We know that for an immersion $j:U \to X$ the restriction functor $j^*:{\cal O}_Xmod \to {\cal O}_umod$ has a left adjoint $j!$. I am looking for some condition to deduce that $j!$ takes its values in Qco(X) that is to be a left adjoint for the functor $j^*:Qco(X) \to Qco(u)$.

$\begingroup$ By the way, it is more interesting to write down the right adjoint of $j^*$ (which exists by general nonsense). For example, you can do this when $j$ is quasicompact, i.e. $U$ is retrocompact in $X$. Then the right adjoint is just $j_*$. $\endgroup$– Martin BrandenburgMay 13, 2012 at 18:47
1 Answer
The restriction functor $\mathrm{Qcoh}(X) \to \mathrm{Qcoh}(U)$ doesn't preserve infinite products in general (which always exist, by the way). Therefore it cannot have a left adjoint.

$\begingroup$ can we add some assumptions on U or X to make this true? (or is U=X the only case when this works?) $\endgroup$ May 13, 2012 at 8:30

$\begingroup$ Well it is true when $U$ is a clopen subset of $X$, because then $\mathrm{Qcoh}(X) = \mathrm{Qcoh}(U) \times \mathrm{Qcoh}(X \setminus U)$. But otherwise probably it's only true in pathological cases. This is already visible in the affine case, say $X=\mathrm{Spec}(A)$ and $U=D(f)$ for some $f \in A$. When does $M \mapsto M_f$ preserve infinite products? Almost never. $\endgroup$ May 13, 2012 at 9:05

$\begingroup$ does the decomposition of QC(X) you described for clopens work for an infinite number of them? (I'm thinking of things like the Hilbert scheme which decomposes according to topological data of the parameterised subscheme, that's a countable disjoint union) $\endgroup$ May 14, 2012 at 9:24

$\begingroup$ Yes, but this is a special case. When $X \setminus U$ is a disjoint union of opens, then it is open ... $\endgroup$ May 14, 2012 at 9:30