\section{The sign and ten-row coefficients}\label{sec:11-top-coefficients}
We prove that $\mathcal C_{(11)}$ and $\mathcal C_{(10,1)}$ vanish by
proving injectivity of the induced compact boundary map on the residual
quotient. We begin with the compact nonseparating injectivity theorem,
stated for every genus at least five, since
Section~\ref{sec:16-main-proof} uses it in the genus induction.

\begin{theorem}[Compact nonseparating injectivity]
\label{thm:compact-injectivity}
For every $g\geq5$ and finite $F$, with the maps of
Definition~\ref{def:compact-boundary-row}, both maps
\[
\xi^*:H^{15}(\Mbar_{g,F})\longrightarrow N_{g,F},
\qquad
\beta|_{C^1_{\rm sep}}:C^1_{\rm sep}\longrightarrow C^2_{g,F}
\]
are injective. They are equivariant morphisms of pure Hodge structures.
\end{theorem}

\begin{proof}
\emph{Lower-degree inputs.}
By \cite[Theorem 2.10]{arbarello1998}, the map
$\xi^*:H^k(\Mbar_{g,F})\to H^k(\Mbar_{g-1,F\cup\{c,d\}})$ is injective
for $k\leq2g-2$ if $g\leq7$ and for $k\leq g+5$ if $g\geq7$; this gives injectivity
on $H^2$ and $H^4$ for every genus at least three.  We also need injectivity of the
$H^{13}$ pullback in genus at least three.  By
Theorem~\ref{thm:cited-thirteen-cohomology}, $H^{13}$ is the direct sum of
the images of the elliptic-tail Gysin maps with domains
\[
 H^{11}(\Mbar_{1,A\cup\{p\}})\otimes
 H^0(\Mbar_{h-1,A^c\cup\{p'\}})\otimes\Lmot,
\]
with the sum over stable allocations $A$ of the ordinary markings; for
$h\ge2$ these sources map injectively by the same theorem.  Normalize a
nonseparating node in such a source.  If it lies on the elliptic
factor, the resulting factor is a stable genus-zero moduli space, so
Theorem~\ref{thm:cited-genus-zero-cycles} gives
$H^{11}(\Mbar_{0,m})=0$ for every stable $m$; hence this pullback component
is zero.  If it lies
on the unit partner, the resulting source is the corresponding elliptic
tail with the two new flags on a genus-$(h-2)$ partner.  The allocations are
distinct on the labelled cover, and the unit pullback has coefficient one.
Thus nonseparating pullback is injective on $H^{13}$ for every
$h\ge3$.  The same argument applies after taking graph invariants: pullback from a finite
labelled quotient is injective, and the loop flags are fixed as an
unordered pair.

The lower-odd vanishing of
Theorem~\ref{thm:cited-low-odd-vanishing}, together with this description of
the degree-eleven elliptic factors, leaves on a separating normalization only
the six degree pairs
\begin{equation}\label{eq:compact-injectivity-six-pairs}
 (15,0),\ (0,15),\ (13,2),\ (2,13),\ (11,4),\ (4,11).
\end{equation}
There is one odd factor in each pair; in the degree-eleven cases it has
genus one.  Different ordered K\"unneth pairs remain distinct under pullback
because pullback preserves the degree on each vertex factor.

\emph{Loop--bridge detectors.}
Let $\Lambda$ be a separating one-edge graph at $(g,F)$, with vertex genera
$u,v$ and ordinary-label partition $F=A\sqcup B$.  Choose one vertex of
positive genus, lower its genus by one, and add two new flags $c,d$ there,
joined to a loop.  Keep the original separating bridge and order the two
edges as $(e_{\rm loop},e_{\rm br})$.  The resulting two-edge graph
$\Gamma$ is stable, and contracting $e_{\rm loop}$ recovers $\Lambda$ while
contracting $e_{\rm br}$ gives the nonseparating one-edge graph at $(g,F)$.
Consequently, for $z\in C^1_{\rm sep}$, the $\Gamma$-coordinate of
$\beta(z,0)$ is
\begin{equation}\label{eq:compact-injectivity-loop-bridge}
 \xi_{e_{\rm loop}}^*z_\Lambda\otimes
 (e_{\rm loop}\wedge e_{\rm br}),
\end{equation}
because the second contraction contributes the zero nonseparating coordinate.
No other separating source coordinate contributes to this labelled target.
The loop and bridge are distinguished, so they cannot be exchanged by a
graph automorphism.  Reversal of the loop flags fixes the edge determinant
and the clutching pullback.  Hence \eqref{eq:compact-injectivity-loop-bridge}
has coefficient $+1$ also after
passing to $\Aut_F(\Gamma)$-invariants.  If $\Lambda$ has a vertex
exchange symmetry, perform the test on the labelled cover and then take
invariants; for a vector $w\otimes1+1\otimes w$, projection to the unchanged
degree-fifteen factor returns $w$.

We now apply this detector to each term of
\eqref{eq:compact-injectivity-six-pairs}.  For a $(15,0)$ term whose
unit factor has positive genus, put the loop on that unit factor.  The unit
pullback is the identity, so \eqref{eq:compact-injectivity-loop-bridge}
is the coefficient itself.  If the unit
factor is rational, stability gives it at least two ordinary labels.  The
degree-fifteen factor then has $n-b+1<n$ flags, where $b\ge2$ is the number
of labels on the rational tail (the bridge flag is included), and the same
detector uses induction on that smaller marking set.

For a $(13,2)$ term, put the loop on the $H^2$ factor when its genus is at
least three and use the $H^2$ injectivity above.  When that genus is at most
two, the $H^{13}$ factor has genus at least $g-2\ge3$; put the loop there and use
the $H^{13}$ injectivity just established.
The argument is unchanged for a rational
$H^2$ partner.  For a $(11,4)$ term, the $H^{11}$ factor is elliptic and the
other factor has genus $g-1\ge4$; putting the loop on the latter and using
the full $H^4$ pullback detects the coefficient.  Exchanging the two named
factor positions gives the reverse pairs in
\eqref{eq:compact-injectivity-six-pairs}.  Thus every separating
K\"unneth coefficient is detected, including the compact $H^4$ factors.

