\section{The eight-row coefficients}\label{sec:13-eight-row-coefficients}

Proposition~\ref{prop:g5-coefficient-nine-two} already gives
$\mathcal C_{(8,3)}=0$. We prove the vanishing of
$\mathcal C_{(8,2,1)}$ and $\mathcal C_{(8,1,1,1)}$ by support tests
on the double-loop coordinate. The restriction to the separating summand is already
injective, so these tests also prove injectivity of the full compact boundary map.

\begin{lemma}\label{lem:eight-row-coefficients-compact-row-injection}
Let $F$ be finite and let $\rho\vdash|F|$. If $M_\rho(q_F)$ is
injective, then $M_\rho(\beta_{5,F})$ is injective.
\end{lemma}

\begin{proof}
Write $C^1_{5,F}=C^1_{\rm sep}\oplus N_{5,F}$. Let
$\pi:C^2_{5,F}\to\mathscr Q_{5,F}$ be the quotient map and put
$p=\operatorname{pr}_{\mathrm{dl}}\pi$. By
Lemma~\ref{lem:g5-double-loop-coordinate},
\[
 p\beta_{5,F}(z,n)=q_F(n),
\]
since $\pi\beta_{5,F}(z,0)=0$ and
$\pi\beta_{5,F}(0,n)=\delta_F(n)$.
If $(z,n)\in\ker M_\rho(\beta_{5,F})$, injectivity of
$M_\rho(q_F)$ gives $n=0$. Theorem~\ref{thm:compact-injectivity}
gives injectivity on the separating summand. Hence $z=0$ as well.
\end{proof}

\begin{lemma}\label{lem:eight-row-coefficients-nine-row-counts}
Let $L$ be a finite marking set.
\begin{enumerate}
\item[(0)] Let $\mathcal D_L=\mathsf D_{4,L}\otimes\Pmot$ and
$\mathcal T_L=\mathsf T_{4,L}\otimes\Pmot$
(Definition~\ref{def:g5-auxiliary-picard-source} with $L$ for $I$).
Let $\pi_9$ project to the partitions of $|L|$
with exactly nine rows. Then the ordinary clutching
maps induce an isomorphism
\[
 \pi_9(\mathcal D_L\oplus\mathcal T_L)
       \xrightarrow{\ \sim\ }\pi_9H^{15}(\Mbar_{4,L}).
\]
\item[(i)] The constituents of $H^{15}(\Mbar_{4,L})$ have at least
nine rows. Those of $H^{15}(\Mbar_{5,L})$ have at least
nine rows for $|L|\leq10$, and at least eight rows for all $L$.
\item[(ii)] For an integer $m\geq9$ and a partition $\theta\vdash m$
with exactly nine rows, put
$\gamma=(\theta_1-1,\ldots,\theta_9-1)$, omitting zero parts,
and $f_\theta(z)=s_\gamma(1,1,z)$. Then
\[
 M_\theta(H^{15}(\Mbar_{4,[m]}))
 \cong\Pmot^{\oplus\left(f_\theta(1)+
                     \sum_{j\geq3}[z^j]f_\theta(z)\right)}.
\]
In particular this group is zero if $\theta_4\geq2$.
\item[(iii)] One has
\[
 M_{(1^9)}(H^{15}(\Mbar_{5,9}))\cong\Pmot,
 \qquad
 M_{(2,1^8)}(H^{15}(\Mbar_{5,10}))\cong\Pmot^4.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
(0) Theorem~\ref{thm:ordinary-image} identifies the ordinary image
with the four summands of Definition~\ref{def:ordinary-summands}.
Theorem~\ref{thm:genus-four-base} identifies that image with compact
cohomology. The characters of $\mathsf W_{4,L}$ and $\mathsf R_{4,L}$
are products of $H$ with sums of complete symmetric functions. Every
term of $H$ has at least ten rows (Notation~\ref{def:common-series}),
and induction cannot decrease the number of rows
(Lemma~\ref{lem:row-bound-calculus}(1)). Thus these two summands have
zero $\pi_9$-projection. The other two summands, with their
realizations and twists restored, are precisely $\mathcal D_L$ and
$\mathcal T_L$. The restriction of the injective map $\Omega_{4,L}$ is
their displayed sum of clutching maps, proving the assertion.

