\section{The geometric base cases}\label{sec:08-geometric-bases}
Projectivization parametrizes one-dimensional subspaces.

We prove that the images of the ordinary one-edge Gysin maps span degree-fifteen
cohomology in genus four for every number of markings, and in genus
five through ten markings. The additional cases beyond the
Chow--K\"unneth bounds are controlled by the loci where marked points
fail to impose independent conditions on the equations of a cover.

Let \(\mathcal H_{k,g,n}\) parametrize smooth connected degree-\(k\)
covers of a projective line with \(n\) distinct ordered source markings,
allowing arbitrary ramification, and let \(\beta_n\) forget the cover.
Let \(\mathcal M_{g,n}^{k}\) denote the locus of gonality at most \(k\).
By \cite[Section 8.2, equation (8.2)]{canninglarson2022}, the
forgetful morphism
$\beta_n':\mathcal H_{k,g,n}\setminus\beta_n^{-1}(\mathcal M_{g,n}^{k-1})
\longrightarrow\mathcal M_{g,n}\setminus\mathcal M_{g,n}^{k-1}$
is proper.
We recall the two evaluation constructions in
\cite[Sections 9.2 and 10.3]{canninglarson2022}.

\begin{definition}[Trigonal and tetragonal evaluation data]
\label{def:external-trigonal-evaluation}
\label{def:external-tetragonal-evaluation}
Let \(k\in\{3,4\}\).
For \(k=3\), let \(\pi:\mathcal P\to\mathcal B\) carry the universal globally
generated rank-two degree-\(g+2\) bundle \(\mathcal E\) on the
universal projective line, and set
\[
 \mathscr S=\mathbb P\mathcal E^\vee,\quad \gamma:\mathscr S\to\mathcal P,\quad
 \mathcal L=\mathcal O_{\mathscr S}(3)\otimes\gamma^*\det\mathcal E^\vee,
 \quad \mathcal U=\gamma_*\mathcal L
   =\operatorname{Sym}^3\mathcal E\otimes\det\mathcal E^\vee.
\]
For \(k=4\), let \(\mathcal B\) parametrize globally generated bundles
\(\mathcal E,\mathcal F\) on a projective-line bundle
\(\pi:\mathcal P\to\mathcal B\), of ranks \(3,2\), both of
degree \(g+3\), together with an isomorphism
\(\det\mathcal E\simeq\det\mathcal F\), and set
\[
 \mathscr S=\mathbb P\mathcal E^\vee,\quad \gamma:\mathscr S\to\mathcal P,\quad
 \mathcal L=\mathcal O_{\mathscr S}(2)\otimes\gamma^*\mathcal F^\vee,
 \quad\mathcal U=\gamma_*\mathcal L
       =\operatorname{Sym}^2\mathcal E\otimes\mathcal F^\vee.
\]
For \(k=3\), restrict to \(\mathcal B'\) where \(\mathcal U\) is relatively
globally generated; for \(k=4\), let
\(\mathcal B'=\mathcal B\setminus\operatorname{Supp}R^1\pi_*\mathcal U\).
Put \(\mathcal X=\pi_*\mathcal U|_{\mathcal B'}\).
For \(T_n=\mathscr S\times_{\mathcal B'}\cdots\times_{\mathcal B'}\mathscr S\),
let \(\eta_i:T_n\to\mathscr S\) and \(\epsilon:T_n\to\mathcal B'\)
be the projections, and set
\[
 e_n:\epsilon^*\mathcal X\longrightarrow
       \bigoplus_{i=1}^n\eta_i^*\mathcal L,
 \qquad T_n^{\mathrm{ev}}=\{e_n\text{ is surjective}\},
 \qquad \mathcal Y_n=e_n^{-1}(0).
\]
For \(k=4\), let \(\mathcal H'_{4,g,n}\) be the open substack over \(\mathcal B'\)
parametrizing smooth covers with distinct markings. For an integer \(f\), let
\(\mathcal H^f_{4,g,n}\subset\mathcal H'_{4,g,n}\) be the locus
where \(\mathcal F\simeq\mathcal O(f_1)\oplus\mathcal O(f_2)\),
\(f_1\le f_2\), has \(f_1\ge f\).
\end{definition}

