# Derived push-forward and pull back.

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 $$j^*j_*F=F\otimes j^*j_*O_D?$$

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I should add a few remarks: 1) This is not the well known $j_*j^*F=F\otimes j_* O_D$ 2) If it is not well known it is probably wrong, but there is always a triangle: $F\otimes O_D(-D)[1]\rightarrow j^*j_*F \rightarrow F$ by Bondal, Orlov "Semiort. decomp." Lemma 3.3 3) We can calculate RHS by restricting and tensoring the resolution $0\rightarrow O(-D)\rightarrow O\rightarrow j_*O_D\rightarrow 0$ As, after restricting to $D$ the map is zero, we see that RHS is a sum of $F$ and $F\otimes O_D(-D)[1]$. In particular it fits to the previous triangle (that maybe does not split). –  Wajcha Oct 17 '12 at 13:50
4) If $F$ is an injective coherent sheaf then LHS and RHS fit into the triangle: $(LvR)HS\rightarrow F\rightarrow F\otimes O_D(-D)[2]$ with the last morphism being zero, as these are injective sheaves in diffrent shifts. 5)There is a morphism from LHS to RHS induced by a morphism from $j_* F$ to $j_*(F\otimes j^*j_*O_D)=j_*F\otimes j_*O_D$, that comes from $O_X\rightarrow j_*O_D$. –  Wajcha Oct 17 '12 at 13:58
6) It is true on the (not nec. full) subcategory $j^*(D(X))$ –  Wajcha Oct 17 '12 at 14:00

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 $j^*j_*F$.