is fixed.

By Lemma \[lem:smoothable\] and the minimality of the very free curve, $C_1$ and $C_4$ are not connected by a single curve $C_2$ or $C_3$ in the last three cases. And the curve $C_2$ (resp. $C_3$) is fixed once two points of the curve is fixed by a bend-and-break type argument.

In the first case, the degree $4$ curve $C_2$ has to meet $\Gamma$. Then the dimension of every irreducible component of the Hom-scheme at $C_2$ is at most $9$. Indeed, if in one irreducible component the deformation of $C_2$ dominates $X$, then that irreducible component has expected dimension $-K_X \cdot C_2+\dim X=8$ since a general deformation is free. Now assume that the deformation is contained in a divisor ( it cannot be contained in a surface since we can choose the two degree $2$ free curves to avoid any finite number of codimension $2$ locus by choosing the two points to be general). Consider the evaluation map $${{\mathbb}{P}}^1 \times {{\mathbb}{P}}^1 \times {{\mathbb}{P}}^1 \times Hom(C_2, X) \to X \times X \times X.$$ The image in $X\times X \times X$ has dimension $9$. So it suffices to show that the map has fiber dimension $3$. If the fiber dimension is 