
$(i^{\prime}i)_{*}(h_{n})$ which satisfies
$\alpha^{\prime\prime}_{*}(i^{\prime}i)_{*}(h_{n})$ = 0, then the middle group
has unique generator $\tau_{*}(h_{n})$ such that
$u_{*}\tau_{*}(h_{n})=(i^{\prime}i)_{*}(h_{n})$. Q.E.D.

Since $u\tau\cdot p=i^{\prime}i\cdot p=0$, then by (9.3.7) we have $\tau\cdot
p=wi^{\prime}i\alpha_{1}$ (uo to nonzero scalar), this is because
$\pi_{q-1}K\cong Z_{p}\\{i^{\prime}i(\alpha_{1})\\}.$ Then, by
$Ext_{A}^{2,p^{n}q+1}(H^{*}K,Z_{p})$ = 0(cf. Prop. 9.7.5(1)) and the Ext exact
sequence induced by (9.3.7) we have
(9.7.8)
$\tau_{*}(a_{0}b_{n-1})=\tau_{*}p_{*}(b_{n-1})=w_{*}(i^{\prime}i)_{*}(\alpha_{1})_{*}(b_{n-1})$

$\qquad=w_{*}(i^{\prime}i)_{*}(h_{0}b_{n-1})\neq 0\in
Ext_{A}^{3,p^{n}q+q}(H^{*}X,Z_{p})$.

Proposition 9.7.9 Let $p\geq 3,n\geq 2$, then

(1) $Ext_{A}^{1,p^{n}q+q-1}(H^{*}K,H^{*}K)$ = 0,
$Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}X)$ = 0.

(2) $Ext_{A}^{1,p^{n}q-q+1}(H^{*}K,H^{*}X)\cong
Z_{p}\\{u^{*}(h_{n})^{\prime}\\}$.

(3) $Ext_{A}^{2,p^{n}q+q}(H^{*}X,Z_{p})\cong
Z_{p}\\{w_{*}(i^{\prime}i)_{*}(h_{0}h_{n})\\}$.

(4) $Ext_{A}^{2,p^{n}q+1}(H^{*}X,H^{*}K)$ = 0.

Proof: (1) Consider the following exact sequence

$\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q+2q}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q+q-1}(H^{*}K,H^{*}K)$

$\qquad\qquad\quad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q+q-1}(H^{*}K,H^{*}M)$
induced by (9.1.2). The right group is zero since
$Ext_{A}^{1,p^{n}q-2}(H^{*}M,H^{*}M)$ = 0,
$Ext_{A}^{1,p^{n}q+q-1}(H^{*}M,H^{*}M)$ = 0 which is obtained by
$Ext_{A}^{1,t}(Z_{p},Z_{p})$ = 0 for $t\equiv-1,-2$ (mod $q$) and
$Ext_{A}^{1,p^{n}q+q+t}(Z_{p},Z_{p})$ = 0 for $t=-1,0,1$. The left group also
is zero since $Ext_{A}^{1,p^{n}q+q-1}(H^{*}M,H^{*}M)$ = 0 and
$Ext_{A}^{1,p^{n}q+2q}(H^{*}M,\\\ H^{*}M)$ = 0 which is obtained by the same
reason as above. Then the middle group is zero.

The second result follows by the following exact sequence

$0=Ext_{A}^{1,p^{n}q+q-1}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle
u^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}X)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
w^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle(\alpha^{\prime\prime})^{*}}}{{\longrightarrow}}$
induced by (9.3.7), where the right group has unique generator
$(h_{n})^{\prime}$ which satisfies
$(\alpha^{\prime\prime})^{*}(h_{n})^{\prime}=(h_{0}h_{n})^{\prime\prime}\neq
0\in Ext_{A}^{2,p^{n}q+q-1}(H^{*}K,H^{*}K)$ ( cf. (9.3.8)).