\begin{proposition}[Canning--Larson {\cite[Section 9.2, Lemmas 9.1--9.2; Section 10.3, (10.9) and Lemma 10.11]{canninglarson2022}}]
\label{prop:external-trigonal-evaluation-open}
With the preceding data, for either \(k\), \(\mathcal X\) is a vector bundle, and
\(\mathcal Y_n|_{T_n^{\mathrm{ev}}}=\ker(e_n|_{T_n^{\mathrm{ev}}})\)
is a vector bundle over \(T_n^{\mathrm{ev}}\).
For \(k=3\), \(\mathcal H_{3,g,n}\) is the open of \(\mathcal Y_n\) on which the
zero curve is smooth and the markings are distinct, and
the base \(\mathcal B'\) has the Chow--K\"unneth generation property.
For \(k=4\), \(\mathcal H'_{4,g,n}\subset\mathcal Y_n\) is open, and
the base \(\mathcal B\) has the Chow--K\"unneth generation property.
\end{proposition}

\begin{lemma}\label{lem:geometric-bases-evaluation-opens}
Let $g\geq2$ and $n\geq0$. Let $\mathscr H_3^{\mathrm{ev}}$ be the
open substack of $\mathcal H_{3,g,n}$ lying over $T_n^{\mathrm{ev}}$ in
Definition~\ref{def:external-trigonal-evaluation} with $k=3$, and let
$\mathscr H_4^{\mathrm{ev}}$ be the open substack of $\mathcal H'_{4,g,n}$
lying over $T_n^{\mathrm{ev}}$ there with $k=4$.
Then both stacks have the Chow--K\"unneth generation property.
\end{lemma}

\begin{proof}
Proposition~\ref{prop:external-trigonal-evaluation-open} gives the property
for the respective bases $\mathcal B'$, for $k=4$ by open
restriction from $\mathcal B$. Each factor of $T_n$ is a projective
bundle over the universal projective line. The factors are taken
over $\mathcal B'$, so their points on the projective line vary
independently. Successive projective bundles and open restriction
give the property for $T_n^{\mathrm{ev}}$ by
Proposition~\ref{prop:cited-chow-kunneth-operations}. The kernel of
$e_n$ is a vector bundle there, and its smooth-curve,
distinct-marking locus is open. Another application of that
proposition proves the assertion.

One may perform these constructions on the finite gerbe of lifts
of the target structure group from $\mathrm{PGL}_2$ to
$\mathrm{SL}_2$, where the universal line is the projectivization
of a rank-two bundle. Such a gerbe has the same rational Chow
groups, also after product with any test stack, and the same
rational cohomology. The assertions therefore descend to the
Hurwitz stacks used here.
\end{proof}

For $g=4,5$, Theorem~\ref{thm:motivic-types} and
Proposition~\ref{prop:primitive-degree-fifteen-quotient} show that
$\mathcal O_{g,n}$ is pure and has only types $(13,2)$ and $(2,13)$.
When applying the cycle-class assertion of
Proposition~\ref{prop:cited-chow-kunneth-operations}, the smooth
separated stacks below admit smooth proper compactifications:
their coarse spaces are quasi-projective, and in characteristic
zero one can compactify and resolve while preserving the given
smooth open.

