Proof. Since [Reference BirkarBir16] has settled the Borisov–Alexeev–Borisov conjecture, [Reference Lehmann, Tanimoto and TschinkelLTT18, Theorem 4.8] proves that the closure of $V$ is a proper closed subset of $X$ . In fact the proof gives a little bit more: every component of $V$ is dominated by a family of subvarieties with $a$ -value higher than $X$ . By [Reference Lehmann, Tanimoto and TschinkelLTT18, Proposition 4.1] every component of $V$ will also have higher $a$ -value than $X$ .◻

Proof. The proof is by induction on the length of the chain. In the base case of one component, the first statement is obvious. For the second, note that $U^{\prime \prime \prime }\subset \overline{M}_{0,3}(X)$ can be identified with an open set in $\text{Mor}(\mathbb{P}^{1},X)$ . By [Reference KollárKol96, Corollary 3.5.4] the evaluation map for the second marked point is flat. This factors through the natural map $U^{\prime \prime \prime }\rightarrow U^{\prime \prime }$ ; since the forgetful map is faithfully flat, we see that the second statement also holds.

We next prove the induction step. The projection from $N$ onto the first $n-1$ factors maps $N$ to a component $Q$ of $U_{1}^{\prime \prime }\times _{X}\cdots \times _{X}U_{r-1}^{\prime \prime }$ . By induction, $Q$ has the two desired properties. Also, since the space of free curves through a fixed point has the expected dimension, the map $U_{r}^{\prime \prime }\rightarrow X$ induced by the first marked point is dominant and flat. Consider the following diagram.

Both projections from $Q\times _{X}U_{r}^{\prime \prime }$ have equidimensional fibers by base change. Furthermore, every component of $Q\times _{X}U_{r}^{\prime \prime }$ has the same dimension (as it parametrizes chains of free curves). Together, this shows that every component of $Q\times _{X}U_{r}^{\prime \prime }$ will dominate $Q$ so long as there is some component which dominates $Q$ . But it is clear that a general chain of free curves in $Q$ can be attached to a free curve in $U_{r}$ , so that the map from $Q\times _{X}U_{r}^{\prime \prime }$ to $Q$ must be dominant for at least one component. Noting that $N$ is a component of $Q\times _{X}U_{r}^{\prime \prime }$ for dimension reasons, we obtain from the induction hypothesis the first statement for $N$ since flatness is stable under base change and composition. The last statement follows by the same logic.◻

Proof. The proof is by induction on the length of the chain. Consider the base case $f:M_{1}^{\prime \prime }\rightarrow X$ . Since reducible curves form a locus of codimension $1$ , every component of a general fiber of $f$ must contain irreducible curves. Since there is a closed subset of $X$ containing every non-free irreducible curve in $M_{1}$ , we see that every component of a general fiber of $F$ must contain free curves. The ability to avoid $Z$ follows from Lemma 5.7.

We now prove the induction step. Via projection $N$ maps into a component $Q\subset M_{1}^{\prime \prime }\times _{X}\cdots \times _{X}M_{r-1}^{\prime \prime }$ . By induction $Q$ satisfies the desired property. By [Reference KollárKol96, II.3.5.4 Corollary and II.3.10.1 Corollary] there is a proper closed subset $Z_{0}\subset X$ such that if $C_{0}$ is a component of a curve parametrized by $M_{r}^{\prime \prime }$ and $C_{0}\not \subset Z_{0}$ then $C_{0}$ is free. Consider the evaluation along the first marked point (of $Q$ ) denoted by $\tilde{f}:Q\times _{X}M_{r}^{\prime \prime }\rightarrow X$ . The fibers of this map are products $F\times _{X}M_{r}^{\prime \prime }$ where $F$ is a fiber of $Q\rightarrow X$ ; choosing the fiber $F$ general with respect to $Z_{0}$ , we see that every component of every fiber will contain a chain of free curves. In particular, this is also true for the map $f:N\rightarrow X$ which is a restriction of $\tilde{f}$ to a component. The ability to avoid $Z$ via the last marked point follows from Lemma 5.7.◻

Proof. By definition the projection $N\rightarrow M_{j}^{\prime \prime }$ is dominant, hence surjective by properness. So we know that $N$ contains a point of

$$\begin{eqnarray}M_{1}^{\prime }\times _{X}M_{2}^{\prime \prime }\times _{X}\cdots \times _{X}M_{j-1}^{\prime \prime }\times _{X}\widetilde{M}_{j}^{\prime \prime }\times _{X}\widehat{M}_{j}^{\prime \prime }\times _{X}M_{j+1}^{\prime \prime }\times _{X}\cdots \times _{X}M_{r-1}^{\prime \prime }\times _{X}M_{r}^{\prime }.\end{eqnarray}$$

If we can show that it contains a point which is a chain of free curves, then since such points are smooth in $\overline{M}_{0,0}(X)$ we can conclude that $N$ will contain an entire component of chains of length $r+1$ .