\emph{The marking induction.}
Fix $g\ge5$ and induct on $n=|F|$.  We first record the base range needed
for the compact boundary map.  For $g=5$, the series $H,G,U$ have least
possible degrees ten, nine, and twelve, respectively
(Notation~\ref{def:common-series}). Hence all four ordinary summands of
Theorem~\ref{thm:ordinary-image} vanish for $n\le8$.
Proposition~\ref{prop:genus-five-initial} then gives
$H^{15}(\Mbar_{5,F})=0$ in that range.
If $(g,n)=(6,n)$ with $n\le5$, or
$(g,n)=(7,n)$ with $n\le3$, the Chow--K\"unneth property in
Proposition~\ref{prop:cited-chow-kunneth-bases} gives
$W_{15}H^{15}(\Mopen_{g,n})=0$.  We check directly that the compact
boundary group is zero in these arities.  A separating boundary divisor with two
positive-genus vertices has strictly smaller vertex genera, and each factor has at most $n+1\le6$ flags;
the established genus-one, genus-two, and genus-three low-arity vanishings
in Theorem~\ref{thm:genus-one},
Proposition~\ref{prop:genus-two-gysin-surjectivity}, and
Proposition~\ref{prop:genus-three-threshold}, together with the ordinary
boundary generation in Theorem~\ref{thm:genus-four-base} and
Proposition~\ref{prop:genus-five-initial}, show that its degree-fifteen
K\"unneth source is zero.  In genera four and five the displayed ordinary
sources have the same minimum elliptic-tail arity, so their surjectivity
also gives compact vanishing for these small numbers of flags.  Indeed, an odd factor of degree at
most nine vanishes by Theorem~\ref{thm:cited-low-odd-vanishing}, while the
elliptic degree-eleven and degree-thirteen sources in
Theorem~\ref{thm:cited-thirteen-cohomology} require at least eleven flags.
For a rational tail, stability gives at least two ordinary labels on that
tail, so the remaining genus-$g$ factor has fewer markings; induction on
the marking number gives the same vanishing.  Thus every one-edge source is
zero.  The localization sequence and
$W_{15}H^{15}(\Mopen_{g,n})=0$ then give
$H^{15}(\Mbar_{g,n})=0$.  In all these initial cases the loop--bridge
detectors above prove the
separating injection simultaneously, starting at the empty
marking set and using the same rational-tail induction.  Hence $\alpha$ is
injective in the exceptional initial ranges.

For all remaining stable $(g,F)$ needed below,
Proposition~\ref{prop:compact-restriction-row} gives injectivity of $\alpha$
whenever $15<2g$ for $|F|\le1$, or $15<2g-2+|F|$ for $|F|\ge2$; for
$g=5$ and $n\ge9$ this is $15<8+n$.

Assume now that $n$ is beyond the initial range and that nonseparating
$H^{15}$ pullback is injective for all smaller marking sets at this genus.
Let $z\in C^1_{\rm sep}$ satisfy $\beta(z,0)=0$.  The loop--bridge tests
above, together with the induction in the rational-unit case, force every
coefficient in \eqref{eq:compact-injectivity-six-pairs} to vanish.
Thus $z=0$, proving injectivity of
$\beta|_{C^1_{\rm sep}}$.

Finally let $x\in H^{15}(\Mbar_{g,F})$ satisfy $\xi^*x=0$.  The nonseparating
coordinate of $\alpha(x)$ is then zero, so $\alpha(x)$ lies in
$C^1_{\rm sep}$.  Since $\beta\alpha=0$ by
Proposition~\ref{prop:compact-restriction-row}, the separating injectivity
just proved gives $\alpha(x)=0$.  Since $\alpha$ is injective (above), $x=0$.

All constructions are defined on labelled covers and commute with relabelling
of $F$.  Pullback along clutching maps and the K\"unneth projections are
morphisms of pure Hodge structures. Tensoring with the determinant lines
preserves this property, so both injections are equivariant morphisms of
pure Hodge structures.
\end{proof}

\begin{proposition}[Residual interface]\label{prop:residual-interface}
Let $(F,\rho)$ be one of
\[
 ([11],(1^{11})),\quad([11],(2,1^9)),\quad
 ([12],(2,2,2,1^6)),\quad([11],(3,1^8)),
\]
with $N_\rho$, $\mathscr Q_\rho$, $J_\rho$, $\delta_\rho$ of
Set-up~\ref{setup:g5-detector-row}.
For $\lambda\in\mathscr R_{11}$ set
\[
 \mathcal B_{\rho,\lambda}=
 \operatorname{Hom}_{\Sn_{11}}
 \left(\operatorname{Res}^{\Sn_{|F|}}_{\Sn_{11}}V_\rho,
 V_{\lambda^T}\right).
\]
\begin{enumerate}
\item The kernel of $\overline\delta_\rho$ is canonically the last
column of the following table, and
the multiplicities of $\Pmot$ are as in its middle columns:
\[
 \begin{array}{c|ccl}
 \rho&J_\rho&N_\rho/J_\rho&\ker\overline\delta_\rho\\\hline
 (1^{11})&\Pmot^{29}&\Pmot^{150}&\mathcal C_{(11)}\\
 (2,1^9)&\Pmot^{37}&\Pmot^{312}&\mathcal C_{(10,1)}\\
 (2,2,2,1^6)&\Pmot^{2}&\Pmot^{30}&
   (\mathcal C_{(9,2)}\otimes\mathcal B_{\rho,(9,2)})
   \oplus(\mathcal C_{(8,3)}\otimes\mathcal B_{\rho,(8,3)})\\
 (3,1^8)&\Pmot^{8}&\Pmot^{197}&\mathcal C_{(9,1,1)}
 \end{array}
\]
Both branching spaces in the third row have dimension one.
\item For the two nine-row partitions put
\[
 \mathcal U_\rho=M_\rho(\mathscr U_F),\qquad
 \mathcal R_\rho=M_\rho(\mathcal R_F^{\mathrm{Pic}})
       \oplus M_\rho(\mathcal R^{\mathrm{par}}_{+,F}),
\]
let $\Phi_\rho:\mathcal R_\rho\to N_\rho$ be the sum of the Picard
and parallel maps, and define the residual map
\[
 \varepsilon_\rho:\mathcal R_\rho\longrightarrow
 \mathscr Q_\rho/\delta_\rho(\mathcal U_\rho),\qquad
 r\longmapsto[\delta_\rho\Phi_\rho(r)].
\]
Then $J_\rho\subseteq\mathcal U_\rho$,
$J_\rho=\ker(\delta_\rho|_{\mathcal U_\rho})$, and
$\delta_\rho(\mathcal U_\rho)\simeq\Pmot^{29}$ and $\Pmot^{174}$ for
$\rho=(2,2,2,1^6)$ and $(3,1^8)$, respectively. Projection to the
residual source induces a canonical isomorphism
$\ker\varepsilon_\rho\cong\ker\overline\delta_\rho$.
\end{enumerate}
\end{proposition}