(i) The genus-four bound and the genus-five bound through ten
markings are Lemma~\ref{lem:marking-supports-row-bounds}(iii)
and~(ii). For larger marking sets, the exact sequence in
Proposition~\ref{prop:genus-five-support} adds induced modules
whose seven initial shapes have at least eight rows, and induction
preserves this bound (Lemma~\ref{lem:row-bound-calculus}(1)). This
proves the remaining bound without using any coefficient vanishing.

(ii) The nine-row coefficients of the genus-four character are those of
\[
 \sum_{k\geq1}s_{(k,1^8)}(EE_{\geq3}+E^2).
\]
Here the first term comes from $E\partial U$: only removing
the bottom corner of its ten-row hook contributes. In the path
term only the nine-row part of $G$ contributes. For a fixed
$\theta$, removing the first column of $\theta/(k,1^8)$
identifies it with $\gamma/(k-1)$. In a semistandard tableau
of $\gamma$ on three ordered letters, the first letter occupies
an initial segment of the first row. Summing over that segment
gives
\[
 \sum_{k\geq1}s_{\theta/(k,1^8)}(1,z)=s_\gamma(1,1,z).
\]
The two horizontal strips contributed by $E^2$ count
$f_\theta(1)$; requiring the second strip to have at least
three boxes counts $\sum_{j\geq3}[z^j]f_\theta(z)$.
If $\theta_4\geq2$, then $\gamma$ has a column of height four,
so no such three-letter tableau exists.

(iii) Through ten markings the nine-row genus-five coefficients are
those of
\[
 \sum_{k\geq1}s_{(k,1^8)}(E^2+e_2[E]).
\]
The nine-row part of the $\mathsf D$ term starts in degree twelve,
and the remaining terms have at least ten rows
(Notation~\ref{def:common-series}). At $(1^9)$ the coefficient is
$1$. At $(2,1^8)$ the $E^2$ contribution is $2+1=3$, and
the $e_2[E]$ contribution is $1$. This proves the two counts.
\end{proof}

\begin{lemma}\label{lem:eight-row-coefficients-double-loop-injection}
Let $F=[11]$ and let $\rho=(3,2,1^6)$ or $(4,1^7)$. With the
labelled flags and maps of Set-up~\ref{setup:g5-detector-row},
the map
\[
 M_\rho(q_F):M_\rho(N_{5,F})\longrightarrow M_\rho(\mathscr D_F)
\]
is injective.
\end{lemma}

\begin{proof}
Work on the $\rho$-isotypic subspace of $N_{5,F}$.
Let $x$ belong to this isotypic subspace of $N_{5,F}$ and
suppose $q_F(x)=0$.

\emph{The old flags on the decorated factor.}
By Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(i),
compact genus-four cohomology has at least nine rows. A
constituent contributing to $M_\rho(N_{5,F})$ differs from
$\rho$ by a horizontal two-strip, since the old flags $c,d$
have ordinary invariants, and such a strip adds at most one
first-column box (Lemma~\ref{lem:row-bound-calculus}(1)); so the
constituent has exactly nine rows.
Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(0) therefore
gives $x$ a unique preimage
\[
 (z_D,z_T)\in \pi_9(\mathcal D_I\oplus\mathcal T_I),
 \qquad (c\,d)^*(z_D,z_T)=(z_D,z_T).
\]
In each nonzero coordinate of $z_D$, at least one of $c,d$
belongs to the genus-one factor. Indeed its full $H^{13}$
module has row bound ten, and nine after fixing its node
(Lemma~\ref{lem:row-bound-calculus}(2),(1)). If neither old flag
were on this factor,
all these rows would be on $F$ labels, excluding $\rho$ after
induction. Similarly, the eleven-row hook on a path center
has at least nine rows after its two nodes are fixed. Thus
every nonzero coordinate of $z_T$ has an old flag on its
center. The forbidden allocations form submodules under
$\Sn_F\times\langle(c\,d)\rangle$ in the full direct sum of
sources. Their $\rho$-projections vanish coordinate by
coordinate, including for the preimage under $\pi_9$.

\emph{The transverse pullback.}
As in the proof of Proposition~\ref{prop:g5-auxiliary-kernel},
$\nu_{ab}^*$ of an ordinary tree class is the sum over positive-genus
vertices of the branch lowering that vertex and placing $a,b$ on it,
with coefficient one.