\begin{theorem}[The genus-four base]\label{thm:genus-four-base}
For every $n\geq0$, the complete ordinary one-edge Gysin map
$\Phi^{15}_{4,n,\mathrm B}$ is surjective, and $\mathcal O_{4,n}=0$.
\end{theorem}

\begin{proof}
\smallskip\noindent\emph{Removing the hyperelliptic locus.}
Put $X=\Mopen_{4,12}$, let $Y\subset X$ be its hyperelliptic locus,
and put $X^\circ=X\setminus Y$. The hyperelliptic stack is smooth:
it is a finite gerbe over the quotient of the space of ten distinct
unordered branch points on $\mathbb P^1$, and adding markings gives
an open substack of a fiber power of the universal curve. Thus
\[
\dim X=21,\qquad \dim Y=(10-3)+12=19.
\]
$Y$ is open in the substack $\mathcal H^{\mathrm{irr}}_{4,12}$ of
\cite[Theorem 6.2]{canninglarson2022}, which has the Chow--K\"unneth
generation property since $12\le 2\cdot4+6$; with
Proposition~\ref{prop:cited-chow-kunneth-operations} this gives
$W_{11}H^{11}(Y)=0$.
Lemma~\ref{lem:geometric-bases-supported-types}(2), applied with
$c=2$, gives an injection
\[
\mathcal O_{4,12}\hookrightarrow W_{15}H^{15}(X^\circ).
\]

\smallskip\noindent\emph{Pullback along the proper trigonal morphism.}
Let $\mathscr T=\mathcal H_{3,4,12}$. A hyperelliptic genus-four
curve cannot carry a degree-three map: together with its
degree-two map it would be birational to a curve of bidegree
$(2,3)$ on $\mathbb P^1\times\mathbb P^1$, of arithmetic genus two.
The degree of the map onto its image is one because it divides both two and three.
Consequently $\beta:\mathscr T\to X^\circ$ is proper.

This map is surjective and generically finite. Indeed, the canonical
model of a nonhyperelliptic genus-four curve has degree six in
$\mathbb P^3$. Since $h^0(2K_C)=9<10$, it lies on a quadric.
The quadric is irreducible, since the curve is nondegenerate.
Since $h^0(3K_C)=15$, there is a cubic containing the curve which
is not a multiple of that quadric: the space of cubics has
dimension twenty, and the multiples of the quadric have dimension
four. The complete intersection of the quadric and cubic has degree six and hence is
the curve. A ruling of the quadric cuts out a degree-three pencil;
if the quadric is a cone, smoothness of the complete intersection
implies that the curve avoids its vertex. Conversely, a divisor of a
degree-three pencil spans a line in the canonical model by
Riemann--Roch. The line lies on the quadric, since its intersection
with that quadric contains a length-three divisor. The pencil
therefore comes from a ruling. A general smooth quadric has two
rulings, so $\beta$ has degree two with the ungerbed convention.

The stack $\mathscr T$ is smooth of dimension twenty-one.
Deformations of a finite map
$\alpha:C\to\mathbb P^1$ are controlled by the torsion cokernel
of $T_C\to\alpha^*T_{\mathbb P^1}$; its first cohomology vanishes
and its length for a trigonal curve is $2g+4$. Quotienting by the
three-dimensional target automorphism group gives unmarked
dimension $2g+1$, and distinct markings add twelve dimensions.
For the proper map $\beta$ between smooth stacks of the same
dimension, Poincar\'e duality and the projection formula give
\[
\beta_*\beta^*=2\operatorname{id}.
\]
Thus the pullback embeds $\mathcal O_{4,12}$ in $H^{15}(\mathscr T)$.

