over $k$. Let $S$ be a smooth prime divisor on $X$ and let $A$ be an ample Cartier divisor on $X$. Assume the following two conditions.

1.  [$K_X+S+A$ is nef.]{}

2.  [$\kappa(S, K_S+A)\neq 0$.]{}

Then, there exists $m_0\in \mathbb Z_{>0}$ such that, for every $m\geq m_0$, the natural restriction map $$H^0(X, m(K_X+S+A))\to H^0(S, m(K_S+A))$$ is surjective.

To show the above two theorems, we use the trace map of Frobenius. This strategy is essentially the same as [@Schwede2 Proposition 5.3] and its proof. Let us see the idea of the proofs. Let $X$ be a smooth projective variety. Let $S$ be a smooth prime divisor on $X$ and let $A$ be an ample Cartier divisor on $X$. Then, for every $e\in\mathbb Z_{>0}$, we obtain the following commutative diagram by using the trace map of Frobenius: [ $$\begin{CD}
H^0(X, K_X+S+p^eA)@>>>H^0(S, K_S+p^eA)@>>>H^1(X, K_X+p^eA)\\
@VVV @VV{{\operatorname{Tr}}}_S^e(A)V\\
H^0(X, K_X+S+A)@>\rho >> H^0(S, K_S+A)
\end{CD}$$]{} where the lower horizontal arrow $\rho$ is the natural restriction map and the upper horizontal sequence is exact. By the Serre vanishing theorem, for large $e\gg 0$, we obtain the vanishing $H^1(X, K_X+p^eA)=0$. Thus, to prove that the restriction map $\rho$ is surjective (resp. a non-zero map), it is sufficient to show 