X_2}$, then $\mathcal{R}_{X\to X_1}$ is surjective.

-   If $(K_X+\Lambda)|_{X_2}$ is very ample and $X_1\cap X_2$ is a line on $X_2$, then $\mathcal{R}_{X\to X_1}$ is surjective and $\Phi_{|K_X+\Lambda)|}$ separates $X_1$ and $X_2$.

-   If each $(K_X+\Lambda)|_{X_i}$ is very ample and $X_1\cap X_2$ is a line on each $X_i$, then $K_X+\Lambda$ is very ample.

\(i) and (ii) is obvious by chasing the diagram.

For (iii), $X_1\cap X_2$ is a line implies $\mathcal{R}_{X_2\to X_1\cap X_2}$ is surjective. Then any section in $H^0(X_1,(K_X+\Lambda)|_{X_1})$ can be extended into a section in $H^0(X,K_X+\Lambda)$. Therefore $\mathcal{R}_{X\to X_1}$ is surjective. Next $\ker \mathcal{R}_{X_2\to X_1\cap X_2}$ is nontrivial and its base part is the line $X_1\cap X_2$. Then $\ker \mathcal{R}_{X\to X_1}$ is nontrivial and its base part is $X_1$. Therefore $\Phi_{|K_X+\Lambda)|}$ separates $X_1$ and $X_2$.

For (iv) we see that $\mathcal{R}_{X\to X_i}$ is surjective and the kernel is nontrivial by (ii). Then we have plenty of sections to separate points and tangents, which implies $K_X+\Lambda$ is very ample.

\[equaldelta\] Let $(X,\Lambda)$ be a log surface as in Lemma \[2comps\]. We assume further $X_1$ is irreducible and $\Delta(X,K_X+\Lambda)= \Delta(X_i,(K_X+\Lambda)|_{X_i})\le 1$ for $i=1,2$. Then

-   if $\Delta(X,K_X+\Lambda)=0$, then $X_1\cap X_2$ is a line on each $X_i$. Moreover, $K_X+\Lambda$ is very ample.