For $\mathcal T_I$, pulling back on the elliptic center is
zero by genus-zero odd vanishing. Pullback on the genus-one
leaf gives a path with leaf genera $(0,2)$; pullback on the
genus-two leaf gives a path with leaf genera $(1,1)$ and
$a,b$ on the lowered leaf. For $\mathcal D_I$, pullback on
the elliptic $H^{13}$ factor is zero for the same reason,
and pullback on its partner gives an elliptic $H^{13}$ tail
with a genus-two partner carrying $a,b$. These are all the
terms. The original leaf genera distinguish the path leaves;
in each resulting positive path, the external labels $a,b$
distinguish the two elliptic leaves. Thus all these graph
groups are trivial. The output is invariant under $(a\,b)$,
with no extra factor from this labelled pullback.

\emph{Eliminating the positive paths.}
Write $r(x)=\nu_{ab}^*x$. By
Lemma~\ref{lem:g5-double-loop-coordinate}, the double-loop
equation is the cohomological equality
\begin{equation}\label{eq:eight-row-coefficients-double-loop}
 q_F(x)=r(x)-s^*r(x)=0.
\end{equation}
Choose a coordinate of $z_T$ and lower its genus-two leaf.
Its positive target path $\Gamma$ has neither $a$ nor $b$
on the center, and its coefficient is the unchanged center
class. Raising the leaf carrying $a,b$ uniquely recovers
the original source and its allocation. Every other positive
path in $r(x)$ has different recovered data. Every positive
path in $s^*r(x)$ has at least one of $a,b$ on its center,
by the old-flag condition, and so is also different.

Theorem~\ref{thm:cited-thirteen-cohomology} in genus one
expresses each $H^{13}$ coefficient as a sum of rational-tail
Gysin classes. Thus every $\mathcal D_I$ term, after its outer
attachment, is supported on a two-edge tree with a rational
vertex. Apply Lemma~\ref{lem:positive-path-detection} to $\Gamma$,
with leaf genera $(1,1)$ and the leaves distinguished by the
external labels $a,b$, and let $O_\Gamma$ be its open. Such terms,
and the rational paths in the pullback, restrict to zero on
$O_\Gamma$ by that lemma.
Restriction of \eqref{eq:eight-row-coefficients-double-loop}
to $O_\Gamma$ and its $(11,2,2)$ self-intersection consequently
forces the selected coefficient to vanish. Applying this
argument to every labelled path gives $z_T=0$.

\emph{Eliminating the elliptic tails.}
Write $z_A$ for the remaining $H^{13}$ coefficient with
elliptic external label set $A\subseteq I$. If some $z_A$
is nonzero, choose one with $|A|$ minimal. It contains an
old flag and neither $a$ nor $b$. In $r(x)$ it gives a tail
with genus-two partner on $J\setminus A$. Apply
Lemma~\ref{lem:minimal-elliptic-tail-detection} with partner genus
$h=2$ and $J=A\sqcup(J\setminus A)$, with its open
$W_A\subseteq\Mbar_{3,J}$: a tail of
type $A'$ can meet $W_A$ only for $A'\subseteq A$.
The proper subsets have zero coefficient by minimality.
Every tail in $s^*r(x)$ contains $a$ or $b$ on its elliptic
side, so none meets $W_A$. The $(13,2)$ self-intersection
of the restriction of \eqref{eq:eight-row-coefficients-double-loop}
is therefore
\[
 -z_A|_{\mathrm{ct}}\otimes\psi_q=0.
\]
The same lemma makes this map injective, a contradiction.
Hence $z_D=0$ as well. Hence $x=0$.
\end{proof}

\begin{lemma}\label{lem:eight-row-coefficients-small-genus-three}
The following multiplicities are direct sums of the indicated
numbers of copies of $\Pmot$:
\[
 \begin{array}{c|c|r}
 |L|&\theta&M_\theta(H^{15}(\Mbar_{3,L}))\\\hline
 10&(2,1^8)&1\\
 11&(3,1^8)&2\\
 11&(2,2,1^7)&2\\
 12&(3,2,1^7)&4
 \end{array}
\]
\end{lemma}

