# Long exact sequence of cohomology and fibration

In an article I'm reading, the author is stating :

$O$ is isomorphic to the complement of a zero section of a line bundle over $X$. We have a long exact sequence of (étale) cohomology associated to the fibration : $$\cdots \to H^{i-2}_c(X) \to H^{i}_c(X) \to H^{i+1}_c(O) \to H^{i-1}_c(X) \to \cdots$$

Is there a result involving a long exact sequence (of cohomology) and the complement of a zero section of a (line) bundle ?

For now, I'm guessing the first and the last term of the sequence above are the cohomology groups of the line bundle : if $Y \to X$ is the linde bundle, then $$H^i_c(Y) = H^{i-2}_c(X).$$ So hypothetically there should be a long exact sequence like : $$\cdots \to H^{i}_c(Y) \to H^{i}_c(X) \to H^{i+1}_c(O) \to H^{i+1}_c(Y) \to \cdots$$ but I don't understand why.

Thank you.

• Apr 23, 2012 at 19:25

Though Mark Grant's comment links to the right answer (what you are looking for is called the Gysin sequence), the Wikipedia page doesn't state it in the form you want. In general, for an open subset $U \subset X$ of an algebraic variety $X$, there is a long exact sequence $$\cdots \to H^i_c(U) \to H^i_c(X) \to H^i_c(X\setminus U) \to H^{i+1}_c(U) \to \cdots$$ of étale cohomology groups. This should be in any reference on étale cohomology.
The analogue in de Rham cohomology is easy to understand, and this is how I usually remember this sequence. A compactly supported differential form on $U$ admits an extension by zero to all of $X$, and the restriction of a compactly supported differential form on $X$ to the closed subset $X\setminus U$ is a compactly supported form on $X \setminus U$. This gives an exact sequence $$0 \to \Omega_c^\bullet(U) \to \Omega_c^\bullet(X) \to \Omega_c^\bullet(X\setminus U) \to 0$$ whose cohomology gives the above long exact sequence.
• If $L \to X$ is the line bundle, then $O$ is an open subset of $L$ and its complement is isomorphic to $X$ (the zero section provides the isomorphism). Apr 24, 2012 at 7:50