\smallskip\noindent\emph{The locus of nonsurjective evaluation.}
For a cover $\alpha:C\to\mathbb P^1$, let $\mathcal E_\alpha$
be its rank-two bundle in
Definition~\ref{def:external-trigonal-evaluation} with $k=3$. Denote its ruled
surface by $\Sigma=\mathbb P(\mathcal E_\alpha^\vee)$ and its
projection by $\gamma_\alpha$. Put
\[
L_\alpha=\mathcal O_\Sigma(3)\otimes
\gamma_\alpha^*\mathcal O_{\mathbb P^1}(-6),
\qquad D=p_1+\cdots+p_{12}.
\]
The curve is a divisor of $L_\alpha$. The canonical bundle of
$\Sigma$ is $\mathcal O_\Sigma(-2)\otimes
\gamma_\alpha^*\mathcal O(4)$, so adjunction gives
\[
\omega_C\simeq
\mathcal O_\Sigma(1)|_C\otimes\alpha^*\mathcal O(-2),
\qquad L_\alpha|_C\simeq\omega_C^{\otimes3}.
\]
Since $H^1(\Sigma,\mathcal O_\Sigma)=0$, the divisor sequence
shows that ambient sections of $L_\alpha$ surject onto
$H^0(C,\omega_C^{\otimes3})$. Also
$H^1(C,\omega_C^{\otimes3})=0$. The cokernel of evaluation at $D$
is therefore
\[
H^1(C,\omega_C^{\otimes3}(-D))
\simeq H^0(C,\mathcal O_C(D)\otimes\omega_C^{-2})^\vee.
\]
The last line bundle has degree zero. It has a section precisely
when it is trivial, and then its space of sections is
one-dimensional. Thus there are exactly two rank strata:
evaluation has rank twelve off the locus
\[
Z=\{(\alpha:C\to\mathbb P^1,p_1,\ldots,p_{12}):D\sim2K_C\},
\]
and rank eleven on $Z$.

Give $Z$ the closed scheme structure obtained by pulling back the
bicanonical section of the relative degree-twelve Picard space
under the Abel map. Over $\mathcal H_{3,4}$ it is the open of the
projective bundle of bicanonical divisors with simple zeros,
followed by the finite \'etale cover ordering those zeros. The
bundle of bicanonical sections has rank
$h^0(C,2K_C)=12-4+1=9$ and vanishing first cohomology.
The degree-twelve Abel map is smooth, since $12>2g-2$.
It follows that $Z$ is smooth and closed, and
\[
\dim Z=9+8=17,\qquad
\operatorname{codim}_{\mathscr T}Z=4.
\]
This locus is the entire complement of the evaluation open.

Lemma~\ref{lem:geometric-bases-evaluation-opens} and
Proposition~\ref{prop:cited-chow-kunneth-operations} give
$W_{15}H^{15}(\mathscr T\setminus Z)=0$.
By Lemma~\ref{lem:geometric-bases-supported-types}(1), applied with
$c=4$, restriction to this open is injective on the image of
$\mathcal O_{4,12}$.
That image is therefore zero. The hyperelliptic localization and
the proper pullback then give $\mathcal O_{4,12}=0$.

\smallskip\noindent\emph{Arbitrary numbers of markings.}
Proposition~\ref{prop:cited-chow-kunneth-bases} gives
$\mathcal O_{4,n}=0$ for $0\leq n\leq11$.
For $n=13,14$, every partition $\lambda\vdash n$ of length at most
four has $\lambda_1\geq4$. Applying
Lemma~\ref{lem:low-degree-symplectic-vanishing} with
$d=15-n\in\{2,1\}$ gives
\[
H^{15-n}(\Mopen_4,\mathbb V_\lambda)=0.
\]
The primitive quotient in
Proposition~\ref{prop:primitive-degree-fifteen-quotient} is thus
zero for these two numbers of markings. Its forgetful span is zero in each case
because the quotient with one fewer marking vanishes. For $n\geq15$, the same
proposition identifies $\mathcal O_{4,n}$ with its forgetful span,
so induction proves vanishing for every $n$.
The right-exact one-edge
sequence of that proposition now proves all the claimed
surjectivities.
\end{proof}

\begin{proposition}[The initial genus-five range]\label{prop:genus-five-initial}
For $0\leq n\leq10$, the complete ordinary one-edge Gysin map
$\Phi^{15}_{5,n,\mathrm B}$ is surjective, and $\mathcal O_{5,n}=0$.
\end{proposition}

\begin{proof}
Proposition~\ref{prop:cited-chow-kunneth-bases} proves the assertion for $n\leq7$.
Fix $n\in\{8,9,10\}$.