\begin{proof}
\emph{The nine-row ordinary source.}
Proposition~\ref{prop:genus-three-generation} supplies ordinary
generation. The complete ordinary source has row bound nine by
Lemma~\ref{lem:row-bound-calculus}(3), and so has its compact image
$H^{15}(\Mbar_{3,L})$ by Lemma~\ref{lem:marking-supports-row-bounds}(iii).
We identify the nine-row coordinates of the source.
On a separating source the degree-thirteen factor
has at least nine rows after fixing its node; the mixed
$H^{11}\otimes H^2$ sectors have at least ten. On the
nonseparating normalization use the independent elliptic-tail
coordinates of Theorem~\ref{thm:cited-thirteen-cohomology}
in genus two. If neither loop flag lies on the inner elliptic
factor, at least ten rows remain. If one does, fixing it
and the inner node leaves at least nine. If both do, write
the hook as the tenth exterior power of the standard
representation and restrict its three distinguished flags: on $a$ external flags
the restricted standard representation is
$\operatorname{Std}_{[a]}\oplus\operatorname{Std}_{[3]}\oplus\mathbb Q$,
and external exterior degree eight is
paired with $\operatorname{sgn}_3\oplus\operatorname{Std}_{[3]}$.
Taking invariants under a transposition eliminates the sign summand and gives a
one-dimensional fixed space in $\operatorname{Std}_{[3]}$; this is the
nine-row two-loop-flag coordinate.

For the nine-row part, the hook on a center with $a$ external
flags contributes $V_{(a-8,1^8)}$, and exchange of its two
attaching flags acts by minus one. The surviving separating
coordinates are elliptic $H^{13}$ tails with genus-two
partners, and the independent inner elliptic-tail coordinates
in higher-genus $H^{13}$ factors. In the latter coordinates
the outer node must lie on the inner elliptic factor.
They therefore recover a centered path together with its
chosen inner leaf. The nonseparating coordinates divide
into the one-loop-flag and two-loop-flag cases just described.

\emph{Eliminating the redundant factorizations.}
Use the face maps of Definition~\ref{def:low-genus-ordinary-row};
their composites with ordinary Gysin are zero by the
normal-crossings row. A parallel-edge graph has selected-edge
stabilizer of order one and target loop group of order two.
Its two signed faces add to twice the transfer of one labelled
clutching. Projection to one of the two inner coordinates
recovers twice the original hook. Thus these faces eliminate the
entire one-loop-flag part.

Next take a bridge with a loop on its decorated elliptic
vertex. Contracting the bridge is the identity on the
two-loop-flag invariant coordinate: its source and target
loop groups both have order two. The other face is minus
one half of the labelled loop Gysin into a separating
coordinate, since its target group is trivial. This is
the graph of a specified map out of the independent
two-loop-flag coordinate, and eliminates that coordinate
without imposing a relation on the remaining separating
part.

A centered path with leaf genera zero and two has trivial
stabilizers. Its faces identify its independent higher-genus
coordinate with its rational-tail Gysin image in the
elliptic $H^{13}$ source. For two elliptic leaves distinguished
by external labels, the faces similarly identify the two
independent inner-tail factorizations. Each resulting
coordinate recovers the center labels, the inner leaf
labels, and the outer leaf labels, so no different path
uses that coordinate. If both elliptic leaves are unmarked,
their exchange acts by minus one on the local nine-row
hook. The two oriented faces add with coefficient two
and kill the sole factorization; the ordinary invariant
path source is zero in exactly this case.

These eliminations leave the elliptic $H^{13}$ tail
source and the ordinary paths with elliptic center and
two elliptic leaves. On the four partitions in the
statement the elliptic $H^{13}$ source is zero. Its nine-row
coefficients are those of
\[
 \sum_{k\geq1}s_{(k,1^8)}EE_{\geq3},
\]
which starts in degree twelve with $s_{(1^9)}h_3$ and has
zero coefficient at $(3,2,1^7)$. The positive-path source
is injective by Lemma~\ref{lem:positive-path-detection}, applied
with leaf genera $(1,1)$: its nonzero nine-row coordinates have
externally distinguished leaves, and the opens $O_\Gamma$ of that
lemma isolate the coefficients. Thus this quotient maps isomorphically
onto the compact components in question. All further
face relations, including those from double-loop graphs,
map to zero in this quotient and impose no further relation.

