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Mar 30, 2017 at 21:09 vote accept user45397
Mar 29, 2017 at 21:03 comment added Dan Petersen Oh, I see. I thought the "usual" Lefschetz theorem had an assumption that $Y$ is smooth, but I see now that it's enough that the singular locus of $X$ is contained in $Y$. Yeah, then I agree with you.
Mar 29, 2017 at 20:59 comment added Mikhail Bondarko I am just saying that if you replace the group $H^k(Y)$ in the Lefschetz theorem by the group $IH_k(Y)$ that is not isomorphic to it then you would get a wrong statement. However, it may happen that $H^k(Y)$ is always isomorphic to $IH_k(Y)$ the range in question. Anyway, the (current) argument of user45397 is not correct.
Mar 29, 2017 at 20:34 comment added Dan Petersen I don't think this answers the question. I would've expected the answer to be "yes" and I don't really see an argument in your answer for why you think the isomorphism fails in general.
Mar 29, 2017 at 12:16 history answered Mikhail Bondarko CC BY-SA 3.0