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Let $X$ be a smooth projective algebraic variety over a field of characteristic zero and let $D$ be a simple normal crossing divisor on $D$. Put

$j: U \hookrightarrow X$

for the inclusion of the complement $U=X-D$ on $X$.

Are the functors $j_\ast$ and $j_!$ exact?

I think the answer is yes.

I would be very grateful if someone could provide me with a proof.

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I assume you're asking about quasicoherent sheaves.

Since the inclusion of $U$ into $X$ is an open immersion, $j_!$ is exact. You can find this in Tag 03DJ.

Since $j$ is an affine morphism, $j_*$ is exact (EGA 2 Corollary 5.2.2). $j$ is affine, because the affine property of morphisms is étale local on the target, and the inclusion of the complement of a principal closed subscheme (such as the locus defined by $x_1 x_2 \cdots x_r = 0$ in affine space) is affine. See e.g., Tag 07ZT

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  • $\begingroup$ $j_!$ does not preserve quasicoherent sheaves... $\endgroup$ – anon Jun 2 '13 at 15:20
  • $\begingroup$ Could you explain Why not? $\endgroup$ – exactfun Jun 2 '13 at 15:30
  • $\begingroup$ I should have said $\mathcal{O}$-modules. $\endgroup$ – S. Carnahan Jun 2 '13 at 23:41
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For a general open immersion $j: U \hookrightarrow X$, $j_!$ is exact (it is left exact and left adjoint to $j^*$), but I would guess that $j_*$ is (in general) only left exact (it is right adjoint to $j^*$).

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  • $\begingroup$ Thanks, Timo? Is that general or you have taken into account the shape of $U$? Is the morphism $j$ affine or something that allows to conclude exactness? $\endgroup$ – exactfun Jun 1 '13 at 18:26
  • $\begingroup$ This holds for $j: U \hookrightarrow X$ an open immersion. An affine morphism is exact. $\endgroup$ – user19475 Jun 1 '13 at 18:31
  • $\begingroup$ So now the question is: is $j$ affine in the situation "complement of a SNCD"? $\endgroup$ – exactfun Jun 1 '13 at 18:35
  • $\begingroup$ I would guess so, since according to [Milne, Étale cohomology], Theorem VI.7.1, the complement in a projective variety of a hypersurface section is affine. $\endgroup$ – user19475 Jun 1 '13 at 18:44
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    $\begingroup$ If the divisor is not ample then it is not such a hyperplane section. For example the complement of a line on a quadric surface is a $\mathbb{P}^1$ bundle over $\mathbb{A}^1$. $\endgroup$ – Jack Huizenga Jun 1 '13 at 19:08

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