\begin{proof}
\emph{The canonical kernels.}
For $|F|=11,12$ we have $16<2\cdot5-2+|F|$.
Proposition~\ref{prop:cited-compact-support-bound} and the identities
in Proposition~\ref{prop:compact-restriction-row} give exactness of
\[
 0\longrightarrow H^{15}(\Mbar_{5,F})
 \xrightarrow{\ \alpha\ }C^1_{5,F}
 \xrightarrow{\ \beta_{5,F}\ }C^2_{5,F}.
\]
For $x\in\ker\delta_F$, there is a separating coordinate $z$ with
$\beta_{5,F}(z,x)=0$. Theorem~\ref{thm:compact-injectivity}
makes $z$ unique. Thus projection to the nonseparating coordinate
identifies $\ker\beta_{5,F}$ with $\ker\delta_F$, and under row
exactness this projection is $\xi_F^*$. Quotienting by its ordinary
image gives
\[
 \ker\overline\delta_\rho
 \cong M_\rho\bigl(H^{15}(\Mbar_{5,F})/\mathcal I_{5,F}\bigr).
\]
Proposition~\ref{prop:genus-five-support} identifies the right side.
At arity eleven this is $\mathcal C_{\rho^T}$, which gives the first,
second, and fourth rows of the table.
At arity twelve, removing either corner of $(2,2,2,1^6)$ gives
\[
 (2,2,1^7)=(9,2)^T,\qquad
 (2,2,2,1^5)=(8,3)^T.
\]
Branching is multiplicity free, giving the two indicated lines.

\emph{The sign and ten-row multiplicities.}
Let $\rho=(1^{11})$ or $(2,1^9)$.
By Lemma~\ref{lem:genus-four-source-multiplicities}, the multiplicities
of $\Pmot$ in $N_\rho$ are $179$ and $349$.
By Theorem~\ref{thm:ordinary-image} at genus five, the genus-five
ordinary series is
\[
 E\partial U+3HE+5HE^2+GE^2+Hh_2[E]+Ge_2[E].
\]
Its six terms contribute, respectively,
\[
\begin{array}{c|rrrrrr}
 (1^{11})&1&6&15&3&2&2\\
 (2,1^9)&1&6&15&9&2&4
\end{array}
\]
by the finite Pieri rule in the proof of
Lemma~\ref{lem:genus-four-source-multiplicities}. In the last two columns one uses
$[h_2[E]]_0=1$, $[h_2[E]]_1=h_1$, $[e_2[E]]_0=0$,
$[e_2[E]]_1=h_1$, and $[e_2[E]]_2=h_2+e_2$.
The ordinary image therefore has multiplicities $29$ and $37$.
Its pullback is injective by Theorem~\ref{thm:compact-injectivity},
so these are the multiplicities in $J_\rho$. Subtracting gives $150$
and $312$. Polarizability splits these subobjects of sums of $\Pmot$,
giving the stated, noncanonical, pure-Hodge isomorphisms.

\emph{The nine-row ordinary pullback.}
Let now $\rho=(2,2,2,1^6)$ or $(3,1^8)$.
Apply Theorem~\ref{thm:ordinary-image} at genus five with ordinary
markings $F$. By the old-absent exclusion in the proof of
Lemma~\ref{lem:genus-four-source-multiplicities}, only centered paths
contribute to $M_\rho(\mathcal I_{5,F})$.
Their leaf genera are $(1,3)$ and $(2,2)$.

For the first pair of genera the leaves are distinguished, giving the
coefficients of $(\sum_{r\ge1}s_{(r,1^8)})E^2$, which are the
$\mathcal T_{2,F}$ entries of
Lemma~\ref{lem:genus-four-source-multiplicities}: one for
$(2,2,2,1^6)$ and six for $(3,1^8)$. For the equal-genus leaves,
their exchange acts by $-1$ on the nine-row center hook, so the
allocation coefficient is its anti-invariant part. In the first case
the only contributing skew is $e_2$, and the split-leaf allocation
has the required sign, giving one copy. In the second case the
horizontal skews of sizes two, one, and zero give $1,1,0$ copies:
the first two come from the difference between putting all remaining
labels on either leaf; the empty allocation has no anti-invariant
part. There are therefore two copies. The injectivity in
Theorem~\ref{thm:ordinary-image} gives ordinary multiplicities
$1+1=2$ and $6+2=8$.

