Cech Cohomology Isomorphism with Singular Cohomology – Algebraic Geometry

Question

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?

Answer 1

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.)