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Sep 20, 2014 at 23:40 vote accept lime
Sep 19, 2014 at 19:37 comment added Piotr Achinger To show that some presheaf $F$ sheafifies to zero is the same as showing that given a section $s\in F(V)$, we can cover $V$ by $V'_i$ such that $s$ maps to zero in $F(V'_i)$. In other words, for every $x\in V$ there is a $V'_x$...
Sep 19, 2014 at 18:49 comment added lime Piotr, why do we need to do this exactly for geometric points $\bar{p}$? Isn't that enough to show that for any $V$ there exists $V'$ such that corresponding cohomology vanishes? Apparently I am missing some obvious conception.
Sep 19, 2014 at 2:52 history answered Piotr Achinger CC BY-SA 3.0