Proof. Let U be the preimage of S in $\mathcal {M}_{g,r+k}(X, \beta )$ under the map $\pi : \overline {\mathcal {M}}_{g,r+k}(X, \beta ) \to \overline {\mathcal {M}}_{g,r}(X, \beta )$ which forgets the last k points. Thus, U parametrizes maps in S together with the choice of k additional points $\{q_{r+i} \}_{i=1}^{k}$ in $C \setminus \{q_i \}_{i=1}^r$ . We also let $\psi : \overline {\mathcal {M}}_{g,r+k}(X, \beta ) \to \overline {\mathcal {M}}_{g,r+k}$ be the forgetful map and let $\phi : \overline {\mathcal {M}}_{g,r+k} \to \overline {\mathcal {M}}_{g,r}$ be the map forgetting the last k markings. Note that the fiber F of $\phi $ over $(C, q_1, \ldots , q_r)$ is irreducible since it is birational to $C^{k}$ . By construction the closure of the image of U under $\psi $ is F. Note that the nonempty fibers of $\psi : U \to F$ are isomorphic to S and thus have dimension k.

Fix a very ample line bundle $\mathcal {L}$ on X. Set $U_{0} = U$ and for $1 \leq i \leq k$ define $U_{i} \subset \mathcal {M}_{g,r+k}(X, \beta )$ inductively by choosing a general section $D_{i}$ of $\mathcal {L}$ and letting $U_{i}$ be the intersection of $U_{i-1}$ with $ev_{r+i}^{-1}(D_{i})$ where $ev_{r+i}$ denotes the evaluation map at the corresponding marked point. In particular we choose $D_{i}$ so that $p_{i} \not \in D_{i}$ . We prove by induction that the dimension of the fiber of $\psi |_{U_{i}}$ over a general point of F is at least $k-i$ .

By induction we know the fiber $V_{i-1}$ of $\psi |_{U_{i-1}}$ over a general point $(C,q_{1},\ldots ,q_{r+k})$ of F has positive dimension. Note that the image of $V_{i-1}$ in $\mathcal {M}_{g,0}(X, \beta )$ does not depend on the choice of marked points $q_{r+i},\ldots ,q_{r+k}$ . Furthermore, this image must have positive dimension since $V_{i-1}$ has positive dimension and parametrizes maps from a fixed marked curve. Thus for a general point of F the image $s(q_{r+i})$ sweeps out a locus of dimension $\geq 1$ in X as we vary $s \in V_{i-1}$ . We conclude that the preimage of the general divisor $D_{i}$ under the evaluation map $ev_{r+i}|_{U_{i-1}}$ meets $V_{i-1}$ . In particular the dimension of the general fiber of $\psi |_{U_{i}}$ is at most one less than the dimension of the general fiber of $\psi |_{U_{i-1}}$ , proving the claim.

Let $\overline {U_{k}}$ be the closure of $U_{k}$ in $\overline {\mathcal {M}}_{g,r+k}(X, \beta )$ . There is an element of $\overline {U_{k}}$ lying over the locus in F where $q_{r+i}$ specializes to $q_i$ for each i. Let $s^{\prime }:(C^{\prime }, q^{\prime }_1, \dots , q^{\prime }_{r+k}) \to X$ be the corresponding stable map. Because $\pi (s^{\prime })$ lies in the closure $\overline {S}$ of S in $\overline {\mathcal {M}}_{g,r}(X,\beta )$ , we know that the stabilization of $(C^{\prime },q^{\prime }_1,\dots ,q^{\prime }_r)$ must be $(C, q_1, \dots , q_r)$ . Thus each $q^{\prime }_{i}$ is contained in a tree of rational curves (which also contains $q^{\prime }_{r+i}$ ) that is contracted by the stabilization map. Likewise, because $\pi (s^{\prime })$ lies in $\overline {S}$ we see that $s^{\prime }(q^{\prime }_i) = p_i$ for all $i \leq k$ . For all $i,j \leq k$ , generality of $D_j$ ensures it is disjoint from $p_i$ . Because $s^{\prime }(q^{\prime }_{r+i})$ must lie in $D_i$ , we see that the tree of rational curves containing $q^{\prime }_{i}$ has to map to a curve in X connecting $p_i$ to $D_i$ ; in particular, some component is not contracted by $s'$ .

Proof. Let $q_1, \ldots , q_k \in C$ be k general points in C. Let $T_i = \{\tilde {s} \in M \mid \tilde {s}(q_i) = s(q_i)\}$ . Since $s \in T_i$ for all i, the intersection $S := \cap _i T_i$ is nonempty. Moreover, as $\operatorname {codim}(T_i, M) \leq \dim X$ , we get $\operatorname {codim}(S, M) \leq k(\dim X)$ . Thus $\dim S \geq k$ .

Because $2g - 2 + k> 0$ , the pointed curve $(C, q_1, \dots , q_k)$ has finite automorphism group, and hence the natural map $\pi : S \to \overline {\mathcal {M}}_{g,k}(X, \beta )$ is generically finite. Apply Lemma 2.1 to a dense locally closed subset of $\pi (S)$ and let $s^{\prime }: (C^{\prime }, q_1^{\prime }, \ldots , q_k^{\prime }) \to X$ be the stable map it identifies. The desired stable map is obtained from $s'$ by forgetting the k marked points and stabilizing.

Proof of Theorem 1.1.

First suppose that our ground field is algebraically closed of characteristic $p> 0$ and that H is $\mathbb {Q}$ -Cartier. After rescaling H we may suppose it is Cartier. We write $i: C' \to X$ for the normalization of C. For $m> 0$ , let $s_m : C^{\prime } \to X$ be the precomposition of i with the $m^{\text {th}}$ iterate of the Frobenius. The dimension $d_{m}$ of $\operatorname {Mor}(C^{\prime },X)$ at $s_m$ satisfies

$$ \begin{align*}d_{m} \geq p^m(-K_X \cdot i_*C^{\prime}) - g\dim X,\end{align*} $$

where g is the genus of $C'$ . Let $k_{m} = \lfloor \frac {d_{m}}{\dim X + 1} \rfloor $ . For large m, Proposition 2.2 allows us to find a deformation of $s_m$ that breaks off $k_{m}$ rational curves through $k_{m}$ general points of $s(C^{\prime })$ . Because H is nef, at least one of these rational curves has H-degree at most

$$ \begin{align*} \frac{H\cdot s_{m*}C^{\prime}}{k_{m}} = \frac{p^m(H \cdot i_*C^{\prime})}{k_{m}} \leq \frac{p^m(H \cdot i_*C^{\prime})}{p^{m}(-K_{X} \cdot i_{*}C^{\prime}) - (g + 1)\dim X}(\dim X + 1). \end{align*} $$

For large enough m the floor of this upper bound is at most $(\dim X + 1)\frac {H \cdot i_*C^{\prime }}{-K_X \cdot i_*C^{\prime }}$ . This proves that a general point of $s(C^{\prime })$ is contained in a rational curve whose H-degree satisfies the desired inequality. Since the existence of such a rational curve through a point is a closed condition, this statement holds for every closed point in $s(C')$ , and in particular for x.

The extension to ample $\mathbb {Q}$ -Cartier divisors in characteristic $0$ uses the spreading out argument of (Reference Miyaoka and MoriMM86, Step 3 of proof of Theorem 5). The extension to nef $\mathbb {R}$ -Cartier divisors follows as in (Reference KollárKol96, Steps 4 and 5 of proof of II.5.8 Theorem).