Let $X$ be a smooth projective variety of dimension $n$ and $i: D\hookrightarrow X$ be a smooth, anti-canonical divisor in $X$. In other words,$[D]+[\omega\_X]=[0]$ where $\omega\_X$ is the canonical bundle of $X$.


Let $i^\*: D^b\_{coh}(X)\to D^b\_{coh}(D)$ and $i\_\*: D^b\_{coh}(D)\to D^b\_{coh}(X)$ be the derived pullback and pushforward functors. By adjunction, for any $
\mathcal{F}\in D^b\_{coh}(X)$, we have a morphism $\mathcal{F}\to i\_\*i^\*\mathcal{F}$.


Now the question is how to complete the above morphism to an exact triangle in $D^b\_{coh}(X)$. In particular, do we have an exact triangle like
$$
\mathcal{F}\to i\_\*i^\*\mathcal{F}\to \mathcal{F}\otimes^L\_{\mathcal{O}\_X}\omega\_X[1]\to \mathcal{F}[1]?
$$


I also wonder if we can solve the same problem if we do not require that the divisor $D$ is anti-canonical.



What is the cone of $\mathcal{F}\to i_*i^*\mathcal{F}$ for a divisor $i: D\hookrightarrow X$?
For any Cartier divisor $i \colon D \hookrightarrow X$ and any $F \in D^b(X)$ there is an exact triangle
$$
F \otimes \mathcal{O}\_X(-D) \to F \to i\_\*i^\*F.
$$
It can be obtained by tensoring the exact sequence
$$
0 \to \mathcal{O}\_X(-D) \to \mathcal{O}\_X \to i\_\*\mathcal{O}\_D \to 0
$$
with $F$ and using the projection formula
$$
F \otimes i\_\*\mathcal{O}\_D \cong i\_\*i^\*F.
$$