\smallskip\noindent\emph{The hyperelliptic and trigonal strata.}
Put $X=\Mopen_{5,n}$, let $Y\subset X$ be the hyperelliptic
locus, and put $X^\circ=X\setminus Y$. The branch-divisor
description gives $\dim Y=9+n$, whereas $\dim X=12+n$.
Thus $Y$ is smooth of codimension three, and
Lemma~\ref{lem:geometric-bases-supported-types}(1), applied with
$c=3$, makes restriction injective on $\mathcal O_{5,n}$.

Let $\mathscr D\subset X^\circ$ be the trigonal locus and put
$V=X^\circ\setminus\mathscr D$. We show that $\mathscr D$ is a
smooth divisor and has the Chow--K\"unneth generation property.
A hyperelliptic genus-five curve has no degree-three map, by the
bidegree-$(2,3)$ argument used above. A nonhyperelliptic
genus-five curve has at most one degree-three pencil: two
inequivalent such pencils would give a birational model of
bidegree $(3,3)$, whose arithmetic genus is four. If the joint map
had degree three onto its image, both maps would factor through the same
projective line and would be equivalent.

The map from the unmarked trigonal Hurwitz stack is unramified
as well. For a cover $\alpha:C\to\mathbb P^1$, put
$A=\alpha^*\mathcal O(1)$. Then $h^0(C,A)=2$ and
$h^0(C,K_C-A)=3$. The basepoint-free pencil sequence identifies
the kernel of multiplication
\[
H^0(C,A)\otimes H^0(C,K_C-A)\longrightarrow H^0(C,K_C)
\]
with $H^0(C,K_C-2A)$. The dimensions of the displayed spaces
are six and five, so this kernel is nonzero. Its line bundle has
degree two, and nonhyperellipticity forces its space of sections
to have dimension at most one. Riemann--Roch therefore gives
\[
h^0(C,K_C-2A)=1,\qquad h^0(C,2A)=3.
\]
Deformations of $\alpha$ with $C$ fixed have tangent space
$H^0(C,\alpha^*T_{\mathbb P^1})=H^0(C,2A)$.
The three-dimensional space of target automorphisms maps
injectively to this space, so the tangent space after quotienting
is zero.

Properness follows from the properness of $\beta_n'$ recalled above.
The map is representable, since an automorphism inducing the
identity on the source induces the identity on the target.
Uniqueness of the pencil makes it universally injective.
A proper, universally injective, unramified morphism is a closed
immersion. It identifies the ungerbed Hurwitz stack with the
reduced trigonal locus. The deformation calculation for trigonal
covers above gives smoothness and dimension $2\cdot5+1=11$.
Adding distinct markings is base change along the smooth
configuration space of the universal curve. It follows that
$\mathscr D$ is smooth of dimension $11+n$, hence is a divisor
in $X^\circ$.

Since $n\leq10<5+7$,
\cite[Section 9.2, Lemmas 9.1--9.2]{canninglarson2022} give the
Chow--K\"unneth generation property for $\mathscr D$.
Proposition~\ref{prop:cited-chow-kunneth-operations} then gives
$W_{13}H^{13}(\mathscr D)=0$, and
Lemma~\ref{lem:geometric-bases-supported-types}(2), applied with
$c=1$, makes restriction injective in weight fifteen.
We have obtained an injection
\[
\mathcal O_{5,n}\hookrightarrow W_{15}H^{15}(V).
\]
Denote its image by $\mathscr V_n$.

