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Let $X$ be a smooth, projective variety and $Y \subset X$ be a hyperplane section (possibly singular) of $X$. Suppose that the dimension of $X$ is $n$. Is it true that for any $k<n-1$, the induced morphism of intersection homologies: $$IH_k(Y) \to IH_k(X)$$ is an isomorphism?

I would think this is true as the version of the Lefschetz hyperplane section theorem I have it states this conclusion under the additional assumption that $Y$ is transverse to every strata of some Whitney stratification of $X$. As $X$ is smooth, in my case, I can take the only non-trivial strata to be $X$ and the other sub-strata to be empty sets. This should answer the above question, but I am not sure if I am making a mistake.

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So, the answer is positive and follows from the "usual" Weak Lefschetz Theorem if you have transversality (and it suffices to assume that $Y$ is smooth). On the other hand, if $Y$ is singular then you surely don't have transversality and probably don't have the isomorphism in question (as far as I remember, the intersection homology of $X$ is isomorphic to its "usual" cohomology, whereas for a singular $Y$ the group $IH_k(Y)$ is "usually smaller" than $H^k(Y)$).

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  • $\begingroup$ 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. $\endgroup$
    – Dan Petersen
    Commented Mar 29, 2017 at 20:34
  • $\begingroup$ 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. $\endgroup$
    – Mikhail Bondarko
    Commented Mar 29, 2017 at 20:59
  • $\begingroup$ 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. $\endgroup$
    – Dan Petersen
    Commented Mar 29, 2017 at 21:03

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