\section{Fixed-site local phase calculus and weighted table}
\label{fixed:sec-local-calculus}

The local calculation isolates the only place where the two phase choices can
interact.  Once the old surface and the unused sites are treated as
spectators, the problem becomes a calculation in one marked one-holed torus,
with the Walker weight kept alongside the supported mapping class.

Work in the attachment-state Boolean chart of
\cref{common:sec-walker-data}.  Fix \(I\subseteq[k]\) and a label
\(i\notin I\), and abbreviate
\[
 X_I:=\Sigma_I^{\mathrm{out}},
 \qquad
 Y_{i,I}:=\Sigma_{I\cup\{i\}}^{\mathrm{out}}.
\]
The old surface, the handles indexed by \(I\), and every unused disk are
spectators.  We use the homology and Lagrangian splittings
\eqref{fixed:eq-attachment-homology}--\eqref{fixed:eq-attachment-lagrangian}
and the Walker convention \eqref{common:eq-walker-product}.  Every local
map below is supported in the \(i\)-th handle neighborhood with its complete
marking and is the identity on all spectators.  Products act from the
right, and every displayed integer is a Walker weight rather than a power
of a mapping class.

The elementary compression body and its literal reverse are
\begin{align}
 \mathcal C_{i\mid I}(c)&:
 X_I\longrightarrow Y_{i,I},
 \label{fixed:eq-elementary-forward-local}\\
 \mathcal R_{i\mid I}^{\mathrm{rev}}(r)&:
 Y_{i,I}\longrightarrow X_I.
 \label{fixed:eq-elementary-reverse-local}
\end{align}
Here \(c,r\in\mathbb Z\) are independently assigned weights.  The
superscript on \(\mathcal R^{\mathrm{rev}}\) records the literal reverse family; it
does not say that
\(\mathcal R^{\mathrm{rev}}(r)\) is the actual dagger of
\(\mathcal C(r)\).

Recall that \(T_\gamma\) denotes the right Dehn twist about \(\gamma\).  On the new
one-holed torus fix the exact right-twist word
\begin{equation}
 \varsigma_i=T_{m_i}T_{\ell_i}T_{m_i},
 \qquad
 (\varsigma_i)_*=
 \begin{pmatrix}0&1\\-1&0\end{pmatrix}.
 \label{fixed:eq-exact-slope-local}
\end{equation}
Let \(\sigma_{i\mid I}\) be its extension by the identity over the old
surface and all spectators, and put
\begin{equation}
 \Theta_{i\mid I}(u):=C(\sigma_{i\mid I},u):
 Y_{i,I}\longrightarrow Y_{i,I},
 \qquad
 \mathcal S_{i\mid I}:=\Theta_{i\mid I}(0).
 \label{fixed:eq-slope-cylinder-local}
\end{equation}
Thus
\begin{equation}
 (\sigma_{i\mid I})_*\langle m_i\rangle
 =\langle\ell_i\rangle,
 \qquad
 (\sigma_{i\mid I})_*^{-1}\langle m_i\rangle
 =\langle\ell_i\rangle,
 \label{fixed:eq-slope-lines-local}
\end{equation}
where signs of individual generators disappear only because these are
Lagrangian lines.

Fix the phase and weight vectors
\[
 \begin{aligned}
  &\epsilon,\eta\in\{0,1\}^k,\\
  &\mathbf c,\mathbf r,\mathbf u^+,\mathbf u^-\in\mathbb Z^k.
 \end{aligned}
\]
The weight vectors are independent.  At the chosen label abbreviate
\[
 \begin{aligned}
  e&:=\epsilon_i,&v&:=\eta_i,&c&:=c_i,&r&:=r_i,\\
  u^+&:=u_i^+,&u^-&:=u_i^-.
 \end{aligned}
\]
For these local data define
\begin{align}
 J_{i\mid I}^{0}(c)&:=\mathcal C_{i\mid I}(c),
 \label{fixed:eq-phase-forward-zero}\\
 J_{i\mid I}^{1}(c,u^+)
 &:=
 \Theta_{i\mid I}(u^+)\mathcal C_{i\mid I}(c),
 \label{fixed:eq-phase-forward-one}\\
 R_{i\mid I}^{0}(r)&:=\mathcal R_{i\mid I}^{\mathrm{rev}}(r),
 \label{fixed:eq-phase-reverse-zero}\\
 R_{i\mid I}^{1}(r,u^-)
 &:=
 \mathcal R_{i\mid I}^{\mathrm{rev}}(r)\Theta_{i\mid I}(u^-).
 \label{fixed:eq-phase-reverse-one}
\end{align}
Every \(J_{i\mid I}^{e}\) has type \(X_I\to Y_{i,I}\), and every
\(R_{i\mid I}^{v}\) has type \(Y_{i,I}\to X_I\).
When \(e=0\), the coordinate \(u^+\) is inactive and is ignored; when
\(v=0\), the coordinate \(u^-\) is inactive and is ignored.  The
superscripts \(e,v\) are the only powers in this notation.  Put
\begin{equation}
 \begin{aligned}
  a_i&:=c+eu^+,&
  b_i'&:=r+vu^-,\\
  \text{that is}\qquad
  a_i&=c_i+\epsilon_i u_i^+,&
  b_i'&=r_i+\eta_i u_i^-.
 \end{aligned}
 \label{fixed:eq-elementary-weight-totals}
\end{equation}

