Let $\varphi:X\to Y$ be a morphism of schemes of relative dimension 1, and $\mathcal{L}' \xrightarrow{g} \mathcal{L}$ an injection of line bundles on $X$.
The sequence $$0\to \mathcal{L}' \xrightarrow{g} \mathcal{L} \to \overline{\mathcal{L}} \to 0$$ induces a long exact sequence $$0 \longrightarrow R^0\varphi_*\mathcal{L}' \longrightarrow R^0\varphi_*\mathcal{L} \longrightarrow R^0\varphi_*\overline{\mathcal{L}}$$ $$\longrightarrow R^1\varphi_*\mathcal{L}' \longrightarrow R^1\varphi_*\mathcal{L} \longrightarrow 0$$
which can be spliced into two short exact sequences of $\mathcal{O}_Y$-modules $$S_0:\; 0 \to R^0\varphi_*\mathcal{L}' \longrightarrow R^0\varphi_*\mathcal{L} \to \mathcal{F} \to 0$$ $$S_1:\; 0 \to \mathcal{G} \to R^1\varphi_*\mathcal{L}' \longrightarrow R^1\varphi_*\mathcal{L} \to 0$$
The term $R^0\varphi_*\overline{\mathcal{L}}$ straddles the sequences $S_0, S_1$, via its submodule $\mathcal{F}$ appearing in $S_0$ or its quotient $\mathcal{G}$ in $S_1$, and if it "falls on one side" then the other side will be an isomorphism.
In other words, $R^0\varphi\overline{\mathcal{L}}$ may be nonzero in the long exact sequence, but either the map entering it or leaving it may be zero, which is equivalent to one of $\mathcal{F}, \mathcal{G}$ being zero. Then one of $S_0, S_1$ would be a short exact sequence with only two nonzero terms, and the map between them must be an isomorphism.
My question is: are there nice conditions under which either one of $S_0, S_1$ will be an isomorphism?
If either
- $\mathcal{L}$ is acyclic for $R\varphi_*$
- $R^0\varphi_*\overline{\mathcal{L}}$ is a simple $\mathcal{O}_Y$-module
then the property holds, but I am looking for something less piecemeal.
My hope is to express the condition as a relation between a divisor $D$ and a line bundle $\mathcal{L}$, so that the map in question is $$\mathcal{L} \longrightarrow \mathcal{L}(D).$$ I am particularly interested in the case where one of $\mathcal{L},\mathcal{L}'$ is $\Omega_{X/Y}$.
