I have a question regarding to the proof given in lemma 1.9 (Bend and Break II) pg 11. given in the book ‘Birational Geometry of Algebraic Varieties’ by János Kollár and Shigefumi Mori.
In the second paragraph we assume that $\tilde{G}: \tilde{S} \rightarrow X$ is a morphism. By fixing an ample divisor $H$ on $X$, and setting $C_0, C_\infty \subset S$ to be sections extending $\{0\} \times D, \{\infty\} \times D$ we have $(\tilde{G}^*H)>0$ and $C_0 \cdot \tilde{G}^*H = C_\infty \cdot \tilde{G}^*H =0$. By the Hodge index theorem, we have $C_0^2 , C_\infty^2 <0$.
The author went on to claim that $\{\tilde{G}^*H, C_0, C_\infty\}$ must be an linearly independent set in the Néron–Severi group of $S$. How do we see this? Given a linear combination $a\tilde{G}^*H + b C_0 + bC_\infty =0$ we easily see that $a=0$. So we are left with $bC_0 + cC_\infty =0$. How do we use the fact that $C_0^2, C_\infty^2 <0$ to show that $b=c=0$ necessarily?
Edit: I think I have figured it out. We have $b(C_0 - C_\infty)=-(b+c) C_\infty$. Then taking product, $-(b+c) C_\infty^2 = b (C_0^2 - C_0C_\infty)$.
We may assume $b \ge 0$ and so the RHS is $<0$. This forces $b+c \le 0$. Now, by the Hodge Index theorem applied to $C_0-C_\infty$, we have $(C_0 - C_\infty)^2 \le 0$.
So $0 \ge b(C_0-C_\infty)^2= -(b+c)(C_0C_\infty -C_\infty^2)$. So $-(b+c) \le 0$. So $b+c \ge 0$. All in all, $b=-c$ and so we have $C_0 = C_\infty$, which cannot be since they are distinct.
Edit: I think I have figured it out. We have $b(C_0 - C_\infty)=-(b+c) C_\infty$. Then taking product, $-(b+c) C_\infty^2 = b (C_0^2 - C_0C_\infty)$.
We may assume $b \ge 0$ and so the RHS is $<0$. This forces $b+c \le 0$. Now, by the Hodge Index theorem applied to $C_0-C_\infty$, we have $(C_0 - C_\infty)^2 \le 0$.
So $0 \ge b(C_0-C_\infty)^2= -(b+c)(C_0C_\infty -C_\infty^2)$. So $-(b+c) \le 0$. So $b+c \ge 0$. All in all, $b=-c$ and so we have $C_0 = C_\infty$, which cannot be since they are distinct.