Pull back these ordinary paths by $\xi_F$. Their two selected edges
remain bridges under every degeneration, so the nonseparating node
being normalized lies inside a vertex factor and cannot coincide with
either bridge. Independent node-smoothing parameters give transverse
base change with coefficient one on the labelled products.
Normalizing the decorated genus-one factor gives zero by
Proposition~\ref{prop:eleven-hook-realizations}. For leaves $(1,3)$,
normalizing the genus-three leaf gives a centered path in
$\mathcal T_{2,F}$. Normalizing the genus-one leaf gives a rational
leaf carrying $c,d$; its inner Gysin is the class $Lw$ in the
degree-thirteen endpoint. Thus this pullback has coordinates
\[
 (Lw,0,0,w,0,0)
\]
in the coordinate order of Proposition~\ref{prop:g5-auxiliary-kernel}.
For leaves $(2,2)$, either normalization gives a centered path in
$\mathcal T_{1,F}$. On the cover ordering the leaves, the two branches
are exchanged; their quotient has this same support and retains the
ordinary normalization. These exhaust the pullback branches because
the vertex factors are compact. Hence $J_\rho\subseteq\mathcal U_\rho$.
Theorem~\ref{thm:compact-injectivity} preserves the two and
eight ordinary copies under $\xi_F^*$.

\emph{The auxiliary kernel.}
Lemma~\ref{lem:g5-double-loop-coordinate} gives
$M_\rho(q_F)=M_\rho(\operatorname{pr}_{\rm dl})\delta_\rho$.
Proposition~\ref{prop:g5-auxiliary-kernel}, together with the cohomological
source isomorphism of Lemma~\ref{lem:genus-four-source-multiplicities},
therefore gives
\[
 J_\rho\subseteq\ker(\delta_\rho|_{\mathcal U_\rho})
 \subseteq\ker(M_\rho(q_F)|_{\mathcal U_\rho}).
\]
The last kernel is the graph of $L$ on $\mathcal T_{2,F}$ together
with $\mathcal T_{1,F}^+$. Its multiplicities are $1+1=2$ and
$6+2=8$. The outer spaces in the displayed inclusions have the same
dimension, so both inclusions are equalities. Subtracting from the
auxiliary multiplicities thirty-one and one hundred eighty-two of
Lemma~\ref{lem:genus-four-source-multiplicities} gives
the stated images. Subtracting $J_\rho$ from the multiplicities
$32$ and $205$ of $N_\rho$ in the same lemma gives $N_\rho/J_\rho$,
by the same polarizability splitting.

\emph{The residual kernels.}
Finally Lemma~\ref{lem:genus-four-source-multiplicities} gives
\[
 N_\rho/J_\rho=(\mathcal U_\rho/J_\rho)
                  \oplus\Phi_\rho(\mathcal R_\rho),
\]
and $\overline\delta_\rho$ maps the first summand isomorphically onto
$\delta_\rho(\mathcal U_\rho)$. Projecting a kernel element onto its
$\mathcal R_\rho$ coordinate is injective. Conversely, if
$\varepsilon_\rho(r)=0$, there is a unique
$u\in\mathcal U_\rho/J_\rho$ satisfying
$\overline\delta_\rho(u)=-\delta_\rho\Phi_\rho(r)$.
Then $u+\Phi_\rho(r)$ is a kernel element.
\end{proof}

\begin{definition}\label{def:g5-old-absent-endpoints}
Let $F=[11]$, $A\subseteq F$ with $|A|=10$ or $11$, and
$B=F\setminus A$, $m=|B|$. Every source in this definition carries
the full coefficient $H^{11}(\Mbar_{1,A\sqcup\{p\}})(-2)$.
For $C\subseteq B$ let $Z_{A,C}$ be the ordinary path of genera
$(1,2,1)$ with allocations $(A,(B\setminus C)\sqcup\{c,d\},C)$.
Let $V_A(\epsilon,\eta)$ be the elliptic tail with genus-three partner divisor
$\epsilon\delta_{\mathrm{irr}}+\eta(\psi_c+\psi_d)$.
For every ordered partition $B=B_1\sqcup B_2$, form the endpoint
path correction $P_A(\epsilon,\eta)$ with coefficient $\eta$ on $(1,1,2)$
paths with the old flags split between the unit vertices,
$-\eta-2\epsilon$ on $(1,2,1)$ paths with the old flags split, and
$2\eta+2\epsilon$ on $(1,1,2)$ paths with both old flags on the middle
vertex. Both split placements occur with equal coefficients.
All other coefficients are zero. Put
$W_A(\epsilon,\eta)=V_A(\epsilon,\eta)-P_A(\epsilon,\eta)$ as a map into $N_{5,F}$.

On $Y=\Mbar_{2,B\sqcup\{q,a,b,c,d\}}$, write
$D_T=\delta_{0,T}$ for a rational-tail divisor and set
\[
 \Psi=\psi_a+\psi_b-\psi_c-\psi_d,\quad \Delta=D_{ab}-D_{cd}.
\]
If $B=\{i\}$ also set
$\Delta_i=D_{abi}-D_{cdi}$ and
$U_i=D_{ai}+D_{bi}-D_{ci}-D_{di}$.
Let $j_A^0:\Mbar_{1,A\sqcup\{p\}}\times Y\to \Mbar_{3,J}$ glue $p,q$,
and define, for each of these test divisors $v$,
\[
 \mathcal B_{A,v}(y)=
 \left(1\otimes\left[u\mapsto\int_Yu\lambda_1\lambda_2\psi_q^{m+2}v\right]\right)
             \operatorname{pr}_{11,4}(j_A^0)^*y.
\]
\end{definition}

