Assume
\(f,g\in P(X; x,y)\) and
\([f]=[g]\text{.}\) Let
\(F\colon I\times I\rightarrow X\) be a path homotopy from
\(f\) to
\(g\text{.}\) Again, using
Corollary 1.15.8, we can find partitions
\begin{align*}
0=s_0 \amp < s_1 < \dots < s_n=1 \\
0=t_0 \amp < t_1 < \dots < t_n=1
\end{align*}
such that for each subrectangle \(R_{k,l}=[s_{k-1}, s_k]\times [t_{l-1},t_l]\) of \(I\times I\text{,}\) we have \(F(R_{k,l})\subseteq W\text{,}\) where \(W=U\) or \(W=V\text{.}\) For each \((k,l)\) with \(1 \leq k\leq n\) and \(0\leq l\leq n\text{,}\) let \(h_{k,l}\) be the horizontal path in \(I\times I\) from \((s_{k-1}, t_l)\) to \((s_k, t_l)\text{;}\) similarly define \(v_{k,l}\) to be the vertical path from \((s_{k}, t_{l-1})\) to \((s_k,t_l)\) for all pairs \((k,l)\) with \(0\leq k\leq n\) and \(1\leq l\leq n\text{.}\) Lastly, define
\begin{align*}
f_{k,l} \amp =F\circ h_{k,l}\\
\gamma_{k,l} \amp =F\circ v_{k,l}
\end{align*}
for all pairs \((k,l)\) for which these expressions make sense. Fix a pair \((k,l)\) with \(1\leq k,l\leq n\text{.}\) The polygonal paths \(h_{k,l-1}*v_{k,l}\) and \(v_{k-1,l}*h_{k,l}\) respectively traverse the bottom-and-right segment and left-and-top segment of the rectangle \(R_{k,l}\text{.}\) Since the rectangle is convex, these paths are homotopic in \(R_{k,l}\text{.}\) Composing with \(F\) gives us a homotopy of \(f_{k,l-1}*\gamma_{k,l}\) and \(\gamma_{k-1,l}*f_{k,l}\) in \(F(R_{k,l})\subseteq W\text{.}\) Using properties of \(\tau\vert_{D_2}\text{,}\) we conclude that
\begin{equation*}
\tau\vert_{D_2}(f_{k,l-1})\tau\vert_{D_2}(\gamma_{k,l})=\tau\vert_{D_2}(\gamma_{k-1,l})\tau\vert_{D_2}(f_{k,l})
\end{equation*}
and hence that
\begin{equation}
\tau\vert_{D_2}(f_{k,l-1})=\tau\vert_{D_2}(\gamma_{k-1,l})\tau\vert_{D_2}(f_{k,l})\tau\vert_{D_2}(\gamma_{k,l})^{-1}\text{.}\tag{2.15.11}
\end{equation}
Let \(f_{l}(s)=F(s,t_l)\) for all \(0\leq j\leq n\text{.}\) So in particular, we have \(f_{0}(s)=F(s,0)=f(s)\) and \(f_n(s)=F(s,1)=g(s)\text{.}\) Note that we have
\begin{equation*}
f_j\circ \phi_k=F\circ h_{k,l}=f_{k,l}\text{.}
\end{equation*}
\begin{align*}
\tau(f_{l-1}) \amp =\prod_{k=1}^n\tau\vert_{D_2}(f_{k,l-1})\\
\amp = \prod_{k=1}^n(\tau\vert_{D_2}(\gamma_{k-1,l})\tau\vert_{D_2}(f_{k,l})\tau\vert_{D_2}(\gamma_{k,l})^{-1}) \amp \knowl{./knowl/xref/eq_conj.html}{\text{(2.15.11)}}\\
\amp = \tau\vert_{D_2}(e_{x})\cdot \prod_{k=1}^n\tau\vert_{D_2}(f_{k,l})\cdot \tau\vert_{D_2}(e_y) \amp \text{(cancellation)}\\
\amp = \prod_{k=1}^n\tau\vert_{D_2}(f_{k,l}) \amp (\tau\vert_{D_2}(e_x)=\tau\vert_{D_2}(e_y)=e)\\
\amp = \tau(f_l)\text{.}
\end{align*}
We conclude that \(\tau(f)=\tau(f_{0})=\tau(f_{n})=\tau(g)\text{,}\) as desired.