# How to compute the Betti numbers of S-D for a surface S and a divisor D?

Let S be a projective non-singular surface and D a Cartier divisor which has a smooth representative. Can the Betti numbers of S-D be represented by the Betti numbers of S and D? In a paper $b_i(S-D)=b_i(S)-b_i(D)$ is used without explanation; however, I cann't prove it and I doubt whether it would hold.

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This is clearly false, think about $i=0$... In general you would study this by means of the long exact sequence for relative cohomology and the isomorphism $H^i(S;D) = H^{4-i}(S-D)^\vee$. – Dan Petersen Mar 19 '12 at 9:07
Using the long exact sequence it is easy to represent the Euler number of S-D by the Euler numbers of S and D, but we still cannot represent the Betti numbers of S-D by the Betti numbers of S and D, right? At least we cannot find a universal formula... – rose Mar 19 '12 at 9:24
What it is true is that $\sum (-1)^i b_i(S-D)=\sum (-1)^i b_i(S) - \sum (-1)^i b_i(D)$. In fact, the topological Euler-Poincaré characteristic is additive, as you can see by taking a triangulation of $S$ that contains a triangulation of $D$ as a sub-triangulation (racall that any smooth, complex projective variety can be triangulated). – Francesco Polizzi Mar 19 '12 at 10:07