(2) Consider the following exact sequence

$\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q+2}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle(j^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q-q+1}(H^{*}K,H^{*}K)$

$\qquad\qquad\quad\stackrel{{\scriptstyle(i^{\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q-q+1}(H^{*}K,H^{*}M)\stackrel{{\scriptstyle\alpha^{*}}}{{\longrightarrow}}$
induced by (9.1.2). The left group is zero since
$Ext_{A}^{1,p^{n}q-q+1}(H^{*}M,H^{*}M)$ = 0 and
$Ext_{A}^{1,p^{n}q+2}(H^{*}M,H^{*}M)$ = 0 which is obtained by
$Ext_{A}^{1,p^{n}q-q+t}(Z_{p},Z_{p})$ = 0 for $t=0,1,2$ and
$Ext_{A}^{1,p^{n}q+t}(Z_{p},Z_{p})$ = 0 ($t=1,2$). The right group also is
zero since $Ext_{A}^{1,p^{n}q-2q}(H^{*}M,H^{*}M)$ = 0 and
$Ext_{A}^{1,p^{n}q-q+1}(H^{*}M,H^{*}M)$ = 0 . Then we have
$Ext_{A}^{1,p^{n}q-q+1}(H^{*}K,H^{*}K)$ = 0.

The desired result can be obtained by the following exact sequence

$\stackrel{{\scriptstyle(\alpha^{\prime\prime})^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle
u^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q-q+1}(H^{*}K,H^{*}X)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
w^{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q-q+1}(H^{*}K,H^{*}K)=0$
induced by (9.3.7), where the left group has unique generator
$(h_{n})^{\prime}$ (cf. Prop. 9.3.6).

(3) Consider the following exact sequence

$Ext_{A}^{1,p^{n}q+1}(H^{*}K,Z_{p})\stackrel{{\scriptstyle\alpha^{\prime\prime}_{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+q}(H^{*}K,Z_{p})\stackrel{{\scriptstyle
w_{*}}}{{\longrightarrow}}$

$\qquad\qquad Ext_{A}^{2,p^{n}q+q}(H^{*}X,Z_{p})\stackrel{{\scriptstyle
u_{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+1}(H^{*}K,Z_{p})=0$
induced by (9.3.7). The right group is zero by Prop. 9.7.5(1) and the left
group has unique generator $(i^{\prime}i)_{*}(h_{0}h_{n})$ since
$Ext_{A}^{2,p^{n}q+q}(Z_{p},Z_{p})\cong Z_{p}\\{h_{0}h_{n}\\}$ and
$Ext_{A}^{2,t}(Z_{p},Z_{p})$ = 0 for $t\equiv-1,-2$ (mod $q$). In addition,
$Ext_{A}^{1,p^{n}q+1}(H^{*}K,\\\ Z_{p})$ = 0, this is because
$Ext_{A}^{1,p^{n}q+1}(H^{*}M,Z_{p})$ = 0 ( cf. the proof of Prop. 9.7.5(1))
and $Ext_{A}^{1,p^{n}q-q}(H^{*}M,Z_{p})$ = 0 ( cf Chapter 5), then $w_{*}$ is
monic so that the result follows.

(4) Consider the following exact sequence

$0=Ext_{A}^{2,p^{n}q+1}(H^{*}K,H^{*}K)\stackrel{{\scriptstyle
w_{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+1}(H^{*}X,H^{*}K)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
u_{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q-q+2}(H^{*}K,H^{*}K)=0$
induced by (9.3.7). The left group is zero as pointed out in the proof of
Prop. 9.7.7(2). The right group also is zero since
$Ext_{A}^{2,p^{n}q+3}(H^{*}K,H^{*}M)$ = 0 and
$Ext_{A}^{2,p^{n}q-q+2}(H^{*}K,H^{*}M)$ = 0 which is obtained by
$Ext_{A}^{2,p^{n}q+t}\\\ (Z_{p},Z_{p})$ = 0 for $t=2,3$ and
$Ext_{A}^{2,p^{n}q-rq+t}(Z_{p},Z_{p})$ = 0 for $r=1,2$ , $t=1,2,3$. Then the
middle group is zero as desired. Q.E.D.

Now we proceed to prove the main Theorem 9.7.4 in this section. The proof will
be done by some argument processing in the Adams resolution (9.2.9) . We first
prove the following Proposition and Lemmas.

