I'm trying to compute the Euler Characteristic of the blowing up of $\mathbb{C}\mathbb{P}^2$ at $n$ points. Does anyone know how could I do this?
Euler Characteristic of a blowing up
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I thought I'd write down a couple of thoughts about how we can obtain the result $\chi_{\rm top}(Bl_n(\mathbb P^2)) = 3 + n$ explained in Nicolas' answer.
Method 1:
Noether's formula for surfaces states that $$\chi(\mathcal O_X)= \tfrac 1 {12}\left( K_X.K_X + \chi_{\rm top}(X)\right)$$ where $\chi(\mathcal O_X)$ denotes the holomorphic Euler characteristic of the structure sheaf, $K_X . K_X$ denotes the self-intersection of the canonical divisor, and $\chi_{\rm top}(X)$ denotes the topological Euler characteristic of the space (with the analytic topology).
We also have the following two facts (both of which are proved in Beauville's Complex Algebraic Surfaces):
As Mohan pointed out, $\chi(\mathcal O_X)$ is a birational invariant, so if $\pi : \widetilde X \to X$ is a blow-up, then $\chi(\mathcal O_{\widetilde X}) = \chi(\mathcal O_X)$.
The self-intersection of the canonical class of the blow-up is related to the self-intersection of the original space by $K_{\widetilde X}.K_{\widetilde X}=\pi^\star K_X . \pi^\star K_X - 1$.
Combining all of these observations, we see that blowing up a surface at a point increases the topological Euler characteristic $\chi_{\rm top}(X)$ by one.
Method 2:
I think this method is really equivalent to Nicolas' answer...
I believe that complex quasi-projective varieties obey an inclusion-exclusion principle (see also here): if $X$ is a complex projective variety and $Y \subset X$ is a closed subvariety, then $$ \chi_{\rm top}(X) = \chi_{\rm top}(X \setminus Y) + \chi_{\rm top} (Y)$$ So if you think of the blow-up operation as removing a point ($Y = \{ \rm pt \}$), then adding a two-sphere ($Y = \mathbb {CP}^1$), this formula tells you that the Euler characteristic increases by one after the blow-up.
I must admit I've never studied a proof of this result, but I believe it is based on triangulation.
It is easy to see that for a complex surface $S$ and $p$ a smooth point, we have $\chi(Bl_{p}S) = \chi(S) +1 $, say using triangulations. This gives $\chi(Bl_{n}(\Bbb{P}^2)) = 3 + n$.
For example, $n=6$ is a cubic hypersurface and you can easily check that you get $9$.