\smallskip\noindent\emph{Balanced tetragonal evaluation.}
By \cite[Lemma 10.9 and Section 10.5]{canninglarson2022},
\[
 \mathcal H^4_{4,5,n}
   =\mathcal H_{4,5,n}\setminus\beta_n^{-1}(\mathcal M_{5,n}^{3}),
\]
and its forgetful morphism to
$\mathcal M_{5,n}\setminus\mathcal M_{5,n}^{3}$ is proper
and surjective; this gives a proper
surjective morphism
\[
\beta_n:\mathcal H^4_{4,5,n}\longrightarrow V.
\]
For any of its covers $\alpha:C\to\mathbb P^1$, let
$\mathcal E_\alpha,\mathcal F_\alpha$ be the bundles in
Definition~\ref{def:external-tetragonal-evaluation} with $k=4$.
The rank-two bundle $\mathcal F_\alpha$ has degree eight and
both summands have degree at least four. Hence
$\mathcal F_\alpha\simeq\mathcal O(4)^{\oplus2}$.
Write
\[
\Sigma=\mathbb P(\mathcal E_\alpha^\vee),\qquad
\gamma_\alpha:\Sigma\to\mathbb P^1,\qquad
N_\alpha=\mathcal O_\Sigma(1)\otimes
                   \gamma_\alpha^*\mathcal O(-2).
\]
The curve is the common zero locus of two sections of
$N_\alpha^2$, forming a complete intersection, and the bundle defining it is
$W_\alpha=(N_\alpha^2)^{\oplus2}$.
Here $\det\mathcal E_\alpha\simeq\mathcal O(8)$, so
\[
\omega_\Sigma\simeq
\mathcal O_\Sigma(-3)\otimes\gamma_\alpha^*\mathcal O(6).
\]
Adjunction gives $\omega_C\simeq N_\alpha|_C$, and consequently
$W_\alpha|_C\simeq(\omega_C^{\otimes2})^{\oplus2}$.

The restriction of ambient sections is surjective. With $I_C$
the ideal sheaf of $C$ in $\Sigma$, the twisted Koszul sequence is
\[
0\longrightarrow N_\alpha^{-2}\longrightarrow
\mathcal O_\Sigma^{\oplus2}\longrightarrow
I_C\otimes N_\alpha^2\longrightarrow0.
\]
Every direct image of $N_\alpha^{-2}$ under $\gamma_\alpha$
vanishes, since its restriction to a projective-plane fiber is
$\mathcal O_{\mathbb P^2}(-2)$, which has no cohomology.
Together with $H^1(\Sigma,\mathcal O_\Sigma)=0$, the sequence
therefore gives $H^1(\Sigma,I_C\otimes N_\alpha^2)=0$.
It follows that
$H^0(\Sigma,N_\alpha^2)\to H^0(C,\omega_C^{\otimes2})$
is surjective. Applying this to both summands, and using
$H^1(C,\omega_C^{\otimes2})=0$, identifies the cokernel of
evaluation at $D=p_1+\cdots+p_n$ with
\[
H^1(C,W_\alpha|_C(-D))
\simeq
\bigl(H^0(C,\mathcal O_C(D)\otimes\omega_C^{-1})^\vee\bigr)^{\oplus2}.
\]

Define the closed locus
\[
Z_n=\{(C,p_1,\ldots,p_n)\in V:
       h^0(C,\mathcal O_C(D)\otimes\omega_C^{-1})>0\},
\qquad V^{\mathrm{ev}}=V\setminus Z_n.
\]
Closedness follows from upper semicontinuity. The condition is
independent of the choice of tetragonal pencil. Therefore the
restriction
\[
\beta_n^{-1}(V^{\mathrm{ev}})\longrightarrow V^{\mathrm{ev}}
\]
is the base change of the proper surjective map $\beta_n$.
Its source is precisely the open substack of the balanced Hurwitz stack
where evaluation is surjective. The balanced Hurwitz stack is
open in $\mathcal H'_{4,5,n}$, by the displayed equality of
\cite[Lemma 10.9]{canninglarson2022}.
Lemma~\ref{lem:geometric-bases-evaluation-opens} and open
restriction thus give the Chow--K\"unneth generation property for
$\beta_n^{-1}(V^{\mathrm{ev}})$. The morphism is representable by
Deligne--Mumford stacks, as before. Proper descent
in Proposition~\ref{prop:cited-chow-kunneth-operations} gives the
property for $V^{\mathrm{ev}}$, and hence
\[
W_{15}H^{15}(V^{\mathrm{ev}})=0.
\]