Proposition 9.7.10 Let $p\geq 5,n\geq 2$,
$(h_{0}h_{n})^{\prime\prime}\in[\Sigma^{p^{n}q+q-1}K,KG_{2}\wedge K]$ be
$d_{1}$-cycle which represents the element
$(h_{0}h_{n})^{\prime\prime}=(\alpha^{\prime\prime})^{*}(h_{n})^{\prime}\in\\\
Ext_{A}^{2,p^{n}q+q-1}(H^{*}K,H^{*}K)$(cf. (9.3.8)), then there exist
$\eta^{\prime\prime}_{n,2}\in[\Sigma^{p^{n}q+q-1}K,\\\ E_{2}\wedge K]$ and
$(\eta^{\prime\prime}_{n,2})_{Y}\in[\Sigma^{p^{n}q+q-1}Y,E_{2}\wedge K]$ such
that $(\bar{b}_{2}\wedge 1_{K})\eta^{\prime\prime}_{n,2}=(\bar{b}_{2}\wedge
1_{K})(\eta^{\prime\prime}_{n,2})_{Y}\cdot
r=(h_{0}h_{n})^{\prime\prime}\in[\Sigma^{p^{n}q+q-1}K,KG_{2}\wedge K]$ where
$r:K\to Y$ is the map in (9.1.4).

Proof : Applying Theorem 9.3.9 to $(\sigma,s,tq)=(h_{n},1,p^{n}q)$, or
applying the mian Theorem B’ and its proof to
$(\sigma,\sigma^{\prime})=(h_{n},b_{n-1}),(s,tq)=(1,p^{n}q)$, we have
$(\bar{c}_{2}\wedge 1_{K})(h_{0}h_{n})^{\prime\prime}=0$,. Them there exists
$\eta^{\prime\prime}_{n,2}\in[\Sigma^{p^{n}q+q-1}K,KG_{2}\wedge K]$ such that
$(\bar{b}_{2}\wedge
1_{K})\eta^{\prime\prime}_{n,2}=(h_{0}h_{n})^{\prime\prime}\in[\Sigma^{p^{n}q+q-1}K,KG_{2}\wedge
K]$. For the second result, note that
$(h_{0}h_{n})^{\prime\prime}i^{\prime}i\in[\Sigma^{p^{n}q+q-1}S,KG_{2}\wedge
K]=0$, this is because $\pi_{p^{n}q+tq-u}KG_{2}\cong
Ext_{A}^{2,p^{n}q+tq-u}(Z_{p},Z_{p})$ = 0 for $t=0,1,u=1,2,3$. Then there is
$(h_{0}h_{n})^{\prime\prime}_{Y}\in[\Sigma^{p^{n}q+q-1}Y,KG_{2}\wedge K]$ such
that $(h_{0}h_{n})^{\prime\prime}=(h_{0}h_{n})^{\prime\prime}_{Y}\cdot r$,
where $r:K\to Y$ is the map in (9.1.4). Then by Theorem 9.3.9 we have
$(\bar{c}_{2}\wedge 1_{K})(h_{0}h_{n})^{\prime\prime}_{Y}\cdot r=0$ and so ,
by the cofibration (9.1.4), $(\bar{c}_{2}\wedge
1_{K})(h_{0}h_{n})^{\prime\prime}_{Y}=f^{\prime}_{0}\epsilon=(1_{E_{2}}\wedge\epsilon\wedge
1_{K})(1_{Y}\wedge f^{\prime}_{0})=0$, for some
$f^{\prime}_{0}\in[\Sigma^{p^{n}q+q}S,E_{3}\wedge K]$, where we use
$\epsilon\wedge 1_{K}=\mu(i^{\prime}i\wedge 1_{K})(\epsilon\wedge 1_{K})=0$
which is obtained by (9.1.26) and the cofibration (9.1.4). Hence, there is
$(\eta^{\prime\prime}_{n,2})_{Y}\in[\Sigma^{p^{n}q+q-1}Y,E_{2}\wedge K]$ such
that $(\bar{b}_{2}\wedge
1_{K})(\eta^{\prime\prime}_{n,2})_{Y}=(h_{0}h_{n})^{\prime\prime}_{Y}\in[\Sigma^{p^{n}q+q-1}Y,KG_{2}\wedge
K]$. Q.E.D.