\begin{lemma}[Exact elementary phase weights]
\label{fixed:lem-elementary-weights}
The actual Walker weights of
\(J_{i\mid I}^{e}(c,u^+)\) and
\(R_{i\mid I}^{v}(r,u^-)\) are \(a_i\) and \(b_i'\),
respectively.
\end{lemma}

\begin{proof}
 The weight is computed on the one active symplectic plane; all old and
unused summands are spectators.  Phase zero has no internal seam, while phase
one transports the new meridian line through the marked half-turn.  Direct-sum
additivity then reduces the Maslov calculation to the displayed local
correction.

There is no internal seam in a phase-zero arrow.  For a phase-one forward
arrow, direct-sum additivity gives
\begin{align}
 &\mu\!\left(
 \lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}},
 (\sigma_{i\mid I})_*^{-1}\lambda_{I\cup\{i\}}
 \right)\notag\\
 &\quad=
 \mu(\lambda_g,\lambda_g,\lambda_g)
 +\sum_{s\in I}
   \mu(\langle m_s\rangle,\langle m_s\rangle,\langle m_s\rangle)
 +\mu(\langle m_i\rangle,\langle m_i\rangle,\langle\ell_i\rangle)
 =0.
 \label{fixed:eq-forward-internal-seam}
\end{align}
For a phase-one reverse arrow the seam is
\begin{align}
 &\mu\!\left(
 (\sigma_{i\mid I})_*\lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}}
 \right)\notag\\
 &\quad=
 \mu(\lambda_g,\lambda_g,\lambda_g)
 +\sum_{s\in I}
   \mu(\langle m_s\rangle,\langle m_s\rangle,\langle m_s\rangle)
 +\mu(\langle\ell_i\rangle,\langle m_i\rangle,\langle m_i\rangle)
 =0.
 \label{fixed:eq-reverse-internal-seam}
\end{align}
Every term vanishes by a repeated pair.  Hence the constituent weights add
with no correction, proving the two formulas.
\end{proof}

\subsection{The localized marked half-turn}

Let \(\delta_{i\mid I}\) be the chosen separating curve in the collar of
the \(i\)-th attached one-holed torus.  Let \(d_{i\mid I}\) be the positive
twist about this curve, extended by the identity on the old surface and all
spectators, and write
\begin{equation}
 V_{d_{i\mid I}}:=C(d_{i\mid I},0).
 \label{fixed:eq-retained-disk-twist}
\end{equation}

\begin{proposition}[Relative half-turn localization]
\label{fixed:prop-halfturn}
For every \(I\subseteq[k]\) and \(i\notin I\), the exact relative half-turn
extends across the compression body, fixes the incoming boundary and every
spectator region, and satisfies
\begin{equation}
 \boxed{
 \mathcal S_{i\mid I}^{\,2}\mathcal C_{i\mid I}(0)
 =\mathcal C_{i\mid I}(0).}
 \tag{H1}\label{fixed:eq-H1}
\end{equation}
The equality is literal in the marked Walker Hom set and has weight zero.
\end{proposition}

\paragraph{The local compression model and its reflected reverse.}
Build the compression body in a product neighborhood of the active site,
attach the handle at the two marked feet, and use the explicit reflection to
identify the reverse side.  The co-core supplies the meridian, while the
joining band supplies the framed longitude, so the boundary map and its
marked germ are determined together.