\emph{Counting the ordinary paths.}
Because exchange of the node flags acts by minus one,
the ordinary unordered leaves contribute $e_2[E]$.
Their nine-row coefficients are therefore those of
\[
 \sum_{k\geq1}s_{(k,1^8)}e_2[E].
\]
In complementary degrees zero through three, $e_2[E]$
has components $0,h_1,h_1^2,h_3+h_1h_2$.
The contributions in increasing order of the hook length
$k$ are respectively $1$, $1+1$, $1+1$, and $1+2+1$
for the four partitions displayed. These give $1,2,2,4$.
\end{proof}

\begin{proposition}\label{prop:g5-coefficient-eight-two-one}
Let $F=[11]$ and $\mu=(3,2,1^6)$.
The map $M_\mu(q_F)$ is injective with source $\Pmot^{26}$.
The map $M_\mu(\beta_{5,F})$ is injective with source $\Pmot^{59}$.
Consequently
\[
 M_\mu(H^{15}(\Mbar_{5,F}))=0.
\]
\end{proposition}

\begin{proof}
\emph{The nonseparating map and its source.}
Lemma~\ref{lem:eight-row-coefficients-double-loop-injection}
proves injectivity on this $\mu$. The horizontal two-strip rule for
the ordinary invariants of the two old flags
(Lemma~\ref{lem:row-bound-calculus}(1)) and the nine-row bound of
Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(i) leave precisely
\[
 \theta_1=(4,2,1^7),\qquad
 \theta_2=(3,3,1^7),\qquad
 \theta_3=(3,2,2,1^6)
\]
as partitions of thirteen contributing to $N_{5,F}$, each
once. The polynomials of
Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(ii) are
\[
 \begin{array}{c|c|r}
 \theta&f_\theta(z)&f_\theta(1)+\sum_{j\geq3}[z^j]f_\theta(z)\\\hline
 \theta_1&3+6z+4z^2+2z^3&17\\
 \theta_2&1+2z+3z^2&6\\
 \theta_3&2z+z^2&3
 \end{array}
\]
Thus $M_\mu(N_{5,F})\cong\Pmot^{26}$.

\emph{The entire separating source.}
By Lemma~\ref{lem:marking-supports-row-bounds}(iv), the mixed
K\"unneth sectors $(13,2)$, $(2,13)$, $(11,4)$, $(4,11)$ of
$C^1_{\rm sep}$ have row bound nine and contribute nothing to
$M_\mu$, and a genus-zero degree-fifteen factor
vanishes by Theorem~\ref{thm:cited-genus-zero-cycles}. Thus
only degree-fifteen factors with unit partners contribute.
A genus-one factor has at most twelve flags and vanishes
by Theorems~\ref{thm:genus-one} and~\ref{thm:motivic-types}.
For genera two, three, and four, the compact factor has
at least nine rows by Lemma~\ref{lem:marking-supports-row-bounds}(iii).
The same bound holds for a genus-five factor here, since
its rational partner leaves at most ten flags
(Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(i)).

Let $b$ be the number of external labels on the unit
partner. Fixing the node on the decorated factor and
then inducing with $h_b$ requires exactly the following
one-corner and horizontal-strip possibilities
(Lemma~\ref{lem:row-bound-calculus}(1)). The last
two columns give the compact multiplicities in genera
three and four:
\[
 \begin{array}{c|c|c|rr}
 b&\text{external shape}&\text{full decorated shape}&g=3&g=4\\\hline
 0&(3,2,1^6)&(3,2,1^7)&4&8\\
 1&(3,1^7)&(3,1^8)&2&6\\
 1&(2,2,1^6)&(2,2,1^7)&2&3\\
 2&(2,1^7)&(2,1^8)&1&3
 \end{array}
\]
Every branching and Pieri coefficient is one. Indeed an
eight-row external shape must be obtained by removing the
bottom singleton of a nine-row decorated shape; interlacing
with $\mu$ gives precisely the four rows of the table.
The genus-three entries were proved in the preceding
lemma. For genus four the respective polynomials
$f_\theta(z)$ are $2+4z+2z^2$, $3+2z+z^2$, $1+2z$,
and $2+z$, giving the displayed entries.

All four genus-two entries are zero. In degrees at most
eleven Theorem~\ref{thm:genus-two} has no nine-row
constituent; in degree twelve its only possible nine-row
term comes from $EGE_{\geq3}$ and is $s_{(1^9)}h_3$.
It has zero coefficient at $(3,2,1^7)$. Thus positive
partners contribute $\Pmot^9$ with decorated genus three
and $\Pmot^{20}$ with decorated genus four. For a rational
partner stability requires $b\geq2$, leaving only the
last row of the table. Its full decorated genus-five
shape is $(2,1^8)$ on ten flags, contributing $\Pmot^4$
by Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(iii).
Consequently the complete source, including all label
allocations, is
\[
 M_\mu(C^1_{5,F})
 \cong\Pmot^{26}\oplus\Pmot^{20}\oplus\Pmot^9\oplus\Pmot^4
 \cong\Pmot^{59}.
\]

