Let f:X→Y be proper birational morphism between two quasi-projective varieties over an algebraically closed field $k$. I am particularly interested in the case where the characteristic of $k$ is positive; Y is singular, X is smooth and both X and Y are not projective.

Let D be a closed subset of Y, let E=f-1(D) and assume that f|X\E:X\E→Y\D is an isomorphism.

Based on results in characteristic 0 (and on results on rigid cohomology) I expect that there is a long exact sequence of étale cohomology groups

Hi(Y,ℚl)→ Hi(X,ℚl)⊕ Hi(D,ℚl)→ Hi(E,ℚl)→ Hi+1(Y,ℚl).

I hoped this exact sequence is well-known and would appear in a standard text, but I had trouble identifying such a text. I can think of a prove using Cox's étale version of tubular neighbourhoods for E in X and D in Y and you might be able to compare them using the Mayer-Vietoris sequences in étale cohomology, but such a proof seems quite involved (you need to define the image of an étale tubular nhd under a proper morphism (which seems non-trivially to me) and check that for certain exact sequences taking direct or projective limits turns out to be an exact functor.)

Before working out the details I would like to ask whether anyone knows a reference for the above sequence or knows a simpler/nicer proof.


Let $V := X\setminus E$ and $U := Y\setminus D$ and $j\colon U \rightarrow Y$ and $k\colon V \rightarrow X$ the inclusions. We have exact sequences $\cdots\rightarrow H^\ast(X,k_!\mathbb Z_\ell)\rightarrow H^\ast(E,\mathbb Z_\ell)\rightarrow H^\ast(X,\mathbb Z_\ell)\rightarrow\cdots$ and a similar one for $Y$, $D$ and $j$. Furthermore, $f$ induces a map of these long exact sequences. A standard diagram chasing (also used for instance to the Mayer-Vietoris sequence from excision) can be used to show that to get the desired exact sequence it is enough to show that $f$ induces an isomorphism $H^\ast(Y,j_!\mathbb Z_\ell)\rightarrow H^\ast(X,k_!\mathbb Z_\ell)$. We have that $H^\ast(X,k_!\mathbb Z_\ell)=H^\ast(Y,Rf_\ast k_!\mathbb Z_\ell)$ but $f$ is proper so that $Rf_\ast k_!\mathbb Z_\ell=f_!k_!\mathbb Z_\ell$. However, $f_!k_!=(fk)_!=(jf')_!=j_!f'_!$, where $f'=f_{|V}$, and as $f'$ is an isomorphism we have $f'_!\mathbb Z_\ell=\mathbb Z_\ell$.

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