\begin{proof}
Build the local compression body from \(X_I\times[0,1]\) by attaching
\[
 H_i=[-1,1]\times D^2
\]
at the two ordered feet belonging to \(d_i\).  Its co-core is
\(\{0\}\times D^2\), with boundary \(m_i\); the chosen joining band and the
product push-off of the core determine the framed longitude \(\ell_i\).
On \(H_i\) use
\[
 (t,z)\longmapsto(-t,\overline z).
\]
It is orientation preserving, exchanges the ordered feet, preserves the
co-core setwise, and reverses both the core and its product-framed
longitude.  Indeed, reflection of the interval coordinate and complex
conjugation of the disk coordinate each reverse orientation, so their
product preserves the orientation of \(H_i\).  On the two attaching
rectangles these coordinates agree with the prescribed exchange of the
feet.  Extend the map over the joining band by the corresponding
half-turn.  Inside the chosen separating collar, interpolate from this
rotation to the identity while keeping the outer edge of the collar fixed,
and take the identity on \(X_I\times\{0\}\).  The construction is therefore
relative to the incoming boundary and to a full outer collar, rather than
merely up to an unmarked isotopy.  Because the active handle neighborhood
is disjoint from every \(T_s\), \(s\in I\), and from every unused disk, the
extension is pointwise the identity on all spectator regions.

Cutting the outgoing one-holed torus along \(m_i\cup\ell_i\) gives the
marked square determined by the feet and band.  Reading its paired edges in
the fixed positive-twist convention gives the exact relative word
\begin{equation}
 (T_{m_i}T_{\ell_i})^3
 =
 (T_{m_i}T_{\ell_i}T_{m_i})^2
 =\sigma_{i\mid I}^{\,2}.
 \label{fixed:eq-relative-halfturn-word}
\end{equation}
This is stronger than its action on homology.  The homology matrix alone
cannot detect a power of the separating twist.  With the cut square and its
boundary marking fixed, any alternative relative lift with the same
homological action would have the form
\(T_{\delta_{i\mid I}}^q\sigma_{i\mid I}^{\,2}\).  Track one oriented,
product-framed arc crossing the separating collar once.  Squaring the
explicit half-turn rotates the normal framing of this arc through
\(+2\pi\), and hence produces exactly the positive twist
\(T_{\delta_{i\mid I}}\).  The square of the alternative lift would instead
be \(T_{\delta_{i\mid I}}^{2q+1}\).  The separating twist has infinite order
in the relative marked mapping-class group~\cite{FarbMargalit2012}, so
comparison with the constructed square forces \(q=0\).  Thus no
boundary-twist power is hidden in the extension, and the outgoing
restriction is the written relative word, not merely a representative with
the same homology action.

It remains to check the Walker lift.  First, the only correction in the
product defining \(\mathcal S_{i\mid I}^{\,2}\) is
\begin{align}
 &\mu\!\left(
 (\sigma_{i\mid I})_*\lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}},
 (\sigma_{i\mid I})_*^{-1}\lambda_{I\cup\{i\}}
 \right)\notag\\
 &\quad=
 \mu(\lambda_g,\lambda_g,\lambda_g)
 +\sum_{s\in I}
   \mu(\langle m_s\rangle,\langle m_s\rangle,\langle m_s\rangle)
 +\mu(\langle\ell_i\rangle,\langle m_i\rangle,\langle\ell_i\rangle)
 =0.
 \label{fixed:eq-halfturn-square-seam}
\end{align}
Thus \(\mathcal S_{i\mid I}^{\,2}\) has weight zero.  Its composition
with the compression body has correction
\begin{align}
 &\mu\!\left(
 \lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}},
 (\sigma_{i\mid I})_*^{-2}\lambda_{I\cup\{i\}}
 \right)\notag\\
 &\quad=
 \mu(\lambda_g,\lambda_g,\lambda_g)
 +\sum_{s\in I}
   \mu(\langle m_s\rangle,\langle m_s\rangle,\langle m_s\rangle)
 +\mu(\langle m_i\rangle,\langle m_i\rangle,\langle m_i\rangle)
 =0,
 \label{fixed:eq-forward-halfturn-local-seam}
\end{align}
because \(\sigma_{i\mid I}^{\,2}\) is \(-1\) on the active plane and the
identity on every other summand.  The ordinary relative equality therefore
lifts with no central shift, proving \eqref{fixed:eq-H1}.
\end{proof}

\begin{proposition}[Fourth power and capping-kernel absorption]
\label{fixed:prop-capping}
The exact localized identities
\begin{equation}
 \boxed{\mathcal S_{i\mid I}^{\,4}=V_{d_{i\mid I}}}
 \tag{H2}\label{fixed:eq-H2}
\end{equation}
and
\begin{equation}
 \boxed{
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)V_{d_{i\mid I}}
 =\mathcal R_{i\mid I}^{\mathrm{rev}}(0)}
 \tag{H3}\label{fixed:eq-H3}
\end{equation}
hold as weight-zero marked Walker morphisms.
\end{proposition}

