free rational curve. Thus the dimension is exactly $-K_X \cdot C_2+\dim X \leq 7$. So for now assume that the deformation of $C_2$ in one irreducible component $U$ is contained in a subvariety $S$. Both $C_1$ and $C_3$ intersect the subvariety $S$ at finitely many points and $C_2$ has to pass through at least two of them. Then $C_2$ does not move once we make the choice of the two points. Otherwise we can deform $C_2$ fixing two points and, by bend-and-break, $C_2$ breaks into a reducible or non-reduced curve. But this cannot happen since $-K_X \cdot C_2\leq 3$, $i(X) \geq 2$ and $-K_X$ is ample. The above analysis shows that the evaluation map $${{\mathbb}{P}}^1 \times {{\mathbb}{P}}^1 \times Hom(C_2, X) \to X\times X$$ has fiber dimension $3$ (coming from reparametrization ) and the image has dimension $\dim S \times 2\leq 3 \times2=6$. Therefore, $$\dim_{[f\vert_{C_2}]}Hom(C_2, X) \leq 7.$$

The estimate \[eq:1\] in Lemma \[lem:smoothing\] shows that at least one of the nodes can be smoothed out. If only one node is smoothed, then we get a reducible curve, which is the union of two irreducible curves each passing through a very general point. Then a general deformation of this new curve 