determine an element $\bar{p}(\gamma)$ in $P_p\operatorname{St}(S)_{q_0,...,q_p}$ together with a $(p+1,q+p+1)$-partition $(i_1,...,i_p,i_{p+1},j_1,...,j_{q+p+1})$ of $n=q+2p+1$, where $q=q_0+...+q_p$. Here $$\begin{aligned}
i_1&=&q_0+1\\
i_2&=&q_0+q_1+3\\
&.&\\
&.&\\
&.&\\
i_p&=&q_0+...+q_{p-1}+2p-1\\
i_{p+1}&=&q_0+...+q_p+2p+1\end{aligned}$$ correspond to the $\mu_i$’s defined in Definition  \[4.2\] and the $j$’s correspond to the complement, that is, $j_1,...,j_{q_0+1},j_{q_0+2},...,j_{q_0+q_1+2},...,j_{q_0+...+q_{p-1}+p},...,j_{q_0+...+q_p+p+1}$, are $0,...,q_0,q_0+2,...,q_0+q_1+2,q_0+q_1+4,...,q_0+...+q_{p-2}+2p-2,...,q_0+...+q_{p-1}+2p,q_0+...+q_{p-1}+2p,...,
q_0+...+q_p+2p$, respectively. Then, in terms of Remark 2 at the end of Section 4, we define $$\bar{p}(\gamma)=(\sigma,\tau,
\bar{\sigma})\in{P_p\text{St}(S)_{q_0...q_p}}\subseteq{S_{q+p}\times{S_{q+p}}\times{S_p}}$$ where $$\begin{aligned}
\bar{\sigma}&=&d_{0...q_0}\circ{\hat{d}_{q_0+1}}\circ{...}
\circ{d_{(q_0+...+q_{p-1}+2p)...(q_0+...+q_p+2p)}}
\circ{\hat{d}_{q+2p+1}}(\gamma)=d_{j_1...j_{q+p+1}}(\gamma)\\
\tau&=&d_{q_0+1}\circ{d_{q_0+q_1+3}}\circ{...}
\circ{d_{q_0+...+q_p+2p+1}}(\gamma)=d_{i_1...i_{p+1}}(\gamma)\\
\sigma&=&\hat{s}_{q_0+...+q_p+p}\circ{s_{(q_0+...+q_p+p-1)...(q_0+...+q_{p-1}+p)}}\circ
\hat{s}_{q_0+...+q_{p-1}+p-1}\circ{...}\circ\\
& &\hat{s}_{q_0+q_1+1}\circ{s_{(q_0+q_1)...(q_0+1)}}\circ
\hat{s}_{q_0}\circ{s_{(q_0-1)...(0)}}(\bar{\sigma})\\
&=&s_{(q+p-1)...(q+p-q_p)}\circ{...}\circ{s_{(q_0+q_1)...(q_0+1)}}\circ{s_{(q_0-1)...(0)}}(\bar{\sigma})\\
&=&\mu_{q_0,...,q_p}(\bar{\sigma}).\end{aligned}$$ Using the above expression for $\bar{\sigma}$ in terms of $d$’s and $\gamma$, we get $$\sigma=s_{(q+p-1)...(q+p-q_p)}\circ{...}\circ{s_{(q_0+q_1)...(q_0+1)}}\circ{s_{(q_0-1)...0}}\circ{d_{j_1...j_{q+p+1}}}(\gamma).$$ Now $\operatorname{St}(S)_{q_0+...+q_p+2p+1}$ contains the simplex $$\begin{aligned}
(s_{j_{q+p+1}...j_1}\circ{}s_{(q+p-1)...(q+p-q_p)}&\circ{...}&\circ{}s_{(q_0+q_1)
...(q_0+1)}
\circ{s_{(q_0-1)...0}}\circ{d_{j_1...j_{q+p+1}}}(\gamma),\\
s_{i_{p+1}...i_1}d_{i_1...i_{p+1}}(\gamma))&=&(s_{j_{q+p+1}...j_1}\sigma,s_{i_{p+1}...i_1}\tau).\\\end{aligned}$$ It follows that $(\sigma,\tau)\in{\operatorname{St}(S)}$ and hence $\bar{p}(\gamma)=(\sigma,\tau,\bar{\sigma})\in{P_p\operatorname{St}(S)_{q_0,...,q_p}}$.

Now $\bar{p}$ is a surjective map: Suppose $(\sigma,\tau,\bar{\sigma})\in{P_p\operatorname{St}(S)_{q_0,...,q_p}}$ and we shall find $\gamma\in{\bar{P}_p}S_{q_0,...,q_p}$ such that $\bar{p}(\gamma)=(\sigma,\tau,\bar{\sigma})$. Thus $(\sigma,\tau,\bar{\sigma})\in{P_p\operatorname{St}(S)_{q_0,...,q_p}}\subset{\delta(S\times{S})}_{q+p}$ is such that $$\pi_1(\sigma,\tau)\in{\text{Im}\{\mu_{q_0,...,q_p}:S_p\to{S_{q+p}}\}}$$ where $\bar{\sigma}\in{S_p}$.

Again use the partition $(p+1,q+p+1)$ as above, put $\gamma=s_{i_{p+1}...i_1}\sigma\in{S_{q+2p+1}}$. Indeed since $$(s_{j_{q+p+1}}\circ{s_{j_q-q_p+p-1}}\circ{...}\circ{}s_{j_{q_0+q_1+2}}\circ{s_{j_{q_0+1}}}\sigma,s_{i_{p+1}...i_1}\tau)$$ is of the required form as in Definition  \[4.2\] and since $$(\sigma,\tau)=(d_{k_1....k_{p+1}}\times{}
d_{i_1...i_{p+1}})(s_{j_{q+p+1}}\circ{s_{j_q-q_p+p-1}}\circ{...}\circ{}s_{j_{q_0+q_1+3}}\circ{s_{j_{q_0+1}}}\sigma,s_{i_{p+1}...i_1}\tau)$$ here $d_{k_1....k_{p+1}}=d_{q_0+1}\circ{}d_{q_0+q_1+3}\circ{...}\circ{}d_{q_0+...+q_{p-1}+2p-1}\circ{}d_{q+2p+1}$. So the $d_K$’s and the $d_I$’s are the same, where $d_I=d_{i_1...i_{p+1}}$.

We have $(\sigma,\tau)\in{\operatorname{St}(S)}$. Hence $\bar{p}(\gamma)\in{P_.\operatorname{St}(S)_.}$.

2\) If $S=K^s$, $K$ simplicial complex then $$\begin{aligned}
P_p\text{St}(K^s)_{q_0,\dots,q_p}=\{(\sigma,\tau)\in{\text{St}}(K^s)_{q+p}\subset{\delta}(K^s\times{K^s})_{q+p}\\
\text{
  }|\text{   }
\pi_1(\sigma,\tau)\in{\text{Im}}\{\mu_{q_0,\dots,q_p}:{K_p}^s\to{{K_{q+p}^s}}\}
\}.\end{aligned}$$

The map $\mu_{q_0,\dots,q_p}:{K_p}^s\to{{K_{q+p}^s}}$ takes $(i_0,...,i_p)$ to $(\underbrace{i_0,\dots,i_0}_{q_0+1-\text{times}},\dots,
\underbrace{i_p,\dots,i_p}_{q_p+1-\text{times}})$. Then $$\begin{aligned}
\sigma&=&(a_{i_0},\dots,a_{i_0},\dots,a_{i_p},\dots,a_{i_p})\in{{K^s}_{q+p}},\\
