$X$ and an ample [$\mathbb{Z}$]{}-divisor $A$ such that

1.  [$|K_X+H+A|$ is basepoint free. ]{}

2.  [The natural restriction map $$H^0(X, K_X+H+A)\to H^0(H, K_H+A)$$ is not surjective. ]{}

Let $X$ be a smooth projective surface and let $A$ be an ample [$\mathbb{Z}$]{}-divisor on $X$ such that $$H^1(X, K_X+A)\neq 0.$$ We can find such a surface by [@Raynaud]. Take a smooth hyperplane section $H$ of $X$ such that $|K_X+H+A|$ is basepoint free and that $$H^1(X, K_X+H+A)=0.$$ Consider the exact sequence $$0\to \mathcal O_X(K_X+A)\to\mathcal O_X(K_X+H+A)\to\mathcal O_H(K_H+A)\to0.$$ Then, we obtain the following exact sequence $$H^0(X, K_X+H+A)\to H^0(H, K_H+A)\to H^1(X, K_X+A)\to 0.$$ Since $H^1(X, K_X+A)\neq 0$, the restriction map is not surjective.

We use the following corollary in Section 8.

\[nonzero-criterion\] Let $X$ be a smooth projective surface. Let $L$ be a [$\mathbb{Z}$]{}-divisor on $X$ such that $$L=C+M$$ where $C$ is a smooth prime divisor and $M$ is a nef and big [$\mathbb{Z}$]{}-divisor such that $M|_C$ is ample. If $H^0(C, K_C+M)\neq 0$, then, the trace map $${{\operatorname{Tr}}}_X^e(L):H^0(X, K_X+p^eL)\to H^0(X, K_X+L)$$ is a non-zero map.

By Lemma \[trace-diagram\], we obtain the following commutative diagram [$$\begin{CD}
H^0(X, K_X+p^eL)@>>> H^0(p^eC, K_{p^eC}+p^eM)@>>> H^1(X, K_X+p^eM)\\
@VV{{\operatorname{Tr}}}_{X}^e(L) V @VV \Psi V\\
H^0(X, K_X+L)@>>> H^0(C, K_C+M).
\end{CD}$$]{} By [@T1 Theorem 2.6], we 