Let $E \subset \mathbb P^2$ be a curve cut out by a homogeneous polynomial of degree 3. This is an elliptic curve and so $H^0(E, \mathscr O) = H^1(E, \mathscr O) \cong \mathbb C$. Now I want to calculate the cohomology of the sheaf $\mathscr O_E(p)$, where $p\in E$.
To this end we use the following short exact sequence $$ 0 \to I_{\{p\}} \to \mathscr O_E \to S_p \to 0, $$
where $I$ and $S$ are the ideal sheaf and skyscraper sheaf at $p$. As $I_{\{p\}} \cong \mathscr O_E(-p)$, tensoring the sequence with $\mathscr O_E(p)$ gives
$$ 0 \to \mathscr O \to \mathscr O(p) \to S_p \to 0$$
which induces the following long exact sequence:
$$ 0 \to H^0(\mathscr O) \to H^0(\mathscr O(p)) \overset{g}{\to} \mathbb C \to H^1(\mathscr O) \to H^1 (\mathscr O(p)) \to 0 \to 0 \to \cdots$$
Now I want to show that $g$ is zero. This should be true as we know $H^1(\mathscr O(p)) = \{0\}$, so $\mathbb C \to H^1(\mathscr O_E)$ is an isomorphism. But I have had a hard time showing that this is true (Indeed, I do not want to use $H^1(\mathscr O(p)) = \{0\}$ here).
Use that $\dim H^0(\mathscr O_E(p))=1$. Exactness says $\ker g=\textrm{im } (H^0(\mathscr O_E)\to H^0(\mathscr O_E(p)))=H^0(\mathscr O_E(p))$.