\paragraph{The two-chain relation and capping.}
The two-chain relation converts the chosen half-turn word into a boundary
twist after taking its fourth power.  The twist bounds the separating disk
that caps the active one-holed torus, so the resulting mapping cylinder is
absorbed by the declared capping kernel.

\begin{proof}
The braid relation for the two curves meeting once and the two-chain
relation in the fixed one-holed-torus marking~\cite{FarbMargalit2012} give
\begin{equation}
 \sigma_{i\mid I}^{\,2}=(T_{m_i}T_{\ell_i})^3,
 \qquad
 \sigma_{i\mid I}^{\,4}
 =(T_{m_i}T_{\ell_i})^6
 =T_{\delta_{i\mid I}}.
 \label{fixed:eq-local-two-chain}
\end{equation}
The first equality is the braid calculation for the three-twist word; the
second is the relative two-chain relation.  The final twist is positive:
the separating curve is unoriented, so the opposite boundary orientations
on its two sides do not invert the ambient right Dehn twist.  Equation
\eqref{fixed:eq-halfturn-square-seam} already shows that the first square
has weight zero.  Squaring it has correction
\begin{align}
 &\mu\!\left(
 (\sigma_{i\mid I})_*^2\lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}},
 (\sigma_{i\mid I})_*^{-2}\lambda_{I\cup\{i\}}
 \right)\notag\\
 &\quad=
 \mu(\lambda_g,\lambda_g,\lambda_g)
 +\sum_{s\in I}
   \mu(\langle m_s\rangle,\langle m_s\rangle,\langle m_s\rangle)
 +\mu(\langle m_i\rangle,\langle m_i\rangle,\langle m_i\rangle)
 =0.
 \label{fixed:eq-local-fourth-power-seam}
\end{align}
Hence \eqref{fixed:eq-local-two-chain} lifts exactly to
\eqref{fixed:eq-H2}.

For \eqref{fixed:eq-H3}, take the positive boundary-parallel twist
\(\tau_i\) in the relative \(k\)-disk surface, with weight zero in the
common pulled-back Walker extension of Proposition
\ref{fixed:prop-common-walker-extension}.  Its support is a small
annulus around the \(i\)-th boundary circle.  Capping at the \(i\)-th site
in the high vertex turns this annular twist into the separating twist,
whereas capping by the original disk in the low vertex makes it isotopic to
the identity relative to the remaining marking.  Thus
\[
 e_{I\cup\{i\}}(\tau_i)=d_{i\mid I},
 \qquad
 e_I(\tau_i)=\operatorname{id}_{X_I}.
\]
The twist extends through the literal reverse compression body: the
separating curve bounds the retained compression disk there, and twisting
a collar of that disk realizes the boundary twist in the three-manifold.
After compression the same twist is supported in the capped disk and is
trivial on the lower boundary.  The extension is supported in the active
handle region and fixes both endpoint markings elsewhere.  The equality of
the underlying relative cobordisms is therefore
\[
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)C(d_{i\mid I},0)
 =
 C(\operatorname{id}_{X_I},0)
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0).
\]
The two Walker interfaces are, respectively,
\begin{align*}
 &\mu\!\left(
 (d_{i\mid I})_*\lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}}
 \right)=0,\\
 &\mu(\lambda_I,\lambda_I,\lambda_I)=0.
\end{align*}
Here \(d_{i\mid I}\) is separating and acts trivially on first homology.
Thus the relative covariance equality has no central correction and is
exactly \eqref{fixed:eq-H3}.  This argument localizes the capping-kernel
mechanism at the typed vertex; it does not import an arbitrary-genus
half-turn identity.
\end{proof}

\begin{proposition}[Inverse and positive reverse absorption]
\label{fixed:prop-reverse-absorption}
One has
\begin{equation}
 \boxed{
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)
 \mathcal S_{i\mid I}^{-2}
 =\mathcal R_{i\mid I}^{\mathrm{rev}}(0)}
 \tag{H4}\label{fixed:eq-H4}
\end{equation}
and
\begin{equation}
 \boxed{
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)
 \mathcal S_{i\mid I}^{2}
 =\mathcal R_{i\mid I}^{\mathrm{rev}}(0).}
 \tag{H5}\label{fixed:eq-H5}
\end{equation}
The same identities hold after assigning an arbitrary weight \(r\) to the
reverse handle.
\end{proposition}

\begin{proof}
 Reverse the actual marked cobordism, rather than rewriting its homology row.
Contravariance reverses the boundary order, and the interface Maslov term is
zero because the transported active line agrees with the declared line at
the seam.  The same argument handles both phase choices.

