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Mar 12 at 13:49 comment added Sergey Guminov I realized that I don't understand one thing: is the functor $f^*$ in this answer the whole inverse image functor, or just the 0-th perverse cohomology? Is there a reason why $f^*$ is $t$-exact in this case (if at all true)?
Feb 20 at 15:53 comment added Will Sawin @SergeyGuminov Corrected, thanks.
Feb 20 at 15:53 history edited Will Sawin CC BY-SA 4.0
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Feb 20 at 15:39 comment added Sergey Guminov I think the computation of $v'$ for the case of $f_{k*}$ is off. It seems that the last component of $1+v'c'$ is $e_k+vce_1$ instead of $e_1+vce_1$. Changing the last component of $v'$ from $vf_k$ to $vf_k-f_1-\ldots-f_{k-1}$ seems to work. Also, there is a minor typo, $F'$ should probably be $E^{k-1}\times F$, not $E^k\times F$. This answer was very helpful, by the way!
Mar 21, 2022 at 16:10 comment added user2520938 Thanks for all your help! I will carefully read this over the next couple of days.
Mar 21, 2022 at 15:01 history answered Will Sawin CC BY-SA 4.0