All functors are derived and all categories are bounded derived categories of coherent sheaves. Suppose that we have got an inclusion of a smooth divisor $j:D\rightarrow X$ in a smooth projective variety. Is it true that $$j^\*j\_\*F=F\otimes j^\*j\_\*O\_D?$$



Derived push-forward and pull back.
The answer is no. The simplest example that I know is $X = P^3$, $D = P^1\times P^1$, $F = O(0,1)$. In this case $j^\*j\_\*O\_D = O\_D \oplus O\_D(-2)[1]$ and hence $F\otimes j^\*j\_\*O\_D = F \oplus F(-2)[1]$, while $j^\*j\_\*F$ fits into a triangle
$$
F(-2)[1] \to j^\*j\_\*F \to F
$$
which is not split. To see this note that $j\_\*F$ has a resolution of the form 
$$
0 \to O\_X(-1)^2 \to O\_X^2 \to j\_\*F \to 0,
$$
which gives a distinguished triangle
$$
O\_D(-1,-1)^2 \to O\_D^2 \to j^\*j\_\*F.
$$
It follows easily from this that $Hom(F,j^\*j\_\*F) = 0$, which shows that $F$ is not a direct summand of $j^\*j\_\*F$.