\begin{lemma}\label{lem:g5-endpoint-matrices}
For the maps in Definition~\ref{def:g5-old-absent-endpoints}, the
$\mathcal B_{A,v}q_F$ tests annihilate the image of every elliptic
$H^{13}$ tail. On the selected endpoint sources they are diagonal
in $A$. For $m=0$, their matrix with rows $(\Psi,\Delta)$ and
columns $(Z_{A,\varnothing},W_A(0,1))$ is
\[
 I_0\begin{pmatrix}20&-72\\12&-24\end{pmatrix}.
\]
For $B=\{i\}$, their matrix with rows
$(\Psi,\Delta,\Delta_i,U_i)$ and columns
$(Z_{A,\varnothing},Z_{A,\{i\}},W_A(1,0),W_A(0,1))$ is
\[
 I_0\begin{pmatrix}
 24&-4&-72&-96\\14&-2&-24&-28\\-2&2&0&4\\4&-4&0&-24
 \end{pmatrix}.
\]
Their determinants are $384I_0^2$ and $12288I_0^4$, respectively.
\end{lemma}

\begin{proof}
\emph{The branch coefficients.}
A nonseparating cut of an ordinary tree lowers one vertex and puts
$a,b$ together there. Putting them on different vertices would create
an alternate path across one of its separating edges. The cut is
transverse to the ordinary edges, with coefficient one. Lowering the
decorated elliptic vertex gives zero odd cohomology in genus zero.
For $Z_{A,C}$, lowering the middle vertex gives a positive path fixed
by pair exchange. Lowering the last vertex gives a rational endpoint.
Consequently, for every elliptic coefficient $z$,
\begin{equation}\label{eq:top-coefficients-endpoint-branch}
 q_FZ_{A,C}(z)=j_{A*}^0\bigl(z\otimes(D_{abC}-D_{cdC})\bigr).
\end{equation}
There is one minus sign from the ordered compact loops.

Let $\Sigma_{xy}$ sum the stable genus-two separating divisors
splitting $x,y$, with complementary descriptions counted once.
The full labelled nonseparating divisor pullback is
\[
 \nu_{ab}^*\delta_{\rm irr}
 =\delta_{\rm irr}^{Y}-\psi_a-\psi_b+\Sigma_{ab}.
\]
The negative terms are the normal class of the selected node;
the sum consists of the branches which become nonseparating after
gluing $a,b$. Thus the coefficients of $q_FV_A(0,1)$ and
$q_FV_A(1,0)$ inside $j_{A*}^0$ are, respectively,
\[
 -\Psi,\qquad -\Psi+\Sigma_{ab}-\Sigma_{cd}.
\]

Positive-path coefficients: on a positive
endpoint path let $n_{\mathrm{old}},n_{\mathrm{new}}$ count the old and new
pair flags on the last unit vertex, and put
$j_{\mathrm{old}}=\mathbf1_{n_{\mathrm{old}}>0}$,
$j_{\mathrm{new}}=\mathbf1_{n_{\mathrm{new}}>0}$. The coefficient of
$z\otimes\psi_1\otimes\psi_2$ for $V_A(\epsilon,\eta)$ is
\[
 -2\epsilon(j_{\mathrm{old}}-j_{\mathrm{new}})
       -\eta(n_{\mathrm{old}}-n_{\mathrm{new}}).
\]
Indeed the irreducible divisor contributes
$12-n_T-\mathbf1_{n_T=1}=12-2\mathbf1_{n_T>0}$ for a pair $T$;
when it is split, both persistent Cartier branches contribute a
normal cotangent. The complete table is
\begin{equation}\label{eq:top-coefficients-positive-table}
\begin{array}{cc|c}
 \text{new pair}&\text{old pair}&\text{coefficient}\\\hline
 \text{middle}&\text{middle}&0\\
 \text{middle}&\text{split}&-\eta-2\epsilon\\
 \text{middle}&\text{last}&-2\eta-2\epsilon\\
 \text{last}&\text{middle}&2\eta+2\epsilon\\
 \text{last}&\text{split}&\eta\\
 \text{last}&\text{last}&0
\end{array}
\end{equation}
Both-split coefficients are zero. The three patterns defining
$P_A(\epsilon,\eta)$, together with their pair exchanges, have
exactly this table. Hence $q_FW_A(\epsilon,\eta)$ has zero full
positive endpoint projection, for each allocation of $B$.

\emph{Compact diagonalization.}
In a common refinement with the test bridge, an ordinary source edge
inside the tested elliptic factor raises its least odd degree from
eleven to at least thirteen. Projection to $(11,4)$ annihilates that term.
An odd component outside that factor also cannot contribute. In the
remaining cases the tested set $A$ is contained in the source's
decorated set $A'$. If $A'\setminus A\ne\varnothing$, the refinement
has a rational factor carrying $q$, these extra flags, and one other
attachment. Its dimension is $|A'\setminus A|-1\le m+1$, including
the exchanged sources. The factor $\psi_q^{m+2}$ restricts to zero on this
compact factor. Further refinements pull back this same zero
cotangent power.

When $A'=A$, distinct nested bridges would enclose an unmarked
rational tree with two exterior attachments, which stability
excludes. The test bridge therefore coincides with the source bridge.
Its normal class is $-\psi_p-\psi_q$. An $H^{13}$ tail still has
elliptic degree thirteen and has zero projection. For an $H^{11}$ coefficient
the matching contribution is
\[
 \operatorname{pr}_{11,4}(j_A^0)^*j_{A*}^0(z\otimes u)
       =-z\otimes\psi_qu.
\]
These arguments apply to the tails and every path in the correction,
and prove diagonalization and the asserted $H^{13}$ annihilation.
The use of the self-intersection formula with an odd coefficient follows
directly from the Thom class of the node smoothing line: its Euler
class is $-\psi_p-\psi_q$, and the Thom identity is linear over the
cohomology of the normalization. Thus it has the same normalization
as the excess bundle in Theorem~\ref{thm:genus-three-excess-cited}.