Actual orientation reversal of \eqref{fixed:eq-H1} reverses the order,
inverts the outgoing cylinder, and negates the zero weight.  It gives
\eqref{fixed:eq-H4}.  Its interface correction is
\[
 \mu\!\left(
 (\sigma_{i\mid I})_*^{-2}\lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}},
 \lambda_{I\cup\{i\}}
 \right)=0
\]
by the same old/spectator/active direct-sum splitting as
\eqref{fixed:eq-forward-halfturn-local-seam}.

Now use the fourth power rather than identifying the half-turn square with
the identity:
\begin{align}
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)\mathcal S_{i\mid I}^{2}
 &=
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)
 \mathcal S_{i\mid I}^{-2}\mathcal S_{i\mid I}^{4}\notag\\
 &=
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)V_{d_{i\mid I}}\notag\\
 &=
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0).
 \label{fixed:eq-positive-absorption-calculation}
\end{align}
All even powers preserve the declared Lagrangian, so the displayed
reassociation introduces no Maslov correction.  The last equality is
\eqref{fixed:eq-H3}.  A central weight shift on the reverse handle passes
unchanged through the same zero-seam calculation, proving the assertion for
arbitrary \(r\).  Neither equation asserts
\(\mathcal S_{i\mid I}^{2}=1\); indeed
\(\mathcal S_{i\mid I}^{4}=V_{d_{i\mid I}}\) retains the nontrivial
separating twist.
\end{proof}

\subsection{Actual elementary source dagger}

Actual source orientation reversal keeps the fixed endpoint Lagrangians
fixed.  It reverses the arrow and negates its Walker weight.  Thus, for the
exact slope cylinder,
\begin{equation}
 \boxed{
 \Theta_{i\mid I}(u)^\dagger
 =C(\sigma_{i\mid I}^{-1},-u).}
 \tag{H6}\label{fixed:eq-H6}
\end{equation}
In particular, the right side is not \(\Theta_{i\mid I}(-u)\).  The two mapping
classes have the same image of the meridian as an unoriented line, but they
are distinct relative marked mapping classes.
The preferred inverse cylinder is nevertheless
\[
 \mathcal S_{i\mid I}^{-1}
 =C(\sigma_{i\mid I}^{-1},0),
\]
because its product with \(\mathcal S_{i\mid I}\) has active correction
\(\mu(\langle\ell_i\rangle,\langle m_i\rangle,
\langle\ell_i\rangle)=0\), while every spectator contribution has a
repeated pair.

\Needspace{4\baselineskip}
\begin{proposition}[Elementary phase-arrow dagger]
\label{fixed:prop-elementary-dagger}
For \(e,v\in\{0,1\}\), with inactive slope coordinates ignored,
\begin{align}
 \boxed{
 \bigl(J_{i\mid I}^{e}(c,u^+)\bigr)^\dagger
 =
 R_{i\mid I}^{e}(-c,-u^+)}
 \tag{H7}\label{fixed:eq-H7}\\
 \boxed{
 \bigl(R_{i\mid I}^{v}(r,u^-)\bigr)^\dagger
 =
 J_{i\mid I}^{v}(-r,-u^-).}
 \tag{H8}\label{fixed:eq-H8}
\end{align}
The weights on the two sides are, respectively, \(-a_i\) and \(-b_i'\).
\end{proposition}

\paragraph{Source reversal and the dagger identities.}
Source reversal is applied only after the elementary compression and reverse
families have been constructed.  The local dagger identities follow from
orientation reversal of the marked product model, with weight negation and
the written factor order retained.

\begin{proof}
For phase zero, actual orientation reversal of the weighted compression
body gives
\[
 \mathcal C_{i\mid I}(c)^\dagger
 =\mathcal R_{i\mid I}^{\mathrm{rev}}(-c),
 \qquad
 \bigl(\mathcal R_{i\mid I}^{\mathrm{rev}}(r)\bigr)^\dagger
 =\mathcal C_{i\mid I}(-r).
\]

