First of all consider $\mathbb{P}^n$ as a complex analytic manifold. In Griffiths and Harris's Principles of Algebraic geometry p.145 it is stated $$ H^2 (\mathbb{P}^n, \bf{\mathbb{Z}}) \cong \mathbb{Z}, $$ that is, the second Cech cohomology group of $\mathbb{P}^n$ over the constant sheaf $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$.
I wanted to check this using the usual algebraic cover of open sets: $\mathcal{U} = \{ U_i \}_{0 \le i \le n}$, where $U_i = \{ x_i \neq 0 \}$, but already for $\mathbb{P}^1$ this fails, meaning that $\mathcal{U}$ is not fine enough for computing Cech cohomology.
Edit: As Rene remarked in the answer below, there is a theorem of Leray which states that given a sheaf $\mathcal{F}$ and a cover $\mathcal{U}= \{ U_i \}$ such that $$ H^p(U_{i_0} \cap \dotsb \cap U_{i_k}, \mathcal{F}) = 0 $$ for all finite intersections of $\mathcal{U}$ and all $p$, then $$ H^p(\mathcal{U},\mathcal{F}) \rightarrow H^p(X, \mathcal{F}) $$ is an isomorphism for all $p$.
This still gives rise to questions:
- Is there a standard, or at least well known, cover of $\mathbb{P}^n$ satisfying this condition?
- Is there any intuition on how to choose such covers?
- Is there any other direct method of seeing the isomorphism $H^2 (\mathbb{P}^n, \bf{\mathbb{Z}}) \cong \mathbb{Z}$ ?