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Jun 14, 2018 at 19:30 comment added Fan Zheng @Malkoun: A HUGE Sorry. What happened is that I opened too many mathoverflow tags and copied the wrong address. LInk has been fixed.
Jun 14, 2018 at 19:26 history edited Fan Zheng CC BY-SA 4.0
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Jun 14, 2018 at 11:09 comment added Malkoun I think this page needs some cleaning up.
Jun 14, 2018 at 10:34 comment added Maxime Ramzi @WillSawin ah ok seeing $L:\mathbf{Psh}(X) \to \mathbf{Psh}(X)$. Since this is a tiny bit subtle it's good to make it clear haha
Jun 14, 2018 at 10:27 comment added Will Sawin @Max I was referring to the sheafificatiin functor from presheaves to presheaves . It's all the same.
Jun 14, 2018 at 9:13 comment added Maxime Ramzi @WillSawin probably the actual reason is that in my "proof" the second $\bigoplus$ is the direct sum in $\mathbf{Sh}(X)$ (I realized my blunder too late) and this does not coincide with the direct sum in $\mathbf{Psh}(X)$ (the inclusion functor does not commute with direct sums, while the sheafification functor does)
Jun 14, 2018 at 7:37 comment added Will Sawin @BrianT Sheafification doesn't commute with direct sum as the definition of sheafification involves products, possibly infinite, and infinite products don't commute with infinite sums.
Jun 14, 2018 at 7:04 comment added Malkoun I did some search, and indeed, some counterexamples can be found for example here: math.stackexchange.com/questions/193816/…
Jun 14, 2018 at 6:29 comment added BrianT Hi Fan Zheng, so can you tell me where the above argument of Max fails : let $F^l$ be sheaves, and $F$ be the functor sheafification. Then $$F(\underset{l}{\bigoplus} F^l) = \underset{l}{\bigoplus} F(F^l) = \underset{l}{\bigoplus} F^l$$
Jun 14, 2018 at 4:33 comment added Malkoun Hi. I followed the link, but could not find the counterexample. Do you mean in Ravi Vakil's notes?
Jun 14, 2018 at 0:13 history answered Fan Zheng CC BY-SA 4.0