Lemma 10.28.11959B582751F

Proof. Near one of the sections, coordinates centered on it identify the blowup with $\operatorname {Bl}_0(\mathbb C^2)\times \Delta $. Since the four centers are disjoint, these charts prove smoothness and the submersion property. The blowup of the projective morphism $\mathbb P^2\times \Delta \to \Delta $ is projective and proper, and each fiber $X_ t$ is the blowup of $\mathbb P^2$ at the four points $p_ i(t)$.

The quadratic forms

vanish along all four sections and therefore lift to sections of $L$. They are nonzero, which proves that $L$ is effective. For $t\ne 0$, their common zeros in $\mathbb P^2$ are exactly the four points. In suitable affine coordinates $(u,v)$ centered at $p_1,p_2,p_3,p_4$, their linear initial terms are

Each pair is linearly independent. Hence, after the exceptional divisors are subtracted, the transformed sections have no common zero on an exceptional curve. The two transformed sections therefore generate $L_ t$, and $L_ t$ has nonnegative degree on every irreducible curve.

On a fiber, $H^2=1$, $H\cdot E_ i=0$, and $E_ i\cdot E_ j=-\delta _{ij}$. Hence $L_ t^2=0$, and $K_{X_ t}=-3H+\sum _ iE_ i$ gives $K_{X_ t}\cdot L_ t=-2$. A nonzero section of $K_{X_ t}+mL_ t$ would have an effective zero divisor whose intersection with the nef line bundle $L_ t$ is $-2<0$, a contradiction. This argument also excludes a nowhere-vanishing section, whose zero divisor would have intersection zero. $\square$