Let $Y$ be a smooth algebraic variety and $i: X\hookrightarrow Y$ be a smooth divisor. We consider the derived functors $i^\*: D^b\_{coh}(Y)\to D^b\_{coh}(X)$ and $i\_\*: D^b\_{coh}(X)\to D^b\_{coh}(Y)$. By adjunction, for any $\mathcal{F}\in D^b\_{coh}(X)$ we have a canonical morphism $i^\*i\_\*\mathcal{F}\to \mathcal{F}$ in $D^b\_{coh}(X)$.


My question is: Can we complete $i^\*i\_\*\mathcal{F}\to \mathcal{F}$ to an exact triangle in $D^b\_{coh}(X)$? I am particularly interest in the case that $Y$ is Calabi-Yau and I guess that in this case the expected exact triangle should be
$$
i^\*i\_\*\mathcal{F}\to \mathcal{F}\to \mathcal{F}\otimes\omega\_X^{-1}
$$
but I cannot prove or disprove it.



How to complete $i^*i_*\mathcal{F}\to \mathcal{F}$ into an exact triangle for a smooth divisor $i: X\hookrightarrow Y$?
The cone of $i^\*i\_\*\mathcal{F} \to \mathcal{F}$ is isomorphic to $\mathcal{F} \otimes \mathcal{O}\_X(-X)[2]$.


EDIT. Let me write an argument for a sheaf $F$. Consider the distinguished triangle
$$
i^\*i\_\*F \to F \to F'.
$$
We need to identify $F'$.
Applying $i\_\*$ we obtain
$$
i\_\*i^\*i\_\*F \to i\_\*F \to i\_\*F'.\tag{\*}
$$
By projection formula and the Koszul resolution of $i\_\*\mathcal{O}\_X$, we have
$$
i\_\*i^\*i\_\*F \cong 
i\_\*F \otimes i\_\*\mathcal{O}\_X \cong
\mathrm{Cone}(i\_\*F \otimes \mathcal{O}\_Y(-X) \to i\_\*F),
$$
and since the map in the right-hand side is obviously zero, we conclude that
$$
i\_\*i^\*i\_\*F \cong 
i\_\*F \oplus i\_\*F \otimes \mathcal{O}\_Y(-X)[1].
$$
Now the first map in $(\*)$ is the projection to the direct summand, hence
$$
i\_\*F' \cong 
i\_\*F \otimes \mathcal{O}\_Y(-X)[2] \cong
i\_\*(F \otimes \mathcal{O}\_X(-X)[2]).
$$
Since $i\_\*$ is fully faithful on the category of sheaves, it follows from this that
$$
F' \cong F \otimes \mathcal{O}\_X(-X)[2].
$$