Since the map $N\rightarrow M_{j}^{\prime \prime }$ is surjective, in particular, for any two-pointed chain of length $2$ in $\widetilde{M}_{j}^{\prime \prime }\times _{X}\widehat{M}_{j}^{\prime \prime }$ there is a curve parametrized by $N$ containing this chain. Since the curves are free, we may choose a chain such that the first and last marked points are general. The fiber of $N$ over this point is a union of components of

$$\begin{eqnarray}G_{1}\subset M_{1}^{\prime }\times _{X}\cdots \times _{X}M_{j-1}^{\prime \prime }\end{eqnarray}$$

under a product with

$$\begin{eqnarray}G_{2}\subset M_{j+1}^{\prime \prime }\times _{X}\cdots \times _{X}M_{r}^{\prime },\end{eqnarray}$$

where $G_{1}$ and $G_{2}$ are the fibers of the last and first marking, respectively. Applying Lemma 5.8, we see that every component of the fiber over this point contains chains of free curves.◻

Proof. Suppose that $f:C\rightarrow X$ denotes our original chain of free curves. It suffices to prove the statement when $\tilde{f}^{\dagger }$ differs from $f^{\dagger }$ by a transposition of two adjacent elements. Suppose that $T_{1}$ and $T_{2}$ are two adjacent components of $C$ . Let $S_{1}$ denote the rest of the chain that attaches to $T_{1}$ , and $S_{2}$ denote the rest of the chain that attaches to $T_{2}$ . After deforming $f$ , we may suppose that the intersection of $T_{1}$ and $T_{2}$ maps to a general point $x$ of $X$ .

Suppose we leave $T_{1}$ and $T_{2}$ fixed, but deform $S_{1}$ and $S_{2}$ , maintaining a point of intersection with $T_{1}$ or $T_{2}$ respectively, so that the specialized curves $S_{1}^{\prime }$ and $S_{2}^{\prime }$ contain the point $x$ . By generality of the situation, the deformed $S_{1}^{\prime }$ and $S_{2}^{\prime }$ are still chains of free curves. The stable curve $g:D\rightarrow X$ corresponding to this deformation looks like a single rational curve $Z$ contracted by $g$ to the point $x$ with four chains of free curves $S_{1}^{\prime },T_{1},T_{2},S_{2}^{\prime }$ attached to $Z$ . A tangent space calculation (as in [Reference TestaTes05, Corollary 1.6]) shows that $g$ is a smooth point of $M$ . However, by a similar argument $g$ is the deformation of a chain of rational curves of the type $S_{1}^{\prime \prime }\cup T_{2}\cup T_{1}\cup S_{2}^{\prime \prime }$ , where $S_{1}^{\prime \prime }$ and $S_{2}^{\prime \prime }$ are deformations of $S_{1}^{\prime }$ and $S_{2}^{\prime }$ .◻

Proof. For degree reasons, there are only finitely many components $M$ of $\overline{M}_{0,0}(X)$ which can be the closure of the image of such a map. Each such $M$ generically parametrizes free curves. Thus, there is a unique component $M^{\prime }$ of $\overline{M}_{0,1}$ lying over $M$ . Let $g:M^{\prime }\rightarrow X$ denote the universal family map, and let $h:\overline{Z}\rightarrow X$ denote the Stein factorization of a resolution of $g$ . If the map $g$ factors rationally through a generically finite dominant map $f:Z\rightarrow X$ , then so does $h$ . Thus for any given component $M$ there can only be finitely many corresponding elements of ${\mathcal{Z}}$ .◻

Proof. Let $\{{M_{\unicode[STIX]{x1D6FD}}\}}_{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6EF}}$ denote the elements of ${\mathcal{N}}$ . For each component, we have a universal family map $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D6FD}}:M_{\unicode[STIX]{x1D6FD}}^{\prime }\rightarrow X$ . We let $e_{\unicode[STIX]{x1D6FD}}$ denote the degree of the Stein factorization of the composition of $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D6FD}}$ with a resolution of singularities of $M_{\unicode[STIX]{x1D6FD}}^{\prime }$ and set $e=\sup _{\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6EF}}e_{\unicode[STIX]{x1D6FD}}$ . Note that since we have included a resolution in the definition, if $e_{\unicode[STIX]{x1D6FD}}=1$ then the general fiber of $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D6FD}}$ is irreducible.