\smallskip\noindent\emph{Canonical and residual-divisor strata.}
Put $k=n-8\in\{0,1,2\}$. The line bundle
$\mathcal O_C(D)\otimes\omega_C^{-1}$ has degree $k$.
On a curve of gonality greater than three, a line bundle of
degree at most two has at most one independent section.
Indeed, two independent sections, after removing their common
zeros, would define a map of degree at most two to $\mathbb P^1$.
Thus evaluation has rank $2n$ off $Z_n$, and rank $2n-2$ on
$Z_n$; no further rank drop occurs.

For $n=8$, the degree is zero, so $Z_8$ is exactly the locus
$D\sim K_C$. Over the unmarked open
$\Mopen_5\setminus\mathcal M_5^3$, its unordered divisors form
the projective bundle of canonical sections, of relative
dimension four. Restricting to simple zeros and ordering the
eight zeros gives an open substack followed by a finite \'etale cover. This
identifies $Z_8$, with its relative linear-system scheme
structure, as a smooth closed substack of $V$.

For $n=9,10$, there is a unique effective divisor $R$ of degree
$k$ such that $D\sim K_C+R$. The Abel map
\[
\operatorname{Sym}^k C\longrightarrow\operatorname{Pic}^k C
\]
is a closed immersion: it is proper and universally injective,
and its tangent kernel at $R$ is
$H^0(C,\mathcal O_C(R))/H^0(C,\mathcal O_C)=0$.
These statements hold relatively over the unmarked open as well.
Since $n>2g-2=8$, the degree-$n$ Abel map is smooth, with fibers
the complete linear systems of dimension $n-5=3+k$.
Its inverse image of
$K_C+\operatorname{Sym}^k C$ is the projective bundle of sections
of $\omega_C(R)$ over the relative symmetric power. The direct
image has constant rank $4+k$, since
\[
H^1(C,\omega_C(R))=H^0(C,\mathcal O_C(-R))^\vee=0.
\]
The simple-zero open, with its zeros ordered, identifies with
$Z_n$. In particular, $Z_n$ is smooth and closed, of relative
dimension $k+(3+k)=3+2k$ over the unmarked open.
The dimensions and codimensions are
\[
\begin{array}{c|c|c|c}
n & \text{failure condition} & \dim Z_n & \operatorname{codim}_V Z_n\\
\hline
8 & D\sim K_C & 12+4=16 & 4\\
9 & D\sim K_C+R,\ \deg R=1 & 12+5=17 & 4\\
10 & D\sim K_C+R,\ \deg R=2 & 12+7=19 & 3
\end{array}
\]

\smallskip\noindent\emph{Exclusion by Hodge type.}
For each of the three values of $n$, write
$c=\operatorname{codim}_V Z_n\in\{4,3\}$.
Lemma~\ref{lem:geometric-bases-supported-types}(1), applied with
this $c\geq3$, makes restriction to $V^{\mathrm{ev}}$ injective on
$\mathscr V_n$.
Its target in weight fifteen is zero by proper descent from the
evaluation open, so $\mathscr V_n=0$.
The injections obtained by removing the hyperelliptic and
trigonal strata give $\mathcal O_{5,n}=0$ for $n=8,9,10$.
The right-exact sequence of
Proposition~\ref{prop:primitive-degree-fifteen-quotient} gives
the asserted one-edge surjectivity throughout this range.
\end{proof}