\emph{The full compact map.}
Injectivity of $M_\mu(q_F)$ and
Lemma~\ref{lem:eight-row-coefficients-compact-row-injection} give
injectivity of $M_\mu(\beta_{5,F})$ on the source just counted.
Since $15<19$, Proposition~\ref{prop:compact-restriction-row}
makes $\alpha$ injective. The identity $\beta\alpha=0$ therefore
gives $M_\mu(H^{15}(\Mbar_{5,F}))=0$.
\end{proof}

\begin{proposition}\label{prop:g5-coefficient-eight-one-one-one}
Let $F=[11]$ and $\mu=(4,1^7)$.
The map $M_\mu(q_F)$ is injective with source $\Pmot^{35}$,
and $M_\mu(\beta_{5,F})$ is injective. Consequently
\[
 M_\mu(H^{15}(\Mbar_{5,F}))=0.
\]
\end{proposition}

\begin{proof}
Lemma~\ref{lem:eight-row-coefficients-double-loop-injection}
gives injectivity of $M_\mu(q_F)$. A horizontal two-strip above
$\mu$ (Lemma~\ref{lem:row-bound-calculus}(1)), with exactly one new
row by the nine-row bound of
Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(i), leaves the
two genus-four shapes
\[
 (5,1^8),\qquad (4,2,1^7),
\]
each with coefficient one. Their polynomials in
Lemma~\ref{lem:eight-row-coefficients-nine-row-counts}(ii) are
$5+4z+3z^2+2z^3+z^4$ and $3+6z+4z^2+2z^3$.
The corresponding multiplicities are $18$ and $17$, so
$M_\mu(N_{5,F})\cong\Pmot^{35}$.

Lemma~\ref{lem:eight-row-coefficients-compact-row-injection} now
gives injectivity of $M_\mu(\beta_{5,F})$ and, as in the proof of
Proposition~\ref{prop:g5-coefficient-eight-two-one},
$M_\mu(H^{15}(\Mbar_{5,F}))=0$.
\end{proof}

\begin{theorem}[Vanishing of the genus-five coefficients]
\label{thm:genus-five-coefficient-vanishings}
For every $\lambda\in\mathscr R_{11}$ one has $\mathcal C_\lambda=0$.
\end{theorem}

\begin{proof}
For $\lambda=(11)$ and $\lambda=(10,1)$ this is
Proposition~\ref{prop:g5-coefficient-eleven-ten}.
Proposition~\ref{prop:g5-coefficient-nine-two} proves
$\mathcal C_{(9,2)}=\mathcal C_{(8,3)}=0$, and
Proposition~\ref{prop:g5-coefficient-nine-one-one} proves $\mathcal C_{(9,1,1)}=0$.
For the two remaining coefficients, the
ordinary genus-five image has at least nine rows by
Theorem~\ref{thm:ordinary-image}. At eleven markings,
Proposition~\ref{prop:genus-five-support} therefore identifies
the compact multiplicities at $(3,2,1^6)$ and $(4,1^7)$ with
$\mathcal C_{(8,2,1)}$ and $\mathcal C_{(8,1,1,1)}$, respectively.
Propositions~\ref{prop:g5-coefficient-eight-two-one} and
\ref{prop:g5-coefficient-eight-one-one-one} make both multiplicities
zero.
\end{proof}

\begin{corollary}[The genus-five base]\label{thm:genus-five-base}
For every $n\geq0$, the complete ordinary one-edge Gysin map
$\Phi^{15}_{5,n,\mathrm B}$ is surjective, and $\mathcal O_{5,n}=0$.
\end{corollary}

\begin{proof}
Proposition~\ref{prop:genus-five-initial} proves the assertion
for $0\leq n\leq10$. For $n\geq11$, the support
isomorphism in Proposition~\ref{prop:genus-five-support}
expresses $\mathcal O_{5,n}$ as the induction of the seven
eleven-marking coefficients.
Theorem~\ref{thm:genus-five-coefficient-vanishings}
makes all seven coefficients zero. Thus $\mathcal O_{5,n}=0$.
The exact sequence of the same support proposition then
identifies the image of the complete ordinary one-edge
map with all of $H^{15}(\Mbar_{5,n})$.
\end{proof}