The proof is by induction on $e$ . First suppose that $e=1$ . Let $M$ be a component of $\overline{M}_{0,0}(X,d)$ satisfying the desired condition. Since a chain of free curves is a smooth point of $M$ , to count such components $M$ it suffices to count all possible components of the parameter space of chains of free curves from ${\mathcal{N}}$ of total degree $d$ . In fact, by applying Lemma 5.11, we may reorder the combinatorial type however we please. Furthermore, for any choice of combinatorial type the parameter space of chains of that type

$$\begin{eqnarray}M_{1}^{\prime }\times _{X}M_{2}^{\prime \prime }\times _{X}\cdots \times _{X}M_{r-1}^{\prime \prime }\times _{X}M_{r}^{\prime }\end{eqnarray}$$

is irreducible since by assumption each degree $e_{i}=1$ . Thus the number of possible $M$ is at most the possible ways of choosing (with replacement and unordered) components of ${\mathcal{N}}$ such that the total degree adds up to $d$ . This count is polynomial in $d$ .

Before continuing with the proof, we make an observation:

Now suppose that $e>1$ . For a dominant generically finite map $g:Z\rightarrow X$ of degree at least $2$ with $Z$ smooth, let ${\mathcal{N}}_{Z}$ denote the subset of ${\mathcal{N}}$ consisting of components $M$ such that the universal map $M^{\prime }\rightarrow X$ factors rationally through $Z$ . Note that the locus where the rational map to $Z$ is not defined must miss the general fiber of $M^{\prime }\rightarrow M$ . Thus, we obtain a family of free curves on $Z$ parametrized by an open subset of $M$ . A deformation calculation shows that the general curve has vanishing intersection with the ramification divisor; in particular, $M$ is birational to a component of $\overline{M}_{0,0}(Z)$ . Note that for any component of ${\mathcal{N}}_{Z}$ the degree of the rational map from the component to $Z$ is strictly smaller than for the corresponding component in ${\mathcal{N}}$ . Consider the corresponding families of rational curves on $Z$ measured with respect to the big and nef divisor $g^{\ast }L$ . By the induction hypothesis, there is a polynomial $P_{Z}(d)$ which gives an upper bound for the number of components of $\overline{M}_{0,0}(Z,d)$ which arise by gluing chains from ${\mathcal{N}}_{Z}$ on $Z$ . Furthermore, by Lemma 5.12 there are only finitely many $Z$ for which ${\mathcal{N}}_{Z}$ is non-empty.

Fix a positive integer $r$ and a dominant generically finite map $f:Z\rightarrow X$ of degree at least $2$ with $Z$ smooth. As we vary over possible choices $M_{i}\in {\mathcal{N}}$ , consider all main components of $M_{1}^{\prime }\times _{X}\cdots \times _{X}M_{k}^{\prime }$ such that there is an integer $b$ where the component $M$ of $\overline{M}_{0,0}(X)$ obtained by gluing the first $b$ curves has $L$ -degree $r$ and has a universal family map which factors rationally through $Z$ , but if we consider the component arising from gluing the first $b+1$ curves, the Stein factorization of a resolution of the universal family map has degree $1$ . We see there are at most $P_{Z}(r)$ possible components $M$ obtained by gluing the first $b$ curves in the chain. Next consider adding one more component. By degree considerations, there can be at most $e\cdot P_{Z}(r)$ components obtained by gluing the first $b+1$ curves, and for any such component the evaluation map to $X$ has generically irreducible fibers. Finally, to add on the remaining components, we may use Lemma 5.11 to reorder the other components arbitrarily. Applying Observation 5.14, we see that the total number of glued components for this choice of $b$ and $Z$ is bounded above by $e\cdot P_{Z}(r)$ times the number of ways to choose (with replacement and unordered) $k-b-1$ components from ${\mathcal{N}}$ .

