117 As in the $k=1$ case, the first term, corresponding to $i(b) = \eset$, makes the all balls in $\beta$ $\cV_1$-small. However, if this were the only term $s$ would not be a chain map, because we have no control over $\restrict{\phi_{\beta}}{x_0 = 0}(\bdy b)$. This necessitates the other terms, which fix the boundary at successively higher codimensions.

   117 As in the $k=1$ case, the first term, corresponding to $i(b) = \eset$, makes the all balls in $\beta$ $\cV_1$-small. However, if this were the only term $s$ would not be a chain map, because we have no control over $\restrict{\phi_{\beta}}{x_0 = 0}(\bdy b)$. This necessitates the other terms, which fix the boundary at successively higher codimensions.

   118 

   118 

   119 It may be useful to look at Figure \ref{fig:erectly-a-tent-badly} to help understand the arrangement. The red, blue and orange $2$-cells there correspond to the $m=0$, $m=1$ and $m=2$ terms respectively, while the $3$-cells (only one of each type is shown) correspond to the terms in the homotopy $h$.

   119 It may be useful to look at Figure \ref{fig:erectly-a-tent-badly} to help understand the arrangement. The red, blue and orange $2$-cells there correspond to the $m=0$, $m=1$ and $m=2$ terms respectively, while the $3$-cells (only one of each type is shown) correspond to the terms in the homotopy $h$.

   120 \begin{figure}[!ht]

   120 \begin{figure}[!ht]

   121 $$\mathfig{0.5}{smallblobs/tent}$$

   121 $$\mathfig{0.5}{smallblobs/tent}$$

   123 \label{fig:erectly-a-tent-badly}

   123 \label{fig:erectly-a-tent-badly}

   124 \end{figure}

   124 \end{figure}

   125 

   125 

   126 Now

   126 Now

   127 \begin{align*}

   127 \begin{align*}