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.



Exact sequence in étale cohomology related with proper birational morphism
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 $f'\_!\mathbb Z\_\ell=\mathbb Z\_\ell$.


