The singular cohomology with integer coefficients of a projective variety is isomorphic to the Čech cohomology of the constant sheaf of integers on this variety.
If the above statement is correct then consider the following example:
Look at $\mathbb{P}^1$ and the standard open cover $U\_0,U\_1$. Then the Čech complex for any sheaf $\mathcal{F}$ on $\mathbb{P}^1$ would be
$$
0 \rightarrow \mathcal{F}(U\_0)\oplus\mathcal{F}(U\_1) \rightarrow \mathcal{F}(U\_0\cap U\_1) \rightarrow 0.
$$
If we consider the constant sheaf of integers this becomes
$$
0 \rightarrow \mathbb{Z}\oplus\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0
$$
with non-trivial map $(a,b)\mapsto a-b$.
But the singular cohomology of $\mathbb{P}^1$ is
$$H^i\mathbb{P}^1=\begin{cases} \mathbb{Z},\quad i=0,2 \\
0,\quad \text{else.}\end{cases}$$
Where am I going wrong?



Čech cohomology is isomorphic to singular cohomology
The Cech complex only computes cohomology if the subspaces have vanishing cohomology themselves: you need
$$
H^i(U\_0; \mathcal{F}) \cong
H^i(U\_1; \mathcal{F}) \cong
H^i(U\_0 \cap U\_1; \mathcal{F}) \cong 0
$$
for all $i > 0$. In the example you've written, this is not true for $U\_0 \cap U\_1$. In general there is actually a Mayer-Vietoris long exact sequence relating the cohomology of $X$, $U\_0$, $U\_1$, and $U\_0 \cap U\_1$, or more generally a Mayer-Vietoris spectral sequence if the cover has more open sets.


(As a remark, you may have seen this constraint pushed under the rug because it is automatically true in special cases. If $X$ is a scheme and the subspaces $U\_0$, $U\_1$, and $U\_0 \cap U\_1$ are affine---the last automatic if $X$ is separated---then Serre's vanishing theorem says that these higher cohomologies vanish whenever $\mathcal{F}$ is a quasicoherent sheaf of $\mathcal{O}\_X$-modules.)