\emph{The compact moments.}
For divisors $u,v$ on $Y$ set
\[
 b_m(u,v)=\int_Y\lambda_1\lambda_2\psi_q^{m+3}uv,
 \qquad R=D_{acd}+D_{bcd}-D_{abc}-D_{abd}.
\]
Positive separating divisors and the irreducible divisor have zero pairing
by Lemma~\ref{lem:g5-basic-hodge-moments}. Rational tails containing
$q$ are annihilated by the displayed power of $\psi_q$. For $m=0$ the
remaining part of $\Sigma_{ab}-\Sigma_{cd}$ is $R$, and
\[
 I_0^{-1}\bigl(b_0(\Psi,\Psi),b_0(\Psi,\Delta)\bigr)=(-68,-20),
 \quad
 I_0^{-1}\bigl(b_0(R,\Psi),b_0(R,\Delta)\bigr)=(-4,-4).
\]
For $m=1$ put
$R_i=D_{acdi}+D_{bcdi}-D_{abci}-D_{abdi}$.
The remaining divisor is $U_i+R+R_i$. The rational factor of each
$R_i$ term has dimension two, and the test supplies just one divisor
there, so its pairings vanish. The needed table is
\[
\begin{array}{c|rrrr}
 I_0^{-1}b_1&\Psi&\Delta&\Delta_i&U_i\\\hline
 \Psi&-92&-24&4&-24\\
 \Delta&-24&-14&2&-4\\
 \Delta_i&4&2&-2&4\\
 U_i&-24&-4&4&-24\\
 R&-4&-4&0&0
\end{array}
\]
By the moments of Lemma~\ref{lem:g5-basic-hodge-moments} and the
stable forgetting identities of
Proposition~\ref{prop:external-stable-forgetting}, for $m=0$ the
repeated and distinct cotangent moments in
$b_0(\Psi,\Psi)$ are $13I_0$ and $30I_0$, giving
$4\cdot13-4\cdot30=-68$. For $m=1$ they are $19I_0$ and $42I_0$,
giving $-92$. A rigid tail contributes its genus-two node cotangent
with a minus sign on self-intersection; distinct tails meet only
when nested or disjoint. In particular
\[
\begin{aligned}
 b_m(D_{ab}-D_{cd},\Psi)&=-4(m+5)I_0,\\
 b_m(D_{ab}-D_{cd},\Delta)&=-2(m+6)I_0,\\
 b_1(\Delta_i,\Psi)&=4I_0,\qquad
 b_1(\Delta_i,\Delta)=2I_0.
\end{aligned}
\]
The last two use the rational factor $\Mbar_{0,4}$, on which
each cotangent and each boundary point has integral one.
The four rigid tails in $U_i$ each give $-6I_0$ against $\Psi$
and on self-intersection, hence both entries $-24I_0$.
Their four disjoint intersections with $\Delta$ give $-4I_0$;
their four nested intersections with $\Delta_i$ give $4I_0$.
The two self-intersections in $\Delta_i^2$ contribute $-I_0$ each.
Each $R$ term has a one-dimensional rational factor: restriction of
$\Psi$ and $\Delta$ gives the entries $-4I_0$, whereas the other
two tests either have incompatible boundary supports or have degree below the dimension of that factor.

The rational-middle branches of $P_A(\epsilon,\eta)$ vanish under
these tests: the power of $\psi_q$ exceeds the rational dimension,
except when all flags are there, in which case the additional test
divisor also lies there and exceeds it. The remaining split-old
branches have coefficient $-R$ for $m=0$ and $-(R+R_i)$ for $m=1$.
After the minus sign from the outer normal class, their test vector is the $R$ row
multiplied by $-\eta-2\epsilon$. Thus the selected $W_A(0,1)$ column
is $(-72,-24)I_0$ for $m=0$. For $m=1$, the two $W$ columns are
\[
 (-72,-24,0,0)I_0,\qquad(-96,-28,4,-24)I_0.
\]
Equation~\eqref{eq:top-coefficients-endpoint-branch} gives the
negative $\Delta$ and $\Delta_i$ rows for the $Z$ columns.
The determinants are
$20(-24)+72\cdot12=384$ and $12288$, in the stated powers of $I_0$;
for the latter, adding twice the third row to the fourth leaves
$(0,0,0,-16)$ and the remaining minor is $-768$.
\end{proof}

\begin{definition}\label{def:g5-selected-separating-sources}
Let $F=[11]$, and retain the ordinary sources and eigenspaces of
Definition~\ref{def:g5-auxiliary-source}.
Let $\mathcal S_e$ be the endpoint-path source with no old flag on
the decorated endpoint, and with either the old flags on different
unit vertices, for both genus orders, or both old flags on the
middle vertex of a $(1,1,2)$ path. Include every marking allocation.
For $\mu=(1^{11})$ or $(2,1^9)$, put
\[
S_{0,\mu}=
M_\mu\!\left(
(\mathcal D_F^+\oplus\mathcal T_{0,F}\oplus\mathcal T_{1,F}^-
 \oplus\mathcal E_F^+\oplus\mathcal V_F^+)^{\langle t\rangle}\right),
\qquad
S_{e,\mu}=M_\mu(\mathcal S_e^{\langle t\rangle}).
\]
Let $R_{0,\mu}$ be the multiplicity source of the two selected
$m=0$ columns and four selected $m=1$ columns of
Lemma~\ref{lem:g5-endpoint-matrices}, over all $A$.
Put $R_{p,\mu}=M_\mu(\mathcal R_F^{\mathrm{Pic}})$.
The sum of the Gysin maps on these four domains is denoted
\[
\Phi_{\mathrm{sep},\mu}:
S_{0,\mu}\oplus S_{e,\mu}\oplus R_{0,\mu}\oplus R_{p,\mu}
\longrightarrow N_\mu.
\]
\end{definition}