For phase one, contravariance and \eqref{fixed:eq-H6} first give the
literal formulas
\begin{align}
 \bigl(\Theta_{i\mid I}(u^+)\mathcal C_{i\mid I}(c)\bigr)^\dagger
 &=
 \mathcal R_{i\mid I}^{\mathrm{rev}}(-c)
 C(\sigma_{i\mid I}^{-1},-u^+),
 \label{fixed:eq-forward-phase-literal-dagger}\\
 \bigl(\mathcal R_{i\mid I}^{\mathrm{rev}}(r)
 \Theta_{i\mid I}(u^-)\bigr)^\dagger
 &=
 C(\sigma_{i\mid I}^{-1},-u^-)
 \mathcal C_{i\mid I}(-r).
 \label{fixed:eq-reverse-phase-literal-dagger}
\end{align}
The proved absorptions, not an equality of slope generators, now give
\begin{align}
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)\mathcal S_{i\mid I}^{-1}
 &=
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)\mathcal S_{i\mid I},
 \label{fixed:eq-reverse-inverse-slope-absorption}\\
 \mathcal S_{i\mid I}^{-1}\mathcal C_{i\mid I}(0)
 &=
 \mathcal S_{i\mid I}\mathcal C_{i\mid I}(0).
 \label{fixed:eq-forward-inverse-slope-absorption}
\end{align}
Indeed, the first follows by multiplying
\(\mathcal R^{\mathrm{rev}}\mathcal S^2=\mathcal R^{\mathrm{rev}}\) on the right by
\(\mathcal S^{-1}\), and the second follows by applying
\(\mathcal S^{-1}\) on the left to
\(\mathcal S^2\mathcal C=\mathcal C\).

Both signs of the slope have zero handle seam.  Explicitly, the active
summand in the first equality is
\[
 \mu(\langle\ell_i\rangle,\langle m_i\rangle,
     \langle m_i\rangle)=0,
\]
because both \(\sigma_{i\mid I}\) and
\(\sigma_{i\mid I}^{-1}\) carry the meridian line to the longitude line.
For the second equality it is
\[
 \mu(\langle m_i\rangle,\langle m_i\rangle,
     \langle\ell_i\rangle)=0.
\]
The old and spectator summands contain three equal entries.  Consequently
central shifts may be restored in
\eqref{fixed:eq-reverse-inverse-slope-absorption} and
\eqref{fixed:eq-forward-inverse-slope-absorption} without changing either
weight.  Equations
\eqref{fixed:eq-forward-phase-literal-dagger} and
\eqref{fixed:eq-reverse-phase-literal-dagger} therefore become the
phase-one instances of \eqref{fixed:eq-H7} and
\eqref{fixed:eq-H8}.  Lemma
\ref{fixed:lem-elementary-weights} shows that the common weights are
\(-c-u^+=-a_i\) and \(-r-u^-=-b_i'\).
\end{proof}

\subsection{The complete weighted same-site table}

Define the preferred local double in its actual endpoint Hom set by
\begin{equation}
 \mathcal D_{i\mid I}(0)
 :=
 R_{i\mid I}^{0}(0)J_{i\mid I}^{0}(0)
 =
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)\mathcal C_{i\mid I}(0)
 \in\operatorname{End}(X_I).
 \tag{L0}\label{fixed:eq-L0}
\end{equation}
For \(n\in\mathbb Z\), \(\mathcal D_{i\mid I}(n)\) denotes the same marked
cobordism with total Walker weight \(n\).  This is a central weight shift
inside \(\operatorname{End}(X_I)\); it is not an operator that has been
identified across different labels or subset vertices.

For the fixed local phases \(e,v\), put
\begin{equation}
 n_i:=a_i+b_i'
 =c+r+eu^++vu^-.
 \tag{L1}\label{fixed:eq-L1}
\end{equation}

\Needspace{4\baselineskip}
\begin{theorem}[Weighted local phase table]
\label{fixed:thm-local-table}
For every \(I\subseteq[k]\), \(i\notin I\), every
\(e,v\in\{0,1\}\), and every active constituent integer weight,
\begin{equation}
 \boxed{
 R_{i\mid I}^{v}(r,u^-)
 J_{i\mid I}^{e}(c,u^+)
 =
 \begin{cases}
  \mathcal D_{i\mid I}(n_i),&e=v,\\
  C(\operatorname{id}_{X_I},n_i),&e\ne v.
 \end{cases}}
 \tag{L2}\label{fixed:eq-L2}
\end{equation}
At preferred weight zero this is
\begin{equation}
 \begin{array}{c|cc}
  &e=0&e=1\\ \hline
  v=0&\mathcal D&1\\
  v=1&1&\mathcal D
 \end{array}.
 \tag{L3}\label{fixed:eq-L3}
\end{equation}
Here the table entries abbreviate
\(\mathcal D=\mathcal D_{i\mid I}(0)\) and
\(1=C(\operatorname{id}_{X_I},0)\).
\end{theorem}

\paragraph{Matching phases in the common seam.}
Transport the low-end and high-end Lagrangian lines into one common seam
calculation.  Matching phases cancel the active slope change and leave the
local double; mismatching phases leave the identity on the spectator surface.
The topology of the matching entry is then determined by the same marked
one-holed-torus model.

