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Let $X$ be a projective variety of dimension $n$ and $D \subset X$ is a proper subvariety. Embed $X$ into a projective space $\mathbb{P}^{3n}$. The following argument implies that $H^i(X,X\backslash D)=0$ for all $i>1$, which is not always true. Unfortunately, I am not able to find the mistake. I am afraid that there is something basic that I am misunderstanding. It will be very helpful if somebody could point out the gap in my understanding. I now give my argument:

We know that for every $i$ there are exact sequences of the form:

$$ H^i(\mathbb{P}^{3n}, X) \to H^i(\mathbb{P}^{3n}) \to H^i(X) \to H^{i+1}(\mathbb{P}^{3n}, X ) \to H^{i+1}(\mathbb{P}^{3n})$$ and

$$ H^i(\mathbb{P}^{3n} \backslash D, X \backslash D) \to H^i(\mathbb{P}^{3n} \backslash D) \to H^i(X \backslash D) \to H^{i+1}(\mathbb{P}^{3n}, X \backslash D ) \to H^{i+1}(\mathbb{P}^{3n} \backslash D)$$

There is a natural morphism of exact sequences from the first exact sequence to the second one. Note that, by Excision theorem $$H^i(\mathbb{P}^{3n}, X) \cong H^i(\mathbb{P}^{3n}\backslash D, X \backslash D)\, \mbox{ and } H^{i+1}(\mathbb{P}^{3n}, X) \cong H^{i+1}(\mathbb{P}^{3n} \backslash D, X\backslash D).$$ By Thom isomorphism as given in Fulton's page 371, we have (note that $\dim(D)<n$) $$H^i(\mathbb{P}^{3n},\mathbb{P}^{3n}\backslash D) \cong H_{6n-i}^{\mathrm{BM}}(D)=0\, \mbox{ for all } 0 \le i \le 2n$$ This implies that the morphism from $H^i(\mathbb{P}^{3n})$ to $H^i(\mathbb{P}^{3n}\backslash D)$ are isomorphisms for all $0 \le i \le 2n$. Finally, using five lemma, we conclude that the morphism from $H^i(X)$ to $H^i(X\backslash D)$ is an isomorphism from all $i>0$. This implies that $H^i(X,X\backslash D)=0$ for all $i>0$.

Where am I making a mistake?

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The hypothesis of the excision theorem that the closure of $D$ is contained in the interior of $X$ is not satisfied, since the interior of $X$ is empty.

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  • $\begingroup$ I see. Is this because X is not an open subscheme in $\mathbb{P}^{3n}$? I thought D is in the interior of X because D is a closed subvariety in X. $\endgroup$
    – user45397
    Commented May 2 at 13:57
  • $\begingroup$ @user45397 Exactly. The interior of a subset is the largest open subset but $X$ itself is not open and in fact does not contain any nonempty open subset. $\endgroup$
    – Will Sawin
    Commented May 2 at 14:12
  • $\begingroup$ Thank you so much for your answer. I understand my mistake now. $\endgroup$
    – user45397
    Commented May 2 at 14:18

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