Let $X$ be a topological space, $A$ a sheaf of (unital and associative but not necessarily commutative) rings on $X$. Suppose $M$ is a simple quasicoherent $A$-module and $U$ an open subset of $X$. Is $M|_U$ a simple $A|_U$-module?
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asked Jul 24, 2017 at 18:41
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1 Answer
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Yes. Restriction is an exact functor split by its right adjoint (pushforward). If $M|_{U}$ has a proper non-zero quotient $N$, then $i_*N$ receives a non-zero and thus injective map from $M$. However, this means that $M|_U$ must map injectively to $N$, which is a contradiction.
answered Jul 24, 2017 at 19:12
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$\begingroup$ What do you mean by split? I haven't seen that term in this context before. $\endgroup$ Commented Jul 24, 2017 at 19:36
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1$\begingroup$ In this case, I mean its right adjoint is a 1-sided inverse. $\endgroup$– Ben Webster ♦Commented Jul 24, 2017 at 20:12