\begin{proof}
\emph{Seam and weight.}
We first calculate the gluing seam independently of the topology.  The
line transported to the high endpoint by the forward phase arrow is
\[
 L_J^e=
 \begin{cases}
  \langle m_i\rangle,&e=0,\\
  \langle\ell_i\rangle,&e=1,
 \end{cases}
\]
and the line inverse-transported from the low endpoint by the reverse phase
arrow is
\[
 L_R^v=
 \begin{cases}
  \langle m_i\rangle,&v=0,\\
  \langle\ell_i\rangle,&v=1.
 \end{cases}
\]
These statements use the exact cylinder
\(\sigma_{i\mid I}\); they are not definitions made from the desired
table.  On the old surface and on each \(s\in I\), the corresponding three
entries are \(\lambda_g,\lambda_g,\lambda_g\) and
\(\langle m_s\rangle,\langle m_s\rangle,\langle m_s\rangle\).
The four active-plane seams are therefore
\begin{equation}
 \begin{array}{c|cc}
 &e=0&e=1\\ \hline
 v=0&
 \mu(\langle m_i\rangle,\langle m_i\rangle,\langle m_i\rangle)&
 \mu(\langle\ell_i\rangle,\langle m_i\rangle,\langle m_i\rangle)\\[2pt]
 v=1&
 \mu(\langle m_i\rangle,\langle m_i\rangle,\langle\ell_i\rangle)&
 \mu(\langle\ell_i\rangle,\langle m_i\rangle,\langle\ell_i\rangle)
 \end{array}
 =0.
 \tag{L4}\label{fixed:eq-L4}
\end{equation}
In particular, the \((e,v)=(1,1)\) seam is the two-slope seam
\[
 \mu(\langle\ell_i\rangle,\langle m_i\rangle,
     \langle\ell_i\rangle)=0;
\]
it is not omitted or replaced by a parity slogan.  Together with
\eqref{fixed:eq-forward-internal-seam} and
\eqref{fixed:eq-reverse-internal-seam}, this proves that every
constituent weight contributes exactly once and that the total is \(n_i\).

\emph{Matching phases.}
It remains to identify the four underlying marked cobordisms.  At preferred
weight zero their right-factor-first words are
\begin{equation}
 \begin{array}{c|cc}
 &e=0&e=1\\ \hline
 v=0&
 \mathcal R^{\mathrm{rev}}\mathcal C&
 \mathcal R^{\mathrm{rev}}\mathcal S\mathcal C\\
 v=1&
 \mathcal R^{\mathrm{rev}}\mathcal S\mathcal C&
 \mathcal R^{\mathrm{rev}}\mathcal S^2\mathcal C,
 \end{array}
 \label{fixed:eq-four-local-words}
\end{equation}
where every symbol in this display carries the fixed tag \(i\mid I\).
The first word is \(\mathcal D_{i\mid I}(0)\).  The last word is the same double,
because either
\(\mathcal S^2\mathcal C=\mathcal C\) by
\eqref{fixed:eq-H1}, or
\(\mathcal R^{\mathrm{rev}}\mathcal S^2=\mathcal R^{\mathrm{rev}}\) by
\eqref{fixed:eq-H5}.  Thus the \((1,1)\) entry is \(\mathcal D\), not the
identity.

\emph{Mismatching phases.}
For either mismatch, the common middle word is
\(\mathcal R^{\mathrm{rev}}\mathcal S\mathcal C\).  We now identify it with the
actual product relative to both endpoint markings.  Before the slope is
inserted, the return compression disk has boundary \(m_i\).  Pulling that
boundary through the exact word
\(T_{m_i}T_{\ell_i}T_{m_i}\) gives the actual product-framed longitude, not
only its homology class, and this curve meets the belt meridian once.  Take
a small neighborhood of the ordered feet, joining band, one-handle, and
return compression disk.  In these coordinates the return attaching curve
crosses the belt circle exactly once and its normal framing is the product
framing.  The standard local cancellation deletes the one-handle and the
return two-handle inside this neighborhood while leaving the complement
and both endpoint collars fixed.

\emph{Excluding a boundary twist.}
This is an exact marked cancellation, not a conclusion from the
intersection number alone: the attaching feet, band, co-core, actual
curves, and product normal have all been retained.  To see that no relative
boundary twist remains, cut the local one-holed torus along
\(m_i\cup\ell_i\).  The written three-twist word carries the marked
meridian edge to the marked longitude edge with the collar framing fixed by
the same crossing arc used in the proof of Proposition
\ref{fixed:prop-halfturn}.  A word differing by
\(T_{\delta_{i\mid I}}^q\) would turn that framed arc through an additional
\(2\pi q\), whereas the explicit word has \(q=0\).  The cancellation is
therefore the product \(X_I\times[0,1]\) with its fixed endpoint
parameterizations.  It follows that
\[
 \mathcal R_{i\mid I}^{\mathrm{rev}}(0)
 \mathcal S_{i\mid I}\mathcal C_{i\mid I}(0)
 =C(\operatorname{id}_{X_I},0).
\]
\emph{Restoring the weight.}
Both mismatch entries use this same actual word; only the label-attached
phase bookkeeping says whether the slope belongs to the forward or reverse
constituent.  Restoring the already calculated weight \(n_i\) proves all
four cases of \eqref{fixed:eq-L2}, and
\eqref{fixed:eq-L3} is its zero-weight specialization.
\end{proof}

