\section{Lifting the pentagon to Walker's extended cobordism category}
\label{sec:walker}

The relative isotopy gives equality of marked mapping classes, but the paper's
category also remembers an integer weight.  This section checks the Maslov
seams in the fixed Lagrangian and lifts the geometric pentagon to weight-zero
Walker cylinders.

Equip \(Y\) with the rational Lagrangian \(\Lambda\) of
\eqref{eq:common-lagrangian-section12}.  For a mapping class \(f\) and
an integer \(m\), write \(C(f,m)\) for its extended mapping cylinder of
weight \(m\).  With the right-factor-first convention, forward
multiplication is
\begin{equation}
 C(g,n)C(f,m)
 =C\!\left(g\compose f,
 n+m-\mu(f_*\Lambda,\Lambda,g_*^{-1}\Lambda)\right).
 \label{eq:walker-forward-law}
\end{equation}
The minus sign and the order of the three Lagrangians are the convention
used here; see \cite[arXiv~v3, Section~3, equation~(2)]{GilmerMasbaum2013}.
We use only this forward
multiplication formula.

Put
\begin{equation}
 H_e=C([h_e],0).
 \label{eq:walker-edge-cylinders}
\end{equation}
By \cref{prop:two-sided-lagrangian}, the three seams encountered in
forward multiplication are
\begin{align}
 \mu(h_{0B*}\Lambda,\Lambda,h_{B\star*}^{-1}\Lambda)&=0,
 \label{eq:walker-left-seam}\\
 \mu(h_{0A*}\Lambda,\Lambda,h_{AAC*}^{-1}\Lambda)&=0,
 \label{eq:walker-right-first-seam}\\
 \mu((h_{AAC}\compose h_{0A})_*\Lambda,
      \Lambda,h_{AC\star*}^{-1}\Lambda)&=0.
 \label{eq:walker-right-second-seam}
\end{align}
Indeed, each of the three entries in every displayed Maslov index is
literally \(\Lambda\): the first entries use forward preservation, the
third entries use inverse preservation, and the composite in
\eqref{eq:walker-right-second-seam} preserves \(\Lambda\) because each of
its factors does.  A Maslov index with two equal entries is zero.

\begin{theorem}[Weight-zero Walker pentagon]
\label{thm:walker-pentagon}
All constituent, seam, and route weights are zero, and
\begin{equation}
 H_{B\star}H_{0B}
 =H_{AC\star}H_{AAC}H_{0A}
 =C([h_{B\star}\compose h_{0B}],0).
 \label{eq:walker-pentagon-section14}
\end{equation}
\end{theorem}

\begin{proof}
Apply \eqref{eq:walker-forward-law} first to the left route.  Using
\eqref{eq:walker-left-seam},
\begin{align*}
 H_{B\star}H_{0B}
 &=C([h_{B\star}],0)C([h_{0B}],0)\\
 &=C([h_{B\star}\compose h_{0B}],0+0-0)\\
 &=C([h_{B\star}\compose h_{0B}],0).
\end{align*}

For the right route, first multiply the two rightmost factors.  Equation
\eqref{eq:walker-right-first-seam} gives
\[
 H_{AAC}H_{0A}
 =C([h_{AAC}\compose h_{0A}],0).
\]
Now multiply on the left by \(H_{AC\star}\).  The seam is exactly
\eqref{eq:walker-right-second-seam}, so
\begin{align*}
 H_{AC\star}H_{AAC}H_{0A}
 &=C([h_{AC\star}],0)
   C([h_{AAC}\compose h_{0A}],0)\\
 &=C([h_{AC\star}\compose h_{AAC}\compose h_{0A}],0+0-0)\\
 &=C([h_{AC\star}\compose h_{AAC}\compose h_{0A}],0).
\end{align*}
Only after these forward multiplications do we use the mapping-class
equality \eqref{eq:mapping-class-pentagon-section13}.  It identifies the
two underlying mapping classes, while the computations above identify
their weights as the same literal integer \(0\).  This proves
\eqref{eq:walker-pentagon-section14}.
\end{proof}

No converse to \eqref{eq:walker-forward-law}, injectivity of the cylinder
assignment, pair-determination theorem, projective quotient, or
weight-forgetting argument is used.