Cover for Cech cohomology of the constant sheaf $\mathbb{Z}$ over $\mathbb{P}^n$ 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 $H^2 (\mathbb{P}^n, \bf{\mathbb{Z}}) \cong \mathbb{Z}$ ?

 A: I hope that this will not be the only answer to this question, but I'd like the point out that the theorem of Leray gives sufficent conditions for 
$$H^n(\mathcal{U}, \mathcal{F})\rightarrow H^n(X,\mathcal{F})$$ to be an isomorphism for all $n$, namely that for all $n$
$$H^n(U_{i_0}\cap \cdots \cap U_{i_k},\mathcal{F})=0$$ for all finite intersection of open sets from the cover. This is not satisfied by the open covering given in the question.
A: 
$\text{}$1. Is there a standard, or at least well known, cover of $\mathbb{P}^n$ satisfying this condition?

Not an explicit one as far as I know.

$\text{}$2. Is there any intuition on how to choose such covers?

Any covering consisting of geodesically convex sets works.

$\text{}$3. Is there any other direct method of seeing the isomorphism $H^2 (\mathbb{P}^n, \bf{\mathbb{Z}}) \cong \mathbb{Z}$?

The simplest approach in my opinion is to use the cell complex structure of $\mathbb{P}^n$: it has exactly one cell in each even dimension $0$, $2$, $\ldots$ , $2n$. Then use the definition of homology for CW complexes, and apply it to this cell decomposition.