\begin{proposition}[Bare local topology and first-homology separation]
\label{fixed:prop-local-topology}
For every \(n\in\mathbb Z\),
\begin{equation}
 \boxed{
 \mathcal D_{i\mid I}(n)
 \cong
 C(\operatorname{id}_{X_I},n)
 \mathbin{\#}(S^1\times S^2,0).}
 \tag{L5}\label{fixed:eq-L5}
\end{equation}
Consequently,
\begin{equation}
 \boxed{
 \mathcal D_{i\mid I}(n)
 \ne C(\operatorname{id}_{X_I},m)
 \quad\text{for all }m,n\in\mathbb Z.}
 \tag{L6}\label{fixed:eq-L6}
\end{equation}
\end{proposition}

\paragraph{The separating-sphere identification of the double.}
Close a regular neighborhood of the active forward--reverse pair by its
separating collar.  The exterior remains a product, while the interior is
the genus-one meridian-to-meridian double, which is \(S^1\times S^2\).
This separates the local topological contribution from the boundary
marking and Walker-weight bookkeeping.

\begin{proof}
\emph{The enclosed local piece.}
Use the chosen separating collar to close the boundary of a regular
neighborhood of the active forward--reverse handle pair.  The resulting
two-sphere encloses only the \(i\)-th handle region because the exterior
product structure is fixed throughout the composite.  Outside that sphere
the bare double
\(\mathcal R_{i\mid I}^{\mathrm{rev}}\mathcal C_{i\mid I}\) is the product on
\(X_I\).  Inside it, the forward and return compression disks have the
same marked meridian \(m_i\).  Equivalently, the two punctured solid tori
are glued meridian to meridian.  This is the genus-one Heegaard description
of \(S^1\times S^2\), rather than the meridian-to-longitude gluing that
would give the cancelling local product.  The incoming and outgoing local
interfaces account for the two removed balls, so the enclosed marked piece is
\(S^1\times S^2\) with two open balls removed.  Gluing it back to the
exterior product is precisely an interior connected sum and gives the
ordinary relative cobordism decomposition in \eqref{fixed:eq-L5}.

\emph{Walker weight.}
Its Walker seam is
\[
 \mu(\langle m_i\rangle,\langle m_i\rangle,
     \langle m_i\rangle)=0,
\]
and the old and spectator terms also have three equal entries.  The
separating sphere has zero first homology and creates no further Maslov
term.  Hence the complete weight \(n\) may be placed on the product
cylinder and the closed summand may be recorded at weight zero, proving the
extended statement \eqref{fixed:eq-L5}.

\emph{Homological separation.}
Let \(g_I=g+|I|\) be the genus of \(X_I\).  The underlying identity cylinder
has
\[
 \beta_1(X_I\times[0,1])=2g_I,
\]
whereas connected sum in the interior with \(S^1\times S^2\) gives
\[
 \beta_1\!\left(
 (X_I\times[0,1])\mathbin{\#}(S^1\times S^2)
 \right)=2g_I+1.
\]
Walker weight does not change the underlying three-manifold.  Therefore no
choice of \(m,n\) can make the two marked extended cobordisms equal, which
proves \eqref{fixed:eq-L6}.
\end{proof}

\begin{remark}[Scope of the local calculation]
\label{fixed:rem-local-scope}
Every equality in this section was proved in a fixed marked Walker Hom set
from the handle model, the exact slope word, relative covariance, and
explicit Maslov calculations.  No target matrix, target faithfulness,
finite-dimensional value, or physical TQFT was used.  The argument proves
no interchange between different sites, no order independence, and no
global connected-sum theorem; those questions require the disjoint-support
calculus of the next section.  It makes no statement for repeated,
overlapping, nested, moved, or relabelled sites, and uses no refined
genus-at-least-three carrier.  It also never identifies
\(\mathcal S_{i\mid I}^2\) with the identity or
\(\sigma_{i\mid I}^{-1}\) with \(\sigma_{i\mid I}\).
\end{remark}