Lemma 9.7.11 Let $p\geq 5,n\geq 2$ and
$(\eta^{\prime\prime}_{n})_{Y}=(\bar{a}_{0}\bar{a}_{1}\wedge
1_{K})(\eta^{\prime\prime}_{n,2})_{Y}\in[\Sigma^{p^{n}q+q-3}Y,K]$ be the map
obtained in Prop. 9.7.10, then $w(\eta^{\prime\prime}_{n})_{Y}\cdot
r=\lambda^{\prime}w(\zeta_{n-1}\wedge
1_{K})+(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}$ for some
$f^{\prime\prime}_{1}\in[\Sigma^{p^{n}q+q+1}K,E_{4}\wedge X]$ and nonzero
$\lambda^{\prime}\in Z_{p}$, where $\zeta_{n-1}\in\pi_{p^{n}q+q-3}S$ is the
element obtained in Theorem 9.5.1 which is represented by $h_{0}b_{n-1}\in
Ext_{A}^{3,p^{n}q+q}(Z_{p},Z_{p})$ in the ASS.

Proof : By Prop. 9.7.10 and (9.3.8)(9.3.7), $(\bar{b}_{2}\wedge
1_{X})(1_{E_{2}}\wedge w)(\eta^{\prime\prime}_{n,2})_{Y}\cdot r$ =
$(1_{KG_{2}}\wedge w)(h_{0}h_{n})^{\prime\prime}=(\bar{b}_{2}\bar{c}_{1}\wedge
1_{X})g^{\prime\prime}$ with
$g^{\prime\prime}\in[\Sigma^{p^{n}q+q-1}K,KG_{1}\wedge X]$, then
(9.7.12) $(1_{E_{2}}\wedge w)(\eta^{\prime\prime}_{n,2})_{Y}\cdot
r=(\bar{c}_{1}\wedge 1_{X})g^{\prime\prime}+(\bar{a}_{2}\wedge
1_{X})f^{\prime\prime}_{0}$
for some $f^{\prime\prime}_{0}\in[\Sigma^{p^{n}q+q}K,E_{3}\wedge X]$. The
$d_{1}$-cycle $(\bar{b}_{3}\wedge
1_{X})f^{\prime\prime}_{0}\in[\Sigma^{p^{n}q+q}K,KG_{3}\wedge X]$ represents
an element in $Ext_{A}^{3,p^{n}q+q}(H^{*}X,H^{*}K)$ and this group has unique
generator $w_{*}[h_{0}b_{n-1}\wedge 1_{K}]$ ( cf. Prop. 9.7.7(2)), then

$(\bar{b}_{3}\wedge
1_{X})f^{\prime\prime}_{0}=\lambda^{\prime}(1_{KG_{3}}\wedge
w)(h_{0}b_{n-1}\wedge 1_{K})+(\bar{b}_{3}\bar{c}_{2}\wedge
1_{X})\tilde{g}_{0}$

$\qquad=\lambda^{\prime}(\bar{b}_{3}\wedge 1_{X})(1_{E_{3}}\wedge
w)(\zeta_{n-1,3}\wedge 1_{K})+(\bar{b}_{3}\bar{c}_{2}\wedge
1_{X})\tilde{g}_{0}$
with $\lambda^{\prime}\in Z_{p}$ and
$\tilde{g}_{0}\in[\Sigma^{p^{n}q+q}K,KG_{2}\wedge X]$, where we use
$(\bar{b}_{3}\wedge 1_{K})(\zeta_{n-1,3}\wedge 1_{K})=h_{0}b_{n-1}\wedge
1_{K}$ (cf. Theorem 9.5.1) . Then
$f^{\prime\prime}_{0}=\lambda^{\prime}(1_{E_{3}}\wedge w)(\zeta_{n-1,3}\wedge