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Let $X$ be a projective complex manifold and $Y\subset X$ be an irreducible hypersurface. If $Y$ is smooth, there is a well known Gysin sequence. However, even if $Y$ is not smooth, a kind of Gysin map can still be devised.

Consider a desingularization $f:\widetilde{Y}\to Y$ and the inclusion $i:Y\hookrightarrow X$. We have two maps

$H^i(Y,\mathbb{Q})\to H^i(\widetilde{Y},\mathbb{Q})\to H^{i+2}(X,\mathbb{Q});$

the first one is $f^*$ and the second one is the Poincare dual to $(i\circ f)^*$ (essentially the Gysin map). I am convinced that the composition does not depend on desingularization, though I do not know a rigorous proof of this.

QUESTION: Is the sequence

$H^i(Y,\mathbb{Q})\to H^{i+2}(X,\mathbb{Q})\to H^{i+2}(X\setminus Y,\mathbb{Q})$

exact (as it is in the smooth case)?

All I know about it is a result of Deligne [Theorie de Hodge III. Publ. Math. IHES 44 (1974) pp. 5–77.; Corollary 8.2.8] that

$H^i(\widetilde{Y},\mathbb{Q})\to H^{i+2}(X,\mathbb{Q})\to H^{i+2}(X\setminus Y,\mathbb{Q})$

is exact, but this is much weaker than what I need.

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If you work with Chow groups rather than cohomology, a version of this exact sequence is given (with much weaker hypotheses than the ones you have specified) in Fulton's Intersection Theory, Proposition 1.8, page 21. This is not quite what you are asking for, but it suggests that something similar is probably true for cohomology. – Charles Staats Aug 10 '12 at 2:04
Thank you. Though I am not sure if it may be helpful. – Alex Gavrilov Aug 10 '12 at 2:51
When $Y$ is smooth, by Poincar\'e duality, the sequence you consider is the dual sequence of the sequence with compact supports $\dots H^{2n-i-2}_c(X\backslash Y,{\bf Q})\to H^{2n-i-2}(X,{\bf Q})\to H^{2n-i-2}(Y,{\bf Q})\to\dots$ This suggests that the "right sequence" without compact supports should be obtained by dualising the above sequence using Grothendieck-Verdier duality (which generalizes Poincar\'e duality to the non-smooth setting). In particular, you would expect the dualising complex to appear on $Y$ (and not just $\bf Q$). – Damian Rössler Aug 10 '12 at 8:02
Perhaps, but I hope that the statement may be true as it is. I do not need the Whole Gysin sequence after all. – Alex Gavrilov Aug 10 '12 at 10:13

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