\begin{proposition}\label{prop:g5-selected-separating-injection}
For $\mu=(1^{11})$ and $(2,1^9)$ the map
\[
 M_\mu(\delta_F)\Phi_{\mathrm{sep},\mu}:
 S_{0,\mu}\oplus S_{e,\mu}\oplus R_{0,\mu}\oplus R_{p,\mu}
 \longrightarrow M_\mu(\mathscr Q_{5,F})
\]
is injective, and its source maps into $M_\mu(B_{\mathrm{sep},F})$
before applying $\delta_F$. The multiplicities of $\Pmot$ in its
four source summands are
\[
 \begin{array}{c|rrrr}
 \mu&S_{0,\mu}&S_{e,\mu}&R_{0,\mu}&R_{p,\mu}\\\hline
 (1^{11})&114&9&6&15\\
 (2,1^9)&253&9&6&32.
 \end{array}
\]
\end{proposition}

\begin{proof}
\emph{Positive support tests.}
Let $\Gamma$ be one of the resulting genus-three paths with three
elliptic vertices, one decorated by $H^{11}$ and two carrying units,
with $L_0$ the full flag set on the decorated vertex and $L_1,L_2$
those on the unit vertices. Refine each unit vertex to a rational
vertex with a self-loop, leaving
the decorated vertex unchanged; call the resulting graph $\widetilde
\Gamma$. Remove from $\Mbar_{3,J}$ the proper clutching images of
all graphs which are not contractions of $\widetilde\Gamma$.
There are finitely many such graphs. The remaining open
$V_\Gamma$ has precisely the strata whose graphs are contractions
of $\widetilde\Gamma$. The inverse image of this
open in the normalization is
\[
 U_\Gamma=\Mopen_{1,L_0}\times\mathscr U_{L_1}\times\mathscr U_{L_2},
\]
where $\mathscr U_L$ is the open of
Lemma~\ref{lem:ordinary-image-cotangent-restrictions}, the smooth or
irreducible one-nodal genus-one curves with distinct smooth markings.
For a centered decoration this is the situation of
Lemma~\ref{lem:positive-path-detection} with leaf genera $(1,1)$; in
every application below the two unit leaves carry different external
label sets. That lemma makes $i_\Gamma:U_\Gamma\to V_\Gamma$ a closed
regular embedding of codimension two, gives the self-intersection with
$(11,2,2)$ component
$z|_{\Mopen_{1,L_0}}\otimes\psi_{q_1}\otimes\psi_{q_2}$, injective in
the odd coefficient $z$, and shows that $V_\Gamma$ meets no other
two-edge tree stratum and no two-edge tree stratum with a rational
vertex.

For an endpoint decoration the same argument applies with the roles
of the vertices changed. A two-edge tree meeting $V_\Gamma$ must
contract the two loops and retain the two bridges, so it is $\Gamma$.
Each bridge is recovered uniquely from its genus and marking
partition: repeated nested bridges of that type would enclose an
unstable unmarked rational tree with two exterior attachments.
The bridge smoothing parameters therefore give a closed regular
codimension-two embedding $i_\Gamma:U_\Gamma\to V_\Gamma$.
We use the labelled cover if a finite graph quotient is present.
Restriction to $V_\Gamma$, pullback by $i_\Gamma$, and projection
to degree eleven on the decorated vertex and degree two on each
unit give a cohomological test. For an endpoint decoration, the
product of the two normal classes has exactly one term in this
degree, namely
\begin{equation}\label{eq:top-coefficients-positive-self-normal}
 z|_{\Mopen_{1,L_0}}\otimes\psi_1\otimes\psi_2.
\end{equation}
Here $\psi_1$ is at the middle side of the first bridge and
$\psi_2$ at the last side of the second, and the coefficient is $+1$;
for a centered decoration one takes the cotangent on each unit side,
as in Lemma~\ref{lem:positive-path-detection}.
Lemma~\ref{lem:elliptic-eleven-open-restriction} makes
restriction of the odd coefficient to $\Mopen_{1,L_0}$ injective, and
Lemma~\ref{lem:ordinary-image-cotangent-restrictions} makes each
cotangent nonzero on $\mathscr U_{L_i}$, so these maps are injective on
the full elliptic coefficient.
All other two-edge trees, including their further degenerations,
miss this open. This defines cohomological test maps on
the chosen Gysin images.

\emph{External Picard coefficients.}
Fix a decorated endpoint set $A$ meeting $\{c,d\}$ and containing
neither new flag, and put $B=I\setminus A$. For each $C\subseteq B$
use the preceding test on the positive path with allocations
\[
 (A,(B\setminus C)\sqcup\{a\},C\sqcup\{b\}).
\]
By Proposition~\ref{prop:g5-auxiliary-kernel}(a), for a coefficient
$z\otimes(\epsilon\delta_{\rm irr}+\sum_{i\in B}d_i\psi_i)$ in
$\mathcal V_F^+$ the $(11,2,2)$ component of this test is
\[
 -z|_{\Mopen_{1,L_0}}\otimes
       \left(10\epsilon+\sum_{i\in C}d_i\right)x\otimes y,
\]
with $x,y$ the nonzero cotangents on the two partial elliptic
factors; taking $C=\varnothing$ and then $C=\{i\}$ recovers every
coefficient, and the other families of $\mathcal A_F^{\rm aux}$ and
the exchanged terms contribute nothing to these tests.
The tests also annihilate the remaining sources under consideration:
an old-absent path or a tail in $R_{0,\mu}$ cannot isolate the
old-present endpoint. Thus the tests recover the entire
$\mathcal V_F^+$ component in a simultaneous relation.