In total, the number of components of $\overline{M}_{0,0}(X)$ containing chains of curves from ${\mathcal{N}}$ of degree $d$ will be bounded above by the sum of the previous bounds as we vary $Z$ and $r$ . Let $Q(k)$ denote the polynomial representing the number of ways to choose $k$ components (with replacement and unordered) from ${\mathcal{N}}$ . Altogether, the number of components is bounded above by the polynomial in $d$ given by

$$\begin{eqnarray}e\cdot Q(d)\cdot \mathop{\sum }_{Z}\mathop{\sum }_{r\leqslant d}P_{Z}(r).\Box\end{eqnarray}$$

Proof. Let $\widetilde{Y}$ be a resolution of $Y$ with map $\widetilde{f}:\widetilde{Y}\rightarrow X$ . By taking the strict transform of the family of rational curves, one obtains a dominant family on $\widetilde{Y}$ which is parametrized by an open subset of a component of $\text{Mor}(\mathbb{P}^{1},\widetilde{Y})$ . Since the dimension of this component is the same on $\widetilde{Y}$ and on $X$ , and equals the expected dimension in both cases, we have

$$\begin{eqnarray}K_{X}\cdot C=K_{\widetilde{Y}}\cdot C\quad \;\Longrightarrow \;\quad (K_{\widetilde{Y}}-\widetilde{f}^{\ast }K_{X})\cdot C=0.\end{eqnarray}$$

Thus the divisor $K_{\widetilde{Y}}+a(X,-K_{X})(-\widetilde{f}^{\ast }K_{X})$ is pseudo-effective but not big.◻

Proof. [Reference ManinMan95, Equation 0.5] shows that there is a positive constant $C$ such that the number of components of free curves of degree $d$ against a fixed big and nef divisor is at most $C^{d}$ . The result follows by combining this equation with Theorem 4.4 and a standard counting argument.◻

Proof. By [Reference Tian and ZongTZ14, Theorem 1.3], $N_{1}(X)_{\mathbb{Z}}$ is spanned by the classes of rational curves $\{C_{i}\}$ . Since $X$ is rationally connected, there is a family of very free curves $C$ such that there is a very free member of the family through any point of $X$ . By gluing sufficiently many of these $C$ onto one of the $C_{i}$ to form a comb, we can deform to get a smooth curve (as in [Reference KollárKol96, II.7.10 Proposition]). It is then clear that these smoothed curves and the class of $C$ together span $N_{1}(X)$ .◻

Proof. Choose a dominant family of rational curves $C$ on $Y$ as in Conjecture 6.13 such that

$$\begin{eqnarray}L\cdot C>\frac{\dim (X)-\dim (Y)}{a(Y,L)-a(X,L)}.\end{eqnarray}$$

By computing the expected dimension on $X$ and on $Y$ one concludes the statement.◻

Proof. Let $b$ denote the difference in dimensions between $F(Y,f^{\ast }L)$ and its image under $f_{\ast }$ . [Reference Hassett, Tanimoto and TschinkelHTT15, Theorem 2.16] shows that that lattice points $F(Y,f^{\ast }L)\cap N_{1}(Y)_{\mathbb{Z}}$ generate a subcone of $F(Y,f^{\ast }L)$ which is full rank. Thus there are a class $\unicode[STIX]{x1D6FC}\in \text{Nef}_{1}(X)_{\mathbb{Z}}$ and a constant $C>0$ such that for sufficiently large integers $m$ there are at least $Cm^{b}$ points of $F(Y,f^{\ast }L)$ mapping to $m\unicode[STIX]{x1D6FC}$ .

For sufficiently large $m$ , Conjecture 6.13 guarantees that for each class $\unicode[STIX]{x1D6FD}$ that pushes forward to $m\unicode[STIX]{x1D6FC}$ there is a component $M_{\unicode[STIX]{x1D6FD}}\subset \text{Mor}(\mathbb{P}^{1},Y,\unicode[STIX]{x1D6FD})$ parametrizing a dominant family of rational curves. For each such $M_{\unicode[STIX]{x1D6FD}}$ , by composing with $f$ we get a dominant family of rational curves on $X$ . At most $\deg f$ different components on $Y$ can get identified to a single component on $X$ , showing that the number of different components on $X$ has the desired asymptotic growth rate.◻