\emph{Positive paths and endpoint remainders.}
Having removed the external Picard component, take all the tests
\eqref{eq:top-coefficients-positive-self-normal} for the selected
centered and endpoint paths.
Proposition~\ref{prop:g5-auxiliary-kernel}(b) then recovers the
$\mathcal E_F^+$, $\mathcal T_{0,F}$ and $\mathcal T_{1,F}^-$
coefficients in turn.
For an old-split allocation with no old flag on the decoration, the
original has its new pair together and its old pair split; its
exchange has the reverse pattern, so raising the
vertex carrying $a,b$ recovers the original source and all its
marking allocations. For the additional endpoint sources with
both old flags on the middle of a $(1,1,2)$ path, the exchanged
target would instead arise from a $(1,2,1)$ path with both old
flags on its last vertex; that source is not selected. Centered
and endpoint decorations are distinguished by their degree-eleven
vertex; when the undecorated graphs agree, the flag threshold
excludes a second nonzero decoration. These cases exhaust
$\mathcal T_{0,F}$, $\mathcal T_{1,F}^-$, $\mathcal E_F^+$, and
$\mathcal S_e$.

The positive tests annihilate all elliptic $H^{13}$ tails by degree.
They also annihilate $R_{0,\mu}$. The $Z$ outputs have rational
endpoints by \eqref{eq:top-coefficients-endpoint-branch}, and the
entire positive endpoint projection of each $W$ is zero by the
six coefficients in \eqref{eq:top-coefficients-positive-table}.
A centered positive target cannot receive an old-absent endpoint
tail, since the other possible degree-eleven vertex has too few
flags. This checks the added corrections on all positive targets.

Now consider a relation in the $M_\mu(q_F)$ images of
$S_{0,\mu}\oplus S_{e,\mu}\oplus R_{0,\mu}$.
The external Picard tests first force its $\mathcal V_F^+$
coefficient to vanish, and the positive tests force its selected path
coefficients to vanish. The remaining relation involves only elliptic
$H^{13}$ tails and $R_{0,\mu}$. The tests in Lemma~\ref{lem:g5-endpoint-matrices}
annihilate the former and recover every coefficient of the latter,
since both displayed determinants are nonzero. Finally, the map in
Proposition~\ref{prop:g5-auxiliary-kernel} is injective on the
selected $\mathcal D_F^+$ summand, so its remaining coefficient
vanishes. We have proved
\begin{equation}\label{eq:top-coefficients-three-source-injection}
 M_\mu(q_F)(\Phi_{\mathrm{sep},\mu}|_{S_{0,\mu}\oplus
 S_{e,\mu}\oplus R_{0,\mu}})\quad\text{is injective}.
\end{equation}

\emph{The full Picard source.}
Each of the first three sources maps into $M_\mu(\mathscr U_F)$:
the $W$ columns are sums of external Picard tails and endpoint
paths. Pass a relation among all four sources in
$M_\mu(\mathscr Q_{5,F})$ to
\[
 M_\mu\bigl(\mathscr Q_{5,F}/\delta_F(\mathscr U_F)\bigr).
\]
Proposition~\ref{prop:g5-picard-injection} forces the $R_{p,\mu}$
coefficient to vanish. Project the remaining relation to the
double-loop coordinate using Lemma~\ref{lem:g5-double-loop-coordinate}.
Equation~\eqref{eq:top-coefficients-three-source-injection} makes
the other three coefficients zero. This proves the asserted
injection into the full compact quotient.

The Gysin image of every selected source also belongs to $B_{\mathrm{sep},F}$.
An $H^{13}$ tail is the image of a defining one-edge map; a Picard tail has
the defining degree-thirteen coefficient $H^{11}\otimes H^2$.
For a two-edge path, glue one bridge first. Its coefficient then
has degree thirteen, and the remaining bridge is separating.
Transitivity of proper Gysin, with the quotient-stack normalization,
identifies this composition with the path class. The same holds
for each linear combination $W$. This proves the image inclusion
with the full compact coefficients.

\emph{The source multiplicities.}
The multiplicities are those of Lemma~\ref{lem:genus-four-source-multiplicities}.
\end{proof}

\begin{proposition}\label{prop:g5-coefficient-eleven-ten}
Let $F=[11]$ and $\mu=(1^{11})$ or $(2,1^9)$. Then the map
$\overline\delta_\mu:N_\mu/J_\mu\to\mathscr Q_\mu$ of
Proposition~\ref{prop:residual-interface} is injective.
Consequently $\mathcal C_{(11)}=\mathcal C_{(10,1)}=0$.
\end{proposition}

\begin{proof}
Let $R_{\mathrm{par},\mu}=M_\mu(\mathcal R^{\mathrm{par}}_{+,F})\cong\Pmot^6,\Pmot^{12}$
(Lemma~\ref{lem:genus-four-source-multiplicities}) and $p_4$ as in
Definition~\ref{def:g5-parallel-source}.

Suppose that a separating source vector $w$ and a parallel vector
$r$ satisfy
\[
 \delta_\mu\bigl(\Phi_{\mathrm{sep},\mu}(w)+M_\mu(p_4)(r)\bigr)=0.
\]
Pass to
$M_\mu(\mathscr Q_{5,F}/\delta_F(B_{\mathrm{sep},F}))$.
The first term vanishes by the image inclusion in
Proposition~\ref{prop:g5-selected-separating-injection}.
The normalized left inverse of
Proposition~\ref{prop:g5-parallel-left-inverse} gives $r=0$.
The separating injection then gives $w=0$. Thus the combined
source injects into the full compact quotient.

The combined multiplicities are
\[
 114+9+6+15+6=150,\qquad
 253+9+6+32+12=312.
\]
Since $\delta_\mu$ vanishes on $J_\mu$, the induced map to
$N_\mu/J_\mu$ is injective.
Proposition~\ref{prop:residual-interface} shows that
these are precisely the dimensions of the two reduced domains.
The combined Gysin images therefore span these quotients. The reduced maps
$\overline\delta_\mu$ are injective, and their
canonical kernels $\mathcal C_{(11)}$ and $\mathcal C_{(10,1)}$
are zero.
\end{proof}